# The light clock and the simple clock

1. Dec 1, 2012

### vinven7

This is probably a simple question but I can't seem find a convincing enough explanation. Many books on introductory Special relativity use what are called "light clocks" to explain Time dilation. I myself learned this from Arthur Beiser's Modern Physics. This is a simple clock that measures time as the length of a clock or rod/speed of light. In a moving reference frame this is bound to change and time dilation is introduced.
However, how can we show that this is applicable for an ORDINARY CLOCK as well? Beiser states that if you had an ordinary clock (say a quartz clock) that does not show time dilation, then the difference between the two - a light clock and an ordinary clock can be used to figure out the ABSOLUTE velocity of the frame without any reference. Can someone explain this to me? I can't seem to follow the logic.
Please let me know if the question is not clear enough, I'll put up a more detailed explanation in that case

2. Dec 1, 2012

### phinds

There is no such thing as an absolute velocity (without reference). Velocity is ONLY meaningful when referenced to something else. You must be talking about some kind of difference in velocity measurements, but again, there is no such thing as absolute velocity.

3. Dec 1, 2012

### Bill_K

Note "IF". This is reductio ad absurdum. Such a clock cannot exist, because it would violate Lorentz invariance. If you had two clocks that, in the same frame, one exhibited time dilation and the other did not, then from t' = γt, you could unambiguously calculate your absolute γ and hence your absolute v.

4. Dec 1, 2012

### phinds

And what would absolute v mean?

5. Dec 1, 2012

### Staff: Mentor

It would mean v relative to the frame where light clocks keep the same time as quartz clocks. This frame would be the one frame where the laws of physics would work without modification, the absolute frame.

6. Dec 1, 2012

### Staff: Mentor

Basically, time dilation of a light clock follows from the second postulate, and then from that time dilation of all other clocks follows from the first postulate. Does that make sense to you?

7. Dec 1, 2012

### phinds

No, I'm not following any of this. Guess I don't know SR very well. Bottom line, however, is that you ARE saying that such a situation does not exist.

8. Dec 1, 2012

### Staff: Mentor

Correct, such a situation does not exist.

The first postulate states that the speed of light is c in all reference frames. You can use that fact to analyze a light clock in two different frames to derive time dilation for a light clock.

The second postulates states that all of the laws of physics are the same in all reference frames. So if the laws of physics ensure that another clock keeps time with a light clock in one frame then that other clock must keep time with the light clock in all frames. It will therefore dilate along with the light clock.

9. Dec 1, 2012

### vinven7

Thanks for all the replies. Let me restate the question thus:
let's say i have an 'ordinary quartz clock' that somehow is not subject to relativity and hence do not show time dilation. I have a light clock, that does show dilation. I take these two and hop on to a space craft and move with some velocity wrt to a reference frame. If my understanding of SR is right, then since both the clocks are in my frame of reference, they should both show the same time to me. I cannot find out that my time has changed unless I compare it with a reference frame, at which point I will find out my velocity as well. It's seems like I need a reference frame to find the velocity anyway. Is there way that I can find out what my velocity is from just inside the craft? I mean to ask if with respect to myself inside the space craft, are the velocities of the two different?

10. Dec 1, 2012

### pervect

Staff Emeritus
Why are you using relativity to calculate the behavior of something that is not subject to relativity??

There would be nothing "ordinary" about a quartz clock that didn't follow the laws of relativity. An "ordinary" quartz clock would be subject to relativity.

If you've got an extra-ordinary clock (quartz or otherwise) that doesn't follow the laws of relativity, it's up to you to figure out what it does. You can't ask what relativity predicts, because you've just got trhough saying that your clock isn't following those laws.

11. Dec 1, 2012

### vinven7

@ pervect. Thanks for your reply. Let me make my question clearer.
The usual description of time dilation is using a light clock - it serves as a proof that shows and calculates how time dilates. But then all text books claim that this extends to all clocks - mechanical, pendulum , quartz and so on, because of the second postulate. I am trying to see if we had a situation where the normal clock did not show time dilation and the light clock did, will it lead to a visible contradiction?
Personally, I know that SR is true and the normal clock will also show dilation, but I am trying to convince myself why this should be so.

12. Dec 1, 2012

### pervect

Staff Emeritus
It's really experiment that shows that relativity is true, not something that can be "proved".

The FAQ on experiments that test relativity is at http://www.edu-observatory.org/physi...periments.html [Broken]

Historically, people initially expected the speed of light in a vacuum to be non-constant. It was a bit of surprise when it turned out that that "c" was universal. It is the observation that "c" is universal that makes light clocks work - if "c" wasn't universal, they wouldn't.

The light clocks is not really intended as a way to test relativity. It's more or less a vessel to try to illustrate some of its consequnces. I.e. the argument goes - If we initially assume relativity is correct, and the speed of light is constant, then we can make the following predictions.

Last edited by a moderator: May 6, 2017
13. Dec 1, 2012

### vinven7

Please stick to my question, this argument is going in some other direction Let me try this again.
1) Granted that the speed of light in vacuum is always 'c' and that it is independent of the observer.
2) This implies that a light clock traveling with some velocity will show a different time than the same light clock that is at rest.
3) How does this imply that an ordinary clock also should saw the same time dilation effect?

People usually simply invoke the second postulate and claim that time has to be the same in all frames so then all clocks should show the same effect. But how do we know that this is a universal effect and not some quirkiness of the light clock? As far as I can comprehend, this only shows that light clocks undergo dilation and not that all clocks undergo the same effect. Do you get my argument now?

14. Dec 2, 2012

### Staff: Mentor

Suppose we have an ordinary clock and a light clock set up so that, on every "tick" of both clocks, the light beam in the light clock and its counterpart in the ordinary clock meet up with each other. For example, suppose we make the ordinary clock using some particle with non-zero rest mass that can bounce between perfectly elastic walls at a constant velocity, and one of the walls has one of the light clock's mirrors on the other side. Then we can verify that the two clocks tick at the same rate by watching the light beam and the particle hit at the same spot on opposite sides of the wall/mirror at the same instant, at every tick.

By the principle of relativity, if the light beam and the particle hit at the same spot on every tick in their mutual rest frame, they must hit at the same spot on every tick in any frame, because each hit is a definite observable event, a coincidence of worldlines (light beam, particle, and wall/mirror), and such events are frame invariant. Hence, the two clocks must tick at the same rate in every frame; so if the light clock appears slowed down to an observer moving relative to it, the ordinary clock must appear slowed down as well.

15. Dec 2, 2012

### A.T.

This comes up pretty often here. Here my answer from a previous thread:

16. Dec 2, 2012

### arindamsinha

Don't take a light clock literally as a physical object. It is only a theoretical artefact, to demonstrate the time dilation effect (of a velocity or gravity) on someone, as observed by someone else assumed not subject to such velocity/gravity. So it does not make sense to compare a light clock (theoretical artefact) with some other type of clock (physical object).

All it shows is that the 'rate of passage of time' differs between observers who are subjected to different velocities/gravities. Clocks local to each of the observers tick at rates dictated by the local 'rate of passage of time' of the observers.

17. Dec 2, 2012

### vinven7

This is an excellent reply and I am now beginning to get the logic.
If I may press you a bit further, several texts claim that the difference in the two clocks - (if there is a difference under the hypothesis that the ordinary clock does not dilate) can be used to measure the velocity of the craft without the need of a reference frame. How does this work? If I am the person in the craft, I do not notice a difference at all, so I cannot make that velocity measurement, is that correct? This difference in time will only be visible to a person standing on the ground. - and if he has access to the data inside the craft, then how are we able to claim that he is making a velocity measurement without a frame of reference?

18. Dec 3, 2012

### A.T.

No. All observers must agree on whether the two clocks stay synced or not. Otherwise you get paradoxes. If some observer would see you and the clocks moving and going out of sync, you would also see them out of sync. So you both would conclude that the clocks must be moving in an absolute sense and could compute the absolute velocity from the clock rate difference. But that would violate the relativity principle.

19. Dec 3, 2012

### harrylin

Hmm no, that there is a disagreement must be agreed by both for a consistent theory.
However, it is not possible to directly compare the rates of two clocks with only two clocks that are in relative motion, since they can only meet for an instant. You need to compare the times at two different instances. More on that you can read in threads on "relativity of simultaneity" which is about clock synchronisation.

Last edited: Dec 3, 2012
20. Dec 3, 2012

### Staff: Mentor

All frames agree if two things happen at the same place and time. So if a light clock and a quartz clock are right together then all frames will agree if they are going at the same rate or not. So, if a quartz clock did not dilate like a light clock then all frames would agree that the quartz clock was running e.g. 25% faster than the light clock, corresponding to v=0.6 in the absolute frame.