# Infinitesimal volume element in different coordinate system

• teddd
In summary, the speaker is discussing a doubt about the transformation of infinitesimal volume elements when performing a coordinate transformation. They provide a formula for the transformation and a concrete example of a cartesian to polar coordinate transformation. They then mention a mistake in their calculation and ask for clarification on the commutation and anticommutation rules for differentials. The conversation ends with the clarification that the anticommutation rule comes from the properties of orientation and algebraic simplification.
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 above$$dxdy=rdrd\theta$$but 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!

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.

$x= r cos(\theta)$, $y= r sin(\theta)$
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:
$$\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ cos(\theta)dr & -r sin(\theta)d\theta) & 0 \\ sin(\theta)dr & r cos(\theta) d\theta & 0 \end{array}\right|= (r cos^2(\theta)+ rsin^2(\theta))drd\theta\vec{k}= r drd\theta\vec{k}$$

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

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

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)

Last edited:
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??

teddd said:
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:
$$(dz)^2 - (dx)^2 - (dy)^2$$​

teddd said:
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...

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

Thanks a lot for your help guys!

teddd said:
My $dr$ and $d\theta$ are the differentials of$$r=\sqrt{x^2+y^2}$$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 $d\theta$.

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).

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.

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

## 1. What is an infinitesimal volume element?

An infinitesimal volume element is a small volume that is used to describe the volume of a three-dimensional object. It is essentially a small cube with sides of infinitesimal lengths, and it is often denoted as dV.

## 2. Why is an infinitesimal volume element important?

An infinitesimal volume element is important because it allows us to describe the volume of complex three-dimensional objects in a precise and accurate manner. It is also used in many mathematical and scientific calculations, particularly in the field of calculus.

## 3. How is an infinitesimal volume element related to different coordinate systems?

The size and shape of an infinitesimal volume element can vary depending on the coordinate system being used. For example, in Cartesian coordinates, the infinitesimal volume element is a cube with sides of dx, dy, and dz. In spherical coordinates, it is a small wedge-shaped volume with sides of dr, dθ, and dφ.

## 4. Can an infinitesimal volume element be used in non-rectangular coordinate systems?

Yes, an infinitesimal volume element can be used in any type of coordinate system, including non-rectangular ones. As long as the coordinate system can describe three-dimensional space, the concept of an infinitesimal volume element can be applied.

## 5. How is an infinitesimal volume element calculated in different coordinate systems?

The formula for calculating an infinitesimal volume element varies depending on the coordinate system being used. In general, it involves taking the product of the differentials of the coordinates. For example, in cylindrical coordinates, the infinitesimal volume element is calculated as dV = r dr dθ dz.

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