# Spacetime / Time Dilation Question

1. Jul 3, 2010

### rede96

As I understand it, anything that is travelling at the speed of light moves through space but not through time.

So if we were able to track a photon from the surface of the Sun, we would say it took about 8 minutes for the photon to travel to Earth. However from the photon's point of view, it reaches the earth at exactly the same time it left the Sun's surface, as there would have been no movement in time.

Therefore, if nothing else is moving through time, I would guess that means from the Photon's point of view, nothing else can be moving through space either.

So in the above example, the Sun and the Earth are in the exact same relative position in space and time during the 8 minute journey from when the photon leaves the Sun until it reaches the Earth and is absorbed.

I would therefore assume that if this is true then the same can be said for light travelling to Earth from any star or source.

So for a photon travelling to Earth from a star that is say 10 light years away, (ignoring gravitational lensing) during the 10 year journey nothing would be moving in space relative to the photon and it would arrive at a detector on Earth at the exact same time it left the star and, both the Earth and the distant star would be in the exact same relative postion.

Is all this correct?

2. Jul 3, 2010

### bcrowell

Staff Emeritus
FAQ: What does the world look like in a frame of reference moving at the speed of light?

This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.)

What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.

3. Jul 4, 2010

### rede96

Thanks for the reply bcrowell, however I was hoping for a simple 'yes' or 'no' answer. :)

I guess what FAQ answer is saying is that there in no point looking at this from a frame of reference that is not possible physically. (Or something like that.)

Excuse my ignorance but to me that is just like saying if a tree falls in the woods and no one is around to hear it, does it make a sound.

Anyway, I agree that maybe I need to look at this a different way. BTW, my goal here is just to try and understand this topic better but using as little Math as I can get away with.

So, if anyone is willing to help me out here I would very much appreciate it.

So here goes...

A spaceship travels from the Sun to Earth at half the speed of light. The pilot of the spaceship times the journey and he notes that it takes 16 minutes.

An observer on the earth also times the journey by using a device that receives and sends signals to the spaceship at the speed of light and calculates the distance of the spaceship at regular time intervals. Finally, the spaceship arrives and the observer notes the same time for travel the same as the pilot, 16 minutes.

All this makes sense as the distance and speed are absolute, so it doesn't matter what frame of reference you are in, it would always take the spaceship 16 minutes to make the journey.

However relativity states that time moves more slowly in a moving frame of reference. Therefore, the pilot has aged less than the observer, even though they both noted the same time for travel.

How can this be, I'm well confused!

4. Jul 4, 2010

### Passionflower

If we assume it takes light to go from the Sun to the Earth exactly 8 minutes and the spaceship goes at half the speed of light the traveler would clock less than 16 minutes due to time dilation. However an observer on the Earth would clock 16 minutes. How much exactly you can calculate by using the Lorentz transformations. However this assumes conditions valid for special relativity, a more complete solution would be to use general relativity but the end result would be practically, but not exactly, the same.

5. Jul 4, 2010

### Staff: Mentor

As Passionflower has already pointed out, earth observers and the pilot will not measure the same time for the trip. At a speed of 0.5c, the time dilation factor (gamma) is about 1.15. If earth observers measure the trip to take 16 minutes, the pilot will measure the trip to take 16/1.15 = 13.9 minutes according to his clock.

6. Jul 4, 2010

### rede96

Thank you for the replies. I think I am getting there, but please bear with me!

So time dilation makes the trip shorter for the pilot, which I am assuming is due time moving slower for the pilot and traveling a shorter distance due to length contraction. So would a photon make the trip in 0 minutes?

Also, if the pilot and the observer had an identical set of candles that burnt one every minute, I take it that the pilot would only burn 13.9 candles but the observer would burn 16 candles during the trip. So the candles on the spaceship must physically burn slower than the ones on Earth. But I thought the laws of physics were the same for all bodies in uniform motion? Why don't candles burn at the same rate?

7. Jul 4, 2010

### Staff: Mentor

According to earth observers, the moving clock of the pilot runs slow. Of course the pilot sees his clock as running normally (and earth clocks running slow) and the distance between sun and earth as being contracted.
Photons do not belong to an inertial frame, so the question has no real meaning. Change the question slightly and you might get something just as good: Imagine the ship moves at 0.99999 times the speed of light. Would the trip be really short according to the pilot? Of course! The faster the ship moves, the greater the length contraction and time dilation.

Everyone will agree that the pilot will burn 13.9 candles during the trip. Earth observers will say that 16 earth candles burned during the trip, but the pilot will say that the earth observers started burning the candles way too early. He'd say that during his trip, only about 12 candles burned on earth.
No, the candles are identical. But observers do see moving clocks (and candles!) run (and burn!) slowly.
They do. Except when in relative motion. Just like any other physical process.

8. Jul 5, 2010

### rede96

Thanks Doc Al, thanks for your time.

I think I am getting there but let me do another scenario to see if this is making sense!

Two space stations are built together in space, each with its own clock. The clocks are synchronised and then the space stations moved 4 light minutes apart in opposite directions, so they are eight light minutes apart in total. Also the clocks are kept in sync.

The pilot on space station A has a meeting on space station B and must be there at 1pm prompt. His spaceship travels at 0.5c so he works out his journey time to be 13.9 minutes and thus arranges to leave his space station at 13.9 minutes to 1pm so he can arrive on time. (Ignoring acceleration, docking etc.)

When he departs he sends a message to space station B (at c) to let them know he as set off. Space Station B gets his message 8 minutes later and because their clocks are in sync and they know his journey time, they know he will arrive 5.9 minutes later as exactly 1pm.

Something tells me that is not quite right but I'll be dammed if I know what!

Also, if the pilot had a space ship that travelled at 0.9999c, what would his journey time be and what time would he need to leave his space station to get to other space station on time?

9. Jul 5, 2010

### Staff: Mentor

OK.

Since the trip takes 16 minutes according to space station clocks, he'd better leave when clock A reads 16 minutes to 1pm.

When they receive the message, they know he's halfway there (since his speed is 0.5c). So they know he'll be there in 8 minutes according to their clocks.

During his journey, his clock will record a travel time of 13.9 minutes.

At such a speed the time dilation factor (gamma) would be about 71 (instead of 1.15). So during his trip his clocks will record 16/71 minutes or about 14 seconds.

10. Jul 5, 2010

### rede96

Ah, I think I get it!

Although the actual journey is 13.9 minutes for the pilot, for all intents and purposes, his journey is 16 minutes for space stations. The pilot just gets the benefit of a 'shorter' trip because of the effects of time dilation.

Same again I assume. The pilot would actually set off approximately 8 minutes to the hour and arrive at 1pm. But he would have only felt a journey time of 14 secs, hence from his point of view, he's jumped ahead in time by about 7.45 minutes. (Give or take a few seconds)

Do the effects of time dilation only happen when travelling through space-time? (i.e. relating to the expansion on the universe.)

11. Jul 5, 2010

### nismaratwork

Where else would you be traveling? Other than that, you seem to have it now. Go Doc Al!

12. Jul 5, 2010

### Staff: Mentor

Right!

Sounds good.

Not sure what you mean. (How do you not 'travel through space-time'?) I'd say that the effects of time dilation must be considered whenever comparing measurements made in frames in relative motion.

13. Jul 5, 2010

### rede96

I thought I'd read or watched something where it was said that the universe was expanding at a rate faster than the speed of light. Obviously, this is allowed because it wasn't that objects were moving through space faster than c, just that the space in between them was expanding at a rate larger than c.

So as they are not actually travelling though space-time, but are still moving away faster than c relative to us, would time dilation still have an effect? (I.e. would their clock be running backwards?)

14. Jul 5, 2010

### nismaratwork

What you're thinking of is actually the expansion of spacetime. Think of an inflating balloon; you're never left with something other than "balloon", but it is expanding. This leads to galactic recession speeds which exceed 'c', but that is a result of an additive property. Galaxies are still not exceeding c within the balloon. Their clocks would not run backwards; in their frame they are still at or below c. Good question however.

15. Jul 5, 2010

### rede96

Ok nismaratwork, thank you for that. I understand your answer but find it really difficult to get my head around spacetime expanding. That'll take a bit more thought.

Anyway, I'll go away and do some more reading. I find this stuff really interesting, albeit very confusing at times!

Thanks to everyone for their help, it is much appreciated.

16. Jul 5, 2010

### TrickyDicky

I think you are referring to space expansion, if spacetime expanded there would be no recession of galaxies since the stretching of time would cancel out the stretching of space.

17. Jul 5, 2010

### nismaratwork

Metric expansion then if you wish.

18. Sep 9, 2010

### MrXavia

The pilot would measure his speed as faster than .5c
but isn't it also valid to say that he is stationary and the rest of the universe moves? after all evey frame of reference is valid.
So to his measurements the clocks on the space stations would be slow and he would arrive early?

How does that work?

19. Sep 9, 2010

### Staff: Mentor

No, the pilot would measure his speed to be 0.5c.
Definitely!
He will see the space station clocks as running slow, but he'll also see them as being out of synch. According to the pilot, the clock at space station B is set ahead of the clock at space station A.

20. Sep 9, 2010

### MrXavia

Are you sure? If by his clock it takes 13.9 minutes to travel 8 light minutes, his calculated speed would be higher than 0.5c.
Not sure I understand that, I thought they were in perfect sych before he left, and both are moving the same relative speed to him, so how can they be out of sych as they are motionless relative to each other?

I really want to understand how this works!

21. Sep 9, 2010

### Staff: Mentor

According to the space stations the distance is 8 light minutes, but according to the pilot the distance is only 8/1.15 = 6.96 light minutes.

Just like measurements of time and distance are frame dependent, so is whether clocks are in synch. Simultaneity and clock synchronization depend on who's doing the observing. According to the space station frame the clocks are in synch, but not according to the pilot.

22. Sep 9, 2010

### MrXavia

Ok so the pilot would see space contract when he sets off from the first station?
Interesting indeed

Would the clocks seem in sync again once he arrives at the second station?

Very interesting.

23. Sep 10, 2010

### Passionflower

Yes, and if he moves a certain way the space station could get closer to him even when he actually moves away from it!.

24. Sep 10, 2010

### bcrowell

Staff Emeritus
Spacetime doesn't expand. Space expands.

Yes.

Not exceeding c relative to what?

In their own frame their velocity is zero.

The universe as a whole doesn't have an expansion speed, but it's true that you can have galaxy A and galaxy B with the distance between them increasing at a rate greater than c. It is even possible observers in A to receive photons from B, even if that rate of increase has always been greater than c: Davis and Lineweaver, Publications of the Astronomical Society of Australia, 21 (2004) 97, msowww.anu.edu.au/~charley/papers/DavisLineweaver04.pdf

There is nothing in relativity that forbids objects distant from one another from moving at >c relative to one another. What can't happen is for one object to zip right by another object at >c. General relativity doesn't even provide an unambiguous way to define the velocity of one object relative to another *distant* object.

It's valid to interpret them as moving relative to one another at >c. It's also valid to think of it as an expansion of space.

This is a general relativity problem, not a special relativity problem, so you can't just analyze all time dilation in terms of the special-relativistic kinematic time dilation. In any case, you will never be able to observe an object's time running backward. As the Doppler shift approaches infinity, you can interpret the object's time as running at a rate closer and closer to zero. Once the Doppler shift hits infinity, the object is beyond an event horizon for you, and you can't observe it.

25. Sep 10, 2010

### nismaratwork

I'm too used to saying spacetime, my error.
Yes, good.
Not exceeding c relative to local to commoving bodies.
My bad, I should say that a commoving observer will note that they are at or below c.