How can we accurately measure the relative velocity between two moving bodies?

[country boy]

I may be missing something, but it seems that the problem here of understanding relative velocity is not a special relativity problem. Can it be treated classically as a first step?

 

[Meir Achuz]

I may be missing something, but it seems that the problem here of understanding relative velocity is not a special relativity problem. Can it be treated classically as a first step?

This is the relativity forum. Ask your question in the appropritate forum.

 

[country boy]

This is the relativity forum. Ask your question in the appropritate forum.

That is why I asked. The original question was posed in the context of relativity, but upon examination it may not be a relativity question. The difficulty in understanding the relative velocity between A and B is not a result of the magnitude of the velocities. It can be dealt with at low velocities.

 

[Meir Achuz]

That is why I asked. The original question was posed in the context of relativity, but upon examination it may not be a relativity question. The difficulty in understanding the relative velocity between A and B is not a result of the magnitude of the velocities. It can be dealt with at low velocities.

What puzzled me is that this is taught in the first week of physics 101.
V_x=u_x-v_x
V_y=u_y-v_y.
How could this lead to a thread with 33 posts?

 

[country boy]

What puzzled me is that this is taught in the first week of physics 101.
V_x=u_x-v_x
V_y=u_y-v_y.
How could this lead to a thread with 33 posts?

34 ... or 35 ... :-)

 

[George Jones]

I may be missing something, but it seems that the problem here of understanding relative velocity is not a special relativity problem. Can it be treated classically as a first step?

For speeds sufficiently high that Lorentz transformations are significantly different than Galilean transformations, as implied by the original post in this thread, then relativity really is needed.

From the the last equation in my previous post (#30),

(1 - V^2)^(-1/2) = (1 - u^2)^(-1/2) (1 - v^2)^(-1/2),

which, after restoring the c's, leads to

V^2 = u^2 + v^2 - (u^2 v^2)/c^2.

Without relativity, the last term disappears. This is to be expected, since the original post sums perpendicular velocities, so, non-relativistically, the speeds satisfy the Pythagorean theorem.

 

[country boy]

V^2 = u^2 + v^2 - (u^2 v^2)/c^2.

Without relativity, the last term disappears. This is to be expected, since the original post sums perpendicular velocities, so, non-relativistically, the speeds satisfy the Pythagorean theorem.

Thanks for the detailed 4-vector derivations you have posted. It is interesting to follow your reasoning. However, the relation above is only correct for the case where A and B were at the origin at the same time. That constraint was not stated in the original post.

In reading the original post and early exchange, it still seems that the conceptual problem here is with the contruction of a relative velocity. Once that is understood, one can move on to the Lorentz trasformations.

 

[Mentz114]

Anant, if you're still following this, look at post 29. Country Boy has given the method.

Start with D^2 = x^2 + y^2, differentiate wrt time and you're there.

 

[George Jones]

Anant, if you're still following this, look at post 29. Country Boy has given the method.

Start with D^2 = x^2 + y^2, differentiate wrt time and you're there.

I don't think that this gives the relative speed. The relative speed is the magnitude of the derivative of the relative position vector, not the derivative of of the magnitude of the relative position vector. Post #29 and the above use the part of the relative velocity that is (edit) parallel to the relative position vector, but neglect the part of the relative velocity that is perpendicular to the relative position vector.

Let $\vec{D}$ be the relative position of B with respect to A. Then,

$$\vec{V} = \dot{\vec{D}} = \dot{D} \hat{D} + D \dot{\hat{D}}.$$

The relative of B with respect to A is

$$\vec{D} = \vec{r_B} - \vec{r_A};$$

differentiating gives

$$\vec{V} = \vec{v_B} - \vec{v_A};$$

dotting this with itself gives

$$\vec{V} \cdot \vec{V} = \left( \vec{v_B} - \vec{v_A} \right) \cdot \left( \vec{v_B} - \vec{v_A} \right).$$

Finally,

$$V^2 = v^2_A + v^2_B,$$

since $\vec{v}_A$ is perpendicular to $\vec{v}_B.$

However, the relation above is only correct for the case where A and B were at the origin at the same time.

The above non-relativistic stuff is: modified by relativity; not dependent on whether the observers go through the spatial origin. I think my relativistic version also is independent of spacetime origin, but I could be wrong. A good check would be a derivation using methods similar to those used in the standard derivation of the sums of parallel and anti-parallel (in C's frame) velocities.

 

[country boy]

Reply to George Jones:

I see the difference in our two approaches to relative speed. [There has been some confusion between speed and velocity in this thread, which I probably added to.] You derive the relative speed between the two reference frames of A and B, which is the same everywhere, while I derive the speed between the moving points A and B. These are, of course, different. I thought from reading the early posts that the latter was what was asked for, but now I believe that your derivation is what is needed when using the Lorentz transformation. The relative speed between A and B, as I defined it, is what the two observers see as they watch each other. It relates to the doppler effect, for instance. But I agree that it is better to approach the problem through the velocity between two moving reference frames. Relativity can then be applied without confusion and everything can be derived, including the doppler effect.

Thanks very much for sticking with me on this.