Question regarding the equivalence principle

In summary: This is due to the fact that the apparent length is a vector quantity, and the velocity vector (the direction and magnitude of the acceleration) is always pointing downwards.
  • #1
Hello,

So I read that a person in a rocket accelerating at 9.8m/s^2 would feel the same pull downwards as a person standing on the Earth's surface.
However, I can think of a few instances where you could tell the difference:

- If you measure g (9.8N/kg) on Earth, you will notice that the value decreases as you go up. On the rocket in remains constant.
- If you have a charged particle, wouldn't a moving rocket make it emit electromagnetic radiation?
 
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  • #2
AdvaitDhingra said:
Hello,

So I read that a person in a rocket accelerating at 9.8m/s^2 would feel the same pull downwards as a person standing on the Earth's surface.
However, I can think of a few instances where you could tell the difference:

- If you measure g (9.8N/kg) on Earth, you will notice that the value decreases as you go up. On the rocket in remains constant.
- If you have a charged particle, wouldn't a moving rocket make it emit electromagnetic radiation?
First, the equivalence principle explicitly talks about local experiments. There is certainly no equivalence to an outside observer of someone on the surface of the Earth and someone accelerating through space.

More to the point, of course constant acceleration is not equivalent to a time-varying gravitational field.

The question about EM radiation is more subtle. If you have a charged particle in the two cases, then it's not going to change over time. In fact, the equivalence principle tells you that the local EM field must be the same in both cases, as measured locally.

If you have an observer outside the rocket, then as before the two situations are not (to them) equivalent. The EM field they measure due to the charged particle may well be different in the two cases.
 
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  • #3
AdvaitDhingra said:
If you measure g (9.8N/kg) on Earth, you will notice that the value decreases as you go up. On the rocket in remains constant.
Actually, the locally measured acceleration decreases up the rocket. If the acceleration were the same the rocket's length would not change viewed from an inertial frame - i.e. it would not length contract.

Understanding this is the key to resolving Bell's spaceships paradox.
 
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  • #4
AdvaitDhingra said:
If you measure g (9.8N/kg) on Earth, you will notice that the value decreases as you go up. On the rocket in remains constant.
As @PeroK mentioned the equivalence principle is a local principle. In other words, it is valid only over distances and times small enough that the curvature of spacetime is negligible. This corresponds to an absence of tidal gravity.

In relativity a uniformly accelerating reference frame in flat spacetime is called Rindler coordinates. Due to time dilation, hovering observers higher up have lower proper acceleration than observers lower down. So even in flat spacetime the value decreases as you go up in the rocket. This effect can actually be used to derive gravitational time dilation over small distances, e.g. in the Pound-Rebka experiment.

However, on Earth, due to tidal effects (curved spacetime), this effect is more pronounced. The acceleration of a hovering observer decreases faster around the Earth than in flat spacetime. This effect is real and can really distinguish between the two, but it is explicitly excluded by the equivalence principle. Where tidal effects are measurable the equivalence principle does not apply.
 
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  • #5
Dale said:
As @PeroK mentioned the equivalence principle is a local principle. In other words, it is valid only over distances and times small enough that the curvature of spacetime is negligible. This corresponds to an absence of tidal gravity.

In relativity a uniformly accelerating reference frame in flat spacetime is called Rindler coordinates. Due to time dilation, hovering observers higher up have lower proper acceleration than observers lower down. So even in flat spacetime the value decreases as you go up in the rocket. This effect can actually be used to derive gravitational time dilation over small distances, e.g. in the Pound-Rebka experiment.

However, on Earth, due to tidal effects (curved spacetime), this effect is more pronounced. The acceleration of a hovering observer decreases faster around the Earth than in flat spacetime. This effect is real and can really distinguish between the two, but it is explicitly excluded by the equivalence principle. Where tidal effects are measurable the equivalence principle does not apply.
Ahh thank you. I was always confused what "local" meant.
 
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  • #6
AdvaitDhingra said:
Ahh thank you. I was always confused what "local" meant.
If you've studied a bit of SR then you may notice a interesting subtlety regarding a "uniformly accelerating" rocket. Let's analyse the rocket from its original rest frame:

If every part of the rocket has identical acceleration (starts to accelerate simultaneously and always has the same velocity - as measured in the original rest frame), then there is clearly no length contraction. The head and tail of the rocket must, by a simple application of kinematic formulae, remain the same distance apart.

This means that from the rocket's perspective it must be stretching! I.e. it's proper length must be increasing.

(To see this just imagine that the accceleration stops at some point. Externally, the rocket still has its original proper length, but as it's now traveling at some non-zero speed, it ought to be length contracted.)

And, if from the rocket's perspective it remains rigid (i.e. retains its original rest length), then the different parts of the rocket must have (slightly) different accelerations as measured externally. As the rocket is slowly length contracting in the orignal rest frame.

You can formalise all this using Rindler coordinates, as above, but it's a good example where you must be careful with seemingly simple assumptions when dealing with relativistic motion.
 
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