# Difference between a clock and a series of radar pulses

1. Dec 15, 2016

### Ken Miller

1. The problem statement, all variables and given/known data

I have been working through “Basic Concepts in Relativity and Early Quantum Theory” by Resnick and Halliday. I've read about and done most of the problems about time-dilation, length-contraction, and Doppler effect. But then I got to problem 2-76, and I’ve been swirling in confusion for several days about the essential difference between a clock and a series of radar pulses.

Problem statement: A radar transmitter T is fixed to a reference frame S’ that is moving to the right with speed v relative to reference frame S. A mechanical time (essentially a clock) in frame S’, having a period \tau_0 (measured in S’), causes transmitter T to emit radar pulses, which travel at the speed of light and are received by R, a receiver fixed in frame S.

(a) What would be the period τ of the timer relative to observer A, who is fixed in frame S?

(b) Show that the receiver R would observe the time interval between pulses arriving from T, not as $\tau$ or $\tau_0$, but as $$\tau_R = \tau_0 \sqrt{\frac {c+v} {c-v}}.$$

(c) Explain why the observer at R measure a different period for the transmitter than does observer A, who is in the same reference frame. [Hint: A clock and a radar pulse are not the same thing.]

2. Relevant equations

Time-dilation: $$\tau = \frac {\tau_0} {\sqrt {1-\frac{v^2}{c^2}} }.$$

Doppler effect (with $angle = 180^o$): $$f = f_0 \sqrt {\frac {1-\beta} {1+\beta} }$$

3. The attempt at a solution
Before I worked on part (a) and before I saw the question to part (c), I used the Doppler-effect equation to solve part (b), and came up with the correct answer. Then I went to part (a), and thought that A should detect the same period as R does. Then when I read part (c), I realized I was wrong about (a), and that I should probably use the time-dilation formula to calculate the period measured by A.

So, I have two questions.

--Am I correct in my answer to part (a)? (i.e. use time-dilation equation)

--What is really puzzling me is why a clock and a radar pulse are not the same thing?!! (By the way, in the hint, I suspect that the authors meant that “a clock and a series of radar pulses....”)

2. Dec 16, 2016

### Staff: Mentor

Yes.

Forget relativity for the moment. Realize that the apparent time between radar pulses is affected by the relative motion. Imagine you are shooting out pulses at a rate of 1 per second. But the observer is moving toward you, thus they would intercept the pulses sooner than 1 per second. (There's a Doppler effect even without relativity.)

3. Dec 17, 2016

### Ken Miller

Thank you for trying to help me understand. I totally get what you've written. But understanding that was never my problem. My problem is that it seems to me that what you say about the radar pulses (in fact pretty much anything you can say about the radar pulses) you can also say about a clock. So, given that the radar pulses are triggered by the clock, why does the radar receiver R measure a different period than observer A (when both R and A are fixed in frame S)?

4. Dec 17, 2016

### nrqed

There is a subtle point here and if it is completely understandable to be confused if the book does not explain this clearly (most books do not explain this and it leads to a lot of confusion when working with special relativity the first time). When they ask what would be the period relative to observer A, who is in S, they do NOT mean what would be the period between the pulses when they reach observer A, which of course would give the same answer as part b.

What they mean is something more subtle: they mean "what is the time interval between the emission of the pulses, as observed in frame S". To know that, observer A cannot simply use the time interval between the reception of the pulses at his position, he would have to do some calculation to eliminate the effect of the source moving away and the speed of light being finite. This is one way, but a more direct way is for A to rely on "local observers" in the frame S who have synchronized their clocks. For this, imagine that there is an observer at each point of A (that means an infinite number of them which is fine in a thought experiment!) so that whenever there is a pulse emitted by the emitter in S', there is one local observer in S that happens to be exactly at that location. Now, every time a pulse is emitted, th local observer in S who happens to be aligned with the emitter records the time on his clock. The advantage of this is that there is no time delay for light to travel from the emitter to the observer as the local observer aligned with the emitter at a given instant is essentially at a zero distance from the emitter.

When at least two local observers in S have recorded their clock time when a pulse was emitted just in front of them, they send the information to A who *only then* can calculate the time interval between the emission of the pulses as measured in S. This number is the answer to question a). It is *this* value that will show the time dilation effect of moving clocks.

This is a different value than the time interval between reception of the pulses by A, which will include both the time dilation factor and the fact that the source emitting a pulse is at a different distance from A every time a pulse is emitted. This is the subject of question b)

5. Dec 17, 2016

### Ken Miller

Oh, OK, I think I've got it. (Thank you very much!) Let me try it in my own words--I would appreciate a confirmation or correction. This is basically an issue of "infinite array of clocks" vs. "single clock." With an infinite array of clocks, there will always be some clock at the source, so time-dilation formula applies. With only one clock, the clock can't be at the source if the there is relative movement, so the Doppler-effect equation applies. Have I got it?
By the way, if I do finally understand it, I believe that that means that the problem statement is certainly not precise, because it refers to "observer at A," which makes it sound like a single clock/observer rather than infinite array. Though I acknowledge that elsewhere the text does explicitly say that the authors mean "infinite array" when they use the term "observer."

6. Dec 17, 2016

### nrqed

Yes, you got it! You understood very quickly, good job. And yes, books (and teachers) are notoriously ambiguous or even misleading when explaining special relativity. If that books mentions an infinite array, they are at least better than most references. When applying the Lorentz transformations, for example, they will often say something like "what is the time interval between these two events as measured by observer A who is in frame S" which is very ambiguous, usually they do not mean the time interval between the two events as witnessed directly by A but rather the time interval between the two events as reported to A by local observers.