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.

The EEP is used to justify that if an observer on the ground shoots a beam of light towards a tower, then when the light reaches the tower, it will be red shifted. This is because of what happens in an accelerating spaceship.

The books seem to say this implies time dilation, but I don't completely see why. Could it not just be like any other doppler effect? If I send a sound wave towards you and you are moving towards me, the frequency you observe will be greater than what I send out, but that doesn't mean your clock ticks slower (assume his speed is non-relativistic). Why does it necessarily imply time dilation?

In the derivation used to derive this redshift, from the light beam on a spaceship, special relativity doesn't even seem to come into play, just newtonian physics is used. Anyway if the physics on the ground is indistinguishable from the physics in the spaceship, would that not mean that at the back of the spaceship a clock ticks slower, rather than just appearing to click slower, relative to someone at the front? Otherwise it would seem to me that the EEP would just say that we can't do any experiment to distinguish the physics between the rocket and the gravitational field, though the physics wouldn't really seem to be the same.

share|improve this question
    
"The EEP is used to justify that if an observer on the ground shoots a beam of light towards a tower, then when the light reaches the tower, it will be red shifted." This phenomenon has also been measured. The first time in 1959. –  dmckee Feb 22 '13 at 8:47
    
Yes I'm fine with that part, my question is with regards to time dilation. –  JLA Feb 22 '13 at 15:49

2 Answers 2

The experiments you describe can all be analyzed in a flat spacetime using SR. We can switch back and forth between an inertial frame A and an accelerated frame B. The equivalence principle says that observers in B see a gravitational field. (That is, we have a gravitational field and a varying gravitational potential, but spacetime is flat.)

Observer Alice in inertial frame A describes the source and receiver as accelerating, so she interprets the observations as a kinematic Doppler effect. Bob in B sees source and receiver as both being at rest, so he interprets the effect as gravitational rather than kinematic. He can call it a gravitational Doppler shift, or he can call it a gravitational time dilation. GR doesn't distinguish between the two.

If you wanted to construct a theory in which the distinction between gravitational Doppler shifts and gravitational time dilation was meaningful, an example of how you could do that would be that your theory could predict that clocks of different types that were rate-matched in one location could become mismatched in rate when moved together to some other gravitational potential. This would clearly be an example of gravitational time dilation, not a Doppler shift, because arguments about Doppler shifts can't explain the difference in behavior between the two clocks. But this theory is not a metric theory and it violates the equivalence principle (because Alice sees no gravitational field and therefore can't explain what's going on). Because GR is a metric theory that incorporates the equivalence principle, the distinction you're trying to make isn't a distinction that GR can make.

So the result is that you can call this effect a kinematic Doppler effect, a gravitational Doppler effect, or gravitational time dilation, and all of these interpretations are equally valid.

share|improve this answer

From emission to reception a wave can not change its frequency if the receptor and the emitter are not moving from or to each other. If they are moving you have doppler red/blueshift. Now the only possibility which provides a redshift without movement is a that the origin has send it with the redshifted frequency. It only appears to be redshifted if we assume that the receptors clock goes as fast as the emitters one. In other words the clock-speed must be lower at the emitters side if we received a signal that has a lower frequency as the same signal would have if the receptor would have emitted it.

share|improve this answer
1  
"From emission to reception a wave can not change its frequency if the receptor and the emitter are not moving from or to each other." Call this statement P. P is not a testable statement. It's very common for relativists to interpret cosmological redshifts by assuming not-P. They choose to think of a light wave that travels over cosmological distances as being stretched to a longer wavelength by cosmological expansion of space. The same relativist may wake up the following morning and decide to assume P, in which case the cosmological redshift is kinematic. Either way is right. –  Ben Crowell May 3 '13 at 18:03
    
Well it is clear by simple information logic that without some kind of infinite buffer the speed of incoming information must be the same like the speed of the outgoing information in any system transferring information. The only exception is movement because it delivers some kind of infinite buffer that is the additional space generated at every time-unit. From an informations point of view there is no difference between cosmological and doppler redshift. –  luis May 4 '13 at 14:12
    
small correction: of course the buffer must only grow as long the information is transfered so it must only be infinite if the information is continuously generated until infinity. So technically speaking it must only grow not be infinite –  luis May 4 '13 at 14:29

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.