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Problem is as follows:

In two dimensions, show that the divergence transforms as a scalar under rotations.

Aim is to determine $\bar{v}_{y}$ and $\bar{v}_{z}$, and show that

$\frac{\partial\bar{v}_{y}}{\partial\bar{y}}$ + $\frac{\partial\bar{v}_{z}}{\partial\bar{z}} = \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{y}}{\partial z}$

I'm not sure at all why the above shows that divergence transforms as a scalar under rotations. It was a part of the question (given as a hint) so I was just trying to solve it without really understanding what I was doing. Any help on clarifying why I do need to show that would be appreciated.

Since this is a rotation in two dimensions (in the $y$ and $z$ axis),

$ \left( {\begin{array}{cc} \bar{v}_{y} \\ \bar{v}_{z} \end{array} } \right) $ = $ \left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right) $ $ \left( {\begin{array}{cc} {v}_{y} \\ v_{z} \end{array} } \right) $.

Expanding this out, I found that $\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$ and $\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$.

Solving for $v_{y}$ and $v_{z}$, by multiplying $\bar{v}_{y}$ and $\bar{v}_{z}$ by $\sin\phi$ and $\cos\phi$, I was able to use the $\sin^{2}\phi + \cos^{2}\phi = 1$ identity to get

$v_{z} = \bar{v}_{y}.\sin\phi + \bar{v}_{z}.\cos\phi$ and

$v_{y} = \bar{v}_{y}.\cos\phi - \bar{v}_{z}.\sin\phi$.

Next, I found the components of the original equation:

The first partial derivative in the equation was determined as

$\frac{\partial \bar{v}_{y}}{\partial \bar{y}} = (\frac{\partial \bar{v}_{y}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial \bar{v}_{y}}{\partial z})(\frac{\partial z}{\partial \bar{y}})$

and the second as

$\frac{\partial \bar{v}_{z}}{\partial \bar{z}} = (\frac{\partial \bar{v}_{z}}{\partial y})(\frac{\partial y}{\partial \bar{z}}) + (\frac{\partial \bar{v}_{z}}{\partial z})(\frac{\partial z}{\partial \bar{z}})$.

At this point, I become stuck because I cannot seem to be able to find $\partial \bar{v}_{y}$ and $\partial \bar{v}_{z}$ with respect to $\partial y$ and $\partial z$.

How to I continue? Thanks in advance.

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Divergence of a vector is a scalar; and a scalar is a constant and doesn't change under rotations, so if you transform all your variables under a rotation and then calculate the divergence of the vector in the new coordinates, the divergence must remain unchanged.

First find $\bar v_y$ and $\bar v_z$ from the matrix transformation relation:

$ \left( {\begin{array}{cc} \bar{v}_{y} \\ \bar{v}_{z} \end{array} } \right) $ = $ \left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right) $ $ \left( {\begin{array}{cc} {v}_{y} \\ v_{z} \end{array} } \right) \to$

$$\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$$ $$\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$$

Then find their derivative w.r.t. $\bar y$ and $\bar z$, as is needed: $$\frac{\partial \bar v_y}{\partial \bar y}=\frac{\partial v_y}{\partial \bar y}\cos \phi+\frac{\partial v_z}{\partial \bar y}\sin \phi \tag{1}$$

expanding the right hand side derivatives $\frac{\partial {v}_{y}}{\partial \bar{y}}$ and $\frac{\partial {v}_{z}}{\partial \bar{y}}$ as:

$$\cases{\frac{\partial {v}_{y}}{\partial \bar{y}} = (\frac{\partial {v}_{y}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial{v}_{y}}{\partial z})(\frac{\partial z}{\partial \bar{y}}) \\ \frac{\partial {v}_{z}}{\partial \bar{y}} = (\frac{\partial {v}_{z}}{\partial y})(\frac{\partial y}{\partial \bar{y}}) + (\frac{\partial{v}_{z}}{\partial z})(\frac{\partial z}{\partial \bar{y}})} \tag{2}$$ we should now find the derivatives $\frac{\partial y}{\partial \bar y}$ and $\frac{\partial z}{\partial \bar y}$. Using the rotation formula in 2D to write $y$ and $z$ in terms of $\bar y$ and $\bar z$, we can find $\frac{\partial y}{\partial \bar y}$ and $\frac{\partial z}{\partial \bar y}$ : $$\cases{ y=\bar y \cos \phi-\bar z \sin \phi\\z=\bar y \sin \phi+ \bar z \cos \phi}\to\cases{\frac{\partial y}{\partial \bar y}=\cos \phi\\ \frac{\partial z}{\partial \bar y}=\sin \phi}\tag{3}$$ now substituting $(3)$ in $(2)$ and then $(2)$ in $(1) $, we will have $$\frac{\partial \bar v_y}{\partial \bar y}=\left(\frac{\partial v_y}{\partial y}\cos\phi+\frac{\partial v_y}{\partial z}\sin\phi \right )\cos\phi + \left( \frac{\partial v_z}{\partial y}\cos\phi+\frac{\partial v_z}{\partial z}\sin\phi \right)\sin \phi$$

doing the same for $\frac{\partial \bar v_z}{\partial \bar z}$, we will arrive at:

$$\frac{\partial \bar v_z}{\partial \bar z}=-\left(-\frac{\partial v_y}{\partial y}\sin\phi+\frac{\partial v_y}{\partial z}\cos\phi \right )\sin\phi + \left(- \frac{\partial v_z}{\partial y}\sin\phi+\frac{\partial v_z}{\partial z}\cos\phi \right)\cos \phi$$

Now just sum up the two terms$\frac{\partial\bar{v}_{y}}{\partial\bar{y}}$ and $\frac{\partial\bar{v}_{z}}{\partial\bar{z}}$ and apply $\sin^{2}\phi + \cos^{2}\phi = 1$ to arrive at the final result: $$\frac{\partial\bar{v}_{y}}{\partial\bar{y}} + \frac{\partial\bar{v}_{z}}{\partial\bar{z}} = \frac{\partial v_{y}}{\partial y} + \frac{\partial v_{y}}{\partial z}$$

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  • $\begingroup$ Thanks, I understand why I am meant to show the equivalence. However, would you mind explaining your working? I'm afraid I'm having difficulty following from when you expanded the r.h.s. derivatives. $\endgroup$ – achacttn Jul 28 '13 at 15:30
  • $\begingroup$ @dlckd I think it is clear now. $\endgroup$ – Mo_ Jul 28 '13 at 17:39
  • $\begingroup$ Rigorously speaking how do we show that a scalar is unchanged under rotation? We cannot operate on it with the rotation matrix and show that it is the same, because that operation is not defined. $\endgroup$ – pmac Aug 30 '18 at 1:51
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Here is an answer from a completely different point of view (=the "geometric view"). For a transformation $\Phi:U\mapsto V$, define the push forward $\Phi_*$ for different types of objects as follows:

  • $(\Phi_* f)(y):=f\circ \Phi^{-1}(y)$ for a function $f:U\mapsto\mathbb R$
  • $(\Phi_* v)(y):=(\Phi'v)\circ \Phi^{-1}(y)$ for a vector field $v:U\mapsto\mathbb R^n$
  • $(\Phi_* p)(y):=(\frac{p}{|\det \Phi'|})\circ \Phi^{-1}(y)$ for a density $p:U\mapsto\mathbb R$

For a density $p$ and a vector field $v$, define the divergence of $v$ relative to $p$ as the function $$ \operatorname{div}_p v:=\frac{1}{p}\sum_{i=1}^n\frac{\partial[pv]_i}{\partial x_i} $$ The usual divergence is then just given by $\operatorname{div}_1 v$ (or more generally by $\operatorname{div}_\alpha v$ for any $\alpha\in\mathbb R$). Now we can prove that $$ \Phi_*(\operatorname{div}_p v)=\operatorname{div}_{(\Phi_*p)} (\Phi_*v) $$ where $\operatorname{div}_p v$ is transformed like a function, $v$ is transformed like a vector field and $p$ is transformed like a density. The most straightforward way to prove this is to first show that $$ \int pdx\;L_v\varphi = -\int pdx\;\varphi \operatorname{div}_p v $$ for arbitrary smooth test-functions $\varphi$ with compact support, and then use the known trivial transformation behavior of the directional derivative $L_v$ (or call it Lie-derivative if you want) and the known transformation behavior of the integral to derive the transformation behavior of the divergence $\operatorname{div}_p v$. A long time ago, I wrote a German text where this is explained in more detail with some pictures, some applications and a bit more context.


The connection to the question at hand is that $\det \Phi'=1$ if $\Phi$ is a rotation, hence $\Phi_*1=1$ if $1$ is considered as a density, and hence $\Phi_*(\operatorname{div}_1 v)=\operatorname{div}_1 (\Phi_*v)$.

More generally, we don't even need $\det \Phi'=1$, it's enough that $\det \Phi'$ is constant, which is satisfied for any linear tranformation $\Phi$.

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At this point, I become stuck because I cannot seem to be able to find $\partial \bar{v}_{y}$ and $\partial \bar{v}_{z}$ with respect to $\partial y$ and $\partial z$.

You already determined earlier that $\bar{v}_{y} = v_{y}.\cos\phi + v_{z}.\sin\phi$ and $\bar{v}_{z} = -v_{y}.\sin\phi + v_{z}.\cos\phi$. Hence

$\frac{\partial \bar{v}_{y}}{\partial y}=\frac{\partial v_{y}}{\partial y}.\cos\phi + \frac{\partial v_{z}}{\partial y}.\sin\phi$

$\frac{\partial \bar{v}_{y}}{\partial z}=\frac{\partial v_{y}}{\partial z}.\cos\phi + \frac{\partial v_{z}}{\partial z}.\sin\phi$

and

$\frac{\partial \bar{v}_{z}}{\partial y} = -\frac{\partial v_{y}}{\partial y}.\sin\phi + \frac{\partial v_{z}}{\partial y}.\cos\phi$

$\frac{\partial \bar{v}_{z}}{\partial z} = -\frac{\partial v_{y}}{\partial z}.\sin\phi + \frac{\partial v_{z}}{\partial z}.\cos\phi$

However, this is boringly obvious, so you probably meant to ask something else instead.


Let's assume that you asked

At this point, I become stuck because I cannot seem to be able to find $\partial \bar{y}$ and $\partial \bar{z}$ with respect to $\partial y$ and $\partial z$.

We have $ \left( {\begin{array}{cc} \bar{y} \\ \bar{z} \end{array} } \right) $ = $ \left( {\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{array} } \right) $ $ \left( {\begin{array}{cc} y \\ z \end{array} } \right) $, hence

$\frac{\partial \bar{y}}{\partial y}=\cos\phi$

$\frac{\partial \bar{y}}{\partial z}=\sin\phi$

and

$\frac{\partial \bar{z}}{\partial y}=-\sin\phi$

$\frac{\partial \bar{z}}{\partial z}=\cos\phi$

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