# Speed Of Light Question

1. Jun 25, 2004

### zoobyshoe

I'm on a space station in deep space. I depart in a rocket and accelerate to .5 c then level off to that speed. No longer accelerating, I now define myself to be stationary. I define the speed of anything I percieve to be in motion as relative to my own stipulated-to-be-unmoving inertial frame

On board I have a Michelson-Morley (or better) apparatus for measuring the speed of light. However, I am not going to measure the speed of any light that originates on my own ship. I am going to measure the speed of a beam of light that comes from the space station I just left.

At a prearranged time the space station sends this beam of light in exactly the same direction I am traveling.

When it catches up to me I measure it to be going:

A. c?

B. .5 c?

c. other ?

Last edited: Jun 25, 2004
2. Jun 25, 2004

### jcsd

A. c

second postulate

3. Jun 25, 2004

### Nereid

Staff Emeritus
How, exactly, do you measure the speed of light from the space station?

4. Jun 25, 2004

### zoobyshoe

I assumed it could be done since people have measured the speed of light from stars.

You're saying it can't be done?

5. Jun 25, 2004

### Nereid

Staff Emeritus
There are several methods by which one could measure the speed of light. Many of those who find SR hard to understand (or accept) feel that part of the 'problem' is the actual method of measuring the (local) speed of light. :tongue2:

In the case of your hypothetical experiment, I was wondering if you were getting at something special about the space station you had left (or would any distant source do)?

6. Jun 25, 2004

### zoobyshoe

Any source would do, actually.

7. Jul 3, 2004

### zoobyshoe

Another Speed Of Light Question

In the above situation (space station, rocket ship) my speed away from the station is .5 c. The speed of the light departing the space station in my direction is c.

If I am, say, 300,000 km away from the station ( a distance selected for convenience) when the light is emitted from the station, what math do I use to determine 1.) how long it will take for the light from the station to catch up to me and 2.) How far away from the station I will be when it does catch up?

8. Jul 3, 2004

### Staff: Mentor

Assuming you are 300,000 km away from the station as measured in your frame, then:
(1) The light must travel 300,000 km to reach you, so T = D/c = 1 sec.
(2) During that 1 sec the station has traveled a distance D = VT = 0.5cT = 150,000 km. So when the light reaches you you'll be a total of 450,000 km from the station.​

9. Jul 3, 2004

### Janus

Staff Emeritus
As measured by you or the space station? If measured by you, the light will reach you in one sec and the station will be 450,000 km away when it reaches you.

If measured by the space station, then the light will reach you in 2 sec and you will be 600,000 km away from the station.

10. Jul 3, 2004

### zoobyshoe

Sorry I wasn't clear.

In the second question I meant to change my definition of myself as being motionless to being in motion at .5c relative to the source of the light - the space station.

So, If I am traveling at .5 c away from the station, and the beam of light is emitted from the station when I am a distance of 300,000 km away, how long will the light take to reach me, and how far away will I be from the station?

11. Jul 3, 2004

### Janus

Staff Emeritus
Doc Al and I both gave you the answer already: 1 sec and 450,000 km, assuming that you are making the measurements.

12. Jul 3, 2004

### zoobyshoe

OK, What math did you use to arrive at the 2 sec, 600,000 km figures?

13. Jul 3, 2004

### Janus

Staff Emeritus
From the space stations perpective, the relative speed between the light it emits and you is .5c. Since it the distance between you and the station is 1 light sec at the time of emission, it takes 2 sec for the light to close the distance. If the light travels for 2 secs, it travels 600,000 km.

14. Jul 4, 2004

### zoobyshoe

OK the space station viewpoint makes sence to me, but I don't understand how there could be a difference from my perspective.

If the light is emitted from the station when I am 300,000 km away from it, it will take the light one second to reach that point. Once that second has elapsed and the light has reached that point, I will no longer be there. It will be one second later, and at .5c, I will be 150,000 km farther away, the light will not have caught up with me yet. In the half second it takes the light to reach that point 150,000km away, I will have traveled yet further, and won't be there when it arrives. And so on, such that I'm thoroughly confused as to why it isn't also 2 seconds 600,000 km viewed from my perspective as well.

15. Jul 4, 2004

### Staff: Mentor

view from the ship

The light travels at speed c with respect to you. And, with respect to you, "that point" is you! The fact that you are also moving with respect to something else (the spaceship) is irrelevant. No matter what speed you are moving, if the light flashes when it's 300,000 km away from you, it will reach you in one second.
In your frame, you don't move. (You are at rest in your own frame.) In the frame of the station you are moving, but that doesn't matter.
Yes, you will be 150,000 km further from the station. And yes the light will have caught up with you!

Just remember that in your frame light travels 300,000 km/sec with respect to you.

16. Jul 4, 2004

### GOD__AM

Hi, I'm new so please excuse me if I'm butting in. This is all very interesting but I'd like to add another observer to the seudo experiment. Assuming what Doc Ai and Janus are saying is correct (as far as I understand they are) then we get the following senario.

Let's say there are 2 space stations 450,000k apart. The rocket leaves the first station (call it "a") as outlined already heading towards space station "b". The light is still turned on when the rocket is 300,000k from space station "a".

Now the observer in the rocket sees the light from space station "a" when he is right beside space station "b" . Space station "b" cannot see the light for another 1/2 second.

Now lets suppose that the observer in the rocket has a mirror set up that aligns the light source from space station "a" with our observer at space station "b" as it passes so that he sees the light reflected from the rocket nearly a 1/2 second before he sees the light from space station "a". :surprise:

Hope I didn't botch that up too bad. It makes perfect sense to me (the experiment not the results that is).

17. Jul 4, 2004

### zoobyshoe

The light is moving at c in its direction of propagation. That is: at any given time, the tip of the beam of light is propagating away from the point in space from which it was emitted, at speed c.

My speed in the same direction is relevant with respect to the tip of that beam of light. The light cannot reach me in one second, because I am moving away from its approaching tip. And: I cannot be aware of the beam of light to calculate anything about its time untill it first catches up with me. Were I able, by magic, to see the tip of the beam of light, I would not see it closing the distance between us at c. The time it takes for the light to reach me from the time I was at 300,000 km from the station must be greater than 1 sec.
If, once the light has caught up with me, I measure its speed, I will clock it to be going at c regardless of my speed in any inertial frame. Jcsd confirmed this above, and I believe this is the agreed result. Einstein used this mysterious fact as his second postulate.

That, however, is a separate issue from the fact that the tip of a beam of light is limited to propagation at c from its point of origin. It is for this reason that the space station could determine that there was a difference of .5 c between my speed, and the speed of the beam of light as it is on its way toward me.

18. Jul 4, 2004

### Staff: Mentor

The light is moving at speed c with respect to the observer.
No. From the viewpoint of the spaceship, the separation distance between the tip of the light beam and the space station increases at only 0.5 c. (It is only from the view of the space station that the light travels at speed c with respect to the space station.)
Not true. You are moving away from the space station, but the light is still moving towards you at speed c. This is what "invariant speed of light" means.
The only way to measure the speed at which the light got to you is to know when and where it started. You'll have to imagine an extended space ship frame--or another space ship moving along with yours but located next to the space station at the very moment the light is emitted. Your sister space ship has its clock synchronized to yours, and it is 300,000 km away from you. When the light reaches you, you note the time. Later you compare notes with the other ship, which tells you the time the light flashed: you will find that the light took 1 second to reach you, according to the synchronized clocks in the two ships.
Exactly right. But in your frame, the light only traveled 300,000 km to get to you. So it took 1 second.
The speed of light will always be c with respect to whatever observer is making the measurement. That does not mean that from your point of view that light cannot separate from a moving object at some speed less than c. Of course it can.

19. Jul 4, 2004

### zoobyshoe

Thanks for your patience, Doc Al. I have never explored this in such detail, and these answers are not what I expected.

If I reverse the direction of the rocket in the set up, then, to be headed toward the station, I take it that a beam of light emitted from the station when I am 300,000 km from the station will also reach me in one second, from my viewpoint, however: I will be only 150,000 km from the station when I detect it?

Last edited: Jul 4, 2004
20. Jul 4, 2004

### plover

Linguistic point:
The phrasing used here (e.g. "headed toward the station") carries the connotation that the rocket is moving and the station is stationary, which is, of course, the familiar setup - vehicles move and destinations don't. However, part of the essence of special relativity is that describing events for a given observer must always presume that observer as being at rest.

Thus while the mindset of a person on the rocket might be "I am approaching the station at .5c", the necessary viewpoint for setting up any special relativistic calculations is "the rocket is at rest, the station is approaching the rocket at .5c".

Once that framework is established, it becomes easier to see that a photon emitted from a station 300k km away takes 1 second to reach the rocket (since the rocket is at rest) no matter the velocity of the station.

My point is that it's easy for intuitive descriptions of a situation to carry assumptions that conflict with the requirements of the formal model.