3
$\begingroup$

I understand the equivalence principle as "The physics in a freely-falling small laboratory is that of special relativity (SR)." But I'm not quite sure why this is equivalent to "One cannot tell whether a laboratory on Earth is not actually in a rocket accelerating at 1 g".

$\endgroup$
2
$\begingroup$

The best entry level introduction to the equivalence principle I have found to date is found here. As it explains in the article, one of the key items that is general not well understood about the equivalence principle and its relationship to the concept of tidal forces.

If one can imagine a freely falling elevator on earth, versus a spaceship flying in empty space, two test particles will actually feel a slight lateral attraction associated caused by the earth's gravitational field effecting the trajectory of the particles.

freely falling

If you had no knowledge of the earth outside the elevator, you would equate the attraction of the two test particles with a gravitational force between the particles.

gravitating test particles

Similarly if you were in the rocket ship and had no knowledge of the outside world and you had two particles that started gravitating towards each other, you might think that they are actually moving under the influence of a third massive body that was outside the spaceship pulling on the particles along two different paths.

Since it would be impossible to tell the difference, you have to conclude that there is an equivalence between those two frames of reference.

Similarly, if there is absolutely no measurable tidal force inside a freely falling elevator, one would have the same spacetime as used in special relativity.

$\endgroup$
  • $\begingroup$ Is it taken here that the gravitational effect of the big red ball in the spatially nonzero experimental volume is expressible by a homogenous/constant gravitational field? I.e. how much does the gravitational field of the red ball change within the cube? Can the effect even point in two opposing directions? $\endgroup$ – Nikolaj-K Jan 31 '13 at 11:31
  • 1
    $\begingroup$ @Hal Swyers I just made this account to say thanks to your answer, but I can't do anything right now because of the new account. $\endgroup$ – DaiTran Feb 9 '13 at 0:37
  • $\begingroup$ Your welcome! If you want, try to upvote again! :-) $\endgroup$ – user11547 Feb 9 '13 at 1:09
  • $\begingroup$ But you can not regard the attraction observed between the blue and green balls to the gravitational force exerted by the blue ball to the green one or vise versa. That is because if the culprit was the gravitational force that the blue ball exerts on the green one (or vise versa), it would depend on the mass of the green and blue balls however the blue and green balls are just following the geodesic path towards the Earth and does not matter the mass neither of the blue or green ball, they two always would follow the same path and therefore always get together at the same rate. $\endgroup$ – Stefano Jun 11 '17 at 15:56
1
$\begingroup$

Say there were two identical rooms with no windows, one on earth and one on a spaceship accelerating at 9.8 m/s^2 (acceleration due to gravity on earth.) You are in one of them, and can conduct any mechanical experiment you like as long as it does not involve peeking outside of the room in any way shape or form.

Unfortunately no matter what you try you will not be able to determine which room your in because experiments in both rooms will yield identical results.

However one room is experiencing gravitational acceleration and the other inertial acceleration. The principle of equivalence tells us that these two things are equal, or as Einstein said "The outcome of any local non-gravitational experiment in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime."

$\endgroup$
  • $\begingroup$ Yes but, my question is with regards to the two formulations being equivalent, because the EEP says something about free falling frames, but if you are in an accelerating spaceship you aren't in a free falling frame. $\endgroup$ – JLA Jan 30 '13 at 22:13
  • $\begingroup$ @JLA: Yes! That's the key point--if you're in a rocket ship sitting on the surface of the Earth, you're not in a freely falling frame, either. Accelerating in empty space = resisting the Earth's acceleration using the ground. $\endgroup$ – Jerry Schirmer Jan 31 '13 at 3:15
1
$\begingroup$

The equivalence principle (the version you mention) means that you cannot tell, locally, whether you are in a freely falling frame or in "outer space", i.e., in a region of space with no gravitational field. This version is the EEP. There is another version, namely, WEP which says that the inertial mass is the same as the gravitational mass. This means that when standing on earth, the force you experience is proportional to your gravitational mass, which is the same mass that you would put in Newton's second law. Thus, if you are in a rocket accelerating at 1g you will experience the same force. Now, from WEP you see that if you throw a ball upwards, the trajectory it follows will be the same on earth and in the rocket. Therefore we can reformulate WEP in terms of freely falling objects: locally, the motion of freely falling particles are the same in a uniformly accelerated frame and in a gravitational field. Of course, WEP doesn't imply EEP because in SR the mass is not unique. But it is easy to see that you can generalize it by imposing SR in the motion of the objects. However, EEP is stronger in the sense that not only the trajectories are the same, but all laws of physics. This is the basis of QFT.

$\endgroup$
  • $\begingroup$ In general relativity, you could say that EEP implies WEP. But actually, the principle doesn't refer to a particular theory. So you could imagine a theory of gravity (not GR) in which freely falling particles rotate when they are in a gravitational field, but follow exactly the same path. Thus, violating EEP but not WEP. This is not the theory that we experience, but there is nothing that prohibits this kind of theories. $\endgroup$ – Prastt Jan 31 '13 at 3:09
1
$\begingroup$

The way it makes sense to me is that if you shoot a ray of light across a room inside an accelerating rocket ship, the ray will fall some distance (relative to the room). Therefore on Earth the ray of light must also fall (ostensibly "due to gravity"), otherwise you would know whether you are in a ship or not based on whether light does or does not fall. And I imagine there would be some kind of physical consequence to that. The following video seems to imply that the consequence would be that inertial mass and gravitational mass would not be the same, which makes intuitive sense to me.

This is a video from "The Mechanical Universe." It's a good overview of the deduction. http://www.youtube.com/watch?v=VbOKxkj0_wc&feature=youtu.be&t=13m24s

$\endgroup$
0
$\begingroup$

I'm not sure if the O.P. was satisfied with the previous answers or not but I feel that there is something that wasn't answered, at least in the way that the question asked for it. The reason why the equivalence between a free falling system and a Special Relativity Rest Frame implies the equivalence between a laboratory on Earth and a rocket accelerating at 1 g is this:

1) First, notice that a laboratory in earth is not the same system as a free falling system, altough both of them feel gravity. The laboratory (or just elevator) on earth is meant to represent a system that feels gravity but holds its place do to some other force, namely, the normal in the ground or the tension in the elevator.

2) Suppose that the first equivalence is right: A free falling system is equivalent to a rest system in Special Relativity. Great, how we get to the second equivalence?

3) Add an upward force to the free falling system, namely, add a tension going upwards to the free falling elevator and you will get an elevator hunging up in earth.

4) According to the first equivalence, if a free falling system was the same as a rest frame, then a free falling system plus a force acting against gravity should be the same as a rest frame with the same force applied.

5) So, an elevator hunging up in Earth is the same as a elevator feeling a force upwards with no gravity.

This is the logical way to go from the first equivalence to the second one, just add the tension in the elevator to go from "free falling system" to "Laboratory on earth" so that the second part of the equivalence goes from "rest frame of Special Relativity" to "rocket going upwards at 1g".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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