High School Confusion over relative velocities and reference frames

[etotheipi]

I will refer to the example given in 'On the electrodynamics of moving bodies' concerning a rod moving in a coordinate system, in which a beam of light is sent from one end of the rod to the other and is then reflected back.

Usually when calculating relative velocities, we may simply consider the object the velocities are relative to to be at rest and go from there, so for example if an object A is at 3ms-1 and object B is at 5 ms-1, the relative velocity of B from A (taking A to be at rest in its frame of reference) is 2ms-1.

In the case of the rod, from the perspective of an observer in the stationary frame, a beam of light of velocity c traveling in the same direction of the rod of velocity v is given a relative velocity to the rod of c - v, in accordance with the relative velocity equation. However, I don't quite understand this in this context since if we consider the rod to be at rest - as we usually do when imagining relative velocities - the speed of light relative to it is c - v, and not c as implied by the axioms of special relativity.

My question is whether, in the stationary rest frame, when we calculate relative velocities are we 'still in' the stationary rest frame? That is, it is quoted in many places that the time taken for the beam to reach the end of the rod is (c-v)/L implying a relative velocity of (c-v) and length of L, however the notion of a relative velocity when we are in another reference frame is slightly confusing to me.

 

[Orodruin]

You need to distinguish two concepts, the relative speed between two objects, which is the speed of one of the objects in the rest frame of the other, and the speed of separation between two objects in some given reference frame. The speed of separation between a light signal and an object moving at speed v in some reference frame is c-v in that reference frame. The speed of light in the object's rest frame is c.

 

[Ibix]

You can take the difference between two velocities as measured in some frame S, yes. It gives you the separation rate between the two objects with those velocities, as measured in S.

This isn't the same as the velocity of one object as measured by a frame, S', in which the other is at rest. According to S, S' is using rulers that are length contracted and clocks that are time dilated and improperly synchronised, so it's no surprise that they measure a different velocity.

So the separation rate between two things may be anywhere between zero and $2c$ (for light pulses traveling in opposite directions). But the relative velocity as measured by one of the objects will always be below $c$.

 

[etotheipi]

Thank you both for your swift responses! This distinction makes a lot of sense.

Would I be right in assuming that, when using the speed of separation concept, it is valid to imagine it in a similar fashion to regular relative velocities (i.e. imagining that one object is not moving and saying that this separation speed is the speed of the other object moving past it), except keeping in mind that we are still in the original reference frame and that one object is not really at rest?

 

[Orodruin]

i.e. imagining that one object is not moving and saying that this separation speed is the speed of the other object moving past it

You need to let go of this notion if you want to look at separation speeds. It is not helpful for your understanding. But in general, yes, the separation velocity is the difference between the velocities of the objects and you can use regular subtraction to find it. When you look at the actual relative velocity, you need to consider relativistic velocity addition.

 

[etotheipi]

You need to let go of this notion if you want to look at separation speeds. It is not helpful for your understanding. But in general, yes, the separation velocity is the difference between the velocities of the objects and you can use regular subtraction to find it. When you look at the actual relative velocity, you need to consider relativistic velocity addition.

Thank you very much, I have a better understanding now