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 October 22nd, 2018, 01:31 PM #11 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 $\displaystyle \big| \frac{\partial (x,y)}{\partial (r, \theta) }\big|=\begin{vmatrix} x_{r} & y_{r}\\ x_{\theta} & y_{\theta} \end{vmatrix}$ $\displaystyle x=r\cos\theta$ $\displaystyle y=r\sin\theta$ $\displaystyle x_{r}=\frac{\partial x}{\partial r}= \cos\theta$ ....... But it looks like you got it on your own. By the way, your first transformation isn't spherical coordinates, whatever it is.
October 22nd, 2018, 09:33 PM   #12
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 Originally Posted by zylo Unintelligible jargon posing as a derivation. Please illustrate in polar coordinates. Your last sentence is not a given, it's what's to be shown.
You wanted to know why the determinant of the Jacobian measures volumes under a change of coordinates. The reason is that Jacobians are just derivatives, hence linear operators, and these are decomposed into invariant subspaces with eigenvalues determining the scaling in each.

Volumes are nothing more than a product of scalings in each dimension so the product of the eigenvalues is what we want. It just happens to be what we call the determinant. If you want more details than this open a book. Carefully covering this material typically takes the better part of a semester of linear algebra so I don't know what to tell you.

I also don't know what you mean by illustrate with polar coordinates. Illustrate what?

Quote:
 Originally Posted by zylo $\displaystyle J=\begin{bmatrix} x_{u} &y_{u} & z_{u}\\ x_{v} &y_{v} & z_{v}\\ x_{w} &y_{w} & z_{w} \end{bmatrix}$ Note $\displaystyle dxdydz\neq Jdudvdw$ but $\displaystyle \int_{V}dxdydz=\int_{V}Jdudvdw$
As written, this equation is certainly not true because the expression $J dudvdw$ is meaningless. $J$ is a linear transformation on $\mathbb{R}^3$ and $dudvdw$ is a linear functional, so composing them makes no sense since the image of $dudvdw$ is a scalar and $J$ acts on vectors.

On the other hand, I think you meant to say that $dxdydz \neq$ det$(J) dudvdw$, but this is also not true because in this case these are most definitely equal. This follows from the definition of the exterior derivative which is defined explicitly by this equality.

Last edited by skipjack; October 22nd, 2018 at 11:51 PM.

 October 23rd, 2018, 05:38 AM #13 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Jacobian can refer to either determinant or matrix, and it is clear from my post#7 what I meant. I do agree I should have used straight brackets to indicate the determinant. This is dxdy in 2d: Given $\displaystyle x=x(u,v), y=y(u,v)$ $\displaystyle dx=x_{u}du+x_{v}dv$ $\displaystyle dy=y_{u}du+y_{v}dv$ $\displaystyle dxdy=x_{u}y_{u}dudu+x_{u}y_{v}dudv+x_{v}y_{u}dudv+ x_{v}y_{v}dvdv$ This is the approach of literally transforming dxdy instead of determining a volume element in uv coordinates. But how do you integrate that? I don't have the slightest idea what you are talking about. You mention a book. What book? It seems to me that if you understood what you were talking about you should have no problem using it to demonstrate the volume transformation which is the subject of this thread, ie dV=Jdxdy in polar coordinates, as used in printed matter of OP. But I guess we're even. You don't have the slightest idea what I am talking about.
October 23rd, 2018, 05:44 AM   #14
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 Originally Posted by zylo You don't have the slightest idea what I am talking about.
I think on that we all agree on this forum.

October 23rd, 2018, 05:53 AM   #15
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 Originally Posted by Micrm@ss I think on that we all agree on this forum.
We? So far I make out you and SDK.

The implication is you understand what SDK is saying. Could you please answer my last post for him?

Oh, I meant to give a reference for 3d volume transformation.

11.4 Computing the Volume Element: the Jacobian

October 23rd, 2018, 06:05 AM   #16
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 Originally Posted by zylo Jacobian can refer to either determinant or matrix, and it is clear from my post#7 what I meant. I do agree I should have used straight brackets to indicate the determinant.
The point is that it certainly was not clear what you meant. As you said, Jacobian is occasionally used by some authors to mean the determinant of the derivative (though this is rare in my experience). My point was that your statement is wrong for either interpretation. $J dudvdw$ is either a meaningless expression, or $dxdydz \neq J dudvdw$ is false.

Quote:
 Originally Posted by zylo It seems to me that if you understood what you were talking about you should have no problem using it to demonstrate the volume transformation which is the subject of this thread, ie dV=Jdxdy in polar coordinates, as used in printed matter of OP. But I guess we're even. You don't have the slightest idea what I am talking about.
Just to be clear, are you asking me if I know how to compute the determinant of this matrix
$\begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}?$

Edit: I think I see your problem. In your 2D example you seem to be treating $dxdy$ as the product of differentials. This is true for the right definition of product but it isn't the usual geometric product. In particular, $dx,dy$ are both linear functionals so it makes no sense to compose them. Your last line where your "FOIL" these is where this mistake becomes apparent. $dxdy$ is a bivector which is formally defined as the wedge product of $dx,dy$. However, it is often misleading to think of it as an object generated by a binary operation on $dx,dy$. It is better to think of it as just some linear functional which has 2 vectors as input (i.e. a linear functional on the wedge product of two vector spaces).

See https://en.wikipedia.org/wiki/Bivector for a more thorough explanation.

Last edited by SDK; October 23rd, 2018 at 06:14 AM.

October 23rd, 2018, 06:31 AM   #17
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 Originally Posted by SDK . Just to be clear, are you asking me if I know how to compute the determinant of this matrix $\begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}?$ See https://en.wikipedia.org/wiki/Bivector for a more thorough explanation.
I am asking if you can derive the expression for volume transformation of coordinates in polar coordinates, or 2d in general, the subject of this thread (see OP), using the formalism you claim to be describing. I suspect your question is the answer.

October 23rd, 2018, 07:34 AM   #18
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 Originally Posted by zylo I am asking if you can derive the expression for volume transformation of coordinates in polar coordinates, or 2d in general, the subject of this thread (see OP), using the formalism you claim to be describing. I suspect your question is the answer.
Sure but I don't see how this helps. The whole point of doing algebra abstractly on vector spaces is to avoid painful computations like this. However, since its only 2-dimensional (i.e. eigenvectors are easy to compute) here is the computation.

1. Pick a point $(r,\theta)$ in the plane and compute the linearization of the polar transformation at this point. This is just the derivative:
$A = \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}$

2. You can compute the eigenvalues explicitly as
$\lambda_1 = \frac{1}{2}(r+1) \cos \theta + \sqrt{(r+1)^2\cos^2 \theta - 4r}, \qquad \lambda_2 = \frac{1}{2}(r+1) \cos \theta - \sqrt{(r+1)^2\cos^2 \theta - 4r}.$

3. With these you can also explicitly compute eigenvectors for each
$\xi_1 = \begin{pmatrix} \frac{-2r \sin \theta}{(r-1) \cos \theta + \sqrt{(r+1)^2 \cos^2 \theta - 4r}} \\ 1 \end{pmatrix}, \qquad \xi_2 = \begin{pmatrix} \frac{-2r \sin \theta}{(r-1) \cos \theta - \sqrt{(r+1)^2 \cos^2 \theta - 4r}} \\ 1 \end{pmatrix}.$
I'll let you check my computations. Notice that $A$ is a real matrix so if the eigenvalues are complex, then each pair of eigenvalues and eigenvectors are conjugates of one another. In particular, both eigenvectors have the same norm. Moreover, eigenvectors are linearly independent which means that $\xi_1,\xi_2$ form a basis for $\mathbb{C}$ (as a 2-dimensional vector space over $\mathbb{R}$).

4. Now, suppose $k\xi_1$ is a unit vector and thus $k \xi_2$ is also a unit vector. Now consider the parallelogram in $\mathbb{R}^2$ whose sides are the vectors $v_1 = ak\xi_1$ and $v_2 = bk\xi_2$ for arbitrary $a,b \in \mathbb{R}$. Since $k\xi_i$ are both unit vectors, this parallelogram has volume $ab \sin \alpha$ where $\alpha$ is the acute planar angle between $\xi_1,\xi_2$. The image of these sides under the action of $A$ is
$A\cdot ak \xi_1 = \lambda_1 a k \xi_1, \qquad A \cdot bk \xi_2 = \lambda_2 b k \xi_2$
so the volume of the parallelogram spanned by $Av_1,Av_2$ is simply $\lambda_1 a \lambda_2 b \sin \alpha$. Hence, the action of $A$ scales the volume of any parallelogram by $\lambda_1 \lambda_2 = \det(A) = r$.

Last edited by SDK; October 23rd, 2018 at 07:38 AM.

 October 23rd, 2018, 09:38 AM #19 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 $\displaystyle \vec{r}=x\vec{i}+y\vec{j}$ is position of point in x,y plane where x=x(u,v) and y=y(u,v)) $\displaystyle d\vec{r}=\vec{r}_{u}du+\vec{r}_{v}dv$ The volume of $\displaystyle \vec{r}_{u}du$ and $\displaystyle \vec{r}_{v}dv$ is $\displaystyle |\vec{r}_{u}du\times\vec{r}_{v}dv|=Jdudv$ You start off with $\displaystyle \begin{vmatrix} dx\\ dy \end{vmatrix}=A \begin{vmatrix} du\\ dv \end{vmatrix}$ where $\displaystyle A =\begin{bmatrix} x_{u} &x_{v} \\ y_{u} & y_{v} \end{bmatrix}$ just as I do. Then you calculate the eigen vectors of A, and find the volume of the normed eigenvecors, multiplied by a and b respectively, by cross product*, just as I did, and come up with the volume $\displaystyle ab\lambda_{1}\lambda_{2}=abJ$ So how do you go from there to the formula for coordinate transformation of "volume" element: dV =Jdudv J is the absolute value of the determinant of the matrix of the coordinate transformation, just to eliminate that straw man. * Fine for 2 and 3 dimensions, but what is volume for 4 dims. Last edited by zylo; October 23rd, 2018 at 09:59 AM. Reason: dV =Jdxdy to Jdudv

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