# I Green's theorem in tensor (GR) notation

1. Apr 3, 2017

### davidge

Hi. I was trying to translate the divergence theorem and the Green's theorem to tensor notation that we use in Relativity. For the divergence theorem, it was easy (please tell me if I'm wrong in the below derivation). I'm using the standard electromagnetic tensor $F_{\mu \nu}$ in place of the individual $E$ and $B$ vector fields. And I haven't written the various integral signs in the multiple integrals below. Since $\vec{\nabla} \cdot \vec{E} \to \partial_{\mu}E^{\mu}$ for a vector field $E$,

$$\int F_{\mu \nu}dx^{\mu}dx^{\nu} = \int \partial_{\rho}F_{\mu \nu} dx^{\mu}dx^{\nu}dx^{\rho} = \int \partial F_{\mu \nu} dx^{\mu}dx^{\nu}$$
The solution is evident.

Now, I'm having trouble in deriving the Green's theorem. I'm not sure if the correct translation for the "curl" of $F_{\mu \nu}$ is $$\epsilon^{\mu \kappa}{}_{\sigma} \epsilon^{\nu \lambda}{}_{\rho} \partial_{\lambda} F_{\mu \nu} = J_{\sigma \rho}$$ ($\epsilon$ = 1, for indices 123, 1 and -1 for even and odd permutation of them and 0 otherwise.) It will become evident below why I inserted that "don't matching" $\kappa$ - index in the previous expression.
An attempt to the Green's equality then would be
$$\int \epsilon^{\nu \lambda}{}_{\rho} dx^{\rho} \int \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} dx^{\sigma} = \int F_{\rho \sigma}dx^{\rho}dx^{\sigma}dx^{\kappa}$$

Now, for this to be true, we must have

$$\epsilon^{\nu \lambda}{}_{\rho} \epsilon^{\mu \kappa}{}_{\sigma} \partial_{\lambda} F_{\mu \nu} = \int F_{\rho \sigma} dx^{\kappa}$$

Is this correct?

Last edited: Apr 3, 2017
2. Apr 3, 2017

### Orodruin

Staff Emeritus
It is not clear to me what your integrals stand for, nor whether they are 3-dimensional, 2-dimensional, over closed surfaces, etc. You will need to be more accurate with your notation. It is also not clear to me what you want to represent with your integrals.

The only integral theorem you will generally need is Stokes' theorem for integration of differential forms.

3. Apr 3, 2017

### davidge

Well, I got motivated by looking the Maxwell's equations. I just want to represent them in the notation we use in General and Special Relativity. As you may have wounded, $E$ and $B$ in my first post represent the electric and magnetic fields.

4. Apr 3, 2017

### davidge

Just realized that what I mean is the Stokes' theorem, not the Green's theorem.

5. Apr 3, 2017

### Orodruin

Staff Emeritus
I am sorry, your notation is still unclear. I am not referring to the curl theorem (aka Stokes' theorem), I am referring to Stokes' theorem for differential forms. A priori, these integral theorems do not have a particular connection to electromagnetism.

6. Apr 3, 2017

### davidge

In the above

one dx = line element
two dx = area element
three dx = volume element

I guess they should be evaluated over closed surfaces to be consistent with the theorems. Not knowing it is part of my thread.

7. Apr 3, 2017

### Orodruin

Staff Emeritus
What worries me is your way of using the differentials and contracting them with the tensors. It is very reminiscent of integrals of differential forms, but it does not seem as if you are familiar with them. The last step of your first relation is not correct - it is even dimensionally inconsistent.

Stokes' theorem for differential forms relates an integral of a p-form over a p-dimensional closed surface to the integral of its exterior derivative over the p+1-dimensional volume enclosed by the surface. All of the integrals cannot be over closed surfaces/volumes.

If you just want the divergence theorem in N dimensions, it is given by
$$\int_V (\partial_\mu J^\mu) dV = \oint_S J^\mu dS_\mu,$$
where $V$ is an N-dimensional volume and $S$ its boundary surface. Furthermore $dV$ is the N-dimensional volume element and $dS_\mu$ the N-1-dimensional surface element. For the curl theorem, it depends a bit by exactly what you consider to be the generalisation of the curl theorem.

8. Apr 3, 2017

### davidge

That's because the integrands in question are inner products, like $\vec{E} \cdot d\vec{A}$. So I thought it would be valid to multiply the correspondent components of the quantities, considering of course that the inner product should be taken between a vector and a covector, with the differential forms being the covectors in question and also considering the metric.

Indeed. Thank you for pointing it out.

Ok. Then can you write down in tensor notation, by considering your definition of the generalisation of the curl, the equations (or give me a link for a website) $$\vec{\nabla} \times \vec{E} = - \partial \vec{B} / \partial t \\ \vec{\nabla} \times \vec{B} = \mu_{0}( \epsilon_{0} \partial \vec{E} / \partial t + \vec{j})$$

Last edited: Apr 3, 2017
9. Apr 3, 2017

### pervect

Staff Emeritus
There are two usual way of writing this in tensor notation. One way starts out with defining the Maxwell tensor, usually written as *F, also known as the "hodges dual" of the Faraday tensor.

$$*F^{\alpha\beta} = \frac{1}{2} \epsilon^{\alpha\beta\gamma\delta}\, F_{\gamma\delta}$$

where $\epsilon^{\alpha\beta\gamma\delta}$ is the Levi-Civita symbol. The Levi-Civita symbol is zero if any of it's components are equal. If the components are all unequal, it has a + sign for even permutations and an odd sign for odd permutations.

Given this defintion, we write both of these equations as the single tensor equation

$$\partial_\alpha (*F^{\alpha\beta}) = 0$$

The other way doesn't need us to define the Maxwell tensor, but uses bracket notation to indicate a tensor anti-symmeterization operator.

$$F_{[\alpha\beta, \gamma]} = 0$$

The comma here represents partial differentiation by $\gamma$, the bracket notation starts to look ugly if one uses $\partial_\gamma$ instead of the comma notation.

If you're not familiar with the bracket notation for tensor anti-symmeterization, it means that one adds together all six permutations of $\alpha, \beta, \gamma$ with even permutations having a plus sign, and odd permutations having a minus sign, and dividing this sum by the number of permutations (six).

A wiki mention of both ways can be found in wiki at <<link>>, these are the homogeneous Maxwell's equations mentioned in the wiki article. They are homogeneous because the right hand side is zero, as opposed to the inhomogeneous equations in which the right hand side is nonzero. The imhogeneous Maxwells equations, in contrast, are the one that relate the E-field and the B-field (or the Faraday tensor in tensor notation) to the charge and current density.

Let me just add here that the curl is only defined in three dimensions, it takes two vectors (or one bi-vector) as an input, and outputs a vector orthogonal to both input vectors. The dual operation as used above maps a bi=vector into another bi-vector, so it works in four dimensions. A rotation in three space would be represented as bi-vector with one component pointing in the time direction, and the other component pointing in the spatial direction of the curl.

Last edited: Apr 3, 2017
10. Apr 3, 2017

### davidge

Thanks @pervect. Why that factor of $\frac{1}{2}$ in the definition of the Maxwell Tensor? And how it looks like when we put it in a integral?

11. Apr 3, 2017

### George Jones

Staff Emeritus
It would be a good exercise to work an example. What is $*F^{01}$?

12. Apr 3, 2017

### davidge

It would be $\frac{1}{2}(F_{2 3} - F_{3 2})$, which corresponds to the anti-symmetric part of $F_{2 3}$. Am I right?

13. Apr 3, 2017

### George Jones

Staff Emeritus
Yes. The Faraday tensor is anti-symmetric, i.e., $F_{\mu \nu} = -F_{\nu \mu}$, so ...

14. Apr 3, 2017

### davidge

Aw, thanks. I understand it now.

15. Apr 3, 2017

### Orodruin

Staff Emeritus
Just to say that this is not the two equations quoted by the OP (it has four components compared to the OP's total of six - also OP's has an inhomogeneous equation). It is the first one together with $\nabla\cdot\vec B = 0$. The other is $\partial_\mu F^{\mu\nu} = J^\nu$, which also contains $\nabla\cdot\vec E = \rho$ (in reasonable units).

16. Apr 4, 2017

### vanhees71

Perhaps we should start with defining the integrals right first. The key is the Levi-Civita tensor, which is given by
$$\epsilon_{\mu \nu \rho \sigma}=\sqrt{-g} \Delta_{\mu \nu \rho \sigma},$$
Where $\Delta_{0123}=1$ and otherwise it's totally antisymmetric under reordering of its indices; $g=\det (g_{\mu \nu})$.

Then let $S$ be a $d$ ($d \leq 4$) dimensional hypersurface in the Lorentzian manifold. Let $x^{\mu}=x^{\mu}(q^1,\ldots,q^d)$ be a parametrization of this hypersurface. Then the integral over an (antisymmmetric) tensor field $F^{\mu_{d+1}\ldots \mu_{4}}$ or rank $4-d$ over this hypersurface is given by
$$\int_S \mathrm{d} \Sigma_{\mu_{d+1}\ldots \mu_4} F^{\mu_{d+1}\ldots \mu_4} =\frac{1}{(4-d)!} \int_D \mathrm{d} q^{1} \cdots \mathrm{d} q^{d} \frac{\partial x^{\mu_1}}{\partial q^1} \cdots \frac{\partial x^{\mu_d}}{\partial q^d} \epsilon_{\mu_1\ldots \mu_d \mu_{d+1} \ldots \mu_4} F^{\mu_{d+1}\ldots \mu_4},$$
and this is a scalar by construction. Here $D \subseteq \mathbb{R}^d$ is the domain of the parametrization of the hypersurface.

CAVEAT: One must be careful concerning the signs with the Levi-Civita tensor. Raising the indices gives
$$\epsilon^{\alpha \beta \gamma \delta} =\sqrt{-g} g^{\mu \alpha} g^{\nu \beta} g^{\rho \gamma} g^{\sigma \delta} \Delta_{\mu \nu \rho \sigma}=\sqrt{-g} \frac{1}{g} \Delta^{\alpha \beta \gamma \delta} = -\frac{1}{\sqrt{-g}} \Delta^{\alpha \beta \gamma \delta},$$
where again $\Delta^{0123}=+1$ and otherwise totally antisymmetric under reordering of the indices. It depends on the book, which sign convention is chosen. In many QFT books in SRT the sign convention of the Levi-Civita tensor is the opposite!

17. Apr 6, 2017

### davidge

Thanks @vanhees71. I have two questions:
1 - What does $d \Sigma_{\mu_{d+1}... \mu_{4}}$ mean? And what name can we give to it?
2 - Why the indices of your tensor $F$ stops at 4 (i.e. $\mu_4$)?

18. Apr 6, 2017

### davidge

I don't understand, because I have written explicitaly all the components of the right hand side and it seems that the integrand vanishes, since $F^{\mu \nu} = -F^{\nu \mu}$. (OOPS, I'm sorry, I did it for the case $d = 2$, not 3 as said in the image below. Every other thing is for d = 2, though.)

Last edited: Apr 6, 2017
19. Apr 7, 2017

### vanhees71

$d \Sigma_{\mu_{d+1}... \mu_{4}}$ is the hypersurface element. It's analogous to the surface-normal vectors in 3D vector calculus: There it's a vector perpendicular to the surface with the length given by the area of the surface element it belongs to.

The indices stop at $\mu_4$ because spacetime is four-dimensional. You can thus have only 1D (curves), 2D, 3D hypersurfaces and 4D volumes.

20. Apr 7, 2017

Thank you