Relativity question

[The Wrath]

Okay, I'm having trouble grasping something. According to relativity, the speed of light will be the same for all observers. I understand why there is no such thing as absolute space or absolute time...but I don't understand how it fits into this example:

Suppose that I am moving at 25% the speed of light, and a spaceship that is 1 light year away is moving towards me at 75% the speed of light. By my point of view, that spaceship reaches me, and I conclude that it has moved 1 light year. By the spaceship's point of view, I have moved 1 light year. But, due to time dilation, much less time has passed for the spaceship than it has for me. Not only that, but the time that it took for that distance to close was less than one year...meaning that a distance of 1 light year had been covered in less than a year, from either point of view, due to time dilation. How does this make sense, in the context of relativity? We both measured the speed of light to be more than it really is.

I guess a simpler example would be to assume that I, myself, am moving at the speed of light...and I cover a distance of one-light year instantaneously, by my own reckoning. So, if I didn't word the above example in a way that makes sense, just use this one.

[DavidV]

In relation to a stationary observer, the time would take slightly more than a year. Very slightly - extra seconds or a minute to a year. If you were measuring the blue shift of the incoming spacecraft from your own, it would appear to be traveling very slightly slower than it actually is. The equation is here: http://en.wikipedia.org/wiki/Time_dilation

As you approach the speed of light, your mass increases exponentially. Even if your spacecraft weighted only as much as you+spacesuit, all the energy humans have ever produced would probably not be enough to sustain sufficient acceleration to reach another star within your lifetime. (E=mc2)

[The Wrath]

My question is about, not a stationary observer, but a moving one. If I, the observer, am moving at speeds close to the speed of light, I would cover an immense distance in what, to me, seems like a very short time. Thus, the distance/time is not going to equal the speed of light, which is supposed to be the same for all observers.

I know that there are many physical limitations to an object actually traveling that speed, but I'm ignoring those for the sake of my question as it relates to time dilation. This should, essentially, work the same for any speed. If I'm traveling 100 mph, I experience a very slight amount of time dilation, but would run into the same problem. If a beam of light came towards me across a distance of a light year, while I drove 100 mph towards it for an entire year, the distance would be covered (by my reckoning of time) a fraction of a second less than a year. Granted, it would be a very, very tiny fraction of a second, but the difference still exists. Thus, dividing the distance (1 light year) by the time it took (very slightly less than 1 year), I would arrive at the speed of light being greater than what it actually is.

[Adrian Hester]

Suppose that I am moving at 25% the speed of light, and a spaceship that is 1 light year away is moving towards me at 75% the speed of light.

The basic question is: 25% and 75% of the speed of light relative to what? Velocities aren't simply additive, and to get the velocity of the other spaceship as measured in your reference frame you have to change frames. The frame you're saying measures you as having a velocity of .25c in turn has a velocity of -0.25c in your frame; in that frame the other spaceship is measured as having a velocity -0.75c, which would be measured as -(16/19)c (more generally, [v1+v2]/{1+v1*v2}, with v measured in units of c) in your reference frame. By the same token, you are measured as having a velocity (16/19)c in the other ship's frame.

By my point of view, that spaceship reaches me, and I conclude that it has moved 1 light year. By the spaceship's point of view, I have moved 1 light year.

After 19/16 years, yes.

But, due to time dilation, much less time has passed for the spaceship than it has for me.

Only when the elapsed time is measured in the third reference frame. You might want to work through a space-time diagram with all three observers taken explicitly into account and, as always, be very careful about simultaneity, which is frame-dependent.

Not only that, but the time that it took for that distance to close was less than one year...meaning that a distance of 1 light year had been covered in less than a year, from either point of view, due to time dilation.

No, from either point of view 19/16 years have passed.

How does this make sense, in the context of relativity? We both measured the speed of light to be more than it really is.

You're confusing yourself by bringing in a third observer without taking that into account in the math and then mingling measurements in his frame, your frame, and the other spaceship's.

[The Wrath]

But, both agree on the distance that was covered. The only disagreement is on who covered the distance. I guess I hadn't taken the time to figure out the math about something moving at .25c or .75c, but if you assume that you are moving at c, you cover immense distances instantaneously. Thus, dividing the distance you cover by the time it took will yield a much higher value for the speed of light, than is the correct value. Actually, in this case, time would equal 0, so the answer would be undefined, but I think you get what I'm saying.

If I've still missed something, then I'll have to sit down with a pencil, paper, and calculator and try to figure this out, because it's driving me insane.

Edited June 19, 2007 by Moose

[Robert J. Kolker]

But, both agree on the distance that was covered. The only disagreement is on who covered the distance. I guess I hadn't taken the time to figure out the math about something moving at .25c or .75c, but if you assume that you are moving at c, you cover immense distances instantaneously. Thus, dividing the distance you cover by the time it took will yield a much higher value for the speed of light, than is the correct value. Actually, in this case, time would equal 0, so the answer would be undefined, but I think you get what I'm saying.

If I've still missed something, then I'll have to sit down with a pencil, paper, and calculator and try to figure this out, because it's driving me insane.

Yes. The distinction between proper time (a technical term) and laboratory time is subtle and not intuitively obvious.

Read -Space Time Physics- by Taylor and Wheeler. Taylor and Wheeler's approach in presenting relativity and invariance will get you over the "simultaneity hump". Try getting the older edition which is actually clearer than the second edition. See also http://physics.nmt.edu/~raymond/classes/ph...ook/node42.html.

Every one (just about) who has studied special relativity has had this difficulty, so you are in good company.

In special relativity velocities do not add. That is a consequence of Galilean Relativity, which is very intuitively appealing but is also empirically incorrect. Once you see why velocities do not add, you will be on your way.

Bob Kolker

[mrocktor]

But, both agree on the distance that was covered. The only disagreement is on who covered the distance.

Almost, but not quite. If there are only the two spaceships, "which one covered the distance" is meaningless. The only speed you could measure is the relative speed between the two ships - there is no way to say one ship is "stopped" and the other "moving", or that one ship is moving at 0.25c and the other at .75c. You can only say they are moving relative to each other at a given speed.

Enter the third observer. Now, first of all, this third observer is no "better" than the other two, you can't say he is "stationary" because there is no coordinate system nailed to the universe. Assuming the third observer sees ship A moving away from him at .25c and ship B moving towards him at .75c we could just as well declare ship A as our reference frame and say the third observer is moving .25c away in the opposite direction!

OBS() A(.25c)-> <---(.75c)B

is equivalent to

<-OBS(.25) A() <----(.84c)B

it is also equivalent to

OBS(.75c)---> A(.842c)----> B()

The .84c is the speed that "A" sees "B" approaching (and "B" sees "A" approaching) and is the 16/19c resulting from the formula Adrian Hester mentioned. Keep in mind that we could choose any of the three as our reference frame:

OBS sees A going away at .25c and B approaching at .75c;

A sees OBS moving away at .25c and B approaching at .84c;

B sees OBS approaching at .75c and A approaching at .84c.

There is no distinction between a "stationary" and a "moving" observer! The distinction is between an accelerating and an inertial observer (one that is not accelerating). All of our observers are assumed to be inertial here (the ships are not accelerating or decelerating).

If I've still missed something, then I'll have to sit down with a pencil, paper, and calculator and try to figure this out, because it's driving me insane.

This site is very helpful.

[Robert J. Kolker]

This site is very helpful.

That is an excellent site. Thank you for the pointer.

Bob Kolker