# General Relativity + Electrodynamics #2

1. Sep 4, 2006

### JustinLevy

Some of the comparisons between gravity (in the weak field limit) and electromagnetism were really interesting (https://www.physicsforums.com/showthread.php?t=80710 ). Unfortunately, that got locked due to someone presenting their pet theory. The topic seems too good to end, so let's hope we can stay on topic this time.

Referring all the way back to:
https://www.physicsforums.com/showpost.php?p=667979&postcount=12
Can you give some suggestions how one would go about figuring out how g and H do transform? I don't really understand why it would be different, as shouldn't a "point charge" (rho = m * delta function) be the same in all reference frames? So if the "charge" is invarient, and the equations are already "equivalent" to Maxwell's equations (so they have Lorentz symmetry), what makes these any different at all? I would expect them to transform the same.

EDIT: Or are we supposed to use relativistic mass density instead of invarient mass density? I guess that would make the "charge" not invarient and then g,H wouldn't transform like E,B.

Last edited: Sep 4, 2006
2. Sep 4, 2006

### pervect

Staff Emeritus
g and H can be written in terms of the Christoffel symbols, so they transform like Christoffel symbols do.

From the Harris reference, "Analogy between General Relativity and Electromagnetism for slowly moving particles in weak fields", for time-independent fields one can write g and H in terms of the Christoffel symbols as

$$\Gamma^\mu{}_{0\beta} = \frac{1}{2 c^2} \left| \begin{array}{cccc} 0 &-2g_x & -2g_y & -2g_z\\ -2 g_x & 0 & -H_z & H_y \\ -2 g_y & H_z & 0 & -H_x \\ -2 g_z & -H_y & H_x & 0 \end{array} \right |$$

Christoffel symbols don't transform as tensors. I found an ugly expression for how they do transform on the Wikipedia

http://en.wikipedia.org/wiki/Christoffel_symbol#Change_of_variable

which, for convenience, and because of the mutability of Wiki, I'll quote below (though I haven't verified it personally)

$$\overline{\Gamma^k {}_{ij}} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r {}_{pq}\, \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial y^i \partial y^j}$$

So the short version is - they transform in a rather complicated and ugly way :-(. I presume the above non-linear transformation can be "linearized", but I haven't seen or worked out the details.

Last edited: Sep 4, 2006
3. Sep 4, 2006

### JustinLevy

Wow, I don't blame you. I'm not sure I'll work out the details either :)

Thanks for such an indepth answer. I'll definitely have to read up on that, and then maybe I'll try "linearizing" the transformation in some approximation. Anyway, thanks again, very much appreciated.

4. Sep 4, 2006

### masudr

I'm sure you guys are smarter than me (and I haven't read all of the previous threads) but I thought I'd just add this.

Christoffel symbols don't transform as tensors for a very good reason. The derivative of a (non-zero rank) tensor does not transform like a tensor. However, when one sees the form of a derivative in a new co-ordinate system, there is a specific, ugly term that remains. These are precisely the Christoffel symbols, and so the covariant derivative definition includes the Christoffel symbol so that the covariant derivative transforms nicely like a tensor.

i.e. the punch line is that christoffel symbols exactly counter the extra ugly terms that appear in the transformation of a derivative; hence christoffel symbols are ugly.

5. Sep 4, 2006

Staff Emeritus

And then instead of sticking the ugly terms on the end of every derivation, you define them to be part of the derivative and give it a new name: covariant derivative, because with it, derivatives of tensors are now covariant. And then you learn about connections...

6. Sep 5, 2006

### JustinLevy

Ouch, I think I am misunderstanding something very basic here. I thought the point of Gravitoelectromagnetism was that in a weak field limit (and slow moving masses) we could consider instead the equivalent "Maxwell's equations" for g,H and the equivalent "lorentz force law" for "charges", all in flat space, instead of solving the full Einstein equations. Is that not the intent?

7. Sep 5, 2006

### masudr

Well, a flat space would have the Minkowski metric for g. So a non-Minkowski metric implies a non-flat space.

8. Sep 5, 2006

### JustinLevy

Here g is a vector, not the metric. Wikipedia instead uses E and B (for g,H) to prevent confusion (and make it "look" more like Maxwell's equations). http://en.wikipedia.org/wiki/Gravitomagnetism Maybe we should adopt that convention for this thread.

I hope I didn't completely misunderstand your comment.

Last edited: Sep 5, 2006
9. Sep 5, 2006

### masudr

I'm sorry, the fault is all mine.

10. Sep 5, 2006

### pervect

Staff Emeritus
Will, SR doesn't have any nonlinear differential equations, but OK

Huh? Assuming that the unclear PT is "parity time", this doesn't have anything to do with SR and GR, but could have something to do with quantum mechanics

Unless you framulate the blogistan, with slithy toves and onomotapoeia.

Sorry, this is just gibberish.

11. Sep 5, 2006

### JustinLevy

Okay, let's just ignore that post and move on.

...

So if we agree E,B (g,H) are fields we are supposed to use in flat space / inertial frames, and all inertial frames are equivalent, then these equations must be valid in all inertial frames (where the assumption of slow moving masses holds true at least). Thus the fields must have lorentz symmetry, and do transform just like the electromagnetic E,B.

Correct?

I think the problem in the beginning was that it was mentioned that the linearized approximation which gives us the gravitoelectomagnetic (GEM) field equations came from the Christoffel symbols. So we thought that the fields would transform in a complicated way as Christoffel symbols do instead of like a tensor does.

But look back at the Christoffel symbol transformation equation. If we are approximating space as flat, we just get:

$$\overline{\Gamma^k {}_{ij}} = \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r {}_{pq}\, \frac{\partial y^k}{\partial x^r} + \frac{\partial y^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial y^i \partial y^j} \approx \frac{\partial x^p}{\partial y^i}\, \frac{\partial x^q}{\partial y^j}\, \Gamma^r {}_{pq}\, \frac{\partial y^k}{\partial x^r}$$

So in this approximation the Christoffel symbol does transform just like a tensor, and so E,B (g,h) do transform like the electromagnetic fields.

Most of these concepts are new to me, so let me know if my arguement is flawed. But it does look to me like my original intuition worked out on this problem.