I Derivative w.r.t. time of relative position in special relativity

[MrBlank]

TL;DR

In special relativity, is the derivative with respect to coordinate time of relative position equal to relative velocity?

In special relativity, is the derivative with respect to coordinate time of relative position equal to relative velocity?

Does it matter if constant velocity is used?

 

[PeterDonis]

is the derivative with respect to coordinate time of relative position equal to relative velocity?

What do you mean by "relative position" and "relative velocity"?

 

[MrBlank]

Suppose that there are two point test masses, A and B. 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. This is true in both classical mechanics and special relativity.

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.

In special relativity, the situation is not so straight forward.

 

[Ibix]

This is true in both classical mechanics and special relativity.

Sort of.

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.

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.

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.

In particular, note that in special relativity the modulus of the separation rate is limited to $2c$ while that of the relative velocity cannot exceed $c$.

 

[PeterDonis]

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.

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.

Ok.

In special relativity, the situation is not so straight forward.

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?

 

[cianfa72]

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 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.

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.

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.

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.

Sure, separately transform $\vec v_A$ and $\vec v_B$ from frame $S$ to $S'$ and evaluate their difference.

 

[PeterDonis]

Sure, separately transform $\vec v_A$ and $\vec v_B$ from frame $S$ to $S'$ and evaluate their difference.

This only works if one of the velocities is zero in $S'$, as @Ibix already pointed out.

 

[cianfa72]

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.

 

[jbriggs444]

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.

The [magnitude of the] relative velocity of A w.r.t. B is an invariant.
The separation rate of A and B is not invariant -- as you have pointed out.

The two can only be numerically equal if one is careful to select a frame in which they are equal -- for instance, the rest frame of B.