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Explanation for that for a particle, $v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$ $l$ is the arch that the particle...
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In the book of Mechanics by Landau, at page 8, it is claimed that
$$v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$$
where $v$ is a the velocity of the particle in the cartesian
coordinates, $l$ is the arch that the particle traces out.
I know that this is completely a abus d’ notation; however, is there any intuitive way of seeing this ?
analysis physics mathematical-physics classical-mechanics euler-lagrange-equation
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
$endgroup$
add a comment |
$begingroup$
In the book of Mechanics by Landau, at page 8, it is claimed that
$$v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$$
where $v$ is a the velocity of the particle in the cartesian
coordinates, $l$ is the arch that the particle traces out.
I know that this is completely a abus d’ notation; however, is there any intuitive way of seeing this ?
analysis physics mathematical-physics classical-mechanics euler-lagrange-equation
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
$endgroup$
$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46
add a comment |
$begingroup$
In the book of Mechanics by Landau, at page 8, it is claimed that
$$v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$$
where $v$ is a the velocity of the particle in the cartesian
coordinates, $l$ is the arch that the particle traces out.
I know that this is completely a abus d’ notation; however, is there any intuitive way of seeing this ?
analysis physics mathematical-physics classical-mechanics euler-lagrange-equation
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
$endgroup$
In the book of Mechanics by Landau, at page 8, it is claimed that
$$v^2 = (dl / dt)^2 = (dl)^2 / (dt)^2,$$
where $v$ is a the velocity of the particle in the cartesian
coordinates, $l$ is the arch that the particle traces out.
I know that this is completely a abus d’ notation; however, is there any intuitive way of seeing this ?
analysis physics mathematical-physics classical-mechanics euler-lagrange-equation
analysis physics mathematical-physics classical-mechanics euler-lagrange-equation
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
edited Jan 3 at 14:17
user593746
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
asked Nov 23 '18 at 12:27
onurcanbektasonurcanbektas
3,37611036
3,37611036
$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46
add a comment |
$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46
$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46
$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46
add a comment |
3 Answers
3
active
oldest
votes
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I think the notation should be interpreted as follows. First note that
$$v^2 dt^2=v^2(dtcdot dt)=(v dt)cdot (v dt).$$
where $cdot$ is the symmetric product of differential forms. Since $v dt=dl$ you get $$v^2 dt^2=dlcdot dl=dl^2.$$
I would say that writing $v^2=frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
$endgroup$
add a comment |
$begingroup$
Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds).
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{mathfrak x}(t)={frak x}(t+dt)-{frak x}(t)$ and the length of the infinitesimal segment as $dl = |d{mathfrak x}|=sqrt{dx^2+dy^2+dz^2}$.
Today we would write $Delta t,Δ{frak x}$ instead of $dt,d{frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
$endgroup$
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
add a comment |
$begingroup$
It works because the multivariable chain rule and the total differential works the same way.
The velocity is defined as
$$vec{v}=frac{mathrm{d} vec{r}}{mathrm{d}t}$$
If we parametrize the position with $vec{q}$, we get that
$$r_i(t)=x_i(vec{q}(t))$$
And with the multivariable chain rule:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$mathrm{d}vec{r}=frac{partial vec{r}}{partial q_i} mathrm{d}q_i$$
And if you formally divide it by $mathrm{d}t$, you get the same result:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
answered Jan 3 at 14:54
BotondBotond
5,6282732
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the notation should be interpreted as follows. First note that
$$v^2 dt^2=v^2(dtcdot dt)=(v dt)cdot (v dt).$$
where $cdot$ is the symmetric product of differential forms. Since $v dt=dl$ you get $$v^2 dt^2=dlcdot dl=dl^2.$$
I would say that writing $v^2=frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
$endgroup$
add a comment |
$begingroup$
I think the notation should be interpreted as follows. First note that
$$v^2 dt^2=v^2(dtcdot dt)=(v dt)cdot (v dt).$$
where $cdot$ is the symmetric product of differential forms. Since $v dt=dl$ you get $$v^2 dt^2=dlcdot dl=dl^2.$$
I would say that writing $v^2=frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
$endgroup$
add a comment |
$begingroup$
I think the notation should be interpreted as follows. First note that
$$v^2 dt^2=v^2(dtcdot dt)=(v dt)cdot (v dt).$$
where $cdot$ is the symmetric product of differential forms. Since $v dt=dl$ you get $$v^2 dt^2=dlcdot dl=dl^2.$$
I would say that writing $v^2=frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
$endgroup$
I think the notation should be interpreted as follows. First note that
$$v^2 dt^2=v^2(dtcdot dt)=(v dt)cdot (v dt).$$
where $cdot$ is the symmetric product of differential forms. Since $v dt=dl$ you get $$v^2 dt^2=dlcdot dl=dl^2.$$
I would say that writing $v^2=frac{(dl)^2}{(dt)^2}$ does not make mathematical sense, and frankly, this looks horrible.
edited Jan 3 at 14:16
answered Nov 23 '18 at 12:59
user593746
add a comment |
add a comment |
$begingroup$
Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds).
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{mathfrak x}(t)={frak x}(t+dt)-{frak x}(t)$ and the length of the infinitesimal segment as $dl = |d{mathfrak x}|=sqrt{dx^2+dy^2+dz^2}$.
Today we would write $Delta t,Δ{frak x}$ instead of $dt,d{frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
$endgroup$
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
add a comment |
$begingroup$
Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds).
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{mathfrak x}(t)={frak x}(t+dt)-{frak x}(t)$ and the length of the infinitesimal segment as $dl = |d{mathfrak x}|=sqrt{dx^2+dy^2+dz^2}$.
Today we would write $Delta t,Δ{frak x}$ instead of $dt,d{frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
$endgroup$
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
add a comment |
$begingroup$
Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds).
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{mathfrak x}(t)={frak x}(t+dt)-{frak x}(t)$ and the length of the infinitesimal segment as $dl = |d{mathfrak x}|=sqrt{dx^2+dy^2+dz^2}$.
Today we would write $Delta t,Δ{frak x}$ instead of $dt,d{frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
$endgroup$
Physicists and mathematicians of that time (19th, early 20th century) had no problem with the informal use of (formal) infinitesimals (read some papers on differential geometry of the time, for instance by Riemann defining manifolds).
Thus there was no problem in decomposing a curve into a collection of infinitesimal segments and to write $d{mathfrak x}(t)={frak x}(t+dt)-{frak x}(t)$ and the length of the infinitesimal segment as $dl = |d{mathfrak x}|=sqrt{dx^2+dy^2+dz^2}$.
Today we would write $Delta t,Δ{frak x}$ instead of $dt,d{frak x}$ etc., insist there is nothing infinitesimal about them (and even that no such thing exists), and on the finiteness of the number of segments, and embed the whole construction into a limit process where the segment length goes to zero. In the end the result is the same, with a little more overhead.
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
answered Jan 3 at 14:58
LutzLLutzL
57.2k42054
57.2k42054
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
add a comment |
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
$begingroup$
Thanks for the clarification.
$endgroup$
– onurcanbektas
Jan 3 at 17:23
add a comment |
$begingroup$
It works because the multivariable chain rule and the total differential works the same way.
The velocity is defined as
$$vec{v}=frac{mathrm{d} vec{r}}{mathrm{d}t}$$
If we parametrize the position with $vec{q}$, we get that
$$r_i(t)=x_i(vec{q}(t))$$
And with the multivariable chain rule:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$mathrm{d}vec{r}=frac{partial vec{r}}{partial q_i} mathrm{d}q_i$$
And if you formally divide it by $mathrm{d}t$, you get the same result:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
answered Jan 3 at 14:54
BotondBotond
5,6282732
$endgroup$
add a comment |
$begingroup$
It works because the multivariable chain rule and the total differential works the same way.
The velocity is defined as
$$vec{v}=frac{mathrm{d} vec{r}}{mathrm{d}t}$$
If we parametrize the position with $vec{q}$, we get that
$$r_i(t)=x_i(vec{q}(t))$$
And with the multivariable chain rule:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$mathrm{d}vec{r}=frac{partial vec{r}}{partial q_i} mathrm{d}q_i$$
And if you formally divide it by $mathrm{d}t$, you get the same result:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
answered Jan 3 at 14:54
BotondBotond
5,6282732
$endgroup$
add a comment |
$begingroup$
It works because the multivariable chain rule and the total differential works the same way.
The velocity is defined as
$$vec{v}=frac{mathrm{d} vec{r}}{mathrm{d}t}$$
If we parametrize the position with $vec{q}$, we get that
$$r_i(t)=x_i(vec{q}(t))$$
And with the multivariable chain rule:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$mathrm{d}vec{r}=frac{partial vec{r}}{partial q_i} mathrm{d}q_i$$
And if you formally divide it by $mathrm{d}t$, you get the same result:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
answered Jan 3 at 14:54
BotondBotond
5,6282732
$endgroup$
It works because the multivariable chain rule and the total differential works the same way.
The velocity is defined as
$$vec{v}=frac{mathrm{d} vec{r}}{mathrm{d}t}$$
If we parametrize the position with $vec{q}$, we get that
$$r_i(t)=x_i(vec{q}(t))$$
And with the multivariable chain rule:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
On the other hand, in a curvilinear coordinate system the differential element is
$$mathrm{d}vec{r}=frac{partial vec{r}}{partial q_i} mathrm{d}q_i$$
And if you formally divide it by $mathrm{d}t$, you get the same result:
$$vec{v}=frac{partial vec{r}}{partial q_i}frac{mathrm{d}q_i}{mathrm{d}t}$$
answered Jan 3 at 14:54
BotondBotond
5,6282732
answered Jan 3 at 14:54
BotondBotond
5,6282732
answered Jan 3 at 14:54
BotondBotond
5,6282732
answered Jan 3 at 14:54
BotondBotond
5,6282732
5,6282732
add a comment |
add a comment |
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$begingroup$
what is the reason for the down vote ?
$endgroup$
– onurcanbektas
Nov 23 '18 at 15:46