1. Nov 2, 2014

### dynawics

Hello all,

I have not studied general relativity only special relativity, I apologize therefore if my questions seem low level. Thank you for your help.

Three questions:

The way in which I learned that the twin paradox is resolved is through the illustration of the way in which the observer in the rocket would necessarily immediately detect a change in the frequency of the regular signals being sent to him by his twin on earth, but, that the twin on earth would not detect a change in frequency until a certain time later (unless the earth was the one which underwent acceleration away from and back towards the rocket). This seems to me to indicate that there is something about the process of acceleration itself which defines any particular frame of reference as a functionally new physical system with respect to its time characteristics. Is my instinct on this issue correct?

Second: Imagine a scenario in which two twins exist on different stationary planets which are known distances apart. One twin hops in his rocket and moves towards the other twin in a straight line. It would seem that each twin would see the other twin's clock move slower. According to the theory of relativity what would happen when the twin in the rocket arrives on the planet with his other twin? Would the flying twin be younger since he was the one that was flying? How could that be if both twins saw each other's clocks moving slowly and there was no change in direction? What if the twins were sending identical regular signals to each other immediately before and during the rocket flight? It seems as if they would both experience exactly the same pattern of frequency changes. Although, I suppose that, indeed, the twin flying in the rocket would, again, experience the changes in frequency before the stationary twin did. It seems as if the twin who hopped into the rocket would witness the other twins clock running slower for longer than the twin on the earth who would only see the rocket twin's clock moving slowly until the regular signal from his transmitter reached the earth, the time during which the rocket twin had already been moving and witnessing the earth twin as having a slow clock. If this be the case, then it seems necessary that the rocket twin would arrive older, not younger.

Third: Lastly, would not any increase in the frequency of signal registered by either twin make the clock of the other twin seem to move faster? For example, if the signal sent out was regularly timed with a clock, every second representing one signal, if you were moving towards the originating point of the signals then it seems as though you would register them faster. Would this kind of increased speed of signal registration cancel out the relativistic slowing of the clock with respect to both twins? Would in fact the clocks seem to maintain the same relative rate to each observer as a result of this combining of opposite effects in the case that the rocket was moving straight towards the other observer? Does special relativity hold only in the case of motion perpendicular to the observer?

Thank you,
Ian

2. Nov 3, 2014

### Staff: Mentor

A frame of reference isn't a "physical system"; it's an abstraction that we use to help us understand how physical systems behave. You don't "switch frames" by changing your state of motion (i.e., accelerating). You change your relationship to other physical things--such as the immediate change the traveling twin observes in the Doppler shift of light signals from the other twin, when he accelerates to turn around. In fact, you can analyze the twin paradox entirely using the Doppler shift that each twin observes in the signals from the other twin, without ever bringing in frames at all; that's basically what the solution you describe amounts to. The Usenet Physics FAQ article on the twin paradox calls this the Doppler Shift Analysis.

Actually, each twin would directly observe a Doppler blueshift in the other twin's light signals--i.e., he would directly observe the other twin's clock ticking faster. (But the twin who hops in his rocket would see the blueshift as soon as he starts traveling, while the twin who stays on his planet would only see it after a time delay.) Each twin would calculate, if he uses a frame of reference in which he is at rest, that the other twin's clock is ticking slower, once he corrects for light travel time; but that's a calculation, not something he observes directly. This point is often glossed over in discussions of the twin paradox, which is a shame, because it's an important one to keep in mind; it has evidently confused you (see further comments below).

He would have aged less. The Doppler shift analysis gives this answer directly, as should be evident from the parenthetical comment I made above. You can also get the answer by doing a calculation in either frame (the rest frame of the planets or the rest frame of the traveling twin), but it will take more effort.

Yes. Your scenario of the two planets and one twin traveling between them illustrates this, as I noted above.

No. The two are not really different phenomena; they're just different aspects of the same underlying physics. See my comments above.

Yes, but the change is the opposite of what you were thinking, because the "running slower" of the other twin's clock is not what is actually observed directly, as I noted above. Furthermore, if you use just the time dilation (the calculation each twin makes that the other's clock is running slower, after correcting for light travel time), you will get the wrong answer, because you are leaving out relativity of simultaneity; in the traveling twin's frame, the other twin's clock runs slower, but the event at which the traveling twin leaves his planet, assuming that this event is at time t = 0 by the traveling twin's clock, is not, in the traveling frame, simultaneous with the event at which the other twin's clock, on the other planet shows t = 0 (although it is in the rest frame of the planets and the other twin).

Certainly not. See above.

Last edited: Nov 3, 2014
3. Nov 3, 2014

### ghwellsjr

When the rocket twin leaves earth, both twins see the other ones clock ticking slower than their own and by the same factor. When the rocket twin turns around, he immediately sees the earth twin's clock ticking faster than his own but the earth twin won't see this same thing happen until a long time later. If the rocket twin travels at the same speed coming back as he did leaving (according to the earth's rest frame), then the slower factor and the faster factor are the inverse of each other. Since it takes him just as long to reach his turnaround point as it does for him to get back, we can average those two factors and this will tell us the ratio of the earth twin's aging to the rocket twin's aging. For example, let's suppose that the rocket twin sees the earth twin's clock ticking at a factor of one-half during the outbound half. Then he will see a factor of two on the way back. The average of 0.5 and 2 is 1.25 so the earth twin ages 1.25 times whatever the rocket twin ages.

Here is a spacetime diagram to illustrate what I have been saying. The earth twin is shown as the thick blue line with dots marking off increments of one year. The thick red line is the rocket twin. The thin lines show yearly signals propagating at the speed of light from each twin to the other one.

Notice how the red twin's clock takes 1.25 times longer to tick out one year while he is traveling than when he is stationary. Time Dilation is only affected by speed, not by acceleration. Also note the relative factors of 0.5 at the beginning and 2 at the end and note their average is the same as the ratio of the twins' aging. Blue ages by 5 years while red ages by 4.

Let's assume that prior to the rocket twin departing from the distant planet toward earth, their clocks have been synchronized according to their mutual rest frame which means that they each see the other ones clock at an earlier time than their own and by the same amount. Let's assume they were three light-years apart. They would each see the other ones clock three years earlier than their own. Then when the rocket twin starts his trip toward earth, he will immediately see the earth twin's clock advancing faster than his own. Eventually, he will see the earth twin's clock go through those three years of time plus whatever length of time it took for him to make the trip (according to the earth frame). Meanwhile, the earth twin won't see the rocket twin leave for earth until three more years go by. Then he will see the rocket twin's clock advance rapidly but for a much shorter amount of time. This results in the earth twin seeing the rocket twin age less than the rocket twin sees of the earth twin aging.

Here is another spacetime diagram to illustrate this scenario:

While the rocket twin is moving toward the earth twin according to the earth's rest frame, it is only the rocket twin's clock that is ticking slowly. So the increased speed of the signals coming toward the earth twin is reduced by the slower ticking clock of the rocket twin. On the other hand, the increased speed of the signals coming toward the rocket twin is increased by the slower moving clock of the rocket twin. The net result is that both twins see the other ones clock ticking faster than their own by exactly the same amount. You can see these effects illustrated in the previous drawing.

4. Nov 3, 2014

### dynawics

Thanks for the response! Now that you have clarified a few things let me ask another question.

Is there a particular velocity at which the increased frequency registered with the parallel observer of that velocity due to the Doppler effect would precisely equal the decreased registered frequency due to the relativistic effect? Could we simply set the relevant equations equal to each other to find this velocity? It seems that if this were the case, an observer detecting a moving object from this direction would not be able to tell we the the object was even moving or not -at least until it became directly observable. But even then the time of the moving system would appear to be in sync with the reference frame of the observer.

5. Nov 3, 2014

### ghwellsjr

No, this never happens. Think of it this way: if there were no Doppler shift, then the stationary observer would be able to see the slowly ticking clock at its actual rate. But the motion of the clock toward the stationary observer will make it appear to tick faster.

6. Nov 3, 2014

### dynawics

I am still stuck on this: Let's say that there is a clock moving at a particular velocity past you, but you only look at it (detect it) as it passes in front of you from left to right. In this case there would be no, or only negligible Doppler effect, and the clock would be seen as ticking slower than the identical clock in your reference frame.

Now, let's say that the effects of relativity don't exist and that someone else with another identical clock is observing the moving clock in parallel such that it is coming straight at them. In this case, due to the Doppler effect the moving clock would seem to tick faster than the identical clock which they had. The moving clock would be seen as speeding up until it reached the same time as the clock of the observer's clock upon arrival.

It seems then that if these two processes are combined in just the right way- one being the decrease in the detected ticking rate of the moving clock due to relativity, and the other being the increase in the detected ticking rate due to the Doppler effect- then we would observe that one could reduce the other to such a precise extent that it would not be possible to detect wether or not something was actually moving towards you, at least on the basis of the detection of the variations in the clock ticking signals.

7. Nov 3, 2014

### Staff: Mentor

No, this is not correct; there would be a Doppler effect. This is called the transverse Doppler effect, and it has no counterpart in non-relativistic physics. It is present in relativity because of time dilation--the moving clock ticks slower.

Yes, but this is not something separate from the Doppler effect. It's just a different way of looking at the same thing.

They're not two processes; they're different ways of looking at the same process (relative motion). Mixing them up is going to lead to confusion.

I suggest taking another look at the spacetime diagrams ghwellsjr gave in post #3, and thinking hard about what they are telling you.

8. Nov 3, 2014

### A.T.

Look at the formula for the relativistic Doppler effect, which already includes the time dilation between source and observer:

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight

Is there any other value for beta (relative speed) than 0, which gives you a Doppler factor of 1 (no
frequency change)?

Last edited: Nov 3, 2014
9. Nov 3, 2014

### A.T.

What dynawics means is spiting the relativistic Doppler effect in two factors:
- classical Doppler
- time dilation

He thinks that since classical Doppler > 1 and time dilation < 1, their product could become 1 at some relative velocity other than 0. But that is not the case, when you look at the
actual formulas: The classical Doppler grows so fast, that the time dilation never reaches the reciprocal value of it. So their product, the relativistic Doppler factor, never becomes 1. It's always > 1 (blue shift) in the case of an approaching source.

Last edited: Nov 3, 2014
10. Nov 5, 2014

### m4r35n357

The doppler effect calculation combines the Lorentz transormation together with the light delay all in one mathematical step, in other words the doppler effect already includes time dilation.

11. Nov 6, 2014

### A.T.

That is correct (as I pointed out in post #8). But you can still separate those factors (as I pointed out in post #9). dynawics asks if they are ever reciprocal (so their product is 1) for a approach velocity > 0. The answer is no, because classical Doppler tends towards infinity faster, than the time dilation tends towards zero (velocity vs. velocity2 dependency).

12. Nov 6, 2014

### m4r35n357

Acknowledged, beg your pardon for that. But after revisiting my notes on the matter (looks like I need to rewrite my notes in a clearer form!) I think I've changed my mind, in a way; I believe there is a viewing angle (for a moving observer) which I think is defined by:

$$\cos(\alpha') = \frac {\gamma - 1} {\gamma v}$$

where the doppler factor ($DF$) is 1. Theoretically, along this circle, you could directly observe time dilation (modulo the difficulty of observing clocks and their rates virtually instantaneously whilst moving at large relative velocities). (light grey circle of dots in the video)

Furthermore, I reckon there is a smaller viewing angle for which $DF = \gamma$, which I think is $\cos(\alpha') = v$ (magenta circle of dots in the video)

In this video the observer first accelerates away from a home station, then returns. It is on this return leg that you will see the two "circles" of dots move into view.

To see all the details of clocks, speedo etc please view in full HD. There is a fuller description of what is happening in the playlist top level: https://www.youtube.com/playlist?list=PLvGnzGhIWTGR-O332xj0sToA0Yk1X7IuI

13. Nov 6, 2014

### A.T.

That makes sense, because the rel. Doppler along the line of sight (that I was considering in post #8 and #9) is always a blue shift for a source approaching you. But the transverse rel. Doppler (at closest approach) is just the kinetic time dilation, a red shift. So when the source is not approaching you directly, there must be an angle where it switches from blue to red shift, for any relative speed of the source, not just a specific one.

Going back to post #6 this seems what dynawics was actually asking about. So he was actually right assuming there is such a combination, if the source doesn't approach you directly.

Last edited: Nov 6, 2014
14. Nov 6, 2014

### m4r35n357

Agreed. I am a little ashamed that I could have answered earlier and better simply by remembering my own work of a couple of months ago! I really should write it up properly, but I've moved onto other areas now . . .

15. Nov 6, 2014

### A.T.

And I considered only different velocities of direct approach, not different viewing angles. (based on post #4, but #6 is more general).

16. Nov 6, 2014

### ghwellsjr

No, he wasn't right. Your posts prior to #13 were all correct. Don't change your mind. He has always been talking about observing a clock coming directly towards an observer. He thinks there is a speed at which an approaching clock appears to be stationary due to the observed rate of ticking on that clock being identical to the observer's clock. He thinks this observation would be the same from far away to close up until the moment the clock reaches him. His arguments involve a clock not directly coming toward an observer but that is not the situation he wants to know about.

17. Nov 6, 2014

### A.T.

I'm not changing my mind. My post where correct for the direct approach.

Doesn't hurt to consider that situation too.

18. Nov 6, 2014

### ghwellsjr

But he is wondering about a particular value of v for which the Doppler cancels out the Time Dilation. What you are saying about the transverse situation is true for all values of v, not just one. It has nothing to do with what he is asking about.

19. Nov 6, 2014

### stevendaryl

Staff Emeritus
Something that's kind of interesting about time dilation and Doppler shift. Even though, in a sense, they are two different effects, there is a way to think of time dilation as being derivable from Doppler shift plus the principle of relativity.

Assuming nonrelativistic Doppler, then you get different Doppler shifts depending on whether the sender is moving or the receiver is moving.

Assume that a sender is sending light signals at a rate of 1 every $T$ seconds (according to his clock). Then the receiver will receive those signals at the rate of 1 every $T'$ seconds (according to his clock). The relationship between $T$ and $T'$ is, classically:

$\frac{T'}{T} = \frac{1}{1-\frac{v}{c}}$

$\frac{T'}{T} = 1+\frac{v}{c}$

So the two cases are observably different, classically. So classically, the Doppler shift can tell you who is "really" moving, in violation of the principle of relativity.

But now, let's not assume that the two clocks are the same. Instead, let's assume that there is an unknown velocity dependent effect on clocks: A moving clock slows by an unknown factor $g$.

Then if we redo the calculation using $g$, then we get the following results:

$\frac{T'}{T} = g \frac{1}{1-\frac{v}{c}}$

$T'$, the measured time according to the receiver, is decreased by the factor $g$.

$\frac{T'}{T} = \frac{1}{g} (1+\frac{v}{c})$
This time, the measured time according to the sender is decreased by the factor $g$.
Then if $g$ is chosen correctly, we can get these two cases to be the same (restoring the principle of relativity):
$g \frac{1}{1-\frac{v}{c}} = \frac{1}{g} (1+\frac{v}{c})$
$\Rightarrow g^2 = (1 - \frac{v}{c})(1+\frac{v}{c})$
$\Rightarrow g = \sqrt{1-\frac{v^2}{c^2}}$