Infinitesimal volume element in different coordinate system

[teddd]

I've already post this, but I've done it in the wrong section!

So here I go again..

I've a doubt on the way the infinitesimal volume element transfoms when performing a coordinate transformation from x^j to x^{j'}
It should change according to dx^1dx^2...dx^n=\frac{\partial (x^1,x^2...x^n)}{\partial (x^{1'}x^{2'}...x^{n'})}dx^{1'}dx^{2'}...dx^{n'}where \frac{\partial (x^1,x^2...x^n)}{\partial (x^{1'}x^{2'}...x^{n'})} is the Jacobian of the transformation.So i tried to do this in a concrete example: the transformation between cartesian x,y to polar r,\theta coordinates.
The jacobian of this transformation is r and so, according to what I've written abovedxdy=rdrd\thetabut since dr=cos\theta dx+sin\theta dy and d\theta=-\frac{1}{r}sin\theta dx+\frac{1}{r}cos\theta dy i get to dV=r(cos\theta dx+sin\theta dy)(-\frac{1}{r}sin\theta dx+\frac{1}{r}cos\theta dy)=(-sin\theta cos\theta dx^2+sin\theta cos\theta dy^2+cos^2\theta dxdy-sin^2\theta dxdy)and this is not equal to dxdy, the volume element in cartesian coordinate, as it should be!

Where am I mistaking?

Thanks!

 

[dalcde]

Are you sure that \mathrm{d}r=\cos\theta\;\mathrm{d}x+\sin\theta\; \mathrm{d}y? It seems to me that
\mathrm{d}r=\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}.
Also, the formula for the angle also seems to be wrong. Please explain your formulas.

 

[HallsofIvy]

x= r cos(\theta), y= r sin(\theta)
so that dx= cos(\theta)dr- r sin(\theta)d\theta and
dy= sin(\theta)dr+ r cos(\theta)d\theta

The differential of area is
dxdy= (cos(\theta)dr- r sin(\theta)dy)\times(sin(\theta)dr+ r cos(\theta)d\theta)
where the "\times is the cross product:
\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ cos(\theta)dr & -r sin(\theta)d\theta) & 0 \\ sin(\theta)dr & r cos(\theta) d\theta & 0 \end{array}\right|= (r cos^2(\theta)+ rsin^2(\theta))drd\theta\vec{k}= r drd\theta\vec{k}

 

[teddd]

My dr and d\theta are the differentials ofr=\sqrt{x^2+y^2}and\theta=arctg\frac{y}{x}

Isn't this the right way to perform the calculation?

 

[Hurkyl]

Two important properties of differential forms:

dx \, dx = 0
dx \, dy = -dy \, dx​

Try doing your original calculation again, now that you know that the product of differential forms don't commute. (In particular, be careful when you expand the product so as not to reorder things!)

EDIT: I guess I really out to also point out that

f \, dx = dx \, f​

where f is real-valued. (or complex-valued)

 

[teddd]

I knew that there was some sort of commutation rule for infinitesimal!

The dx^{n}=0 with n>1 propriety is ok (they re higher order terms in a taylor expansion of the function, so I guess they're trascurable by definition); but where does the anticommutation rule come from??

 

[Hurkyl]

The dx^{n}=0 with n>1 propriety is ok (they re higher order terms in a taylor expansion of the function, so I guess they're trascurable by definition); but where does the anticommutation rule come from??

Geometrically, it deals with orientation.

Algebraically, it's the same rule. Let z = x+y, and simplify this expression in two different ways:

(dz)^2 - (dx)^2 - (dy)^2​

 

[Hurkyl]

The dx^{n}=0 with n>1 propriety is ok (they re higher order terms in a taylor expansion of the function, so I guess they're trascurable by definition); but where does the anticommutation rule come from??

Geometrically, it deals with orientation.

Algebraically, it's the same rule. Let z = x+y. Then:

(dz)^2 = (dx + dy)^2​

and so...

 

[teddd]

So becaouse (dz)^2 must be zero, we have dxdy=-dydx !

Thanks a lot for your help guys!

 

[HallsofIvy]

My dr and d\theta are the differentials ofr=\sqrt{x^2+y^2}and\theta=arctg\frac{y}{x}

Isn't this the right way to perform the calculation?

Since it is dxdy you want to replace, it make more sense to me to calculate dx and dy in terms of dr and d\theta.

 

[dalcde]

Isn't \mathrm{d}y\mathrm{d}x equal to 0? (I haven't really learned the rules but it seems more intuitive that way).

 

[HallsofIvy]

No, it isn't. It is, however, a "second order differential" so that while "dx" and "dy" separately are differentials of length, "dxdy" is a differential of area.

 

[teddd]

You're absolutely right HallsofIvy; but I wanted to check the rule by doing the backwards calculation!