Mixing
Scalar dot product with directional derivative
$begingroup$
I'm dealing with a problem of the form
$phi cdot left[(A cdot nabla)Bright]$
in which $phi$ is a scalar function and both $A$ and $B$ are vector functions.
Can I simply "distribute" the dot product with scalar function inside the expression to give
$ (C cdot nabla)B$
in which $C = phi cdot A$
Is this correct? I often get confused working with vector calculus operators.
On the side...
It seems that the parenthesis are important when reading the wikipedia identities. For example, would it be true that
- $A times (nabla times B) neq (A times nabla) times B$
Is there a "trick" to understanding how to manage the parenthesis?
vector-analysis
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
$endgroup$
add a comment |
$begingroup$
I'm dealing with a problem of the form
$phi cdot left[(A cdot nabla)Bright]$
in which $phi$ is a scalar function and both $A$ and $B$ are vector functions.
Can I simply "distribute" the dot product with scalar function inside the expression to give
$ (C cdot nabla)B$
in which $C = phi cdot A$
Is this correct? I often get confused working with vector calculus operators.
On the side...
It seems that the parenthesis are important when reading the wikipedia identities. For example, would it be true that
- $A times (nabla times B) neq (A times nabla) times B$
Is there a "trick" to understanding how to manage the parenthesis?
vector-analysis
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
$endgroup$
$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32
add a comment |
$begingroup$
I'm dealing with a problem of the form
$phi cdot left[(A cdot nabla)Bright]$
in which $phi$ is a scalar function and both $A$ and $B$ are vector functions.
Can I simply "distribute" the dot product with scalar function inside the expression to give
$ (C cdot nabla)B$
in which $C = phi cdot A$
Is this correct? I often get confused working with vector calculus operators.
On the side...
It seems that the parenthesis are important when reading the wikipedia identities. For example, would it be true that
- $A times (nabla times B) neq (A times nabla) times B$
Is there a "trick" to understanding how to manage the parenthesis?
vector-analysis
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
$endgroup$
I'm dealing with a problem of the form
$phi cdot left[(A cdot nabla)Bright]$
in which $phi$ is a scalar function and both $A$ and $B$ are vector functions.
Can I simply "distribute" the dot product with scalar function inside the expression to give
$ (C cdot nabla)B$
in which $C = phi cdot A$
Is this correct? I often get confused working with vector calculus operators.
On the side...
It seems that the parenthesis are important when reading the wikipedia identities. For example, would it be true that
- $A times (nabla times B) neq (A times nabla) times B$
Is there a "trick" to understanding how to manage the parenthesis?
vector-analysis
vector-analysis
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
asked Jan 11 at 2:21
ThatsRightJackThatsRightJack
350114
350114
$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32
add a comment |
$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32
$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32
$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32
add a comment |
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$begingroup$
For your first question, yes. If you find it always confusing, you may write down each component, e.g., $Acdotnabla=sum_iA_ifrac{partial}{partial x_i}$, and see if your proposal works. For your second question, well, $Atimes(nablatimes B)$ means $A$ cross the curl of $B$, which is a cross product between two vectors. By contrast, $(Atimesnabla)times B$ simply means some modified curl, if you would like to call it, of $B$ - there is no cross product anymore.
$endgroup$
– hypernova
Jan 11 at 2:32