Is energy conserved in Minkowski space with a time-varying electric field?

[DavidAlan]

A friend and I had an interesting thought and would like to know if it has any consequences.

It is a well known fact that a time-varying electric field is non-conservative, it has a time-dependent Hamiltonian, blah, blah, blah, blah. I'll give this a standard treatment to set up the punchline,

Suppose you have some charged particle in this field, you traverse a closed loop and your net work is nonzero.
\oint_{\partial S}\mathbf{E}\cdot d\mathbf{l}\neq0.

However, while energy is NOT conserved, we have Maxwell's equations telling us that
\mathbf{E}=-\nabla\phi-\partial_{0}\mathbf{A}.

Now, if we define an analogous gradient operator in Minkowski space; I think this is just a covariant derivative with the Minkowski metric, then Stoke's theorem applied to a closed path readily shows that the time-varying electric field IS conservative (mathematically speaking) in Minkowski space. The closed loop integral should be zero because if we integrate with this metric, the gradient we have defined appears on the RHS of Maxwell's equation above. Mind you, you must consider the electric field as being the gradient of the four-potential.

Regardless, energy cannot be conserved. But something, fictitious or not, IS conserved. Does anyone know of any consequences of this?

If I recall, E=mc^2 is merely a consequence of Lorentz invariance. Would one expect to get this out of this song and dance? I think this would make for an interesting problem, and I cannot think of a way to work it out.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Phrak]

A friend and I had an interesting thought and would like to know if it has any consequences.

It is a well known fact that a time-varying electric field is non-conservative, it has a time-dependent Hamiltonian, blah, blah, blah, blah. I'll give this a standard treatment to set up the punchline,

Suppose you have some charged particle in this field, you traverse a closed loop and your net work is nonzero.
\oint_{\partial S}\mathbf{E}\cdot d\mathbf{l}\neq0.

However, while energy is NOT conserved, we have Maxwell's equations telling us that
\mathbf{E}=-\nabla\phi-\partial_{0}\mathbf{A}.

Now, if we define an analogous gradient operator in Minkowski space; ...

Yes, we can form an analogous gradient--or differential operator in Minkowski space, and even better, over generalized coordinates within special relativity. You should take a quick look at the Faraday tensor with lower indices in the wikipedia article, here.

http://en.wikipedia.org/wiki/Electromagnetic_tensor"

Also in this article,

F_{\mu \nu} = \frac{\partial A_\mu}{\partial x^\nu} - \frac{\partial A_\nu}{\partial x^\mu}

Here, A is four dimensional. A_0 is \phi. Zero indexes the temporal part of a four dimensional potential that units A and \phi into a four dimensional vector. Indices 1 through 3 are the usual magnetic potentials with lower indices.

...I think this is just a covariant derivative with the Minkowski metric, then Stoke's theorem applied to a closed path readily shows that the time-varying electric field IS conservative (mathematically speaking) in Minkowski space.

As you correctly surmised, these are covariant derivates of E as well as B.

The closed loop integral should be zero because if we integrate with this metric, the gradient we have defined appears on the RHS of Maxwell's equation above. Mind you, you must consider the electric field as being the gradient of the four-potential.

I'm not sure how you think that the integral of E dot dl over a closed loop should be zero after substitution of E with its derivates in A, however.

In fact, I don't know how it should go. And thank you for the provocative quesiton. What do you get after substitution?

Regardless, energy cannot be conserved. But something, fictitious or not, IS conserved. Does anyone know of any consequences of this?

If I recall, E=mc^2 is merely a consequence of Lorentz invariance. Would one expect to get this out of this song and dance? I think this would make for an interesting problem, and I cannot think of a way to work it out.