What Two Speed Measurements Do Observers in Relative Motion Always Agree On?

[Lillyotv]

What two speed measurements do two observers in relative motion always agree on?

In my opinion, one will be relative velocity...but can't figure out the other..?
Any ideas?

Also, photons of light have zero mass. How is it possible that they momentum?

Cuz Momentum=mv...so if m is zero then how come they have momentum?...I thought abt this and researched...I think the reason is that photons don't have particle properties...kindly elaborate why exactly?

 

[snoopies622]

The other speed that they will agree on -- of course -- is the speed of light.

Photons have no "rest mass". If they did, they would require infinite energy to move at the speed of light. The equation 'momentum=mv' is from classical mechanics which is no longer regarded as exactly correct. Both relativity and quantum mechanics forced some modifications on to it. Look up 'relativistic momentum' for more detail.

 

[Coughlan]

speed of light is the only absolute I can think of. I can see where you might think relative motion but what if the "motion" were "quick" enough to cause time dilation. Well then distance / time would become a quagmire. So my vote is for the speed of light only!

 

[Doc Al]

It has nothing to do with time dilation. Lillyotv is correct. The relative speeds will be the same: The speed of B relative to A is the same as the speed of A relative to B.

 

[Lillyotv]

It has nothing to do with time dilation. Lillyotv is correct. The relative speeds will be the same: The speed of B relative to A is the same as the speed of A relative to B.

So, it will be 2 speeds...just relative to each other..now that makes sense...thanks!

 

[Lojzek]

It has nothing to do with time dilation. Lillyotv is correct. The relative speeds will be the same: The speed of B relative to A is the same as the speed of A relative to B.

I read a book where this was assumed without any proof. Is it possible to prove this?

 

[Doc Al]

I read a book where this was assumed without any proof. Is it possible to prove this?

One way is to use the relativistic addition of velocities:
V_{a/c} = \frac{V_{a/b} + V_{b/c}}{1 + (V_{a/b} V_{b/c})/c^2}

In this case:
V_{a/a} = \frac{V_{a/b} + V_{b/a}}{1 + (V_{a/b} V_{b/a})/c^2}

Since:
V_{a/a} = 0

Therefore:
V_{a/b} = - V_{b/a}

Make sense?

 

[robphy]

I read a book where this was assumed without any proof. Is it possible to prove this?

If you think geometrically, it may be obvious.
All particle 4-velocities are, in spacetime, unit-vectors whose tips lie on the unit future-hyperboloid (analogous to a sphere in Euclidean space).

Relative-speed is simply [the speed of light times] the hyperbolic-tangent of the unsigned "angle" intercepted by the two 4-velocities being considered. (This "angle" is essentially an arc-length on the unit-hyperboloid.)

This is the geometrical picture underlying Doc Al's post.

 

[Lojzek]

The problem is that last two answers are based on validity of Lorentz transformation. That book I mentioned included derivation of Lorentz transformation from the assumption that two observers see each other traveling with equal, but opposite speeds: v(A,B)=-v(B,A). The assumption was needed to obtain equation L(v)*L(-v)=Identity.

So we must either prove the statement in question without using Lorentz transformation
or derive Lorentz transformation without this assumption!

 

[robphy]

The problem is that last two answers are based on validity of Lorentz transformation. That book I mentioned included derivation of Lorentz transformation from the assumption that two observers see each other traveling with equal, but opposite speeds: v(A,B)=-v(B,A). The assumption was needed to obtain equation L(v)*L(-v)=Identity.

So we must either prove the statement in question without using Lorentz transformation
or derive Lorentz transformation without this assumption!

It's actually more general than that.
The analogous argument holds for the Galilean transformations as well.