Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is Einsteins Equivalence theorem in General Relativity correct? It seems to me that it neglects the fact that gravitational acceleration depends upon separation distance squared, thus neglecting the effect of tidal forces.

For example, as I sit on earth, I experience the affect of earth's gravity; Although the acceleration on my head is slightly less than the acceleration acting on my feet. If I make the claim that my frame is equivalent to me being in a space ship traveling at g, doesn't that mean my whole body is accelerating at g uniformly? This is contrary to the previous statement though.

share|improve this question
3  
The equivalence principle is a "local" principle, which means it only applies when considering an arbitrarily small area of space. Within such an arbitrarily small area, the gravitational acceleration, g, will be constant. –  Mew Mar 1 '13 at 2:35
    
Einstein's theory of general relativity is then kind of like the local equivalence principle being integrated over all of space. In other words, general relativity does apply to all of space, even though the equivalence principle is a local principle. –  Mew Mar 1 '13 at 2:36
    
Is this evidence to support the idea that space is Quantized? –  Cactus BAMF Mar 1 '13 at 2:37
1  
No. Integration of xdx = 0.5x^2, is the sum of x multiplied by arbitrarily small elements of "dx". The fact I can choose arbitrarily small values of dx, doesn't mean x is quantised. –  Mew Mar 1 '13 at 2:39
add comment

3 Answers

up vote 5 down vote accepted

The equivalence principle, as stated correctly by Einstein, says that these two situations are equivalent:

  • An uniformly accelerating observer in the absence of a gravitational field
  • A free falling observer in an uniform gravitational field

So, as you noted, this does not apply to the gravitational field of the Earth. Imagine you are in an elevator, free falling towards the Earth. You could let go of two pens - what you would see is that the two pens would come closer to each other, since each of them would be falling towards the centre of the Earth. You, as an observer, could then say with certainty that you are in a gravitational field.

In a hypothetical uniform gravitational field no experiment could reveal, whether you are in a gravitational field or not. Einstein took this thought experiment as a motivation for the development of General Relativity. In the mathematical construction of General Relativity, the equivalence principle does not play an important role.

share|improve this answer
    
Well, EP + metric continuity + the spectral theorem is what allows you to be arbitrarily accurate in treating physics over a small enough region as in SR, doing away with gravity, so I wouldn't say it's not important. It's just that reducing to local SR physics is not always the thing one wants to do in GR. –  Chris White Mar 1 '13 at 14:11
add comment

Quite simply you are right. Tidal forces are not subject to the equivalence principle.

If you are an accelerating reference frame, then you can represent space with the Rindler coordinates. The bottom line is that, no, of course this can't possibly be equivalent to the field you see around a body like Earth. It just can't even represent the same universe.

Rindler coordinates contain the parabolic motion (actually hyperbolic) motion you're familiar with in a constant gravitational field. But here's the more interesting thing about the Wikipedia article I linked to, pay close attention to this wording:

the Rindler coordinate chart is an important and useful coordinate chart representing part of flat spacetime, also called the Minkowski vacuum.

Now, there's something called Minkowski spacetime. It's GR-talk for "flat space". Notice that Rindler coordinates don't replace the Minkowski spacetime. Rindler coordinates still are Minkowski spacetime. A transformation (made possible by the equivalence principle) gets you from one to the next.

The question refers to gravity due to a spherically symmetric object. This "metric" (like a map of spacetime) that corresponds to this is the Schwarzschild metric. This metric is fundamentally different from the (flat) Minkowski spacetime. That's not true going from Rindler to Minkowski, they are the same.

The equivalence principle allows a coordinate transformation between reference frames that do or don't claim to have a gravitational field at their point in space. So there's no universal answer to "what is the gravitational field here?" However, there is a certain kind of invariant topology to spacetime that absolutely doesn't change between reference frames. This topology basically corresponds to tidal forces. Tidal forces gives rise to spacetime curvature, but a field alone does not.

I think of it like this: gravitational field is like the "slope" of space. You can change your body direction and see a different slope! However, tidal forces are like the "curve" or space. It doesn't matter which way you're pointed, you must agree on the presence of curvature.

share|improve this answer
    
So in short: the OP is right about tides, but wrong about what the equivalence principle asserts. –  Chris White Mar 1 '13 at 3:01
    
@ChrisWhite Have to be careful with wording, because obviously the OP was wrong because they arrived at a contradiction, which led them to ask the question. I guess you pretty much summed up that contradiction. Maybe I should be explicit by wording it like the OP was correct to perceive a contradiction. –  AlanSE Mar 1 '13 at 3:04
add comment

There are three forms of equivalence principle in common use (1):

Weak equivalence principle (WEP): If an uncharged test body is placed at an initial event in space–time and given an initial velocity there, then its subsequent trajectory will be independent of its internal structure and composition.

Einstein equivalence principle (EEP): (i) WEP is valid, (ii) the outcome of any local nongravitational test experiment is independent of the velocity of the freely falling apparatus [local Lorentz invariance or (LLI)] and (iii) the outcome of any local nongravitational test experiment is independent of where and when in the universe it is performed [local position invariance (LPI)].

Strong equivalence principle (SEP): (i) WEP is valid for self-gravitating bodies as well as for test bodies, (ii) the outcome of any local test experiment is independent of the velocity of the freely falling apparatus [local lorentz invariance (LLI)] and (iii) the outcome of any local test experiment is independent of where and when in the universe it is performed [local position invariance (LPI)].

General relativity satisfies all three, but note that in every case the principle refers to local experiments. This means that curvature, i.e. tidal effects, are assumed negligible. Tidal effects are real phenomena that cannot be removed by any choice of frame - this is the essence of spacetime curvature. So yes, you are correct.

See also here for experimental tests of the equivalence principles.

  1. SOTIRIOU, T. P., LIBERATI, S., & FARAONI, V. (2008). THEORY OF GRAVITATION THEORIES: A NO-PROGRESS REPORT. International Journal of Modern Physics D, 17(03 & 04), 399. doi:10.1142/S0218271808012097
share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.