Relative Velocity of two particles

[Doc Al]

- Point 1 - I am very aware that scenario #2 is different than scenario #1. Is the 264,705.88 km measurement in S' for BEA where E is moving away at 0.6c from B in S' (B as the origin) and A is moving away from E at 0.6c in S (E as the origin of S) hence requiring the relativistic velocity addition (where B is the origin of this frame of reference) correct?

Yes, that's correct. According to B, A travels 264,705.88 km in 1 second. (I thought I had stated as much in my comment on scenario #2.)

I am aware that 1 second in the S' FR is NOT the same as 1 sec in the original S with E being the origin of that FR.

OK.

- Point 2 - Is BE, again from B as the origin of the FR S' and E being Earth "moving away" at 0.6c, 144,000 km.

No. According to B, E moves 180,000 km in 1 second.

Again, I am aware that the 144,000 km in S' (B as the origin of the FR S') is not the same as 144,000 km in S (E as the origin of S.)

OK, but all distances mentioned in scenario #2 are measured in B's frame (S').

- Point 3 - Is that difference calculated of 120,705.88 calculated as shown because this was all being done from FR S' correct in S'?

You'll have to recalculate that difference, since you had the wrong BE distance. The difference in distance is due to the fact that the Earth and object A have different speeds (of course).

- Point 4 - Is the time of 1 second from B to A correct in S'?

Not sure what this question means, since that's the premise upon which scenario #2 is based. So the answer is: Of course.

- Point 5 - If Point 4 is correct (1 second in S') how would I split the time from B to E and from E to A in S' (B the origin)?

I don't understand what you mean. It takes 1 second (according to B) for E to travel the distance BE and for A to travel the distance BA.

 

[stevmg]

Doc Al -

I am freakin' lost! In the twin paradox question, when the Earth "moved" and the twin in the rocket ship was held still, we cut the distance in the S' (rocket ship as origin) by 1/gamma.

I am just lost...

Need a site where I can bore the folks with really elementary questions. Know of any?

That's all for now

 

[jtbell]

In the twin paradox question, when the Earth "moved" and the twin in the rocket ship was held still, we cut the distance in the S' (rocket ship as origin) by 1/gamma.

I assume you are referring to this exchange:

- Point 2 - Is BE, again from B as the origin of the FR S' and E being Earth "moving away" at 0.6c, 144,000 km.

No. According to B, E moves 180,000 km in 1 second.

In order to use the length-contraction formula, the length must be the distance between two objects (or parts of the same object) that are at rest with respect to each other, i.e. they are at rest in the same inertial reference frame. They have a fixed (constant) distance between them in any inertial reference frame, but different distances in different frames, of course.

Consider the most basic application of length contraction, to the length of a uniformly moving rod. Clearly the two ends of the rod satisfy this condition.

In the twin paradox, the traveling twin's origin and destination (turnaround point) are assumed to be at rest with respect to each other, or practically so. So they also meet this condition.

In the situation you're discussing in this thread, the Earth and object B do not meet this condition, because B is moving away from the Earth at 0.6c.

You cannot apply the simple length-contraction equation to the distance BE, because that distance changes with time, in either reference frame. In order to measure the distance between B and E, you have to (conceptually at least) measure the positions of B and E at the same time, and then subtract one from the other. But if an observer in S does this, then those measurements are not simultaneous in S', and vice versa. This is relativity of simultaneity.

 

[Fredrik]

Need a site where I can bore the folks with really elementary questions. Know of any?

There's no better place than this one. If those elementary questions are "textbook-style questions", you should put them in the homework section (even though they aren't really homework). Otherwise, just ask them in the relativity forum, or the quantum physics forum, or whatever's appropriate for the question you want to ask.

 

[stevmg]

There's no better place than this one. If those elementary questions are "textbook-style questions", you should put them in the homework section (even though they aren't really homework). Otherwise, just ask them in the relativity forum, or the quantum physics forum, or whatever's appropriate for the question you want to ask.

Thank you. I was trying to be polite and not clutter up the forum for you with my elemental questions. If you don't mind answering them, I will proceed, but slowly so that I do not clutter PF with my trivia.

Again, thank you.

 

[Dale]

I will proceed, but slowly so that I do not clutter PF with my trivia.

You should realize that even the questions you feel are trivial are probably questions that lots of other people have or will have, so asking those trivial questions will help so that the answers will show up for them also. I think you should go ahead and not hesitate.

 

[stevmg]

Again, DaleSpam, Fredrik, starthaus and others, thank you for encouaging me to proceed with the elementary questions. By doing so more and more things become clearer.

 

[stevmg]

Then, Doc Al, may we start again.

Assume a point E in space ("Earth.") Assume a particle flies to the left at -0.6c and a second particle to the right at +0.6c. The left point where the first particle is after one second in S (E as the origin) will be labelled B and is 0.6 lt-sec to the left of E and the right point where the second particle is after 1 second is labelled A at 0.6 lt-sec to the right of E.

With S as the frame of reference, B and A are 1.2 lt-sec apart. Let us keep these measurements constant, i.e., independent of time.

If we use a second frame of reference S' which the origin is coincident with the origin of S at t = 0 in S and this S' moves to the left at -0.6c, it will travel with the left particle. Thus x' = 0 in S' no matter how long we wait to measure but clearly at 1 second in S the position of S' is -0.6 lt-sec. but in S', the position of E (the origin in S) is to the right of B (which is coincident with the origin of S'.)

In S, E is 0.6 lt-sec to the right of B. In S, A is 1.2 lt-sec to the right of B.

We will measure in S' B'E' and B'A'. Thus, we will measure at t₁' = t₂'

Thus, B'E' in S' is BE/gamma

gamma = 1([SQRT(1 - v²/c²) = 1/SQRT[1 - 0.6²] = 1/0.8 = 1.25

B'E' = BE/gamma = 0.6*0.8 = 0.48 lt-sec

and

B'A' = BA/gamma = 1.2*0.8 = 0.96 lt-sec (both when t₁' = t₂')

In S', to travel from B' to E' will take 0.48 lt-sec/0.6c = 0.8 sec

In S', to travel from B' to A' will take 0.96 lt-sec/0.6c = 1.6 sec

Are my assumptions correct? Are my calculations correct?

 

[Doc Al]

Then, Doc Al, may we start again.

Good. Permit me to rephrase things a bit as we go along.

Assume a point E in space ("Earth.") Assume a particle flies to the left at -0.6c and a second particle to the right at +0.6c. The left point where the first particle is after one second in S (E as the origin) will be labelled B and is 0.6 lt-sec to the left of E and the right point where the second particle is after 1 second is labelled A at 0.6 lt-sec to the right of E.

With S as the frame of reference, B and A are 1.2 lt-sec apart. Let us keep these measurements constant, i.e., independent of time.

For clarity, let's say that we have markers floating in space (space stations or buoys floating along with the Earth). Marker B is located 0.6 lt-sec to the left of E; marker A is located 0.6 lt-sec to the right of E. (All distances measured in frame S.)

So, after 1 second in S, the left going particle passes marker B and the right going particle passes marker A. OK?

If we use a second frame of reference S' which the origin is coincident with the origin of S at t = 0 in S and this S' moves to the left at -0.6c, it will travel with the left particle. Thus x' = 0 in S' no matter how long we wait to measure but clearly at 1 second in S the position of S' is -0.6 lt-sec. but in S', the position of E (the origin in S) is to the right of B (which is coincident with the origin of S'.)

The origin of frame S' is attached to the left moving particle.

In S, E is 0.6 lt-sec to the right of B. In S, A is 1.2 lt-sec to the right of B.

OK.

We will measure in S' B'E' and B'A'. Thus, we will measure at t₁' = t₂'

I don't know what you mean by B'E' and B'A'. I assume you mean that you will measure the distance between BE and BA according to frame S'. I assume that by t₁' = t₂' you mean to measure the positions of those markers at the same time. Good!

Thus, B'E' in S' is BE/gamma

gamma = 1([SQRT(1 - v²/c²) = 1/SQRT[1 - 0.6²] = 1/0.8 = 1.25

B'E' = BE/gamma = 0.6*0.8 = 0.48 lt-sec

OK. According to S', the distance BE is 0.6 lt-sec/gamma. Good!

and

B'A' = BA/gamma = 1.2*0.8 = 0.96 lt-sec (both when t₁' = t₂')

According to S', the distance BA is 1.2 lt-sec/gamma. Good!

In S', to travel from B' to E' will take 0.48 lt-sec/0.6c = 0.8 sec

OK. According to frame S', it takes the particle 0.8 sec to travel from E to B. Good!

In S', to travel from B' to A' will take 0.96 lt-sec/0.6c = 1.6 sec

I'm not sure what you mean here. What's traveling from B to A?

Are my assumptions correct? Are my calculations correct?

See my comments.

 

[stevmg]

I'm not sure what you mean here. What's traveling from B to A?

See my comments.

B' and A' are the corresponding points for B and A in S'

Mistake - B' to A' (to illustrate the time dilation.)

So, given enough patience, I do get it correctly.

Now the problem is I can't remember what I was thinking about when I posited this problem!
Go figure...

Steve G