Gravitational Analog to Maxwell's Eq.

[kcdodd]

To start, I came across this wikipedia entry on a gravitational analog to maxwell's equations: http://en.wikipedia.org/wiki/Gravitomagnetism

Restating the listed equations:

$$\nabla \cdot \mathbf {E_g} + 4\pi G \rho_g = 0$$

$$\nabla \cdot \mathbf {B_g} = 0$$

$$\nabla \times \mathbf {E_g} + \frac{\partial \mathbf {B_g}}{\partial t} = 0$$

$$\nabla \times \mathbf {B_g} - \frac{1}{c^2}\frac{\partial \mathbf {E_g}}{\partial t} + \frac{4\pi G}{c^2}\mathbf{J_g}= 0$$

Now, they also list the lorentz force analog (which I will rewrite as the force density here):

$$\mathbf{F} = \rho_g \mathbf{E_g} + \mathbf{J_g}\times 2\mathbf{B_g}$$

That factor of two in front of the magnetic analog is what my question is all about.

When I follow a procedure to derive momentum conservation by manipulating the above equations in the same way done for maxwells equations, I arrive at this:

$$\frac{1}{4\pi G} \frac{\partial}{\partial t}(\mathbf{E_g}\times\mathbf{B_g}) + \frac{1}{4\pi G}(\nabla\frac{1}{2}E_g^2 - (\mathbf{E_g}\cdot\nabla)\mathbf{E_g} - (\nabla \cdot \mathbf{E_g})\mathbf{E_g}) + \frac{c^2}{4\pi G}(\nabla\frac{1}{2}B_g^2 - (\mathbf{B_g}\cdot\nabla)\mathbf{B_g} - (\nabla \cdot \mathbf{B_g})\mathbf{B_g}) - \rho_g\mathbf{E_g} - \mathbf{J_g}\times\mathbf{B_g} = 0$$

Where the two messy parts I would naively identify as the divergence of the stress tensor of the gravity field. And the source terms as the Lorentz force analog. However, I do not get a factor of 2 in front of the JxB term, which bothers me.

And, also for completeness, for energy conservation I got:

$$\frac{c^2}{4\pi G}\nabla\cdot (\mathbf{E_g}\times \mathbf{B_g}) + \frac{1}{8\pi G}\frac{\partial}{\partial t}(E_g^2 + c^2B_g^2) - \mathbf{J_g}\cdot\mathbf{E_g} = 0$$

Both look pretty much the same as for EM, except the minus signs in front of the sources, and the constants of course. However, the absence of the 2 in the Lorentz force bothers me. I can manipulate the derivation to get a two there, however that throws off the identification of the stress tensor because the derivatives do not factor, and a factor of 2 appears in the constants as well. EG

$$\frac{1}{2\pi G} \frac{\partial}{\partial t}(\mathbf{E_g}\times\mathbf{B_g}) + \frac{1}{2\pi G}(\nabla\frac{1}{2}E_g^2 - (\mathbf{E_g}\cdot\nabla)\mathbf{E_g} - \frac{1}{2}(\nabla \cdot \mathbf{E_g})\mathbf{E_g}) + \frac{c^2}{2\pi G}(\nabla\frac{1}{2}B_g^2 - (\mathbf{B_g}\cdot\nabla)\mathbf{B_g} - (\nabla \cdot \mathbf{B_g})\mathbf{B_g}) - \rho_g\mathbf{E_g} - \mathbf{J_g}\times2\mathbf{B_g} = 0$$

Any help on understanding the problem will be much appreciated.

 

[bcrowell]

The WP article you linked to seems to answer your question, at "In some literature..."

-Ben

 

[kcdodd]

If I scale the fields by 1/2 as they suggest to get rid of the 2, then I get 1/2 there when doing the derivation. I don't think its that simple.

 

[GiftOfPlasma]

I was actually recently looking at the same WP article.

I also found this disertation interesting: http://idea.library.drexel.edu/bitstream/1860/1123/1/Medina_Jairzinho.pdf"