# Quick question on relativity

1. Jul 13, 2012

### Veni2K

If I syncronise my clock with a spaceship just before it goes flying off and wait a random period of time for it to fly past again so I can get a glimpse of its clock, according to relativity I'll notice that its time will be running slower, is that right? So for me on earth, it would have only have been gone say 10 minutes by my clock, but by its own clock, it might say only 7 minutes?

But what about the man on the spaceship, through his window he might see me on earth whizzing away from him and assume i'm the one moving. Wont he predict that because I'm moving at fast speeds the next time we see eachothers clocks, mine will be running slower. So what does he make of this when, after 7 minutes on his clock, we renedvous again and he sees mine reads 10?

Thanks,
Veni

2. Jul 13, 2012

### Staff: Mentor

To get the spaceship to come back so so that it can "fly past again", the spaceship must accelerate or otherwise change its speed. This situation is described by the "twin paradox" (which isn't really a paradox). Google for "twin paradox", and you'll find a bunch of good references. Read them, and if that doesn't explain this situation for you, come back here with your followup questions.

BTW... The rocket clock will be the slow clock... Rocketman will age three minutes less than you back on earth does.

3. Jul 13, 2012

### ghwellsjr

This is the classic Twin Paradox. Didn't you know that?

4. Jul 13, 2012

### Veni2K

Thanks guys for the quick responses, I'm new to relativity and according to the internet the twin paradox highlights the fact that one of the twins would be older than the other, so I get the meaning of the paradox. My question is on the same sort of track but slightly different. I'm assuming that each person, me and the spaceman, both see the same thing happening to the other (slowing down of time). My question is, if the same thing happens to both of us (I see time for him slow, he sees time for me slow) how can one clock be faster than the other when we see eachother again, and whose one will be faster? Wouldnt the same thing happening to both of us mean both clocks retard by the same amount, Hopefully you understand what I mean, or can spot if I have an error in the logic of the question i'm asking.

Thanks,
Veni

5. Jul 13, 2012

### Staff: Mentor

If you don't bring the spaceship back, so it keeps on moving away from you at a given speed while the space guy sees himself at rest and you moving away from him at that speed.... Yes, you will both see the other's clock running slow.

But there's no contradiction here. You can make yourself think that there is one by overlooking the relativity of simultaneity, but if you carefully work through the relativistic effects, everything is consistent.
(This particular case, in which we never bring the two clocks back together so don't go through the twin paradox explanation, is most easily analyzed using the Lorentz transforms directly instead of using the time dilation formula which is derived from these transforms - the Lorentz transforms make it much easier to keep track of the relativity of simultaneity, keep bogus notions of "at the same time" from sneaking in)

6. Jul 13, 2012

### PAllen

If you take literally your question about what you 'see', the resolution is very simple. The rocket turnaround immediately starts seeing the home clock going faster than theirs, and it does so for the whole trip home. The home twin sees the rocket clock going slow until the rocket is close to them - that is, until the rocket's image from the turnaround point arrives at the home twin. Then the home twin also sees the rocket clock going fast - but only for a short time. Thus, each one ends up in perfect agreement about how much the other's clock has advanced compared to theirs. Each agrees the home clock has advanced further.

The asymmetry is in the Doppler effect, which is caused by the asymmetry of the rocket physically turning around - they see blue shift/ faster home clock time immediately.

Last edited: Jul 13, 2012
7. Jul 13, 2012

### ghwellsjr

8. Jul 13, 2012

### TSny

The asymmetry in the Doppler frequency is a fun way to analyze the twin paradox. I think I first saw it in the book Conceptual Physics by Paul Hewitt.

Of course, it’s worthwhile to do the same analysis for a pre-relativity scenario where the earth is at rest in the “aether”. There would still be a similar asymmetry where the turn-around twin immediately notices a shift in frequency of the earth twin’s signals, but the earth twin doesn’t see a shift in the traveling twin’s signals until much later. But, in this case, they both agree that they have aged the same amount for the whole trip. That is, each twin will have received the same total number of signals from the other twin at the end of the trip.

The difference in the results for the relativistic and prerelativistic scenarios is of course due to the fact that the Doppler shift formula in relativity is different than in pre-relativity.

9. Jul 13, 2012

### Underwood

I think its because the one who turns around says that the one who stayed home gets older very fast when he turns around. The home one says the one who turns around doesn't get any older when he turns around.

10. Jul 13, 2012

### Staff: Mentor

That's a valid way of looking at it, but you have to be very careful with that explanation. There's a pitfall hidden in that phrase "gets older very fast" - fast compared to what?

11. Jul 13, 2012

### PAllen

This is a good suggestion. One thing to point out is that for relativity, the amount of doppler shift and signal speed are symmetric. That is, when either sees blue shift or fast clocks, each sees the same amount of blue shift and the same clock speed up relative to theirs (they see each shift for different amounts of their time, but the shifts appear identical). Further, signal speed is symmetric - invariance of c.

For any classical analog, there must be an asymmetry in signal speed. There must also be an asymmetry signal reception rates. Consider two different classical analogs:

1) Assume light behaves like sound. Then the traveling twin sees slower signal arrival rates during slow phase, and faster signal arrival rates during fast phase, compared to home twin. The reason is that for the traveling twin, the separation is growing/shrinking, and the signal speeds are slower/faster. The latter effect does not occur for the home twin. It then follows that they can both end up with the same total number of signals sent and received, despite the home twin having a longer (but less extreme) slow phase and a shorter (but less extreme) fast phase.

2) Assume light behaves like bullets. Here, each sees asymmetry in signal speed, but it is identical for each. The critical asymmetry here is that a bullet shot just after turnaround will overtake and pass a bullet shot just before turnaround (by the traveling twin). Thus, the home twin is continuing to get slow bullets for a while after starting to get fast bullets. The traveling twin sees no such thing. This allows the signal counts to match.

Thus the elegant symmetry of doppler and signal speed in relativity requires the difference in aging.

Last edited: Jul 13, 2012
12. Jul 15, 2012

### Vandam

Hi Veni,

I think your question is about reciprocity, not specific Twin paradox. I show reciprocity on a (loedel) diagram.

For pink observer his pink watch shows 10 minutes simultaneous with blue watch at 7 minutes.
For blue observer his blue watch shows 7 minutes simultaneous with pink watch at 4,9 minutes.
For blue observer his blue watch shows 10 minutes simultaneous with pink watch at 7 minutes.
Be carefull to interpret reciprocity correctly.

The world of simultaneous events for both observers is different. That's relativity of simultanaeity. Event blue watch at 7 and event pink watch at 4,9 are simultaneous for blue observer (they happen in one blue world if sim events), but those two events do not happen simultaneously for pink (i.e. not both in one pink world of sim events; they happen one before/after the other)

Last edited: Jul 15, 2012
13. Jul 16, 2012

### ghwellsjr

Nice Loedel diagram (I think), but it doesn't apply to Veni's scenario and your conclusions are wrong. Veni, the pink earth observer, is correct. When the blue spaceship observer's clock reads 7 minutes, it will be simultaneous with the pink earth observer's (Veni's) clock reading 10 minutes, not less than 5 minutes as you assert. This is not a reciprocal situation and Veni did not ask about a reciprocal situation.

I challenge you to read the scenario that Veni described, a classic Twin Paradox, and make a Loedel diagram to correctly depict what is happening. Of course you will have to fill in some details that Veni didn't provide, such as, that the spaceship travels away from earth in a straight line for 5 minutes (according to the earth's frame) and then turns around and travels back to the earth at the same speed for another 5 minutes OR you could do it the way Einstein proposed where the spaceship travels in a large circle returning to the earth after 10 minutes.

14. Jul 16, 2012

### PAllen

One part of the above description is incorrect. For the 'waves in medium' analog, though the traveler on return sees faster signal speed, the blueshift signal arrival rate are less than the blueshift portion of the home twin's history. The reason is that the traveling twin gets at most all of the home twin's signals during the return half of their trip. Meanwhile, the signals from the traveling twin can be bunched up as the returning twin nearly keeps up with prior signals. Thus, the home twin sees less extreme redshift, and more extreme blue shift (for a very short time), compared to the traveling twin.

15. Jul 17, 2012

### harrylin

Hi, apparently you did not realize that in your Original Post you described a "Twin paradox" scenario - which is just what you now say that you are not after. I think that you confound two very different cases (and thus there's indeed an error in your question):

1. The case of two inertial reference systems, with what is called "mutual time dilation" - according to measurements with the one system, clocks in the other are slow and vice versa. In this scenario clocks can only meet once, they can never meet again for a close comparison and people have to estimate distant simultaneous clock readings.

2. The "Twin paradox" scenario, in which a traveler goes away and returns, and finds that his clock is behind on clocks on Earth. In this case the traveler is not at rest in an inertial reference system while in approximation the stay-at-homes are at rest in one; the paradox relates to how the traveler will experience events.

I guess that you already don't (at least, not really, in view of the mix-up) understand case 1 - which is supposedly the simpler case. However, to understand it you need to be familiar with "relativity of simultaneity" (you can search for it). I'll stop here, for your feedback.

Last edited: Jul 17, 2012
16. Jul 17, 2012

### ghwellsjr

Since Veni hasn't come back to answer my questions or fill in the details of his scenario, I will make some assumptions and then describe what is happening in this classic Twin Paradox scenario.

We will ignore effects of gravity and assume that the earth is not rotating or revolving around the sun. Veni remains inertial on the earth with a clock that starts out synchronized with a similar clock on the spaceship. The spaceship takes off and follows a circular "orbit" around the earth such that it flies over Veni every ten minutes. Veni observes that the time on the spaceship has only advanced seven minutes after each flyby. This is a time dilation of 1/0.7. From this we can determine that the speed of the spacecraft is 0.714143c or 133 thousand miles per second and the radius of the "orbit" is 12.7 million miles. I put "orbit" in quotes because it is not a gravitational orbit but rather a trajectory that requires constant firing of the rockets to keep the spaceship from flying off into space.

First we will consider this scenario from an inertial frame in which Veni is at rest because that is how Veni described the scenario in the first place. In this frame, since Veni is always at rest, he has no time dilation. Also in this frame, since the spaceship is always traveling at 0.714143c, the time dilation factor is γ or 1/0.7 which explains why the time on the spaceship clock is 7 minutes when Vini's clock reads 10 minutes as the spaceship flies overhead. This pattern repeats as the spaceship flies overhead every 10 minutes, the spaceship clock advances by another 7 minutes.

Now we will consider the same scenario from another inertial frame but this time one in which the spaceship is at rest for the brief moment that it is flying over Vini's head except that in this frame, it's the earth and Vini that are flying under the spaceship in the opposite direction at 0.714143c or 133,000 miles per second. As such, the spaceship's clock is not time dilated (during the brief flyunder) but rather it is Vini's clock that is always time dilated by a factor of 0.7. However, as soon as Vini and the earth pass under the momentarily stationary spaceship, it has to "take off" and zoom around far under the earth to catch up to and pass Vini and the earth and then circle back up and be ready for Vini and the earth to pass under it once more, seven minutes later (according to the spaceship's clock, not according to the coordinate time of the frame). In order for the spaceship to do this, it has to accelerate to a speed of much more than 0.714143c and during this time, it's clock will be running much slower than Vini's clock so that when it gets ready for the next flyunder, it's clock will be reading seven minutes later and Vini's clock will be reading ten minutes later.

So the answer to Vini's question (if he will allow my expansion of the details of his scenario) is that Special Relativity says that each observer will determine that the other ones clock is running slower than his own during the brief moment that they are passing each other but at other times, we have to look at a particular frame to see whose clock is running slower but all frames will conclude that the spaceship's clock advances by seven minutes for every ten minutes of Vini's clock.

And if Vini won't allow my expansion of the details, then he needs to provide whatever details suit him.

Last edited: Jul 17, 2012
17. Jul 19, 2012

### Vandam

O.K. What I wrote is not wrong, but does not apply to Veni's post. I thought it was about reciprocity. Sorry about that.