Area Differential in Cartesian and Polar Coordinates

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Summary:

How do we get rdrdt from dxdy?

Main Question or Discussion Point

The area differential ##dA## in Cartesian coordinates is ##dxdy##.
The area differential ##dA## in polar coordinates is ##r dr d\theta##.
How do we get from one to the other and prove that ##dxdy## is indeed equal to ##r dr d\theta##?

##dxdy=r dr d\theta##
The trigonometric functions are used to obtain the conversions ##x=r \cos \theta ## and ##y=r \sin \theta ## . If we differentiate both with respect to the polar angle ##\theta##,
$$
dx d\theta = \cos \theta dr d\theta -r \sin \theta\; , \;dy d\theta = \sin \theta dr d\theta +r \cos \theta
$$
The differentials then become
$$
dx= \cos \theta dr -r \sin \theta d\theta \; , \;dy = \sin \theta dr + r \cos \theta dr \theta
$$
which to me does not make any sense.
It seems as though finding ##dr d\theta ## is what we must do, but how? What am I missing?
How do we prove that indeed ...
$$
dx dy = r dr d\theta
$$
Is it in Griffiths? On what page?
 
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Answers and Replies

Zap
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O yea, you have to partially differentiate one coordinate with respect to the other.
That's not in Griffiths!
I totally forgot.
Thanks.
That was driving me crazy.
That method was always a great mystery to me, because the professor just kind of told us to think about how it is correct 🤔.
What if we thought too much about it and it no longer makes any sense 🤔?
I was thinking about taking a partial derivative, but it didn't "click."
Griffths has a God awful method for converting coordinate differentials.
I suggest everyone not use that method.
He actually brings out the old ##\Delta x## thing and takes the limit.
Hideous.
 
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WWGD
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O yea, you have to partially differentiate one coordinate with respect to the other.
That's not in Griffiths!
I totally forgot.
Thanks.
That was driving me crazy.
That method was always a great mystery to me, because the professor just kind of told us to think about how it is correct 🤔.
What if we thought too much about it and it no longer makes any sense 🤔?
I was thinking about taking a partial derivative, but it didn't "click."
Griffths has a God awful method for converting coordinate differentials.
I suggest everyone not use that method.
He actually brings out the old ##\Delta x## thing and takes the limit.
Hideous.
You use the total derivatives dx,dy, etc. aka exterior derivative and then sub-in.
 
Zap
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56
Actually, I'm still not getting it. The wiki page shows this:
{\displaystyle dV=dx\;dy\;dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}dr\;d\theta \;dz=r\;dr\;d\theta \;dz}
,
but it leaves a lot to the imagination.
I was thinking of changing coordinates, not differentials. When changing coordinates you can do the $$\hat x = \frac{ \frac{\partial y(x)}{\partial x} }{ |\frac{ \partial y(x)}{ \partial x } |}$$ thingy to find the unit vectors. I guess we are still changing coordinates, but to find the differentials, so maybe the method is similar.

I think someone messed with the OP, because it's not really legible anymore. I was just asking how to get from ##dxdy## to ##rdrd\theta##.
 
fresh_42
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Actually, I'm still not getting it. The wiki page shows this:
{\displaystyle dV=dx\;dy\;dz=\det {\frac {\partial (x,y,z)}{\partial (r,\theta ,z)}}dr\;d\theta \;dz=r\;dr\;d\theta \;dz}
,
but it leaves a lot to the imagination.
I was thinking of changing coordinates, not differentials. When changing coordinates you can do the $$\hat x = \frac{ \frac{\partial y(x)}{\partial x} }{ |\frac{ \partial y(x)}{ \partial x } |}$$ thingy to find the unit vectors. I guess we are still changing coordinates, but to find the differentials, so maybe the method is similar.

I think someone messed with the OP, because it's not really legible anymore. I was just asking how to get from ##dxdy## to ##rdrd\theta##.
I guess that someone was you together with a software bug. The MathJax code sometimes gets lost if you mix it with BB-code like "center" in the edit mode. Anyway, the calculation goes:
\begin{align*}
x&=r \cos \varphi \; , \; y= r\sin \varphi \\ &\Longrightarrow \\
\det J &= \left| \dfrac{\partial (x,y)}{\partial (r,\varphi)} \right| = \left| \begin{bmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial \varphi} \\ \dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial \varphi} \end{bmatrix} \right| = \left| \begin{bmatrix} \cos \varphi & -r \sin \varphi \\ \sin \varphi & r \cos \varphi \end{bmatrix} \right| =r\cos^2 \varphi +r \sin^2 \varphi = r \\
&\Longrightarrow \\
dA&= dx \,dy = \left(\det J \right) \,dr \, d\varphi = r \,dr \,d\varphi
\end{align*}
 
Zap
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Weird. One might think it is better just to memorize it.
I guess I'm confused on this part ##dxdy=(detJ)drdφ##.
I'm not really sure where the det J is coming from. This question has been asked a few times on google, so I will continue reading up on it, but I'm still not entirely getting it.
 
WWGD
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Weird. One might think it is better just to memorize it.
I guess I'm confused on this part ##dxdy=(detJ)drdφ##.
I'm not really sure where the det J is coming from. This question has been asked a few times on google, so I will continue reading up on it, but I'm still not entirely getting it.
Sort of properties of multilinear maps come down to, or simplify into, the DetJ statement.
 
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Zap
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Why is ## dxdy = (cos\theta dr - rsin\theta d\theta)(rcos \theta d\theta + sin\theta dr) ## not allowed? Is it because ##dx## and ##dy## are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like ##(dr)^2##.
 
fresh_42
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I guess I'm confused on this part ##dxdy=(\det J)drd\varphi##.
Try to convert a basis ##\{\,(1,0),(1,4)\,\}## into ##\{\,(2,5),(-1,7)\,\}##. The functional determinant is the factor which adjusts the volumes. If you measure in inch instead of centimeters, you apply a scaling factor, too, don't you? Now what about square inch and square centimeters?
 
fresh_42
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Why is dxdy=(cosθdr−rsinθdθ)(rcosθdθ+sinθdr)dxdy=(cosθdr−rsinθdθ)(rcosθdθ+sinθdr) not allowed? Is it because dxdx and dydy are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like (dr)2(dr)2.
It is allowed. You just used the wrong multiplication. If you insist on differential forms, then we have
\begin{align*}
dx \wedge dy &= (\cos \varphi \,dr -r\sin \varphi \, d\varphi) \wedge (\sin \varphi \, dr + r \cos \varphi \, d\varphi)\\
&= (\cos \varphi )\,dr \wedge (r\cos \varphi \, d\varphi) - (r \sin \varphi \, d \varphi) \wedge (\sin \varphi \, dr)\\
&= r\cos^2 \varphi \,dr \wedge d \varphi - r\sin^2 \varphi \,d\varphi \wedge dr \\
&= (r\cos^2 \varphi + r\sin^2\varphi )\, dr \wedge d\varphi \\
&=r \,dr \wedge d\varphi
\end{align*}
 
lurflurf
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Why is ## dxdy = (cos\theta dr - rsin\theta d\theta)(rcos \theta d\theta + sin\theta dr) ## not allowed? Is it because ##dx## and ##dy## are actually vectors and we have to do some kind of vector product? Or is it allowed, but not very useful, since it produces weird stuff like ##(dr)^2##.
You can do it that way. There is an antisymmetric product so
drdr=0 dθdθ=0 drdθ=-dθdr

dxdy=rdrdθ

The determinant give the ratio of the differentials
 
Zap
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Wow. I always wished that physics could be taught by mathematicians.
There's a lot for me to digest here, but I am learning.
 

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