# More light questions

1. Nov 17, 2005

### SigurRos

Hey, I was reading through a few of these posts, and something crossed my mind:

Say there was a space ship travelling away from earth at the speed of light. To observers on earth, the ship would travel at c, and to observers on the ship, earth would be traveling at c. Suppose that the ship sent a beam of light back to earth. To observers on the ship, the light would travel towards earth at c, but since earth was moving away at c, wouldnt the light never reach earth? Similarly, since the ship is moving away from earth at c, and the light is moving toward earth at c, wouldnt the beam of light have to appear to be unmoving, since if it was moving away from the ship at c (or at any speed), the relative speed of the beam with respect to the ship would be c + c(or any other positive velocity), and this would violate relativity?
I confused myself with these thoughts, so I was wondering if anyone could explain it to me.
Thanks

2. Nov 17, 2005

### JesseM

Relativity says it's impossible for a massive object like a space ship to travel at exactly the speed of light, although it can get arbitrarily close from our perspective. Particles moving at c do not have their own inertial rest frames--the Lorentz transformation equations for translating between different frames' coordinate systems give infinities if you try to plug in v=c.

3. Nov 17, 2005

### SigurRos

OK then, suppose the ship was extremely close to the speed of light, say .999999*c. would the light have to move away from the ship at .000001*c?

4. Nov 17, 2005

### Staff: Mentor

No, the light would still move at speed c, from the point of view of an observer on the earth, or any other observer, for that matter.

5. Nov 17, 2005

### JesseM

Nope, the ship would measure the light to be traveling at c relative to itself. It may help to remember that each observer calculates speed in terms of distance/time as measured on their own set of rulers and clocks, and of course each observer sees the other's rulers shrunk and the other's clocks slowed down, and if each observer has a set of clocks at different locations which are synchronized in his own frame, other observers will see these clocks as out-of-sync (different reference frames disagree about 'simultaneity', so two events that happen at the same time-coordinate in one frame can happen at different time-coordinates in another). Because of this, velocity addition doesn't work the same way in relativity that it does in classical mechanics--in classical mechanics, if I see a car driving by my at u=30 mph, and he sees a car driving by him in the same direction at v=20 mph, then to figure out how fast the second car is driving relative to me, I'd just use the formula u + v = 50 mph. But in relativity, the correct formula would instead by $$(u + v)/(1 + uv/c^2)$$ -- see this page for more on this. So, suppose I observe a ship moving away from me at 0.9c, and an observer on the ship sees a light beam moving in the same direction at velocity c relative to him, and I want to know how fast the beam is moving relative to me--in this case I'd set u=0.9c, v=c, and conclude that the velocity of the beam relative to me is (1.9c)/(1 + 0.9) = c.

6. Nov 17, 2005

### SigurRos

The wouldnt this violate relativity, since the ship and the beam of light would appear to be moving away from each other at a speed greater than c?

7. Nov 17, 2005

### dgoodpasture2005

If you're traveling away from Earth... and you shine the light back at earth, and not in FRONT of the direction you are traveling... then i see no problem here.

[Quote SigurRos]Then wouldnt this violate relativity, since the ship and the beam of light would appear to be moving away from each other at a speed greater than c?

Last edited: Nov 17, 2005
8. Nov 17, 2005

### JesseM

My calculation was assuming the ship and the light beam were moving in the same direction, but if they were instead moving in opposite directions, then in my frame the distance between them could indeed be increasing at a rate faster that 1 light year per year. But relativity only says that no individual object can move faster than c in any given frame, not that the distance between two objects can't increase faster than c in a given frame. And despite the fact that I see the distance between the ship and the light beam increasing faster than c, in the ship's own rest frame the light beam is only moving away from it at c.

9. Nov 17, 2005

### RogerAshford

I think we might be losing sight of the fact that velocities in excess of C are impossible from the perspective of the OBSERVER. Sure, you can see two photons going in opposite directions, each moving at c - giving you a "calculated" relative velocity between them of 2c (from your point of view). However, from the perspective (= inertial frame of reference) of each of the photons, the other is travelling at c.

Going back to your original question, and assuming that a spaceship CAN travel at the speed of light (which it could if it had zero mass!), then, yes, the spaceship pilot would see that the light beam would never reach Earth, whereas Earthlings would see that the light beam WOULD reach Earth after a time interval of of D/c, where D is the distance (as measured from Earth) between the earth & the spaceship when the latter sent back the beam of light. There is no paradox here, because observers on the spaceship would see that time on Earth was standing still (time dilation), so they would never see Earth time advance to the instant when (according to observers on Earth) the light beam reached it. It all boils down to the relativistic effect that is at the bottom of all of its predictions: simultaneity is relative.

10. Nov 17, 2005

### Staff: Mentor

Put another way, two observers will never measure their speed between them to be greater than C. If one moves away from the earth at .99c and another moves away in the other direction at .99c, they cannot just add their speeds together and come up with 1.998c as the relative speed of the other. They must add them using the equation for relativistic velocity, otherwise the speeds won't match what they measure by bouncing a laser off each other.

11. Nov 17, 2005

### JesseM

As I said in my first post on this thread, photons don't have their own reference frame--the lorentz transformation doesn't work for v=c, and anyway giving photons their own inertial frame would violate the basic postulate that the laws of physics should look the same in all inertial frames (because the laws of electromagnetism don't allow electromagnetic waves to be at rest). So, there is no answer to the question about what speed photons have from the perspective of another photon.