Everytime I think I understand relativity I manage to think up a question that I haven’t a clue as to the answer. I know I’m missing something, but what? This ones seems deceptively simple, but I don’t know. I just can’t seem to get my head around it.
No object can travel faster than c. An object can travel at .55c but not c. Now imagine 3 objects and three reference frames. Let’s call the objects North, South, and Stationary. Object North is traveling at .55c relative to object Stationary. All well and good so far. Just one reference frame. Here’s what I can’t get: Object South is also traveling at .55c relative to Object Stationary, but in the exact opposite direction of Object North. There are no fixed reference frames in relativity, so why aren’t Objects North and South travelling at 1.1c relative to each other?
Someone who knows what they are talking about will be laong eventaully, but in the meantime I’ll point out that you can’t just add velocities in General Relativity the way you can in Newtonian Relativity. It works for low speeds but the closer you get to lightspeeds the less valid it is to do so. So while out two objects may be trvaelling away form each other at half lightspeeds their combined velocity will be considerably less than lightspeed.
I tried to understand why once, but my head started aching.
I think the problem is that the question is phrased wrongly, my take is this.(preparing to be shot down in flames) North and South may be regarded as travelling at 1.1c relative to each other from Stationary’s point of view but that is just a deduction not a measurement. In fact if North measures South’s speed it doesn’t get 1.1c it always comes to less than c.
Speed is distance over time; the faster you are travelling (relative to something else), the slower time gets (relative to the time experienced by that something), so the distance over time calculation by which you calculate your speed still works out below c.
Because that’s not how velocities add in Einsteinian relativity. It is how they add in Galilean relativity, but there’s a correction in Einstein’s theory. If I remember correctly the resultant is… v + u/(1+uv/c[sup]2[/sup])
Actually, there’s something wrong with what I posted above, because two bodies travelling away from each other can each declare themselves, or the other one to be stationary, or can declare each other to be moving apart at equal velocities, but it can’t be the case that they both experience time more slowly than each other. What have I done wrong?
Right. The thing is that all of them will measure the speed of the other two relative to them as less than the speed of light. As you say, they could add the measured speeds of the other two together and come up with a sum greater than the speed of light but that is meaningless.
Actually, they can (and do) experience time more slowly than the other, more or less. It’s just that relativity mucks about with your notions of simultaneity as well. For example, if you’re zipping past me at 0.6 c, and we set our stopwatches to start right at we pass each other, my clock will read 0:05 at the same time yours reads 0:04 in my frame of reference. But in your frame of reference, when your clock reads 0:04, mine will read 0:03.2, since events that are simultaneous in my frame aren’t simultaneous in yours.
OK, but what happens if the two bodies each move apart at 0.6c, turn around and move back to meet in the middle? - from their own points of view, they should each haveen been experiencing time slower than the other, but the same thing has happened to both of them, so what happens when they compare clocks?
Whenever an object changes it’s velocity (either speed or direction) it changes frame of reference. Mathochist’s formula actually applies to (inertial) frames of references, not objects.
In your example, there are actually four frames of references involved[ol]
[li]coincident with the first body out-bound []coincident with the first body in-bound []coincident with the second body out-bound [*]coincident with the second body in-bound [/ol]Any calculations will have to use all four to get the correct answer.[/li]
By the way, this is how to resolve the famous Twin’s Paradox. The traveling twin switches their frame of reference.
Neither experiences time any more slowly. Each one sees the other’s clock as running slow. That is, my wristwatch always keeps the same time to me, but if you zoom by it looks like it’s running slow to you. At the same time, yours looks to me like it’s running slow. It’s all in what each observer sees, not in what “changes”.