# Geometric Version of Maxwell Equation related to Tensor Dual

In summary: In the expression you gave, the first and last indices are contracted, but I need three indices for the middle one.In summary, In MTW, exercise 3.15 says that if F is the EM field tensor, then ##\nabla \cdot *F## is a geometric, frame-independent version of the Maxwell equation ##F_{\alpha,\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0##. The problem asks me to demonstrate that this reduces to the Maxwell equation.

## Homework Statement

From Misner, Thorne and Wheeler's text Gravitation (MTW), exercise 3.15:
Show that, if F is the EM field tensor, then ##\nabla \cdot *F## is a geometric, frame-independent version of the Maxwell equation ##F_{\alpha,\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0##
(and similarly for the current equations)
##*F## is the dual of ##F##

## Homework Equations

The Maxwell equation ##F_{\alpha,\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0##
The dual of a vector ##J## is defined ##*J_{\alpha\beta\gamma}=J^\mu \epsilon_{\mu\alpha\beta\gamma}##
where ##\epsilon## is the Levi-Civita symbol.
The dual of a second-rank antisymmetric tensor F is defined by ##*F_{\alpha\beta}=F^{\mu\nu}\epsilon_{\mu\nu\alpha\beta}##
The dual of a third-rank antisymmetric tensor ##B## is defined by ##*B_\alpha=B^{\lambda\mu\nu}\epsilon_{\lambda\mu\nu\alpha}##

## The Attempt at a Solution

##(\nabla \cdot *F)_\beta=(*F)_{\alpha\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}##
##=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu,\alpha}##
In the second expression, the first ##\alpha## should be in the upper position. In the last expression, the final ##,\alpha## should be in the lower position. But I don't know how to get Physics Forum to accept that. On my local machine I declare \usepackage{tensor} and say F\indices{...} but that doesn't seem to work on the forum.

I know that the Maxwell equation as given above is actually 64 equations, mostly redundant. I In MTW ##F## has two indices for the components, which leads me to think ##*F## is second-rank. But when I take the divergence of the second-rank tensor I get a vector as shown in my attempt at a solution. This seems to yield four equations rather than 64.

In equation (3.3) of MTW, it has the expression ##F(u)## which makes me think F is a first-rank tensor. I am unclear what is the rank of ##F## in this problem.

##(\nabla \cdot *F)_\beta=(*F)^{\alpha}{}_{\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}##
Code:
(\nabla \cdot *F)_\beta=(*F)_{\alpha\beta,     \alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}
(\nabla \cdot *F)_\beta=(*F)^{\alpha}{}_{\beta,\alpha}=(\eta^{\alpha\gamma}F^{\mu\nu}\epsilon_{\mu\nu\gamma\beta})_{,\alpha}
##=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu}{}_{,\alpha}##
Code:
=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu,\alpha}
=\eta^{\alpha\gamma}\epsilon_{\mu\nu\gamma\beta}F^{\mu\nu}{}_{,\alpha}

Since ##F_{ab}=\nabla_a A_b - \nabla_b A_a##, it's a two-form.
In n-dimensional space[time], its Hodge dual ##*F## has (n-2) indices. So, in (3+1)-dimensions, its Hodge dual also has 2 indices.

In Eq. (3.3), ##d{\bf p}/{dt}=e{\bf F}({\bf u})## is a 4-vector equation for the Lorentz Force.
In abstract index notation, this is ##d{\bf p}_b/{dt}=e{\bf F}_{ab}{\bf u}^a##

Look at Box 3.3 on "Index Gymnastics" on page 85.

Thanks for the help on Latex coding of tensor indices. Much appreciated!

Still wondering, though, about ##\nabla \cdot *F##. If ##*F## has two indices, shouldn't ##\nabla \cdot *F## have just one index? I seem to need three.

Thanks for the help on Latex coding of tensor indices. Much appreciated!

Still wondering, though, about ##\nabla \cdot *F##. If ##*F## has two indices, shouldn't ##\nabla \cdot *F## have just one index? I seem to need three.
What do you mean by "I seem to need three"? In your expression given in your original post in the "attempt of a solution", what you wrote has only one uncontracted index (the alpha's are contracted).

The problem asks me to demonstrate that it reduces to maxwells equation ##F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0##.

That has three free indices.

The problem asks me to demonstrate that it reduces to maxwells equation ##F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0##.

That has three free indices.

Have you seen "Eq. 3.49b"
MTW said:
Show that Maxwell's "magnetic" equations
##F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha}=0##
can be rewritten in the form
##F_{[\alpha\beta,\gamma]}=0##.

Have you seen Exercise 3.13 on the Levi-Civita tensor ?
https://en.wikipedia.org/wiki/Levi-Civita_symbol

In n-dimensions, the dual of a 3-form (with 3 indices) has (n-3) indices.

(Probably overkill for this problem... but may be a useful reference: https://www.physicsforums.com/threads/algebraic-proofs-of-levi-civita-symbol-identities.919269/ )

Yes, I worked exercise 3.13.

My problem is that ##\nabla \cdot *F=0## seems to produce a formula with only one free index, whereas the exercise asks me to demonstrate that it reduces to an equation with three indices, i.e. ##F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}=0##

I seem to be missing your point, I am sorry.

Can you form the dual of ##F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\alpha,\beta}##?

##*(F_{\alpha\beta,\gamma}+F_{\beta\gamma,\alpha}+F_{\gamma\beta,\alpha})^\mu=\frac{1}{2}F_{\alpha\beta,\gamma}\epsilon^{\alpha\beta\gamma\mu}##
##=-\frac{1}{2}F_{\alpha\beta,\gamma}\epsilon^{\alpha\beta\mu\gamma}=-\frac{1}{2}\frac{\partial}{\partial x^\gamma}F_{\alpha\beta}\epsilon^{\alpha\beta\mu\gamma}=-\frac{1}{2}(\nabla \cdot *F)^\mu=0##

ok that seems to be it. Thanks.

## 1. What is the Geometric Version of Maxwell Equation related to Tensor Dual?

The Geometric Version of Maxwell Equation related to Tensor Dual is a mathematical equation that describes the relationship between the electromagnetic field and its sources in a geometrically invariant way. It is an alternative formulation of Maxwell's equations using tensor calculus.

## 2. How is the Geometric Version of Maxwell Equation related to Tensor Dual different from the traditional form of Maxwell's equations?

The traditional form of Maxwell's equations is based on vector calculus, while the Geometric Version utilizes tensor calculus. This allows for a more concise and elegant representation of the equations, making it easier to apply in different coordinate systems and geometries.

## 3. What is the significance of using tensors in the Geometric Version of Maxwell Equation related to Tensor Dual?

Tensors allow for a more general and covariant formulation of the equations, which means that they remain unchanged under coordinate transformations. This makes them more suitable for use in different reference frames and geometries.

## 4. How is the Geometric Version of Maxwell Equation related to Tensor Dual used in theoretical physics?

The Geometric Version of Maxwell Equation related to Tensor Dual is used in theoretical physics to describe electromagnetic phenomena in a more fundamental and elegant way. It is particularly useful in the study of general relativity and the unification of electromagnetism with other fundamental forces.

## 5. Are there any real-world applications of the Geometric Version of Maxwell Equation related to Tensor Dual?

Yes, the Geometric Version of Maxwell Equation related to Tensor Dual has been applied in various areas such as high-energy physics, cosmology, and quantum gravity. It has also been used in the development of new technologies, such as the theory of electromagnetism in curved spacetime for the design of advanced space propulsion systems.

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