# Formula for non-linear differential forms

1. May 14, 2005

### kleinwolf

Who has any litterature about non-linear differential forms, especially for example if

I would like to compute the following :

$$(dx\wedge dy)(dx\wedge dy\wedge dz)$$

is it equal to $$(dx)^2\wedge (dy)^2\wedge dz$$ ??

2. May 14, 2005

### dextercioby

Technically

$$dx \wedge dx =0$$

.So you're asking for something new.

Daniel.

3. May 14, 2005

### quetzalcoatl9

I am assuming that by this you mean the tensor product?

$$(dx\wedge dy)\otimes(dx\wedge dy\wedge dz)$$

where:

$$dx\wedge dy = dx\otimes dy - dy\otimes dx$$

$$(dx\wedge dy) \wedge dz = (dx\otimes dy - dy\otimes dx)\otimes dz - dz\otimes (dx\otimes dy - dy\otimes dx) = dx\otimes dy \otimes dz - dy \otimes dx \otimes dz - dz \otimes dx \otimes dy + dz\otimes dy \otimes dx$$

lets call $\alpha = dx\wedge dy$ and $\beta = dx\wedge dy \wedge dz$, so

$$\beta(\vec{u}, \vec{v}, \vec{w}) = dx(\vec{u}) dy(\vec{v})dz(\vec{w}) - dy(\vec{u})dx(\vec{v})dz(\vec{w}) - dz(\vec{u})dx(\vec{v})dy(\vec{w}) + dz(\vec{u})dy(\vec{v})dx(\vec{w})$$

is this what you are looking for? if so, then you would need to compute $\alpha \otimes \beta$, keeping the order of the terms correct so that when you plug in your vectors everything works out.

Last edited: May 14, 2005
4. May 14, 2005

### quetzalcoatl9

i believe that by saying $(dx)^2$ that he is using diff. forms shorthand for $dx \otimes dx$ (that is, evaluating $dx$ on a vector twice and multiplying the two real number results together), which is not zero.

5. May 15, 2005

### kleinwolf

In fact i'm just coming to this when I want to compute the area of a manifold, that I found in the thread : https://www.physicsforums.com/showthread.php?t=67268 [/URL]

It's just for example : $$area=\sqrt{(dx\wedge dy)^2+(dx\wedge dz)^2+(dy\wedge dz)^2}$$
this contains forms like : $$(dx\wedge dy)(dx\wedge dy)$$....

Then suppose I want to compute the exterior derivative of area, I get terms of the style : $$(dx\wedge dy)d(dx\wedge dy)$$...

Or can we not say that $$d(f^2)=2fdf$$ ...

Last edited by a moderator: Apr 21, 2017
6. May 15, 2005

### quetzalcoatl9

heres what i did.

i expanded out $(dx)^2\wedge (dy)^2 \wedge dz$ completely, as was done above. the terms did not appear to match.

i then evaluated both $(dx\wedge dy)(dx\wedge dy\wedge dz)$ and $(dx)^2\wedge (dy)^2 \wedge dz$ against 3 vectors $\vec{u}, \vec{v}, \vec{w}$. So, for example:

$$(dx\wedge dy) (\vec{u}, \vec{v}) = (dx\otimes dy - dy\otimes dx) (\vec{u}, \vec{v}) = dx(\vec{u})dy(\vec{v}) - dy(\vec{u})dx(\vec{v}) = u_x v_y - u_y v_x$$

where $u_x$ is the component of $\vec{u}$ in the "x" direction, etc. I did this for the entire expansion, it took around 10 minutes. if i had to verify some non-linear form, this is probably how i would do it. there maybe some easier way, but im not aware of it.

unless i made a mistake, i found the two forms were not equal.

here it is:

$$(dx\otimes dx \wedge dy\otimes dy)\wedge dz = dx\otimes dx \otimes dy \otimes dy \otimes dz - dy \otimes dy \otimes dx \otimes dx \otimes dz - dz\otimes dx \otimes dx \otimes dy \otimes dy + dz\otimes dy \otimes dy \otimes dx \otimes dx)$$

evaluated on the vectors, gives:

$$u_x u_x v_y v_y w_z - u_y u_y v_x v_x w_z - u_z v_x v_x w_y w_y + u_z v_y v_y w_x w_x$$

which is not what i got when expanding the original form in question and evaluating.

Last edited: May 15, 2005
7. May 15, 2005

### kleinwolf

Ok, I think I begin to see what is about...my problem is that I need to compute

$$\int_Md(area)=\int_M\frac{(dx\wedge dy)d(dx\wedge dy)+...}{area}$$

so that I have a 5-form as numerator, and a 2-form at the denominator....but the thing is like if I need to compute

$$\int f(x,y,z)\frac{dx^2dy^2dz}{\sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2}}$$

Whatever it is, even if the notation is right, I have no idea what it means and how to compute it....

ANother question is : is $$\int_0^1f(x)dx^2$$ the double primitive of f(x) evaluated at 1 minus 0 ??

Since I'm in the notation problems, what in fact is the notation

$$\frac{d^2f}{dx^2}$$ this has nothing to do with $$d^2=0$$
??

Last edited: May 15, 2005
8. May 15, 2005

### quetzalcoatl9

No, because with the exterior derivative, you aren't taking the derivative with respect to any particular variable.

$$d(f) = \frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy$$
$$d(df) = \frac{\partial{f^2}}{\partial{y}\partial{x}}dydx + \frac{\partial{f^2}}{\partial{x}\partial{y}}dxdy = 0$$

since partial derivatives commute and $dxdy = -dydx$

back to your original problem: why don't you just compute the exterior derivatives of all of your coordinates, plug them all into the area equation (wedging them and squaring them when appropriate - hopefully after all of the calculations most of your terms will drop out - and if not, given that your coordinates are in terms of trig functions maybe some identity could be used to simply) and then evaluate the integral?

i realize that this will involve alot of calculation, but such is the nature of your problem i think.

p.s. - are you working on something in theoretical physics?

9. May 15, 2005

### kleinwolf

Ok, so I need to make a distinction between exterior derivative d and differentiation d..

$$f(x,y)$$
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$

exterior : $$d^2f=d(df)=\frac{\partial^2 f}{\partial x^2}dx\wedge dx+\frac{\partial^2f}{\partial x\partial y}dy\wedge dx+\frac{\partial^2 f}{\partial y\partial x}dx\wedge dy+\frac{\partial^2 f}{\partial y^2}dy\wedge dy}=0$$

differential : $$d^2f=\frac{\partial^2 f}{\partial x^2}dx^2+2\frac{\partial^2f}{\partial x\partial y}dxdy+\frac{\partial^2 f}{\partial y^2}dy^2$$

the latter being not always 0....

For your proposition that's exactly what I started to do, when I realized I'll fall on something like

$$\int_0^\pi\int_0^{2\pi}\int_0^Rf(r,\theta,\phi)\frac{rdr^2d\theta^2d\phi+....}{\sqrt{r^2dr^2d\theta^2+...}}$$

(like some previous post)....which I don't know how to integrate....

But anyhow for my originial problem (I want to know if the boundary of a 3d manifold is empty), it just suffices that the integral vanishes, so I could forget the denominator with a square root, and integrate only over the numerator...the problem is apparently that I need to double integrate if I find something like dr^2...?

I don't work on anything particular.

10. May 16, 2005

### quetzalcoatl9

I would consider the exterior derivative to be a generalized derivative...which we can use to do calculus when the manifold is the standard $R^3[/tex] and also when the manifold is something more complicated. so in that sense, there is nothing different about the exterior derivative and the normal derivative from calculus. the 2nd differentiation that you have doesn't make any sense to me. If you were integrating [itex]dxdy$ versus $dydx[/tex] you would get a different answer. we can't just say that [itex]dxdy=dydx$

not only that, if one was trying to integrate with respect to another coordinate system then there would be a change of sign due to the Jacobi determinant:

$$\int\int_R f(x,y) dx dy = \int\int_G f(x(u,v), y(u,v)) |\frac{\partial{(x,y)}}{\partial{(u,v)}}| du dv = -\int\int_R f(x,y) dy dx$$

11. May 16, 2005

### Doodle Bob

Last edited by a moderator: Apr 21, 2017
12. May 16, 2005

### kleinwolf

Well it depends which product you consider :

$$dx\wedge dy=(-1)^{mn}dy\wedge dx$$ with the wedge product of m and n differential forms.

$$dxdy=dydx$$ with the normal product of differentials

Just look at the transformation : x'=y, y=x'...then the determinant of the Jacobian is -1, but we have not to forget to take the absolute value : look :

$$\int_0^1(\int_0^1dx)dy=\int_0^1(\int_0^1dy)dx=1$$

and not -1 due to the Jacobi determinant.....so that absolute value is needed.

Bascially the area element of a sphere is $$r^2\sin(\theta)d\phi d\theta$$...which is not linear in $$\theta$$....

Last edited: May 16, 2005
13. May 16, 2005

### quetzalcoatl9

it just so happened to work out in that example, but this is not the case in general. try something like

$$\int_0^2\int_0^1 x dx dy = 1$$

versus

$$\int_0^2\int_0^1 x dy dx = 2$$

they're not the same. there is no change of sign because we are only working in one coordinate system - we are not coordinate independent in this case.

the absolute value was thrown in there when they taught us elementary calculus to hide the fact that the orientation matters.

14. May 16, 2005

### kleinwolf

Well I dont' understand your calculation....(the boundary of the variable is linekd to that variable..if you make a change of var, then the boundaries change too) :

$$\int_0^2\int_0^1xdxdy=\int_0^1\int_0^2xdydx=1$$

that's why the notation $$\int_0^2dy\int_0^1dx x$$ is clearer, because we know which variable which boundary has.

Or if you want : $$\int_0^2(\int_0^1xdx)dy=\int_0^1(\int_0^2xdy)dx=\int_0^1x(\int_0^2dy)dx$$

I think the problem you want to point out is that if the surface we integrate on is not orientable, then the calculation of the area is not defined (or maybe not signed)...I don't know

Last edited: May 16, 2005
15. May 16, 2005

### quetzalcoatl9

ok, ok, yes you are right about the boundaries. but still:

$$dx dy = |\frac{\partial{(x,y)}}{\partial{(u,v)}}| du dv$$

must be true if you want to claim coordinate independence, and therefore the order of appearance of the differentials matters.

for example: let $x = u+v$ and $y = u^2 - v[/tex] $$d(x) = du + dv$$ $$d(y) = 2udu - dv$$ $$dxdy = (du+dv)(2udu - dv) = 2ududu - dudv + 2udvdu - dvdv = 0 - dudv + 2udvdu - 0 = (-1-2u)dudv$$ which is the same as the terms one would get using the more familiar calculus integral. changing the order of [itex]dxdy$ would obviously change the sign. there is no difference between the exterior calculus and normal calculus, the exterior calculus is more general.