# Confusion about reading of clock as seen from different inertial systems

Suppose a clock, located at point $x'$ in the inertial frame $S'$, registers two events $t_1'$ and $t_2'$. Let $\Delta t'=t_2'-t_1'$. The same two events will be registered by two different (synchronized in their rest frame) clocks, located in two different points of the inertial frame $S$ (which moves with a certain speed relative to $S'$). Let $\Delta t=t_2-t_1$ the elapsed time between these two events as seen from $S$. From the theory of relativity we know that $\Delta t' = \sqrt{1-\beta^2}\Delta t$.

At first glance, the time dilation formula looks asymmetrical. We know that all of the clocks in $S$ are synchronized. Thus, it is tempting to assume that $\Delta t$, which is calculated by two different clocks in $S$, can be calculated by one clock in $S$. But then we have a situation where two perfect and equivalent clocks - one from $S$ and one from $S'$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

My textbook says that in order to resolve this "paradox", one must realize that clocks which are synchronized in one inertial frame need not be synchronized in another inertial frame. That is - the clocks "belonging" to the system $S$ aren't showing the same time when seen by an observer in $S'$.

However I don't quite understand this argument. Why do we care what the clocks of $S$ show to an observer in $S'$? As far as I understand, the time interval $\Delta t$ in the formula above is the time elapsed according to the clocks in $S$ as seen by an observer in $S$. In other words - the formula connects what an observer in $S'$ sees to what an observer in $S$ sees = two different observers, not one.

• The situation is symmetrical. For the unprimed observer to measure the elapsed time of the primed observer's moving clock, he has to use two different but synchronized clocks. Vice versa for the primed observer if he wants to measure the unprimed observer's clocks. Each will measure the other's clock ticking more slowly. The key to the apparent paradox is realizing that "same time" measurements made by one observer are not "same time" measurements according to the other observer. I suggest Mermin's "It's about Time: Understanding Einstein's Relativity". – Samuel Weir Jan 3 '16 at 21:03
• There's a subtle difference. To the observer in S', the two events occur at the same location. To the observer in S, the two events occur at two different locations. So the same formula used in going from S' to S cannot be applied in going from S to S'. – Chet Miller Jan 3 '16 at 21:12
• @ChesterMiller - this does not answer OP's question. – cth Jan 4 '16 at 9:13

Perhaps you might find it helpful to consider an analogy involving distances instead of time. Suppose in your coordinate system your x axis is horizontal, but mine slopes down to the right somewhat, and that your frame of reference is moving to the right while mine is still.

Suppose you measure the height at which which two events occur at a certain point along your x axis, and you find a height difference dh between them. In my reference frame the the second event will occur further along my x-axis than the first (as my axis is moving relative to yours), and since my x-axis is sloping down compared with yours I will find that the second event is higher than I would had my x-axis been parallel to yours, so I find a different value for dh. The effect is nothing to do with how I measure the height, or whether I use one ruler for both measurements or two- it arises from the fact that the events are spread along my axis, and my axis is at a different slope to yours.

Exactly the same situation arises with simultaneity, except with height (ie a measurement in the y axis) replaced by time (a measurement in the t axis). You measure the times of two events at one point on your spatial axis. I measure them at two points along mine, and mine is sloping relative to yours so I get different values for dt (whether I use one clock/ruler or two).

Your question is not about relativity in particular so much as it's about perceived asymmetry in general. So here's an example that captures exactly the same kind of "asymmetry".

Stand on the equator, facing South. The South Pole is roughly 6000 miles in front of you and the North Pole is roughly 18000 miles in front of you. (That is, you'd have to walk 18,000 miles in the forward direction to make it 3/4 of the way around the globe to the North Pole.)

I'm standing next to you, facing North. In my coordinate system, the North Pole is 6000 miles in front and the South Pole is 18,000 miles ahead.

Our measurement systems are equivalent in the sense that they make exactly the same predictions for the consequences of any particular journey. For example, if we both go due South, we agree that we'll get to the South Pole in 6000 miles, though you'll describe that journey as walking forward and I'll describe it as walking backward.

So according to you, the South Pole is closer (in the forward direction) than the North. According to me, the opposite is true.

Here we have a situation where two perfect and equivalent measurements are no longer equivalent (they give different values for the distance from where we're standing to the South Pole). But this contradicts the equivalence of our measurement systems!

Or does it?

At first glance, the time dilation formula looks asymmetrical.

You need to use the complete Lorentz transformation for time to see if the formula is symmetric. In your question, let $$\Delta x^\prime=0$$, Lorentz transformation implies:

$$\Delta t=\frac{\Delta t^\prime+(v/c^2)\Delta x^\prime}{\sqrt{1-\beta^2}}=\frac{\Delta t^\prime+(v/c^2)×0}{\sqrt{1-\beta^2}}=\frac{\Delta t^\prime}{\sqrt{1-\beta^2}} \space ,$$

which is your result. However, Lorentz transformation for time in $$S^\prime$$, yields:

$$\Delta t^\prime=\frac{\Delta t-(v/c^2)\Delta x}{\sqrt{1-\beta^2}} \space ,$$

Remember that in, $$S$$, we have $$\Delta x=v\Delta t$$, which implies:

$$\Delta t^\prime=\frac{\Delta t-(v/c^2)\Delta x}{\sqrt{1-\beta^2}}=\frac{\Delta t-(v^2/c^2)\Delta t}{\sqrt{1-\beta^2}}=\sqrt{1-\beta^2}\Delta t \space ,$$

This equation is similar to the first one, which shows the symmetry in relativity.

But then we have a situation where two perfect and equivalent clocks - one from $$S$$ and one from $$S′$$ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

I think you have mistaken time rates (fast or slow) for the elapsed time (readings of the clocks). If you consider the first one, according to the symmetry, each observer measures, say, the angular velocity of the other clock slower (one says $$\omega^\prime/\omega=\gamma$$, and the other one deduces $$\omega/\omega^\prime=\gamma$$ instead.), however, both observers agree that the readings of the clocks ($$\Delta t$$s) are similar in both frames as I showed through the equations above. This similarity is due to clocks synchronization. That is, if two clocks starts ticking simultaneously in one frame; from the viewpoint of other frame, one of the clocks starts ticking earlier than the other one does despite each observer sees the other clock slower.

I hope that I've understood your question correctly.

But then we have a situation where two perfect and equivalent clocks - one from S and one from S′ - are no longer equivalent (one of them is slower than another). But this contradicts the equivalence of all inertial frames of reference.

Whenever you say something, like ...(times) are no longer equivalent/one of them is slower than another.... You automatically put yourself into one of the inertial frames, here it is frame S.

On the flip side, the statement is still true if you are within the context of frame S'. You would still say the other clock seems slower than yours, within the context this S' frame. Therefore, the symmetry holds, in that sense.

To generalize, whenever you describe something in the language of the theory of special relativity, unless it is something invariance like the spacetime interval, be aware that most statement comes automatically with a certain inertial frame.