The discussion revolves around the relationship between the derivative with respect to coordinate time of relative position and relative velocity in the context of special relativity. Participants explore the implications of this relationship, particularly focusing on the differences between classical mechanics and special relativity, and the effects of frame selection on measurements of velocity and separation rate.
Participants express differing views on the relationship between separation rate and relative velocity, with some agreeing that the two can only be equal in specific frames, while others emphasize the complexities introduced by special relativity. The discussion remains unresolved regarding the implications of these differences.
Participants note that the relationship between separation rate and relative velocity is not straightforward in special relativity, and that definitions may vary depending on the frame of reference. There are also mentions of the limitations of applying classical mechanics principles directly to relativistic scenarios.
What do you mean by "relative position" and "relative velocity"?MrBlank said:is the derivative with respect to coordinate time of relative position equal to relative velocity?
Sort of.MrBlank said:This is true in both classical mechanics and special relativity.
Ok.MrBlank said:The position of mass A relative to mass B is given by subtracting the positiin vector for mass B from the position vector for mass A.
Ok.MrBlank said:In classical mechanics, the velocity of mass A relative to mass B is given by subtracting the velocity of mass B from the velocity of mass A. If the derivative with respect to time of the relative position is taken then the result is the relative velocity.
Indeed. But if you recognize this, you presumably already know that the answer to the question you asked is "no". So what are you actually asking?MrBlank said:In special relativity, the situation is not so straight forward.
The separation rate of A and B is defined as ##\vec v_A - \vec v_B## regardless of the frame used. It is just a matter of definition.Ibix said:Using a frame ##S##, you can measure the 3-velocities, ##\vec v_A## and ##\vec v_B## of your two masses. In that frame you are free to subtract those two vectors to get ##\vec V=\vec v_A-\vec v_B##, and this is the relevant quantity if you want to know how long until they collide (assuming they are going to) according to frame ##S##. I usually call ##\vec V## the separation rate of A and B in ##S##. I'm not aware of an "official" name for the value.
The above is another kind of definition: namely the definition of relative velocity of A w.r.t. B. Basically that definition implies picking a special frame: the rest frame of B.Ibix said:You may then transform to the frame ##S'## where B is at rest. In this frame A has velocity ##\vec v'_A##. This is the velocity that B measures A to have, and this is usually called the relative velocity of A with respect to B. Note that it is equal to the separation rate in ##S'##, but this is a special case where one of the velocities is zero.
Sure, separately transform ##\vec v_A## and ##\vec v_B## from frame ##S## to ##S'## and evaluate their difference.Ibix said:In Newtonian physics the separation rate in any frame is equal, and hence equal to the relative velocity. In relativity this is not the case because velocity composition is not a simple addition.
This only works if one of the velocities is zero in ##S'##, as @Ibix already pointed out.cianfa72 said:Sure, separately transform ##\vec v_A## and ##\vec v_B## from frame ##S## to ##S'## and evaluate their difference.
The [magnitude of the] relative velocity of A w.r.t. B is an invariant.cianfa72 said:Sorry maybe I wasn't clear. My point was indeed that in relativity the separation rate of A and B is not frame invariant. It is numerically equal to the relative velocity of A w.r.t. B (which is uniquely determinated) only in the "implied" rest frame of B.