I What are the consequences to electric and magnetic in GR?

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1. Feb 14, 2019 at 2:28 AM

Boltzman Oscillation

So I just learned that in general relativity Magnatic and Electric fields are dependent on the observer. What are some consequences as a result?

2. Feb 14, 2019 at 2:38 AM

Orodruin

Staff Emeritus
This is true in special relativity as well. Also since electromagnetism is incompatible with Galilei transformations, that really means that there is no theory where this does not happen.

3. Feb 14, 2019 at 2:47 AM

Boltzman Oscillation

Electromagnetism is incompatible with Galilei transformations? I just got into the subject so I dint know that. I wont learn anymore relativity until a year or two, unfortunately. What do you mean there is no theory where it does not happen?

4. Feb 14, 2019 at 2:51 AM

Orodruin

Staff Emeritus
Maxwell's theory of electromagnetism is inherently relativistic. It is not Galilei invariant as it would single out a particular frame, the frame where the speed of light is c. However, it is Lorentz invariant. This inconsistency is what led to the development of special relativity in the first place. Einstein's 1905 paper was titled "On the electrodynamics of moving bodies" (but in German).

5. Feb 14, 2019 at 3:05 AM

Boltzman Oscillation

Lorentz invariant means x^2 + y^2 + z^2 - (ct)^2 = s = s' right? What exactly does this tell us? That a position of a particle is also determined by c as opposed to just the xyz positions? This makes me excited for special relativity. I am trying to self learn quantum mechanics but the math is kicking my arse. I dont know if I could take the time to learn special relativity alongside my courses at the moment.

6. Feb 14, 2019 at 3:40 AM

Ibix

If I'm on a train, I don't actually need to factor that into my calculations if I want to do calculations about what would happen if I throw a ball. I can just treat the carriage as stationary and do standard ballistics. Or, if I like making life complicated, I can treat the carriage as moving at 60mph - and do standard ballistics again. I just add 60 to all the initial velocities and add 60t to the positions at time t, and the maths just works.

This is because the laws of (Newtonian) physics are Galilean invariant. Mathematically, in any of the equations, you can replace $x$ with $x'=x-vt$, where $v$ is a velocity in the x direction, and it all just works.

Electromagnetism doesn't work like that. You can describe a charge moving past a magnet or a magnet moving past a charge. But you cannot transform one solution into the other using Galilean invariance. Ether theories of increasing complexity were an attempt to fix that, but experiments eventually led Einstein to the realisation that electromagnetism didn't work if you replaced $x$ by $x-vt$, but did if you replaced $x$ with $x'=\gamma(x-vt)$ and $t'=\gamma(t-vx/c^2)$, where $\gamma=1/\sqrt{1-v^2/c^2}$ - the Lorentz transforms.

Crucially, he also realised that Newton's laws of physics were very slightly inaccurate. He wrote down laws of physics - conservation of momentum and energy and the like - that were the same under the Lorentz transforms, not the Galilean one. So you can still treat the train as moving or stationary as you like - but just adding 60 to the velocities turns out to get you an ever so slightly wrong answer. You'll never notice at everyday speeds, but the point about electromagnetism - and light in particular - is that changes propagate at around the speed of light. So "accurate enough for everyday speeds" doesn't cut it.

Hope that's helpful...

7. Feb 14, 2019 at 5:19 AM

Boltzman Oscillation

Yes, it is insightful. Thank you very much.

8. Feb 14, 2019 at 1:10 PM

pervect

Staff Emeritus
A very short answer. If we assume Maxwell's equations are true in any frame of reference, we find that the speed of light is a constant, "c" in any frame of reference and in any direction. This is of course compatible with special relativity. However, it is not compatible with the Gallilean transform, which would basically demand that the speed of light could only be "c" in all directions (i.e. isotropic) in one particular special frame of reference.

The last statement is a consequence of the fact according to the Gallilean transform, velocities must add linearly.

Thus if we assume that the speed of light is isotropic and equal to "c" in one special frame of reference, in a frame moving at some velocity v relative to this special frame, the Gallilean transform requires that the velocity of light in the direction of motion be v-c and it must be equal to v+c in the opposite direction. So it cannot be isotropic in the moving frame if the Gallilean transform is correct.

In special relativity, velocities add according to the relativistic velocity addition law

$$\beta_t =\frac{ \beta_1 + \beta_2 }{ \left( 1 + \beta_1 \, \beta_2 \right) }$$

where $\beta=v/c$

which allows the speed of light to be isotropic and equal to "c" in all frames of reference no matter how they are moving.

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