# Time dilation in gravitational field

1. Jun 9, 2013

### tonychinnery

A friend of mine posed the following conundrum: If you have 2 identical clocks one at the top and one at the bottom of a tower (on the earth), then for an observer at the top, the clock at the bottom appears to go more slowly than the clock next to him. This one can deduce from the redshift of a beam of light traveling upwards in Earth's gravity. Einstein considered the light to be a continuous wavetrain (why this is allowed I am not quite sure as he himself was a proponent of Planck's quantum theory). By waiting long enough the reading of the lower clock can lag behind by any arbitrary amount. If after a certain time the guy at the top goes down and brings up the lower clock and puts it next to his clock, there will be a difference in the time readings, I understand. But what if the hands of the two (mechanical) clocks are linked by a light lever? Without gravity the two clocks run together, so the linking bar is doing nothing. What happens to the bar when I accelerate the whole thing (equivalent to gravity)? I found myself unable to answer my friend. Help requested

2. Jun 9, 2013

### Staff: Mentor

Check out this entry from the FAQ at the top of this forum: https://www.physicsforums.com/showthread.php?t=536289 [Broken]. You've constructed a variant of that problem by expecting that the presence of the lever will require the hands of both clocks to change position at the same time. They don't, and something in the system must be eventually break.

Last edited by a moderator: May 6, 2017
3. Jun 9, 2013

### tonychinnery

Here is a variant of the problem: a pendulum (if you like activated by a spring, not by gravity) long enough to reach near the ground fixed to the top of the tower. If time runs more slowly at the bottom of the tower then over a large time interval, the bottom of the pendulum will have carried out less oscillations than the top. How is this possible?

4. Jun 9, 2013

### tonychinnery

I don't think the thread Nugatory quotes is relevant. Of course if you push one end of a bar there is a time lag before the other end moves. But in this problem the time difference between the two clocks can get as big as you like by waiting long enough.

5. Jun 9, 2013

### Staff: Mentor

If you mess around with mechanical clocks (adding levers, or whatever), you can change the speed of them. That is independent of any relativistic effects.

No, the clock would have some fixed frequency which depends on the whole setup and its position.

6. Jun 9, 2013

### jartsa

The stronger clock dictates the pace.

If the clocks are identical and have no relativistic moving parts, then the "upper" clock wins.

If clocks are identical and have some relativistic moving parts, then the pace is mostly dictated by the "lower" clock.

7. Jun 9, 2013

### tonychinnery

I've got it now. Light traveling up the tower from the lower clock undergoes a redshift, so the lower clock appears to be ticking faster to an observer at the bottom of the tower than it does to the guy at the top. If the guy at the top then sends back the beam to the bottom it undergoes a blueshift and arrives at the bottom at the same frequency it started out. Both clocks are synchronized to the light beam, but to move together, the lower clock has to be synchronized to a slightly higher frequency (if I'm right)

8. Jun 9, 2013

### Staff: Mentor

Yes indeed, and the relevance is in the fallacy of assuming a rigid connection between two points to establish a definition of simultaneity. If the time is represented by the position of the hands of the two clocks (as in your first post) then something will break - perhaps the lever, or perhaps as Jartsa says, the "stronger" clock will win - so that the two clocks cannot both operate normally and be coerced by a mechanical linkage into remaining synchronized.

[Edit: I hadn't seen your post #7 when I wrote the above. Yes, it's a lot easier to understand what's going on with a light clock, because then we don't have to deal with our preconceptions about how rigid bodies such as the lever in the initial formulation of the problem will behave]

[second edit: Thought experiments involving distributed mechanical relativistic systems can be quite confusing, although also quite interesting. It's probably best to start with Ehrenfest's paradox, Bell's spaceship paradox, and Born rigidity.... Get these down cold before trying new ones]

Last edited: Jun 9, 2013
9. Jun 9, 2013

### Staff: Mentor

A "time interval" is the interval between two points in time: for example, I can say that the interval between "the clock's hour hand pointed straight up" and "the clock's hour hand pointed straight out the right" is three hours. So before we can talk about the number of oscillations in a given time interval, we have to choose the two points that define the ends of that interval - and because the clock at the bottom of the tower doesn't agree with the one at the top, we can't just say something like "OK, both of you start counting oscillations at 1:00, stop counting oscillations at 2:00" and expect both observers to be counting the same oscillations over the same hour.

10. Jun 9, 2013

### tonychinnery

Yes but over a long time period (say, years. Very boring but its only a thought experiment) the number of oscillations of the pendulum as seen, say by a third party standing at a distance, must be the same at the top and the bottom if its one pendulum.

11. Jun 9, 2013

### Staff: Mentor

For that, you don't need to wait for a long period of time. The third party watching at a distance is using his clock to define the interval over which he is counting the oscillations, so of course he ends up counting the same number of oscillations of the top and at the bottom, across any interval of time he pleases.

Yes, the distant observer records X swings of the pendulum in an hour, the guy at the bottom of the potential well reports Y swings of the bottom of the pendulum in an hour,and the guy at the top of the tower reports Z swings of the top of the pendulum in an hour. The discrepancy comes from them having made different choices about when the hour started and when it stopped, not anything weird in the pendulum itself.

12. Jun 10, 2013

### tonychinnery

Thanks for this help. there is one final point that is puzzling me. In his treatment of the subject (I think in 1911) Einstein considers light as a continuous wavetrain (as in the classical model of the vibrating ether) to explain time dilation in a gravitational field, whereas he himself (in his famous 1905 paper on the photoelectric effect) knew that light is emitted in packets ('lumps' Feynman called them). What is the justification for treating light as a continuous wavetrain?

13. Jun 10, 2013

### Staff: Mentor

The lumpiness of light is only apparent at atomic scales. In any macroscopic situation, the number of photons involved is so large and the individual photons are so small that we can ignore the lumpiness and think in terms of continuous electrical and magnetic fields described by Maxwell's equations of classical electrodynamics.

An analogy: we treat water as a fluid with properties like density and volume and pressure, even though we know that it is really composed of tiny lumps called molecules.

14. Jun 10, 2013

### tonychinnery

Excuse me, but I don't think that explanation is quite good enough. I remember that our fluid mechanics teacher (way back in the 70s) told us that the fact that we could treat fluids such as water using a continuous mathematical model was still not fully understood, so you can't explain one phenomenon using another even more complex phenomenon which is not even fully understood. For its as though the photons are queuing up behind each other to form a train, whereas in fact they are being emitted by electrons at random times.

15. Jun 10, 2013

### Staff: Mentor

That bit about water as a fluid was an analogy not an explanation - if it doesn't help you understand, we can forget about it, try some other analogy.

Photons do not queue up "to form a train", and just about any intuition you have about their behavior is going to be misleading. It's also irrelevant for present purposes, as Maxwell's electrodynamics is quite adequate for understanding the behavior of light anywhere except at subatomic scales - we can even calculate c from Maxwell's equations. The quantum mechanical description of light as photons did not replace the Maxwellian description of light as a wave, it supplemented it.

16. Jun 10, 2013

### tonychinnery

Yes but Maxwell was working at a time when light was considered to be the vibration of an 'ether', something that Einstein had helped to supersede in his 1905 paper on the basic principles of special relativity, as well as his paper on the photoelectric effect. In the ether theory its easy to see how the vibrating ether could behave just as an elastic solid transmits sound. I was wondering how Einstein (or anyone) could justify using this classical model when he had already decided that light was really individual wave packets that behaved like particles (each packet is entirely absorbed by one electron in the photoelectric effect, instead of being spread out in space as a classical wave would be).

17. Jun 10, 2013

### Staff: Mentor

Many wave packets together can form a continuous stream of electromagnetic waves, where you can easily count the cycles both at the source and the receiver. This is done with radio waves all the time, and lasers give something similar, too.

18. Jun 10, 2013

### Bill_K

tonychinnery, You need to reread the quote by Nugatory, which is right on.

A beam of light is not the bunch of little bullets that you seem to think it is.

19. Jun 10, 2013

### Staff: Mentor

Einstein's work on SR did get rid of the need for the aether, but it did not in any way modify Maxwell's equations. In fact, if anything it strengthened Maxwell's equations. Prior to Einstein it was thought that Maxwell's equations only applied in one reference frame (the aether frame), and afterwards it was understood that they apply in all frames.

The mathematics of quantum mechanics justifies that. If you take Quantum Electrodynamics and take the appropriate limit, you recover Maxwell's equations. So, as long as the situation is well described by the classical limit then you can justify using Maxwell.

20. Jun 10, 2013

### tonychinnery

Radio waves are formed when many electrons move together backwards and forwards in a wire. Laser light is coherent as well, but did not exist in Einstein's time. An ordinary tungsten bulb is emitting photons in a random incoherent way, and yet Einstein treats it as a wavetrain.