General Relativity as a Theory of Gravity Explained

[davidge]

Suppose there is a charged particle far enough of any mass so that there is no gravitational interaction between the particle and any other body. The trajectory of the particle in space-time would appear to us like this

(we are at the origin of our coordinate system).

Consider that at $t=t_0$ a EM wave arrives in the region our particle is in. The particle will accelerate and its trajectory through space-time would appear to us like this

(I'm not being rigorous on the shape of the trajectory. This is just an illustration made by hand at Paint software. The main point is that this is a curved trajectory.)

Since this is a curved trajectory, Minkowski metric will not be correct to describe it anymore. We therefore would have to use another metric to describe it.

It's all okay, for Einstein's equations predicts it: the electromagnetic wave has energy and the momentum-energy tensor will not vanish.

Now by the above analysis, the Einstein's theory does not seem to be a theory of gravity, but a theory of energy, because it states that energy is responsible for space-time trajectories to be curved and, mathematically speaking, for space-time to be curved. Furthermore, in the above example we did not need to mention gravity anywhere.

So in what sense general relativity is a theory of gravity?

 

[scottdave]

Your example depends on a charged particle interacting with an electromagnetic wave. What about uncharged particles?

 

[davidge]

Your example depends on a charged particle interacting with an electromagnetic wave. What about uncharged particles?

This was just an example. I've chosen to talk about a charged particle, because then we won't need the presence of a massive body to accelerate the particle.

 

[Dale]

Since this is a curved trajectory, Minkowski metric will not be correct to describe it anymore

I don't understand this comment. You certainly can describe a curved trajectory in flat spacetime. You can even choose coordinates where the curved path is at a constant coordinate position. The metric is flat, regardless of a curved worldline or even curved coordinates.

 

[davidge]

You certainly can describe a curved trajectory in flat spacetime. You can even choose coordinates where the curved path is at a constant coordinate position. The metric is flat, regardless of a curved worldline or even curved coordinates.

Ok. But in the example above, Einstein's equations would give a "non flat" metric.

 

[Nugatory]

Since this is a curved trajectory, Minkowski metric will not be correct to describe it anymore. We therefore would have to use another metric to describe it.

The Minkowski metric works just fine in this case. The particle experiences proper acceleration when the wave passes by, so its trajectory is not a geodesic and neither requires nor is explained by curvature.

@scottdave was trying to steer you in the right direction in his post above. In the presence of an electromagnetic field the trajectory of a neutral particle reflects the effects of spacetime curvature (including the tiny contribution from the stress-energy of the electromagnetic field). The trajectory of a charged particle is dominated by the electromagnetic force which has nothing to do with curvature.

 

[Nugatory]

Ok. But in the example above, Einstein's equations would give a "non flat" metric.

If you include the stress-energy of the electromagnetic field, yes. But this non-flatness is negligible, completely undetectable by observing the trajectories of neutral particles.

 

[PeterDonis]

I've chosen to talk about a charged particle, because then we won't need the presence of a massive body to accelerate the particle.

In other words, you have introduced a non-gravitational interaction, and then you wonder why the particle's changed trajectory isn't due to gravity?

Einstein's equations predicts it: the electromagnetic wave has energy and the momentum-energy tensor will not vanish.

Einstein's equations predict what?

They predict that there is nonzero stress-energy in this spacetime, and therefore it is not perfectly flat, yes.

They do not predict that your charged particle will have the curved trajectory it has solely due to the presence of nonzero stress-energy. (If the particle were neutral, as others have pointed out, its trajectory would indeed be predicted solely by the stress-energy present--but in that case its trajectory would be a geodesic, i.e., it would not be curved.) The charged particle's trajectory requires Maxwell's Equations to predict, as well as Einstein's Equation (and as others have pointed out, the effect of Einstein's Equation, i.e., of spacetime curvature, on the trajectory will be miniscule compared to the effect of Maxwell's Equations).

 

[Dale]

Ok. But in the example above, Einstein's equations would give a "non flat" metric.

Why? Is the particle super massive or is the EM wave ultra energetic? I certainly didn't get that impression from your description. Your description makes it sound flat.

 

[davidge]

@scottdave was trying to steer you in the right direction in his post above. In the presence of an electromagnetic field the trajectory of a neutral particle reflects the effects of spacetime curvature (including the tiny contribution from the stress-energy of the electromagnetic field). The trajectory of a charged particle is dominated by the electromagnetic force which has nothing to do with curvature.

Oh, I see now.

If you include the stress-energy of the electromagnetic field, yes. But this non-flatness is negligible, completely undetectable by observing the trajectories of neutral particles.

Ok

In other words, you have introduced a non-gravitational interaction, and then you wonder why the particle's changed trajectory isn't due to gravity?

They do not predict that your charged particle will have the curved trajectory it has solely due to the presence of nonzero stress-energy. (If the particle were neutral, as others have pointed out, its trajectory would indeed be predicted solely by the stress-energy present--but in that case its trajectory would be a geodesic, i.e., it would not be curved.) The charged particle's trajectory requires Maxwell's Equations to predict, as well as Einstein's Equation (and as others have pointed out, the effect of Einstein's Equation, i.e., of spacetime curvature, on the trajectory will be miniscule compared to the effect of Maxwell's Equations).

What do you mean here is that to describe the curved trajectory of the charged particle also requires using Maxwell's Equations?

Why? Is the particle super massive or is the EM wave ultra energetic? I certainly didn't get that impression from your description. Your description makes it sound flat.

Because I thought any amount of energy would be sufficient for the metric not to be Minkowskian.

 

[PeterDonis]

What do you mean here is that to describe the curved trajectory of the charged particle also requires using Maxwell's Equations?

Because the EM wave affects the charged particle's motion due to the electric and magnetic fields associated with the wave. Those effects are governed by Maxwell's Equations. Spacetime being flat or curved is irrelevant to those effects; they aren't due to spacetime geometry. They are due to the EM fields exerting a non-gravitational force on the charged particle. So if you try to predict the charged particle's trajectory just using Einstein's Equations, you will get the wrong answer.

 

[davidge]

Because the EM wave affects the charged particle's motion due to the electric and magnetic fields associated with the wave. Those effects are governed by Maxwell's Equations. Spacetime being flat or curved is irrelevant to those effects; they aren't due to spacetime geometry. They are due to the EM fields exerting a non-gravitational force on the charged particle. So if you try to predict the charged particle's trajectory just using Einstein's Equations, you will get the wrong answer.

Thanks.

So in the example I gave, the trajectory of the particle would be curved due to the EM wave, but the metric still would be the Minkowskian metric?

 

[PeterDonis]

So in the example I gave, the trajectory of the particle would be curved due to the EM wave, but the metric still would be the Minkowskian metric?

How different the metric would be from Minkowski would depend on how intense the EM wave was. If you want a rough order of magnitude, try calculating the energy density of a typical EM wave and comparing it with the energy density of something like the Earth which is known to cause significant spacetime curvature. (The average energy density of the Earth is just its average mass density times $c^2$. The average energy density of an EM wave is the square of its electric or magnetic field strength, times the appropriate physical constant--in SI units this would be $\epsilon_0$ for the electric field.)

 

[davidge]

Thanks.