To Frame My Frame…(Frame II : Einstein) -3. The Trick

December 8, 2005

Now, we are ready to ask what does relativity say about the question – Can moving clocks run slow ? Let us start by assuming that a moving clock runs slower by a factor $\gamma$. What does it mean ? It means when one second is past, the moving clock shows a time $\frac{1}{\gamma}$ seconds. Note that if $\gamma$ turns out to be less than one, then the moving clock would be actually running fast !

The question is of course, what is the value of $\gamma$ ? Our normal intuition says that $\gamma$ should be equal to one(moving clocks run at the same rate as clocks at rest), if relativity is true. But as we have seen, this problem might not be as simple as it appears. But one thing is clear,$\gamma$ cannot be arbitrary- moving clocks can run slow, but the fact that they should not be able to find out whether they are moving should place some condition on $\gamma$. The question is what is that condition ?

To answer this,Let us start by writing down this question in terms of the funny-looking time-like vectors. Assume a clock A moving with a velocity $\vec{v}=v\hat{e}_x$ along x direction is running slow by a factor of $\gamma$, and another clock B at rest with respect to the absolute space. Consider again two explosions which occur one after another. We also assume they happen in such a way that A sees the first explosion then moves with the velocity $\vec{v}$ for a second along x (as seen by B) and then sees the second explosion.

So, according to B,these two explosions are separated by a time duration of one second and a net displacement $\vec{v}=v\hat{e}_x$, or in the notation I introduced before, they are separated by $\hat{e}_t+v\hat{e}_x$. But what would A see ? In this one second, his clock would be showing $\frac{1}{\gamma}$ seconds. And according to him, he did not move at all. So, he would say that, the only thing he did for going from one explosion to another is just to wait for $\frac{1}{\gamma}$ seconds. So, according to him, the two events are separated by $\frac{1}{\gamma}\hat{e}_t'$ where we have put a prime to show that it is waiting as seen by a moving observer.

So, now we want to write down an expression to convey the fact that to go from one explosion to another you can either wait for $\frac{1}{\gamma}$ seconds as seen by A or wait for one second and move $v$ metres as seen by B.Since, these two ways of going from one explosion to another are equivalent and or merely two different ways of going from one explosion to another, we write

\begin{displaymath}\frac{1}{\gamma}\hat{e}_t' = \hat{e}_t + v \hat{e}_x \end{displaymath} (1)

If the principle of relativity is right, equivalently we can treat A to be at rest, imagine two other explosions now seen by B and repeat the whole argument above.Or, we can just interchange all the primed things(measured by A) with all the un-primed things(measured by B).

\begin{displaymath}\frac{1}{\gamma'}\hat{e}_t = \hat{e}_t' + v' \hat{e}_x' \end{displaymath}

where $\gamma'$ is how much B’s clock has slowed down according to A and $v'$ is the velocity of B as seen by A.

Now, to proceed we have to make some further assumptions. These are assumptions precisely because I don’t know how to prove them ! Probably, the only way is to show you that the results that come out of these assumptions agree with whatever we see in the real world. But, even without that, I hope, you would agree with me that these assumptions are ’sensible’.

One, let me assume $\gamma$ and $v'$ to be functions of $v$. What I am saying is this – If you give me the velocity of a moving clock A with respect to clock B at rest, that is sufficient to determine a) how much does A is slow as seen by B and b) the velocity of B as seen by A. Note that, by principle of relativity, $\gamma'$ and $v'$ should be related in the same way as $\gamma$ and $v$ are related,i.e, If $\gamma=f(v)$ then $\gamma'=f(v')$. Similarly, if $v'=g(v)$ then $v=g(v') $.

Two, let me assume that the function $\gamma$ depends only on the speed with which the clock is moving, but not on the direction along which the clock moves. In particular, assume $\gamma(-v)=\gamma(v)$. Well, you will hear people say, of course, that should be true because of what is called as the principle of isotropy. (This is basically the idea that there is no way the Newtonian absolute space distinguishes between two observers rotated with respect to each other by a constant angle. In that sense, the principle of isotropy is completely analogous to the principle of relativity.) But, we have to be a bit more careful in going from the principle of isotropy to the statement that clocks moving at different directions with the same speed run at the same rate.

We have already seen, for example, that principle of relativity alone (which asserts the equivalence of two observers moving with respect to each other by a uniform velocity) DOESNOT imply that the rate of a moving clock should be same as a clock at rest. Similarly, principle of isotropy alone DOESNOT imply that the rate of a clock moving towards the left should be same as one that is moving towards the right with the same speed. So, you have to add some more assumption than just the principle of isotropy.And the necessary assumption is basically this- Any two observers rotated with respect to each other should agree on the time intervals. 1

Now, let me come to my third assumption. Let me assume that $v'=-v$ . i.e, If A sees B move with a velocity $v$ then B sees A moving with a velocity $-v$. In fact, this in my opinion is the most non-trivial assumption among the three. I have some ideas on how to replace it with a weaker assumption, and if you have any idea you are of course welcome to mail me ! But, to discuss these ideas might take us a bit too far from the topic under discussion. So, just take this as one of the assumptions – again sensible,but non-trivial.

Now we are ready. We have $v'=-v$ and $\gamma'=\gamma(v')=\gamma(-v)=\gamma$. Substituting this into the previous equation, we get

\begin{displaymath}\frac{1}{\gamma}\hat{e}_t = \hat{e}_t' - v \hat{e}_x' \end{displaymath} (2)

Now,we can substitute Eqn.(% latex2html id marker 333 $\ref{et}$) in Eqn.(% latex2html id marker 335 $\ref{et'}$) and solve for $\hat{e}_x'$ to get

\begin{displaymath}\hat{e}_x' = \gamma \hat{e}_x + \frac{\gamma}{v}\left(1-\frac{1}{\gamma^2}\right) \hat{e}_t \end{displaymath} (3)

Which is quite a weird relation between $\hat{e}_x'$ , $\hat{e}_x$ and $\hat{e}_t$ which should hold if the moving clocks run slow and still don’t violate the principle of relativity…
(continued here )


This is a stronger statement than just saying two observers are equivalent. For example, If one observers x-axis is rotated with respect to the other at an angle, then these two observers would not agree upon the x-component of the velocity of the particle. And, this disagreement to the question is there despite the fact that both the observers are equivalent !Given this fact, it’s quite non-trivial to assume that they do agree upon the time intervals.It’s a sensible thing to assume. But, we should acknowledge the fact that it is a new assumption over and above just the principle of isotropy. The non-trivial nature of this assumption is clear if you consider these facts

  1. Two observers moving at uniform velocity with respect to each other are equivalent(The principle of relativity). But, they don’t agree on the time intervals if $\gamma$ turns out to be anything other than unity.They also don’t agree upon the velocity of a ball.

  2. Two observers rotated with respect to each other by an angle are equivalent(The principle of isotropy). And they do agree upon the time intervals according to the assumption above. But, they don’t agree upon the components of the velocity of a ball.

If at all anything, this just shows that we have to be careful about arguments which go like “Because the two observers are equivalent, they should agree on blah blah.” In fact, if you read about Reichenbach’s arguments on relativity, you would see that there is a way of formulating relativity in which two observers rotated with respect to each other DON’T agree on the time intervals, whereas the principle of isotropy is well and fine ! But, these are really micro-details and I repeat- We would get into it only if we have to.


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