Identity for divergence of vector product
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I'm reviewing this paper and they state the following "vector identity":
$ nabla cdot (textbf{u}textbf{v}) = nabla textbf{u} cdot textbf{v} + (nabla cdot textbf{v})textbf{u}$
I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $textbf{u}$ and $textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?
EDIT
After some discussion, it seems that $textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.
vector-analysis
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add a comment |
$begingroup$
I'm reviewing this paper and they state the following "vector identity":
$ nabla cdot (textbf{u}textbf{v}) = nabla textbf{u} cdot textbf{v} + (nabla cdot textbf{v})textbf{u}$
I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $textbf{u}$ and $textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?
EDIT
After some discussion, it seems that $textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.
vector-analysis
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$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Duncan
Jan 1 at 4:13
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Which one are you referring to? It's not immediately obvious to me.
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– ThatsRightJack
Jan 1 at 4:21
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Vector dot product
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– Duncan
Jan 1 at 4:32
$begingroup$
I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
$endgroup$
– ThatsRightJack
Jan 1 at 4:44
add a comment |
$begingroup$
I'm reviewing this paper and they state the following "vector identity":
$ nabla cdot (textbf{u}textbf{v}) = nabla textbf{u} cdot textbf{v} + (nabla cdot textbf{v})textbf{u}$
I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $textbf{u}$ and $textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?
EDIT
After some discussion, it seems that $textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.
vector-analysis
$endgroup$
I'm reviewing this paper and they state the following "vector identity":
$ nabla cdot (textbf{u}textbf{v}) = nabla textbf{u} cdot textbf{v} + (nabla cdot textbf{v})textbf{u}$
I tried to find some material online about this identity but didn't have any luck. I'm pretty sure $textbf{u}$ and $textbf{v}$ are both vectors, as they're both bold face in the paper. Can someone help to provide more information on this identity? Is it valid? Can you give the proof?
EDIT
After some discussion, it seems that $textbf{u}$ should be interpreted as a scalar (remove boldface), not a vector. I will clarify this with the original source and follow up if there is any further amendments. If it is a scalar, the answer below is correct.
vector-analysis
vector-analysis
edited Jan 1 at 6:35
ThatsRightJack
asked Jan 1 at 4:02
ThatsRightJackThatsRightJack
352114
352114
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Duncan
Jan 1 at 4:13
$begingroup$
Which one are you referring to? It's not immediately obvious to me.
$endgroup$
– ThatsRightJack
Jan 1 at 4:21
$begingroup$
Vector dot product
$endgroup$
– Duncan
Jan 1 at 4:32
$begingroup$
I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
$endgroup$
– ThatsRightJack
Jan 1 at 4:44
add a comment |
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Duncan
Jan 1 at 4:13
$begingroup$
Which one are you referring to? It's not immediately obvious to me.
$endgroup$
– ThatsRightJack
Jan 1 at 4:21
$begingroup$
Vector dot product
$endgroup$
– Duncan
Jan 1 at 4:32
$begingroup$
I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
$endgroup$
– ThatsRightJack
Jan 1 at 4:44
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Duncan
Jan 1 at 4:13
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Duncan
Jan 1 at 4:13
$begingroup$
Which one are you referring to? It's not immediately obvious to me.
$endgroup$
– ThatsRightJack
Jan 1 at 4:21
$begingroup$
Which one are you referring to? It's not immediately obvious to me.
$endgroup$
– ThatsRightJack
Jan 1 at 4:21
$begingroup$
Vector dot product
$endgroup$
– Duncan
Jan 1 at 4:32
$begingroup$
Vector dot product
$endgroup$
– Duncan
Jan 1 at 4:32
$begingroup$
I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
$endgroup$
– ThatsRightJack
Jan 1 at 4:44
$begingroup$
I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
$endgroup$
– ThatsRightJack
Jan 1 at 4:44
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This is, fundamentally, just another case of the product rule for derivatives.
For the notation to make sense ($nabla u$ and $nablacdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $nabla u$ is the gradient of $u$, and $nablacdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.
To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$):
begin{align*}nablacdot (uv)quad &?quad nabla ucdot v + (nablacdot v) u\
frac{partial (uA)}{partial x}+frac{partial (uB)}{partial y}+frac{partial (uC)}{partial z}quad &?quad left(frac{partial u}{partial x},frac{partial u}{partial y},frac{partial u}{partial z}right)cdot (A,B,C) + left(frac{partial A}{partial x}+frac{partial B}{partial y}+frac{partial C}{partial z}right)u\
ufrac{partial A}{partial x}+Afrac{partial u}{partial x}+ufrac{partial B}{partial y}+Bfrac{partial u}{partial y}+ufrac{partial C}{partial z}+Cfrac{partial u}{partial z} &= Afrac{partial u}{partial x}+Bfrac{partial u}{partial y}+Cfrac{partial u}{partial z}+ufrac{partial A}{partial x}+ufrac{partial B}{partial y}+ufrac{partial C}{partial z} end{align*}
Not too bad there. It's just all the same terms rearranged a bit.
There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule:
$$nablacdot (uv) = d(uwedge v) = duwedge v + (-1)^0 uwedge dv = nabla ucdot v + u(nabla cdot v)$$
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$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
|
show 2 more comments
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Using Einstein notation, we can see it more easily:
$$
nabla cdot (u textbf{v}) = partial_i (uv_i) = partial_iu cdot v_i + ucdotpartial_iv_i = nabla ucdot textbf{v} + u(nabla cdot textbf{v}).
$$
$endgroup$
add a comment |
$begingroup$
For this to make sense $u$ is a scalar function and $mathbf{v} = (v_1, ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$frac{partial}{partial x_j}(umathbf{v})_i = frac{partial}{partial x_j}(uv_i) = frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}$$
Therefore, the Jacobi matrix is given by
$$nabla(umathbf{v}) = left[frac{partial}{partial x_j}(umathbf{v})_iright]_{1le i,jle n} = left[frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}right]_{1le i,jle n} = mathbf{v}otimes (nabla u) + u (nablamathbf{v})$$
The divergence is obtained by taking the trace:
$$nabla cdot (umathbf{v}) = operatorname{Tr}left(nabla(umathbf{v})right) = operatorname{Tr}(mathbf{v}otimes (nabla u)) + u operatorname{Tr}(nablamathbf{v}) = mathbf{v}cdot (nabla u) + u(nablacdot mathbf{v})$$
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
This is, fundamentally, just another case of the product rule for derivatives.
For the notation to make sense ($nabla u$ and $nablacdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $nabla u$ is the gradient of $u$, and $nablacdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.
To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$):
begin{align*}nablacdot (uv)quad &?quad nabla ucdot v + (nablacdot v) u\
frac{partial (uA)}{partial x}+frac{partial (uB)}{partial y}+frac{partial (uC)}{partial z}quad &?quad left(frac{partial u}{partial x},frac{partial u}{partial y},frac{partial u}{partial z}right)cdot (A,B,C) + left(frac{partial A}{partial x}+frac{partial B}{partial y}+frac{partial C}{partial z}right)u\
ufrac{partial A}{partial x}+Afrac{partial u}{partial x}+ufrac{partial B}{partial y}+Bfrac{partial u}{partial y}+ufrac{partial C}{partial z}+Cfrac{partial u}{partial z} &= Afrac{partial u}{partial x}+Bfrac{partial u}{partial y}+Cfrac{partial u}{partial z}+ufrac{partial A}{partial x}+ufrac{partial B}{partial y}+ufrac{partial C}{partial z} end{align*}
Not too bad there. It's just all the same terms rearranged a bit.
There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule:
$$nablacdot (uv) = d(uwedge v) = duwedge v + (-1)^0 uwedge dv = nabla ucdot v + u(nabla cdot v)$$
$endgroup$
$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
|
show 2 more comments
$begingroup$
This is, fundamentally, just another case of the product rule for derivatives.
For the notation to make sense ($nabla u$ and $nablacdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $nabla u$ is the gradient of $u$, and $nablacdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.
To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$):
begin{align*}nablacdot (uv)quad &?quad nabla ucdot v + (nablacdot v) u\
frac{partial (uA)}{partial x}+frac{partial (uB)}{partial y}+frac{partial (uC)}{partial z}quad &?quad left(frac{partial u}{partial x},frac{partial u}{partial y},frac{partial u}{partial z}right)cdot (A,B,C) + left(frac{partial A}{partial x}+frac{partial B}{partial y}+frac{partial C}{partial z}right)u\
ufrac{partial A}{partial x}+Afrac{partial u}{partial x}+ufrac{partial B}{partial y}+Bfrac{partial u}{partial y}+ufrac{partial C}{partial z}+Cfrac{partial u}{partial z} &= Afrac{partial u}{partial x}+Bfrac{partial u}{partial y}+Cfrac{partial u}{partial z}+ufrac{partial A}{partial x}+ufrac{partial B}{partial y}+ufrac{partial C}{partial z} end{align*}
Not too bad there. It's just all the same terms rearranged a bit.
There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule:
$$nablacdot (uv) = d(uwedge v) = duwedge v + (-1)^0 uwedge dv = nabla ucdot v + u(nabla cdot v)$$
$endgroup$
$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
|
show 2 more comments
$begingroup$
This is, fundamentally, just another case of the product rule for derivatives.
For the notation to make sense ($nabla u$ and $nablacdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $nabla u$ is the gradient of $u$, and $nablacdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.
To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$):
begin{align*}nablacdot (uv)quad &?quad nabla ucdot v + (nablacdot v) u\
frac{partial (uA)}{partial x}+frac{partial (uB)}{partial y}+frac{partial (uC)}{partial z}quad &?quad left(frac{partial u}{partial x},frac{partial u}{partial y},frac{partial u}{partial z}right)cdot (A,B,C) + left(frac{partial A}{partial x}+frac{partial B}{partial y}+frac{partial C}{partial z}right)u\
ufrac{partial A}{partial x}+Afrac{partial u}{partial x}+ufrac{partial B}{partial y}+Bfrac{partial u}{partial y}+ufrac{partial C}{partial z}+Cfrac{partial u}{partial z} &= Afrac{partial u}{partial x}+Bfrac{partial u}{partial y}+Cfrac{partial u}{partial z}+ufrac{partial A}{partial x}+ufrac{partial B}{partial y}+ufrac{partial C}{partial z} end{align*}
Not too bad there. It's just all the same terms rearranged a bit.
There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule:
$$nablacdot (uv) = d(uwedge v) = duwedge v + (-1)^0 uwedge dv = nabla ucdot v + u(nabla cdot v)$$
$endgroup$
This is, fundamentally, just another case of the product rule for derivatives.
For the notation to make sense ($nabla u$ and $nablacdot v$ defined), $u$ should be a scalar-valued function of a vector variable, and $v$ should be a vector-valued function of that variable. Then $nabla u$ is the gradient of $u$, and $nablacdot v$ is the divergence of $v$. The two terms on the right are both scalars - the first is the dot product of the vector-valued gradient of $u$ and the vector-valued function $v$, while the second is the product of the scalar-valued divergence of $v$ and the scalar-valued function $u$.
To prove it, we just go down to components. I'll assume three dimensions here, although it works in more generality (Let $A,B,C$ be the components of $v$; $v(x,y,z)=(A(x,y,z),B(x,y,z),C(x,y,z))$):
begin{align*}nablacdot (uv)quad &?quad nabla ucdot v + (nablacdot v) u\
frac{partial (uA)}{partial x}+frac{partial (uB)}{partial y}+frac{partial (uC)}{partial z}quad &?quad left(frac{partial u}{partial x},frac{partial u}{partial y},frac{partial u}{partial z}right)cdot (A,B,C) + left(frac{partial A}{partial x}+frac{partial B}{partial y}+frac{partial C}{partial z}right)u\
ufrac{partial A}{partial x}+Afrac{partial u}{partial x}+ufrac{partial B}{partial y}+Bfrac{partial u}{partial y}+ufrac{partial C}{partial z}+Cfrac{partial u}{partial z} &= Afrac{partial u}{partial x}+Bfrac{partial u}{partial y}+Cfrac{partial u}{partial z}+ufrac{partial A}{partial x}+ufrac{partial B}{partial y}+ufrac{partial C}{partial z} end{align*}
Not too bad there. It's just all the same terms rearranged a bit.
There's another way to look at this. If you go on with mathematics long enough, you'll encounter the language of manifolds and differential forms, which provides a unifying framework for all of these vector calculus notions. In particular, both the gradient and the divergence are examples of the exterior derivative, and the exterior derivative has a product rule. Here, working in $n$ dimensions, we have $u$ as a $0$-form and $v$ as an $(n-1)$-form. The gradient takes $0$-forms to $1$-forms, the divergence takes $(n-1)$-forms to $n$-forms, and the dot product between $1$-forms and $(n-1)$ forms is the wedge product, taking us to a $n$-form. Then we just apply the exterior derivative's product rule:
$$nablacdot (uv) = d(uwedge v) = duwedge v + (-1)^0 uwedge dv = nabla ucdot v + u(nabla cdot v)$$
edited Jan 1 at 5:25
answered Jan 1 at 5:03
jmerryjmerry
16.5k11633
16.5k11633
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Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
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– ThatsRightJack
Jan 1 at 5:21
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Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
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– ThatsRightJack
Jan 1 at 5:21
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Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
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– jmerry
Jan 1 at 5:24
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OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
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– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
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– jmerry
Jan 1 at 6:08
|
show 2 more comments
$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Let me just clarify this part: "u should be a scalar-valued function of a vector variable, and v should be a vector-valued function of that variable"...are you saying that "u" must be a scalar and "v" must be a vector? That would imply that my interpretation of the paper notation is wrong (they use 2 bold letters, so I assumed they were both vectors).
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Then you say "The two terms on the left are both scalars..."? Should say "right" not "left"?
$endgroup$
– ThatsRightJack
Jan 1 at 5:21
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
Yes, your interpretation was wrong. $nablacdot$ is the divergence, which applies to a vector. $nabla$ without the dot is the gradient, which applies to a scalar. That's just what the standard notation means. Other comment - argh, directional dyslexia. I meant the right. Time to edit.
$endgroup$
– jmerry
Jan 1 at 5:24
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
OK, I don't disagree with what you wrote, but I let me throw something at you. In the "material/total" derivative, you often see $(textbf{u}cdot nabla) textbf{v}$. I would read this as a vector, dotted with the gradient of a vector, which is a tensor 2? If I'm right in interpreting it this way, then the gradient operator doesn't necessarily operate on a scalar? I'm just worried that the author did intend that "u" and "v" be vectors.
$endgroup$
– ThatsRightJack
Jan 1 at 5:50
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
$begingroup$
What that notation reads as to me? A directional derivative. $u$ determines a direction by applying multipliers to the partial derivatives, and then we apply that to $v$. It's not the gradient operator at all; it takes a slice of $v$ as a function of one variable, and differentiates that.
$endgroup$
– jmerry
Jan 1 at 6:08
|
show 2 more comments
$begingroup$
Using Einstein notation, we can see it more easily:
$$
nabla cdot (u textbf{v}) = partial_i (uv_i) = partial_iu cdot v_i + ucdotpartial_iv_i = nabla ucdot textbf{v} + u(nabla cdot textbf{v}).
$$
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add a comment |
$begingroup$
Using Einstein notation, we can see it more easily:
$$
nabla cdot (u textbf{v}) = partial_i (uv_i) = partial_iu cdot v_i + ucdotpartial_iv_i = nabla ucdot textbf{v} + u(nabla cdot textbf{v}).
$$
$endgroup$
add a comment |
$begingroup$
Using Einstein notation, we can see it more easily:
$$
nabla cdot (u textbf{v}) = partial_i (uv_i) = partial_iu cdot v_i + ucdotpartial_iv_i = nabla ucdot textbf{v} + u(nabla cdot textbf{v}).
$$
$endgroup$
Using Einstein notation, we can see it more easily:
$$
nabla cdot (u textbf{v}) = partial_i (uv_i) = partial_iu cdot v_i + ucdotpartial_iv_i = nabla ucdot textbf{v} + u(nabla cdot textbf{v}).
$$
answered Jan 1 at 13:27
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
$begingroup$
For this to make sense $u$ is a scalar function and $mathbf{v} = (v_1, ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$frac{partial}{partial x_j}(umathbf{v})_i = frac{partial}{partial x_j}(uv_i) = frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}$$
Therefore, the Jacobi matrix is given by
$$nabla(umathbf{v}) = left[frac{partial}{partial x_j}(umathbf{v})_iright]_{1le i,jle n} = left[frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}right]_{1le i,jle n} = mathbf{v}otimes (nabla u) + u (nablamathbf{v})$$
The divergence is obtained by taking the trace:
$$nabla cdot (umathbf{v}) = operatorname{Tr}left(nabla(umathbf{v})right) = operatorname{Tr}(mathbf{v}otimes (nabla u)) + u operatorname{Tr}(nablamathbf{v}) = mathbf{v}cdot (nabla u) + u(nablacdot mathbf{v})$$
$endgroup$
add a comment |
$begingroup$
For this to make sense $u$ is a scalar function and $mathbf{v} = (v_1, ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$frac{partial}{partial x_j}(umathbf{v})_i = frac{partial}{partial x_j}(uv_i) = frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}$$
Therefore, the Jacobi matrix is given by
$$nabla(umathbf{v}) = left[frac{partial}{partial x_j}(umathbf{v})_iright]_{1le i,jle n} = left[frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}right]_{1le i,jle n} = mathbf{v}otimes (nabla u) + u (nablamathbf{v})$$
The divergence is obtained by taking the trace:
$$nabla cdot (umathbf{v}) = operatorname{Tr}left(nabla(umathbf{v})right) = operatorname{Tr}(mathbf{v}otimes (nabla u)) + u operatorname{Tr}(nablamathbf{v}) = mathbf{v}cdot (nabla u) + u(nablacdot mathbf{v})$$
$endgroup$
add a comment |
$begingroup$
For this to make sense $u$ is a scalar function and $mathbf{v} = (v_1, ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$frac{partial}{partial x_j}(umathbf{v})_i = frac{partial}{partial x_j}(uv_i) = frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}$$
Therefore, the Jacobi matrix is given by
$$nabla(umathbf{v}) = left[frac{partial}{partial x_j}(umathbf{v})_iright]_{1le i,jle n} = left[frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}right]_{1le i,jle n} = mathbf{v}otimes (nabla u) + u (nablamathbf{v})$$
The divergence is obtained by taking the trace:
$$nabla cdot (umathbf{v}) = operatorname{Tr}left(nabla(umathbf{v})right) = operatorname{Tr}(mathbf{v}otimes (nabla u)) + u operatorname{Tr}(nablamathbf{v}) = mathbf{v}cdot (nabla u) + u(nablacdot mathbf{v})$$
$endgroup$
For this to make sense $u$ is a scalar function and $mathbf{v} = (v_1, ldots, v_n)$ is a vector function.
We can calculate the partial derivatives using the chain rule:
$$frac{partial}{partial x_j}(umathbf{v})_i = frac{partial}{partial x_j}(uv_i) = frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}$$
Therefore, the Jacobi matrix is given by
$$nabla(umathbf{v}) = left[frac{partial}{partial x_j}(umathbf{v})_iright]_{1le i,jle n} = left[frac{partial u}{partial x_j}v_i + ufrac{partial v_i}{partial x_j}right]_{1le i,jle n} = mathbf{v}otimes (nabla u) + u (nablamathbf{v})$$
The divergence is obtained by taking the trace:
$$nabla cdot (umathbf{v}) = operatorname{Tr}left(nabla(umathbf{v})right) = operatorname{Tr}(mathbf{v}otimes (nabla u)) + u operatorname{Tr}(nablamathbf{v}) = mathbf{v}cdot (nabla u) + u(nablacdot mathbf{v})$$
answered Jan 1 at 13:18
mechanodroidmechanodroid
29k62648
29k62648
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/…
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– Duncan
Jan 1 at 4:13
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Which one are you referring to? It's not immediately obvious to me.
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– ThatsRightJack
Jan 1 at 4:21
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Vector dot product
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– Duncan
Jan 1 at 4:32
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I'm sorry, but I don't see how that is the same identity I placed in the question. Can you show me how the two are equal?
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– ThatsRightJack
Jan 1 at 4:44