# Moving Clocks Run Slow?

1. Jul 7, 2011

### Master J

I have just started reading Landau and Ligarbagez' fantastic work Classical Theory of Fields. It has got me thinking about some simple scenarios involving moving clocks....

Moving clocks run slower - but which clock is moving?

A spaceship passes the earth say at a constant velocity, the time that passes on the spaceship is less according to the people on earth, than the time thats passed on earth. Yet, can the people on the ship not say that according to them, the time that has passed on earth is less than the time on the ship, since in their frame, the ship is standing still and all else is passing it by??

2. Jul 7, 2011

### Mike_Fontenot

Yes. And they are BOTH correct!

Mike Fontenot

3. Jul 7, 2011

### bcrowell

Staff Emeritus
4. Jul 7, 2011

5. Jul 8, 2011

### Master J

Cheers for the input guys...I'll have a good read of that!

So, if BOTH are correct....let us examine the following (using the twins now!):

The spaceship with the twin returns to Earth. To do this he must undergo non-inertial motion, ie. acceleration, to change his velocity back towards the Earth.

Now, the twin on Earth did not undergo acceleration - he just stood still watching his clock until his sibling arrived home. And so, there is an asymmetry between the 2 twins.

Yet, why can we not take the viewpoint of the traveling twin? If he is considered to be at rest, then the Earth travels away from him, turns around and comes back.

I still cant see how when we compare clocks when the traveling twin is home, his clock is slower. Why isn't his brother who stayed on Earth's clock slower, for the reason I just outlined above?

6. Jul 8, 2011

### bcrowell

Staff Emeritus
It isn't a matter of opinion who accelerates and who doesn't. The traveling twin can measure the accelerations on an accelerometer.

7. Jul 8, 2011

### Master J

Ah excellent, I didn't think of that!!! :)

So, simply because the traveling twin changes inertial frames, this means that it is not valid from his viewpoint to say that the twin on Earth's clock is slower.

8. Jul 8, 2011

### Staff: Mentor

Yes. Without even bringing the Twin Paradox into play, we can make a consistent picture out of this by applying the concept of relativity of simultaneity. Suppose we have two planets at "rest", and two spaceships moving at the same constant velocity with respect to the planets. Observers on the two planets synchronize their two clocks, and observers on the two spaceships synchronize their two clocks similarly. According to the observers on the two planets, the "spaceship clocks" are not synchronized, and according to the observers on the spaceships, the "planet clocks" are not synchronized.

The post linked below includes a diagram that shows the results of a numerical example that I calculated a few years ago. It doesn't show the actual calculation; if you want the details, I can expand on them.

https://www.physicsforums.com/showpost.php?p=1980817&postcount=8

Last edited: Jul 10, 2011
9. Jul 8, 2011

### ZealScience

In this post, you had ignored that every time objects accelerate, there is force exerted which makes them different.

In fact there is also certain ambiguity involved in this "non-inertial reference frame", that brings problem to SR. That's why there was GR, which is more general theory of space time.

10. Jul 8, 2011

### bcrowell

Staff Emeritus
Special relativity is just like Newtonian mechanics in this respect. It has inertial and noninertial frames of reference: http://www.lightandmatter.com/html_books/lm/ch04/ch04.html#Section4.5 [Broken]

Last edited by a moderator: May 5, 2017
11. Jul 8, 2011

### Mike_Fontenot

The traveler's "viewpoint" ISN'T invalid. Both of their "viewpoints" are equally valid, even though they disagree with one another about the correspondence between their ages, during almost all of the trip. That's very strange, but it's just the way special relativity works. It ISN'T inconsistent, even though it sounds like it is.

During segments of his trip for which he isn't accelerating, the traveler DOES conclude that the home twin is ageing more slowly than he himself is. But during his periods of acceleration, he may correctly conclude that the home twin is ageing much faster than he himself is. Or, he may correctly conclude that the home twin is ageing more slowly than he himself is. He may even correctly conclude that the home twin is getting younger. Which of these possibilities actually occurs, in any particular case, depends on the direction and magnitude of his acceleration, and also on their separation.

In the standard twin "paradox" scenario (where the traveler accelerates in the direction TOWARD the home twin), he will conclude that the home twin gets much older during his acceleration, and that's how they can both end up agreeing about whose TOTAL ageing was greater, when they are reunited, even though for most of his trip, he correctly concluded that the home twin was ageing more slowly than he himself was.

https://www.physicsforums.com/showpost.php?p=2934906&postcount=7

https://www.physicsforums.com/showpost.php?p=2923277&postcount=1

Mike Fontenot

Last edited: Jul 8, 2011
12. Jul 8, 2011

### Staff: Mentor

It is not that his viewpoint is invalid. It is simply non-inertial, meaning that the laws of physics are not the "standard" laws. There are fictitious forces, c may be non-constant, etc. The mistake is not in using the traveling twin's viewpoint, but rather in treating it as though it were inertial when it is not.

13. Jul 8, 2011

### Master J

Thanks guys...this has really helped me out.

One last thing I have been wondering....if the traveling twin CAN say that the Earth clock is running slow only in his constant velocity journeys to and from the Earth - is there a formula for the contribution to his clock that the acceleration gives so that his clock does not now predict the Earth's as slower?

14. Jul 8, 2011

### Mike_Fontenot

Yeah, it's the "CADO equation" ... it's explained (and its use is demonstrated) in those links that I gave you in my last post.

Mike Fontenot

15. Jul 9, 2011

### ctxyz

This is not a very well posed question but I will try to recast it and answer it. From the perspective of the traveling twin (the one that accelerates), the results are exactly the same as the results of the calculations done from the perspective of the "stay at home" twin, see HERE

16. Jul 9, 2011

### bcrowell

Staff Emeritus
Just to make sure the OP understands, "CADO equation" is your name for a result you derived. It's not terminology that anyone else uses.

Yes, there is a standard way of calculating this. First off, it's not necessarily accurate to think of the effect as arising only from the acceleration. In the twin paradox scenario, there is actually no way to determine unambiguously whether the time distortion happens at a particular point in the trip (e.g., during the accelerations), accumulates over the whole trip, etc. This is because when the twins are apart, simultaneity is undefined, so there is no way to determine unambiguously what the reading tA on the first twin's clock is at the same time when the reading tB on the other twin's clock has a certain value. The only thing that makes sense to try to predict and measure is the time on their clocks when they're reunited.

If you know calculus, the way to find the time on the clock is to calculate this integral: $s=\int \sqrt{dt^2-dx^2}$, where t and x are the time and position measured in some coordinate system, and the integral is evaluated along the path taken by the clock.

If you don't know calculus, it's still possible to do some nice examples. If the clock's path consists of constant-velocity segments, then you can do a sum like this:
$$\sqrt{\Delta t_1^2-\Delta x_1^2}+\sqrt{\Delta t_2^2-\Delta x_2^2}+\ldots \qquad ,$$
where 1 is the first segment, 2 is the second segment, etc. This is sufficient to handle an example of the twin paradox in which one twin stays at home and the other twin travels away at constant speed and then comes back at constant speed.

17. Jul 9, 2011

### ctxyz

I had a quick look at his "CADO" formulas, they look like a bunch of nonsense. Does this forum allow self-promotion of fringe physics?

18. Jul 9, 2011

### bcrowell

Staff Emeritus
No, it doesn't. I haven't looked at them carefully myself. I'll take a look now.

19. Jul 9, 2011

### ctxyz

Thank you, please do, it is totally fringe, the posts should be removed.

20. Jul 9, 2011

### bcrowell

Staff Emeritus
Here is what it looks like to me. Mike_Fontenot seems to dedicate a lot of his time on PF to trying to get people to pay attention to something he derived that he calls the "CADO equation." I don't have anything in principle against someone doing repeated cut-and-paste posts of their own didactic material, because I do that sometimes myself if there's a FAQ and I think my own answer is the best thing since sliced bread. What bugs me about Mike_Fontenot is that in my opinion his posts are not helpful, it's difficult to tell whether they're correct, and they may violate our rules against "new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion." Taking these points one at a time:

(1) His posts about CADO aren't helpful. He posts his equation over and over, regardless of whether it's relevant. He doesn't define the variables carefully, and when people challenge him to do so, he doesn't seem willing or able to.

(2) It doesn't look correct to me, but it's hard to tell because he doesn't define his variables. He posts links to his web site, where he gives a numerical example involving constant acceleration, but since the numerical example is given without defining his terms or his assumptions, it doesn't clarify what he's claiming. His work was published in Physics Essays 12, 629 (1999): http://physicsessays.org/resource/1/phesem/v12/i4/p629_s1?isAuthorized=no [Broken] The article is behind a paywall, and I'm not inclined to pay \$15 to see it. As far as I can tell, his equation just looks like he took the Lorentz transformation equation $t'=\gamma t-v\gamma x$ (in units with c=1) and rewrote it as $T'=T-vx$ (with no gammas), where T is what he calls the "current age of a distant object." I suppose this equation can be defined to be true, since he never seems to state the definition of the variables T and T' or how they relate to observation.

(3) Physics Essays does not appear to be on the list of journals at thomson.com that we say are OK as academic resources https://www.physicsforums.com/showpost.php?p=2269439&postcount=2 It is a peer-reviewed AIP journal, but probably very low in quality and impact factor. We could get lawyerly about the exact wording of the rule I quoted above, or how crucial it is that Physics Essays isn't on the thomson.com list. But I think it is undeniably true that CADO is "not part of current professional mainstream scientific discussion."

If Mike_Fontenot really wants to be helpful, the thing to do would be to post a reprint of the paper on arxiv or on his own web site. Otherwise there is no way for anyone to evaluate whether CADO is anything more than what it appears to be: fringe stuff that a low-quality journal let slip through peer review.

Last edited by a moderator: May 5, 2017