Mixing
Can we divide a vector by another vector? How about this: $a = vdv/dx?$
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My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$
It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
kinematics acceleration vectors differentiation mathematics
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
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show 2 more comments
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My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$
It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
kinematics acceleration vectors differentiation mathematics
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
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Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
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– Chair
Jan 25 at 14:53
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Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
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– BPP
Jan 25 at 16:12
1
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Related: physics.stackexchange.com/q/14082/2451
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– Qmechanic♦
Jan 25 at 18:23
3
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Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
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– Michael Freimann
Jan 25 at 18:29
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
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– π times e
Jan 26 at 4:33
|
show 2 more comments
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My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$
It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
kinematics acceleration vectors differentiation mathematics
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
$endgroup$
My physics teacher told us that we can’t divide vectors, that vector division has no physical meaning or significance. How about this: $$a = vdv/dx.$$
It says acceleration vector equals velocity (as a function of $x$) times $dv$ ‘divided’ by $dx$.
Here both $dv$ and $dx$ are vectors. How do I make sense of it? Because vector division doesn’t exist in physics right?
kinematics acceleration vectors differentiation mathematics
kinematics acceleration vectors differentiation mathematics
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
edited Jan 25 at 18:26
Qmechanic♦
106k121961223
106k121961223
asked Jan 25 at 14:38
π times eπ times e
16710
asked Jan 25 at 14:38
π times eπ times e
16710
asked Jan 25 at 14:38
π times eπ times e
16710
16710
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Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
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– Chair
Jan 25 at 14:53
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Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
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– BPP
Jan 25 at 16:12
1
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Related: physics.stackexchange.com/q/14082/2451
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– Qmechanic♦
Jan 25 at 18:23
3
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Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
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– Michael Freimann
Jan 25 at 18:29
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
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– π times e
Jan 26 at 4:33
|
show 2 more comments
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Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
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– Chair
Jan 25 at 14:53
$begingroup$
Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
$endgroup$
– BPP
Jan 25 at 16:12
1
$begingroup$
Related: physics.stackexchange.com/q/14082/2451
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– Qmechanic♦
Jan 25 at 18:23
3
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Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
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– Michael Freimann
Jan 25 at 18:29
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
$endgroup$
– π times e
Jan 26 at 4:33
$begingroup$
Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
$endgroup$
– Chair
Jan 25 at 14:53
$begingroup$
Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
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– Chair
Jan 25 at 14:53
$begingroup$
Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
$endgroup$
– BPP
Jan 25 at 16:12
$begingroup$
Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
$endgroup$
– BPP
Jan 25 at 16:12
1
1
$begingroup$
Related: physics.stackexchange.com/q/14082/2451
$endgroup$
– Qmechanic♦
Jan 25 at 18:23
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Related: physics.stackexchange.com/q/14082/2451
$endgroup$
– Qmechanic♦
Jan 25 at 18:23
3
3
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Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
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– Michael Freimann
Jan 25 at 18:29
$begingroup$
Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
$endgroup$
– Michael Freimann
Jan 25 at 18:29
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
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– π times e
Jan 26 at 4:33
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
$endgroup$
– π times e
Jan 26 at 4:33
|
show 2 more comments
7 Answers
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Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=vfrac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $vec{a}=vec{v}frac{dvec{v}}{dvec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $vec{v}(vec{x})$, there are four ways to define differentiation:
- differentiation by a scalar parameter $t$, which is defined for an $n$-dimensional vector as:
$$frac{dvec{v}(vec{x},t)}{dt}=leftlangle frac{d v_1(vec{x},t)}{d t},...,frac{d v_n(vec{x},t)}{d t}rightrangle$$
- the divergence, which is defined for an $n$-dimensional vector as:
$$nabla cdot vec{v}(vec{x})=sum_{i=0}^n frac{partial v_i(vec{x})}{partial x_i}$$
- the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$nabla timesvec{v}(vec{x})=leftlanglefrac{partial v_y(vec{x})}{partial z}-frac{partial v_z(vec{x})}{partial y},frac{partial v_z(vec{x})}{partial x}-frac{partial v_x(vec{x})}{partial z},frac{partial v_x(vec{x})}{partial y}-frac{partial v_y(vec{x})}{partial x}rightrangle$$
- the Jacobian matrix $frac{dvec{v}(vec{x})}{dvec{x}}=J$, whose entries $J_{ij}$ are given by
$$J_{ij}=frac{partial v_i(vec{x})}{partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $vec{v}$:
begin{align}
vec{a}(vec{x})&=frac{dvec{v}(vec{x})}{dt}\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}frac{partial x_i}{partial t},...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}frac{partial x_i}{partial t}rightrangle\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}v_i,...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}v_irightrangle\
&=(vec{v}cdotnabla)vec{v}
end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
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Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
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– π times e
Jan 26 at 4:38
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The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$:
$$
a = frac{dv}{dt} = frac{dv}{dx} frac{dx}{dt} = frac{dv}{dx} v.
$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $vec{v}$ be a function of $x$, $y$, and $z$, since $vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have
$$
a_x = frac{dv_x}{dt} = frac{partial v_x}{partial x} frac{dx}{dt} + frac{partial v_x}{partial y} frac{dy}{dt} + frac{partial v_x}{partial z} frac{dz}{dt} \= frac{partial v_x}{partial x} v_x + frac{partial v_x}{partial y} v_y + frac{partial v_x}{partial z} v_z.
$$
As you can see, we're never "dividing by" the entire vector $vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
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Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
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– π times e
Jan 26 at 4:29
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I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
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– π times e
Jan 26 at 4:31
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@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
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– Michael Seifert
Jan 26 at 14:42
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Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
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– π times e
Jan 26 at 14:51
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I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
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– π times e
Jan 26 at 14:53
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Well, we know how to multiply vectors, via the dot product $vec a cdot vec b$. So we could define a division through:
$$
frac{vec a}{vec b} doteqdot vec a cdot frac{1}{vec b}.
$$
Now we need to define what is meant by $ frac{1}{vec b}$. Well, what about:
$$
frac{1}{vec b} doteqdot frac{vec b}{|vec b|^2}.
$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$
frac{vec a}{vec b} = frac{vec a cdot vec b}{|vec b|^2} = frac{|a|}{|b|}costheta.
$$
This division satisfies what you would expect:
$$
frac{vec a}{vec a} = 1,
$$
but different from usual division, the inverse of $vec a$ isn't unique.
answered Jan 25 at 16:22
TotofofoTotofofo
565
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Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
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– Paul Sinclair
Jan 25 at 17:59
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First off, a key thing to know is that $frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $frac{partial y}{partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $nablavec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
- There is exactly 1 x such that A = Bx
- There is exactly 1 y such that A = yB
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
- 1 dimension is the "normal" division we are used to.
- 2 dimensions is complex division -- division over complex numbers
- 4 dimensions is quaternion division -- division over quaternions (which is very convenient for working with rotations, it turns out)
- 8 dimensions is octonion division
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
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$dy/dx$ is the limit of a series of fractions where one number is divided by another.
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– John McVirgo
Jan 25 at 23:30
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@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
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– Cort Ammon
Jan 26 at 0:04
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While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
- In 2 dimensions ($mathbb R^2$), we could treat any vector $(x,y)$ as a complex number $x+iy$ where $i=sqrt{-1}$, and then use ordinary division of complex numbers (see Cort Ammon's answer, which beat me to this idea)
- In 3 dimensions ($mathbb R^3$) we could use the vector cross product. Define $vec v/vec w$ as the vector $vec p$ such that $vec ptimesvec w=vec v$. Hmmm, but on second thoughts this requires $vec v$ to be orthogonal to $vec w$ so does not seem very useful. If $vec v=vec 0$ the result is not unique as any vector parallel to $vec w$ will suffice. (Perhaps this approach could be generalised by taking only the component of $vec v$ which is orthogonal to $vec w$?)
- In $n$ dimensions, Totofofo's answer defines $vec w^{-1}$ as the vector pointing in the same direction but with inverse magnitude, and then uses the dot product $vec vcdotvec w^{-1}$ to return a number. This requires $vec wnevec 0$.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
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A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
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A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where
s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
answered Jan 25 at 17:03
bobbob
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This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
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– probably_someone
Jan 25 at 17:19
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True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
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– bob
Jan 25 at 17:25
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I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
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– enumaris
Jan 25 at 18:01
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Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
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– bob
Jan 25 at 18:07
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@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
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– enumaris
Jan 25 at 18:10
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7 Answers
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Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=vfrac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $vec{a}=vec{v}frac{dvec{v}}{dvec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $vec{v}(vec{x})$, there are four ways to define differentiation:
- differentiation by a scalar parameter $t$, which is defined for an $n$-dimensional vector as:
$$frac{dvec{v}(vec{x},t)}{dt}=leftlangle frac{d v_1(vec{x},t)}{d t},...,frac{d v_n(vec{x},t)}{d t}rightrangle$$
- the divergence, which is defined for an $n$-dimensional vector as:
$$nabla cdot vec{v}(vec{x})=sum_{i=0}^n frac{partial v_i(vec{x})}{partial x_i}$$
- the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$nabla timesvec{v}(vec{x})=leftlanglefrac{partial v_y(vec{x})}{partial z}-frac{partial v_z(vec{x})}{partial y},frac{partial v_z(vec{x})}{partial x}-frac{partial v_x(vec{x})}{partial z},frac{partial v_x(vec{x})}{partial y}-frac{partial v_y(vec{x})}{partial x}rightrangle$$
- the Jacobian matrix $frac{dvec{v}(vec{x})}{dvec{x}}=J$, whose entries $J_{ij}$ are given by
$$J_{ij}=frac{partial v_i(vec{x})}{partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $vec{v}$:
begin{align}
vec{a}(vec{x})&=frac{dvec{v}(vec{x})}{dt}\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}frac{partial x_i}{partial t},...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}frac{partial x_i}{partial t}rightrangle\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}v_i,...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}v_irightrangle\
&=(vec{v}cdotnabla)vec{v}
end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
$endgroup$
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
add a comment |
$begingroup$
Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=vfrac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $vec{a}=vec{v}frac{dvec{v}}{dvec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $vec{v}(vec{x})$, there are four ways to define differentiation:
- differentiation by a scalar parameter $t$, which is defined for an $n$-dimensional vector as:
$$frac{dvec{v}(vec{x},t)}{dt}=leftlangle frac{d v_1(vec{x},t)}{d t},...,frac{d v_n(vec{x},t)}{d t}rightrangle$$
- the divergence, which is defined for an $n$-dimensional vector as:
$$nabla cdot vec{v}(vec{x})=sum_{i=0}^n frac{partial v_i(vec{x})}{partial x_i}$$
- the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$nabla timesvec{v}(vec{x})=leftlanglefrac{partial v_y(vec{x})}{partial z}-frac{partial v_z(vec{x})}{partial y},frac{partial v_z(vec{x})}{partial x}-frac{partial v_x(vec{x})}{partial z},frac{partial v_x(vec{x})}{partial y}-frac{partial v_y(vec{x})}{partial x}rightrangle$$
- the Jacobian matrix $frac{dvec{v}(vec{x})}{dvec{x}}=J$, whose entries $J_{ij}$ are given by
$$J_{ij}=frac{partial v_i(vec{x})}{partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $vec{v}$:
begin{align}
vec{a}(vec{x})&=frac{dvec{v}(vec{x})}{dt}\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}frac{partial x_i}{partial t},...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}frac{partial x_i}{partial t}rightrangle\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}v_i,...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}v_irightrangle\
&=(vec{v}cdotnabla)vec{v}
end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
$endgroup$
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
add a comment |
$begingroup$
Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=vfrac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $vec{a}=vec{v}frac{dvec{v}}{dvec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $vec{v}(vec{x})$, there are four ways to define differentiation:
- differentiation by a scalar parameter $t$, which is defined for an $n$-dimensional vector as:
$$frac{dvec{v}(vec{x},t)}{dt}=leftlangle frac{d v_1(vec{x},t)}{d t},...,frac{d v_n(vec{x},t)}{d t}rightrangle$$
- the divergence, which is defined for an $n$-dimensional vector as:
$$nabla cdot vec{v}(vec{x})=sum_{i=0}^n frac{partial v_i(vec{x})}{partial x_i}$$
- the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$nabla timesvec{v}(vec{x})=leftlanglefrac{partial v_y(vec{x})}{partial z}-frac{partial v_z(vec{x})}{partial y},frac{partial v_z(vec{x})}{partial x}-frac{partial v_x(vec{x})}{partial z},frac{partial v_x(vec{x})}{partial y}-frac{partial v_y(vec{x})}{partial x}rightrangle$$
- the Jacobian matrix $frac{dvec{v}(vec{x})}{dvec{x}}=J$, whose entries $J_{ij}$ are given by
$$J_{ij}=frac{partial v_i(vec{x})}{partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $vec{v}$:
begin{align}
vec{a}(vec{x})&=frac{dvec{v}(vec{x})}{dt}\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}frac{partial x_i}{partial t},...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}frac{partial x_i}{partial t}rightrangle\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}v_i,...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}v_irightrangle\
&=(vec{v}cdotnabla)vec{v}
end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
$endgroup$
Your expression doesn't make sense for (non-trivial) vectors, only for scalars (which are one-dimensional, a.k.a. "trivial" vectors).
The expression $a=vfrac{dv}{dx}$ only makes sense if $v$ and $x$ are both functions from one dimension to one dimension. You will never see the expression $vec{a}=vec{v}frac{dvec{v}}{dvec{x}}$, because such an operation is not defined.
In general, multiplication of vectors is only well-defined in specific circumstances. There are three commonly-used multiplications of vectors: the dot product (which returns a scalar), the cross product (which returns a vector), and the tensor product (which returns a matrix). The cross product, in addition, only makes sense for three-dimensional and seven-dimensional vectors (the reason for this has to do with the existence of certain extensions of complex numbers called the quaternions and the octonions).
Division of two vectors is in general not defined, because it's not possible to undo either type of vector multiplication. For example, for any given vector, there are an infinite number of other vectors whose dot product with that vector will be zero (namely, all vectors that are perpendicular to it), and similarly, there are an infinite number of other vectors whose cross product with that vector will be zero (namely, all vectors that are parallel to it). Defining an inverse would require a one-to-one correspondence between the input and output of this multiplication, so defining division is not strictly possible.
That said, defining differentiation of vectors is different, namely because differentiation does not involve dividing two vectors. For a vector which is a function of space $vec{v}(vec{x})$, there are four ways to define differentiation:
- differentiation by a scalar parameter $t$, which is defined for an $n$-dimensional vector as:
$$frac{dvec{v}(vec{x},t)}{dt}=leftlangle frac{d v_1(vec{x},t)}{d t},...,frac{d v_n(vec{x},t)}{d t}rightrangle$$
- the divergence, which is defined for an $n$-dimensional vector as:
$$nabla cdot vec{v}(vec{x})=sum_{i=0}^n frac{partial v_i(vec{x})}{partial x_i}$$
- the curl, which is defined for three-dimensional and seven-dimensional vectors, and, in three dimensions, in Cartesian coordinates, is:
$$nabla timesvec{v}(vec{x})=leftlanglefrac{partial v_y(vec{x})}{partial z}-frac{partial v_z(vec{x})}{partial y},frac{partial v_z(vec{x})}{partial x}-frac{partial v_x(vec{x})}{partial z},frac{partial v_x(vec{x})}{partial y}-frac{partial v_y(vec{x})}{partial x}rightrangle$$
- the Jacobian matrix $frac{dvec{v}(vec{x})}{dvec{x}}=J$, whose entries $J_{ij}$ are given by
$$J_{ij}=frac{partial v_i(vec{x})}{partial x_j}$$
As you can see, none of these involve dividing a vector by a vector; in fact, all of them involve multiplying a vector by another vector, namely, the vector of partial derivative operators $nabla$. For the divergence, this is a dot product; for the curl, this is a cross product; and for the Jacobian matrix, this is a tensor product.
So how do we rework the one-dimensional expression for acceleration so that it makes sense in any number of dimensions? The key is to start with the basic definition, assuming, as you did, that there is no explicit time dependence for $vec{v}$:
begin{align}
vec{a}(vec{x})&=frac{dvec{v}(vec{x})}{dt}\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}frac{partial x_i}{partial t},...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}frac{partial x_i}{partial t}rightrangle\
&=leftlangle sum_{i=1}^n frac{partial v_1(vec{x})}{partial x_i}v_i,...,sum_{i=1}^nfrac{partial v_n(vec{x})}{partial x_i}v_irightrangle\
&=(vec{v}cdotnabla)vec{v}
end{align}
There is nothing wrong with differentiating a vector, since it doesn't involve division by a vector.
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
answered Jan 25 at 15:40
probably_someoneprobably_someone
18.1k12959
18.1k12959
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
add a comment |
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
$begingroup$
Thanks a lot for such a thorough explanation. I'm at school so I can't understand the math part of your explanation. But I think I understood the gist. Well I've a question : The formula a = vdv/dx shows that acceleration is variable, what that means is, it is continuously changing (different values at different positions). Does it mean the the acceleration (and position) is only changing in magnitude and not in direction? You have stated in your explanation that acceleration in this form holds only for one dimensional motion. I don't want to create another thread for it, because it's related
$endgroup$
– π times e
Jan 26 at 4:38
add a comment |
$begingroup$
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$:
$$
a = frac{dv}{dt} = frac{dv}{dx} frac{dx}{dt} = frac{dv}{dx} v.
$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $vec{v}$ be a function of $x$, $y$, and $z$, since $vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have
$$
a_x = frac{dv_x}{dt} = frac{partial v_x}{partial x} frac{dx}{dt} + frac{partial v_x}{partial y} frac{dy}{dt} + frac{partial v_x}{partial z} frac{dz}{dt} \= frac{partial v_x}{partial x} v_x + frac{partial v_x}{partial y} v_y + frac{partial v_x}{partial z} v_z.
$$
As you can see, we're never "dividing by" the entire vector $vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
$endgroup$
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$:
$$
a = frac{dv}{dt} = frac{dv}{dx} frac{dx}{dt} = frac{dv}{dx} v.
$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $vec{v}$ be a function of $x$, $y$, and $z$, since $vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have
$$
a_x = frac{dv_x}{dt} = frac{partial v_x}{partial x} frac{dx}{dt} + frac{partial v_x}{partial y} frac{dy}{dt} + frac{partial v_x}{partial z} frac{dz}{dt} \= frac{partial v_x}{partial x} v_x + frac{partial v_x}{partial y} v_y + frac{partial v_x}{partial z} v_z.
$$
As you can see, we're never "dividing by" the entire vector $vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
$endgroup$
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$:
$$
a = frac{dv}{dt} = frac{dv}{dx} frac{dx}{dt} = frac{dv}{dx} v.
$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $vec{v}$ be a function of $x$, $y$, and $z$, since $vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have
$$
a_x = frac{dv_x}{dt} = frac{partial v_x}{partial x} frac{dx}{dt} + frac{partial v_x}{partial y} frac{dy}{dt} + frac{partial v_x}{partial z} frac{dz}{dt} \= frac{partial v_x}{partial x} v_x + frac{partial v_x}{partial y} v_y + frac{partial v_x}{partial z} v_z.
$$
As you can see, we're never "dividing by" the entire vector $vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
$endgroup$
The statement $a = v (dv/dx)$ only holds in that form for one-dimensional motion, where the quantities $v$ and $x$ are just numbers rather than vectors. It follows from the chain rule, if we view $v$ as a function of $x$ instead of as a function of $t$:
$$
a = frac{dv}{dt} = frac{dv}{dx} frac{dx}{dt} = frac{dv}{dx} v.
$$
If you're doing 2-D or 3-D motion, you can still do something similar, but you have to let $vec{v}$ be a function of $x$, $y$, and $z$, since $vec{v}$ can change as each of these quantities changes. This means that you need to use multi-variable calculus to write out an equivalent statement. For example, we have
$$
a_x = frac{dv_x}{dt} = frac{partial v_x}{partial x} frac{dx}{dt} + frac{partial v_x}{partial y} frac{dy}{dt} + frac{partial v_x}{partial z} frac{dz}{dt} \= frac{partial v_x}{partial x} v_x + frac{partial v_x}{partial y} v_y + frac{partial v_x}{partial z} v_z.
$$
As you can see, we're never "dividing by" the entire vector $vec{x}$ when we take these derivatives; we only ever "divide by" its components $x$, $y$, or $z$, which is a valid mathematical operation.
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
answered Jan 25 at 14:54
Michael SeifertMichael Seifert
15.6k22858
15.6k22858
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
Thanks. I'm still at school so I didn't understand the last part. But I understood that the formula a = vdv/dx holds for one-dimensional motion. In case of an n-dimentional motion, I have to treat it as n 1-dimentional motion.
$endgroup$
– π times e
Jan 26 at 4:29
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
I've another question. The formula a = vdv/dx clearly shows that acceleration is variable, means it is continuously changing (different values at different positions). Does it mean the the acceleration (and position as well) is only changing in magnitude and not in direction? Because you said earlier that acceleration in this form holds only for one dimensional motion?
$endgroup$
– π times e
Jan 26 at 4:31
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
@πtimese: We're only talking about 1D motion here. However, the sign of $a$ can be positive or negative, depending on the signs of $v$ and $dv/dx$; and this might qualify as "changing direction". Does that answer your question?
$endgroup$
– Michael Seifert
Jan 26 at 14:42
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
Sorry, I should’ve been more specific. Yeah, what I meant to ask is, does acceleration in this form (a = vdv/dx) mean only its magnitude is changing, and its direction is not changing? Other than the fact that its direction can reverse?
$endgroup$
– π times e
Jan 26 at 14:51
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
$begingroup$
I mean, if the body is going north, either it continues to move north, or it stops and starts moving south. It can’t go east, or west, or at an angle, right?
$endgroup$
– π times e
Jan 26 at 14:53
|
show 2 more comments
$begingroup$
Well, we know how to multiply vectors, via the dot product $vec a cdot vec b$. So we could define a division through:
$$
frac{vec a}{vec b} doteqdot vec a cdot frac{1}{vec b}.
$$
Now we need to define what is meant by $ frac{1}{vec b}$. Well, what about:
$$
frac{1}{vec b} doteqdot frac{vec b}{|vec b|^2}.
$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$
frac{vec a}{vec b} = frac{vec a cdot vec b}{|vec b|^2} = frac{|a|}{|b|}costheta.
$$
This division satisfies what you would expect:
$$
frac{vec a}{vec a} = 1,
$$
but different from usual division, the inverse of $vec a$ isn't unique.
answered Jan 25 at 16:22
TotofofoTotofofo
565
$endgroup$
1
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
add a comment |
$begingroup$
Well, we know how to multiply vectors, via the dot product $vec a cdot vec b$. So we could define a division through:
$$
frac{vec a}{vec b} doteqdot vec a cdot frac{1}{vec b}.
$$
Now we need to define what is meant by $ frac{1}{vec b}$. Well, what about:
$$
frac{1}{vec b} doteqdot frac{vec b}{|vec b|^2}.
$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$
frac{vec a}{vec b} = frac{vec a cdot vec b}{|vec b|^2} = frac{|a|}{|b|}costheta.
$$
This division satisfies what you would expect:
$$
frac{vec a}{vec a} = 1,
$$
but different from usual division, the inverse of $vec a$ isn't unique.
answered Jan 25 at 16:22
TotofofoTotofofo
565
$endgroup$
1
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
add a comment |
$begingroup$
Well, we know how to multiply vectors, via the dot product $vec a cdot vec b$. So we could define a division through:
$$
frac{vec a}{vec b} doteqdot vec a cdot frac{1}{vec b}.
$$
Now we need to define what is meant by $ frac{1}{vec b}$. Well, what about:
$$
frac{1}{vec b} doteqdot frac{vec b}{|vec b|^2}.
$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$
frac{vec a}{vec b} = frac{vec a cdot vec b}{|vec b|^2} = frac{|a|}{|b|}costheta.
$$
This division satisfies what you would expect:
$$
frac{vec a}{vec a} = 1,
$$
but different from usual division, the inverse of $vec a$ isn't unique.
answered Jan 25 at 16:22
TotofofoTotofofo
565
$endgroup$
Well, we know how to multiply vectors, via the dot product $vec a cdot vec b$. So we could define a division through:
$$
frac{vec a}{vec b} doteqdot vec a cdot frac{1}{vec b}.
$$
Now we need to define what is meant by $ frac{1}{vec b}$. Well, what about:
$$
frac{1}{vec b} doteqdot frac{vec b}{|vec b|^2}.
$$
This vector inversion gives a new vector with same direction, but reciprocal magnitude. In fact it's how we define an inversion for other things like the special conformal transformation. Seems like as good a definition as any. Then division of vectors becomes:
$$
frac{vec a}{vec b} = frac{vec a cdot vec b}{|vec b|^2} = frac{|a|}{|b|}costheta.
$$
This division satisfies what you would expect:
$$
frac{vec a}{vec a} = 1,
$$
but different from usual division, the inverse of $vec a$ isn't unique.
answered Jan 25 at 16:22
TotofofoTotofofo
565
answered Jan 25 at 16:22
TotofofoTotofofo
565
answered Jan 25 at 16:22
TotofofoTotofofo
565
answered Jan 25 at 16:22
TotofofoTotofofo
565
565
1
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
add a comment |
1
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
1
1
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
$begingroup$
Since you just defined a specific formula for the "inverse" for $vec b$, it is a little confusing to say later on that the inverse of $vec a$ isn't unique. It might be clearer to point out instead that $vec v = dfrac 1{vec a}$ is not the only solution to $vec a cdot vec v = 1$.
$endgroup$
– Paul Sinclair
Jan 25 at 17:59
add a comment |
$begingroup$
First off, a key thing to know is that $frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $frac{partial y}{partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $nablavec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
- There is exactly 1 x such that A = Bx
- There is exactly 1 y such that A = yB
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
- 1 dimension is the "normal" division we are used to.
- 2 dimensions is complex division -- division over complex numbers
- 4 dimensions is quaternion division -- division over quaternions (which is very convenient for working with rotations, it turns out)
- 8 dimensions is octonion division
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
$endgroup$
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
add a comment |
$begingroup$
First off, a key thing to know is that $frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $frac{partial y}{partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $nablavec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
- There is exactly 1 x such that A = Bx
- There is exactly 1 y such that A = yB
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
- 1 dimension is the "normal" division we are used to.
- 2 dimensions is complex division -- division over complex numbers
- 4 dimensions is quaternion division -- division over quaternions (which is very convenient for working with rotations, it turns out)
- 8 dimensions is octonion division
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
$endgroup$
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
add a comment |
$begingroup$
First off, a key thing to know is that $frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $frac{partial y}{partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $nablavec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
- There is exactly 1 x such that A = Bx
- There is exactly 1 y such that A = yB
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
- 1 dimension is the "normal" division we are used to.
- 2 dimensions is complex division -- division over complex numbers
- 4 dimensions is quaternion division -- division over quaternions (which is very convenient for working with rotations, it turns out)
- 8 dimensions is octonion division
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
$endgroup$
First off, a key thing to know is that $frac{dy}{dx}$ is not actually a division. It is the symbol representing taking the derivative of y with respect to x. The symbol looks like a division because that proved convenient. There's a remarkable number of cases where if you treat it as if it were a division, you get the right answer. However, it does not always work that way, so you can't actually think of it as a division.
As others have pointed out, in multivariable cases, we actually switch the symbol to $frac{partial y}{partial x}$, reminding ourselves that the multi-variable case has a few extra things you have to track. You can't quite use the same old equations. We'll also use symbols liks $nablavec v$ to capture concepts like that.
There is, however, a side conversation that I find very fascinating which is that of real division algebras. These are operations over fields of real numbers (where +, - and * do the operations you expect) where we can meaningfully define division. Mathematicians are very pedantic about this, so they define that to mean that for any A and non-zero B
- There is exactly 1 x such that A = Bx
- There is exactly 1 y such that A = yB
One can see that any reasonable meaning of division will have to be layered on top of such a pattern.
So if we consider these real fields, like (x), (x, y), (x, y, z), (x, y, z, w), etc. we can try to find ways to define them such that those two division rules make sense. As it turns out, due to a rather clever little theorem, division can only work in 1, 2, 4, and 8 dimensional cases (or infinite dimensional... but that's another story entierly).
- 1 dimension is the "normal" division we are used to.
- 2 dimensions is complex division -- division over complex numbers
- 4 dimensions is quaternion division -- division over quaternions (which is very convenient for working with rotations, it turns out)
- 8 dimensions is octonion division
And that's it. Those are literally the only finite-dimensional real fields which can admit division.
This means your 3d vector not only doesn't have a division operator, but in fact there is no possible way 3 real numbers can be put together to create a meaningful division algebra. It actually cannot be done!
You can show that you can define a 2d vector division, but that it must be the same (isometric, technically) as complex division, so you can ask yourself whether it would make sense to think of your vector as a complex number or not. If it doesn't make sense, then there wont be a way to make division make sense either.
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
answered Jan 25 at 19:25
Cort AmmonCort Ammon
23.6k34779
23.6k34779
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
add a comment |
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
$dy/dx$ is the limit of a series of fractions where one number is divided by another.
$endgroup$
– John McVirgo
Jan 25 at 23:30
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
$begingroup$
@JohnMcVirgo Yes, indeed, it is the limit of a series of fractions. And the joy of Libnitz' notation is that you can treat it like division in many cases, and it will happen to work out. However, later on, this will fail spectacularly, especially in multivariable calculus.
$endgroup$
– Cort Ammon
Jan 26 at 0:04
add a comment |
$begingroup$
While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
- In 2 dimensions ($mathbb R^2$), we could treat any vector $(x,y)$ as a complex number $x+iy$ where $i=sqrt{-1}$, and then use ordinary division of complex numbers (see Cort Ammon's answer, which beat me to this idea)
- In 3 dimensions ($mathbb R^3$) we could use the vector cross product. Define $vec v/vec w$ as the vector $vec p$ such that $vec ptimesvec w=vec v$. Hmmm, but on second thoughts this requires $vec v$ to be orthogonal to $vec w$ so does not seem very useful. If $vec v=vec 0$ the result is not unique as any vector parallel to $vec w$ will suffice. (Perhaps this approach could be generalised by taking only the component of $vec v$ which is orthogonal to $vec w$?)
- In $n$ dimensions, Totofofo's answer defines $vec w^{-1}$ as the vector pointing in the same direction but with inverse magnitude, and then uses the dot product $vec vcdotvec w^{-1}$ to return a number. This requires $vec wnevec 0$.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
$endgroup$
add a comment |
$begingroup$
While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
- In 2 dimensions ($mathbb R^2$), we could treat any vector $(x,y)$ as a complex number $x+iy$ where $i=sqrt{-1}$, and then use ordinary division of complex numbers (see Cort Ammon's answer, which beat me to this idea)
- In 3 dimensions ($mathbb R^3$) we could use the vector cross product. Define $vec v/vec w$ as the vector $vec p$ such that $vec ptimesvec w=vec v$. Hmmm, but on second thoughts this requires $vec v$ to be orthogonal to $vec w$ so does not seem very useful. If $vec v=vec 0$ the result is not unique as any vector parallel to $vec w$ will suffice. (Perhaps this approach could be generalised by taking only the component of $vec v$ which is orthogonal to $vec w$?)
- In $n$ dimensions, Totofofo's answer defines $vec w^{-1}$ as the vector pointing in the same direction but with inverse magnitude, and then uses the dot product $vec vcdotvec w^{-1}$ to return a number. This requires $vec wnevec 0$.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
$endgroup$
add a comment |
$begingroup$
While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
- In 2 dimensions ($mathbb R^2$), we could treat any vector $(x,y)$ as a complex number $x+iy$ where $i=sqrt{-1}$, and then use ordinary division of complex numbers (see Cort Ammon's answer, which beat me to this idea)
- In 3 dimensions ($mathbb R^3$) we could use the vector cross product. Define $vec v/vec w$ as the vector $vec p$ such that $vec ptimesvec w=vec v$. Hmmm, but on second thoughts this requires $vec v$ to be orthogonal to $vec w$ so does not seem very useful. If $vec v=vec 0$ the result is not unique as any vector parallel to $vec w$ will suffice. (Perhaps this approach could be generalised by taking only the component of $vec v$ which is orthogonal to $vec w$?)
- In $n$ dimensions, Totofofo's answer defines $vec w^{-1}$ as the vector pointing in the same direction but with inverse magnitude, and then uses the dot product $vec vcdotvec w^{-1}$ to return a number. This requires $vec wnevec 0$.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
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While I concur there is no way to divide vectors, generally speaking, it is fun to think of creative possibilities.
- In 2 dimensions ($mathbb R^2$), we could treat any vector $(x,y)$ as a complex number $x+iy$ where $i=sqrt{-1}$, and then use ordinary division of complex numbers (see Cort Ammon's answer, which beat me to this idea)
- In 3 dimensions ($mathbb R^3$) we could use the vector cross product. Define $vec v/vec w$ as the vector $vec p$ such that $vec ptimesvec w=vec v$. Hmmm, but on second thoughts this requires $vec v$ to be orthogonal to $vec w$ so does not seem very useful. If $vec v=vec 0$ the result is not unique as any vector parallel to $vec w$ will suffice. (Perhaps this approach could be generalised by taking only the component of $vec v$ which is orthogonal to $vec w$?)
- In $n$ dimensions, Totofofo's answer defines $vec w^{-1}$ as the vector pointing in the same direction but with inverse magnitude, and then uses the dot product $vec vcdotvec w^{-1}$ to return a number. This requires $vec wnevec 0$.
The concepts of inverses and uniqueness etc are studied in "abstract algebra", an area of mathematics often introduced at 2nd year university level. For example a structure with some sort of "addition" and "multiplication" is called a ring. A nicer type of ring where every non-zero element has an inverse is called a field. (This is very imprecise, I encourage people to look up the specific details.) Our first example is a field. Example 2 does not have a multiplicative identity.
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
answered Jan 30 at 1:41
Colin MacLaurinColin MacLaurin
417111
417111
add a comment |
add a comment |
$begingroup$
A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
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add a comment |
$begingroup$
A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
$endgroup$
add a comment |
$begingroup$
A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
$endgroup$
A simple-minded way of seeing this is as follows: consider a linear function A that changes a vector x into another vector y: y = Ax. If you "divide by x", you get: A = y/x. So, we could say that the result of dividing vector y by vector x is a more complicated object, i.e., a tensor: the matrix A.
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
answered Jan 30 at 12:26
Raul SimonRaul Simon
11
11
add a comment |
add a comment |
$begingroup$
A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where
s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
answered Jan 25 at 17:03
bobbob
1216
$endgroup$
1
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This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
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– probably_someone
Jan 25 at 17:19
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True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
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– bob
Jan 25 at 17:25
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I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
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– enumaris
Jan 25 at 18:01
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Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
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– bob
Jan 25 at 18:07
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
|
show 1 more comment
$begingroup$
A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where
s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
answered Jan 25 at 17:03
bobbob
1216
$endgroup$
1
$begingroup$
This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
$endgroup$
– probably_someone
Jan 25 at 17:19
$begingroup$
True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
$endgroup$
– bob
Jan 25 at 17:25
$begingroup$
I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
$endgroup$
– enumaris
Jan 25 at 18:01
$begingroup$
Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
$endgroup$
– bob
Jan 25 at 18:07
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
|
show 1 more comment
$begingroup$
A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where
s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
answered Jan 25 at 17:03
bobbob
1216
$endgroup$
A natural way to define vector division is as the inverse of vector-scalar multiplication. Since you can multiply a scalar s by a vector v to obtain a vector w, where
s * v = w
...then you could define w/v = s.
It's not earth-shattering, but could be useful in some contexts. It would have whatever physical interpretation that the inverse of vector-scalar multiplication would have.
answered Jan 25 at 17:03
bobbob
1216
answered Jan 25 at 17:03
bobbob
1216
answered Jan 25 at 17:03
bobbob
1216
answered Jan 25 at 17:03
bobbob
1216
1216
1
$begingroup$
This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
$endgroup$
– probably_someone
Jan 25 at 17:19
$begingroup$
True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
$endgroup$
– bob
Jan 25 at 17:25
$begingroup$
I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
$endgroup$
– enumaris
Jan 25 at 18:01
$begingroup$
Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
$endgroup$
– bob
Jan 25 at 18:07
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
|
show 1 more comment
1
$begingroup$
This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
$endgroup$
– probably_someone
Jan 25 at 17:19
$begingroup$
True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
$endgroup$
– bob
Jan 25 at 17:25
$begingroup$
I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
$endgroup$
– enumaris
Jan 25 at 18:01
$begingroup$
Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
$endgroup$
– bob
Jan 25 at 18:07
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
1
1
$begingroup$
This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
$endgroup$
– probably_someone
Jan 25 at 17:19
$begingroup$
This operation is not defined for vectors which aren't parallel. If all of the vectors you're working with are parallel, then your problem is effectively one-dimensional and you should just use scalars anyway.
$endgroup$
– probably_someone
Jan 25 at 17:19
$begingroup$
True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
$endgroup$
– bob
Jan 25 at 17:25
$begingroup$
True, and probably why I've never seen anyone do this. I guess if for some reason you wanted to solve for the scalar ratio between two vectors? It's really a stretch. Unless you relax the definition somehow to allow an error term so that becomes some measure of how parallel two vectors are. That could be interesting (and probably exists).
$endgroup$
– bob
Jan 25 at 17:25
$begingroup$
I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
$endgroup$
– enumaris
Jan 25 at 18:01
$begingroup$
I have downvoted this answer because I believe it serves no purpose other than to confuse the reader.
$endgroup$
– enumaris
Jan 25 at 18:01
$begingroup$
Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
$endgroup$
– bob
Jan 25 at 18:07
$begingroup$
Huh. Thanks for giving a reason, I appreciate that. Certainly the purpose isn't to confuse the reader, but you mean that it serves no purpose and is just confusing? I would agree that it serves pretty much no purpose, and does require additional thought to unpack, so I can't argue strongly against that. I guess it was just a response to the teacher that technically she is wrong on both counts. Which I think has some value because teachers can often be dogmatic in the vein of Lies to Children, and as a result they can wind up confusing or stifling some of the most creative students.
$endgroup$
– bob
Jan 25 at 18:07
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
$begingroup$
@bob right, I am certainly not suggesting that it was your intention to confuse the reader, only that the answer, as it stands, serves no other purpose. So yes, it can also be rephrased as "the answer serves no purpose and is just confusing". EDIT: I saw your edit, and I don't agree with it. The "answer" you provided does not show that the teacher is wrong. Nobody defines vector division the way you do since it's not useful. There are many other answers/comments in this thread showing why vector division is not well defined.
$endgroup$
– enumaris
Jan 25 at 18:10
|
show 1 more comment
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$begingroup$
Related: physics.stackexchange.com/q/111652 However, the argument specific to your example of acceleration is somewhat different, so I believe it isn't a duplicate.
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– Chair
Jan 25 at 14:53
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Not if you want the result of the division to be unique. For example $(3,0)cdot (2,4)=(1,1)cdot (2,4)=6$ but $(3,0)neq (1,1)$.
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– BPP
Jan 25 at 16:12
1
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Related: physics.stackexchange.com/q/14082/2451
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– Qmechanic♦
Jan 25 at 18:23
3
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Please note that $dx/dy$ is not a fraction, but a rather unfortunate way of writing $frac{d}{dx}y$
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– Michael Freimann
Jan 25 at 18:29
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@Chair I visited that thread before I posted my question. It didn't help me, so I decided to go ahead with it.
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– π times e
Jan 26 at 4:33