Relative Velocity of two particles

[atyy]

If A is a massive particle traveling inertially at v relative to inertial frame S, and P is a photon traveling at c relative to S, what is the speed of the photon relative to A?

 

[Matterwave]

c

 

[starthaus]

If we must have define a term for this concept, I would prefer it to be something like "separation rate in frame C" or something like that. I can't think of a more misleading term than "relative speed"/"relative velocity".

The terms I use are "closing speed" and "separation speed".

 

[starthaus]

If A is a massive particle traveling inertially at v relative to inertial frame S, and P is a photon traveling at c relative to S, what is the speed of the photon relative to A?

The photon travels at c wrt any frame, including the one comoving with A.

 

[atyy]

c

And what is the velocity of A relative to the photon?

And what is the relative velocity of A and the photon?

 

[starthaus]

And what is the velocity of A relative to the photon?

And what is the relative velocity of A and the photon?

There is no way to associate frames of reference to photons. So, the answer is "seven" :-)

 

[stevmg]

If u_x were positive and v_x were negative then without a relativistic correction the sepatartion speeds were u_x - v_x. As stated, basic physics. Writinig u_x and v_x as a fraction of c, it would be possible for this quantitiy to be >c. To wit, u_x = 0.8 and v_x were -0.6c, the speed of separation would be 1.4c. I know this was posited as in relation to the common frame of reference but is it possible to have a separation speed of >c?

I didn't know that was possible.

 

[starthaus]

If u_x were positive and v_x were negative then without a relativistic correction the sepatartion speeds were u_x - v_x. As stated, basic physics. Writinig u_x and v_x as a fraction of c, it would be possible for this quantitiy to be >c. To wit, u_x = 0.8 and v_x were -0.6c, the speed of separation would be 1.4c. I know this was posited as in relation to the common frame of reference but is it possible to have a separation speed of >c?

I didn't know that was possible.

Closing speed vcan be as large as 2c

 

[stevmg]

Closing or separation speed has often come up and been explained in this forum although I cannot immediately give a link. The closing/separation speed of two material objects can approach 2c. But of course their velocity relative to each other is less than c as measured in either of their rest frames and no material object moves faster than c.

Matheinste.

Never thought of it that way. Using that piece of info:

If one object A goes east from the Earth into space (don't take into account the 1000 mph of the Earth rotation near the equator) at 0.6c and a second object B goes west into space at 0.6c each object will be 180,000 km away from the Earth in opposite directions after 1 second or 360,000 km which makes the separation speed 360,000 km in 1 sec or 1.2c. Yet, of course, the separation speed relative to each other is 1.2/(1+.6*.6) = 1.2/1.36 = 0.8824c = 264,706km/sec. (I'm using 300K km/sec as c and not the exact 299000+ whatever it really is.)

That is weird!

OK, sports fans, then how far are they (A and B) "really" apart after 1 second? 360,000 km or 264,706 km?

 

[jtbell]

That depends on your definition of "really."

 

[Dale]

I acknowledge that there is some support for your point of view, so it's arguably inappropriate to call it wrong. That came as a surprise to me

Same here. I have always used the term "relative velocity" to refer to the velocity measured in the rest frame (v<c) and the term "closing velocity" or "separation velocity" to refer to the difference in velocities as measured in another frame (v<2c).

I suspect the OP was looking for the first. It is too bad that we have found yet another piece of unclear terminology.

 

[stevmg]

I suspect the OP was looking for the first. It is too bad that we have found yet another piece of unclear terminology.

language is NOT precise and sometimes, no matter what language you speak, it takes two or three passes at a subject to convey what is really meant...

Any of you married? Ever try talking to a woman? 41+ years now and I do have the final word, "Yes, Dear."

I am not being flippant here. On a smaller scale that is what we run into in even technical jargon.

 

[Ich]

I do have the final word, "Yes, Dear."

Yes, in reply to a list of very precise instructions. Maybe we need more women in science.

 

[stevmg]

That depends on your definition of "really."

You win, jtbell, you caught me.

1) If reality means from S (where the Earth is the frame of reference (FR), then point B and A are 360,000 km apart. Let us say the path BEA is 360,000 km.

2) If reality means from S' (where B is the origin of the new FR) the path BEA is only 264,705.88 km (again, I'm using the exact 300k km/sec).

3) In S' path BE should be not 180,000 km but 144,000 km. (v = 0.6c, gamma = 0.8, L' = gamma*L

4) Again, in S' path EA should be 264,705.88 - 144,000 = 120,705.88 km. The reason for EA being shorter than BE is that EA is moving more quickly relative to B than is BE.

Now, I don't have any fellow students and I don't have any professors or "on-line" classrooms. You folks are all that (unless you can direct me to such a source.)

So, until then I would appreciate if after looking at my italicized comments 1 through 4 would you say they are correct?

My prior comment regarding the XX gender is not pejorative as I am sure they would say the same about men and both of us would be correct. Another example is talking to your kids. Insanity is inherited as you do get it from your children.

 

[yuiop]

If two relativistic particles are traveling with speeds 'u' &'v' ,how to calculate the relative velocity?

If A is traveling at velocity u relative to C and B is traveling at velocity v relative to C, then the relative velocity $V_{AB}$ of A relative to B is:

$$V_{AB} = \frac{u-v}{(1-uv/c^2)}$$

and the relative velocity $v_{BA}$ of B relative to A is:

$$V_{BA} = \frac{v-u}{(1-vu/c^2)}$$

Example 1
If according to C, A is traveling to the right at 0.9c and B is traveling to the left at -0.6c and the speed of light is c=1, then the magnitude of the relative velocity is (0.9+0.6)/(1+0.9*0.6) = 0.974c. The relative velocity of A according to B is +0.974c and the relative velocity of B according to A is -0.974c. The magnitude of the relative velocity is always less than the speed of light c.

The "separation velocity" of A and B according to C is 0.9c+0.6c = 1.5c.

Example 2
If according to C, A is traveling to the right at 0.9c and B is traveling to the right at 0.6c and the speed of light is c=1, then the relative velocity of A according to B is (0.9-0.6)/(1-0.9*0.6) = 0.6521c and the relative velocity of B according to A is (0.6-0.9)/(1+0.6*0.9) = -0.6521c.

The "closing velocity" of A and B according to C is 0.9c-0.6c = 0.3c.

All observers agree on the magnitude of the relative velocity of two particles.

See http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html

 

[stevmg]

If A is traveling at velocity u relative to C and B is traveling at velocity v relative to C, then the relative velocity $V_{AB}$ of A relative to B is:

$$V_{AB} = \frac{u-v}{(1-uv/c^2)}$$

and the relative velocity $v_{BA}$ of B relative to A is:

$$V_{BA} = \frac{v-u}{(1-vu/c^2)}$$

Example 1
If according to C, A is traveling to the right at 0.9c and B is traveling to the left at -0.6c and the speed of light is c=1, then the magnitude of the relative velocity is (0.9+0.6)/(1+0.9*0.6) = 0.974c. The relative velocity of A according to B is +0.974c and the relative velocity of B according to A is -0.974c. The magnitude of the relative velocity is always less than the speed of light c.

The "separation velocity" of A and B according to C is 0.9c+0.6c = 1.5c.

Example 2
If according to C, A is traveling to the right at 0.9c and B is traveling to the right at 0.6c and the speed of light is c=1, then the relative velocity of A according to B is (0.9-0.6)/(1-0.9*0.6) = 0.6521c and the relative velocity of B according to A is (0.6-0.9)/(1+0.6*0.9) = -0.6521c.

The "closing velocity" of A and B according to C is 0.9c-0.6c = 0.3c.

All observers agree on the magnitude of the relative velocity of two particles.

See http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/velocity.html

I may be off base and I am new to this (as stated, I have no classmates, no professor, no online students with me except for this forum) but I have not seen this explained as succinctly as all you folks have done (starthaus, matheinstei, kev) when one uses the old fashioned Galilean concepts (separation and closing velocities which are always betwwen -2c and + 2c - which I didn't know until you folks pointed it out) versus the relative velocities as explained in SR even by Einstein. Thanks to you I have it now but if someone could corroborate my calculations from above that would "ice" it.

 

[starthaus]

I may be off base and I am new to this (as stated, I have no classmates, no professor, no online students with me except for this forum) but I have not seen this explained as succinctly as all you folks have done (starthaus, matheinstei, kev) when one uses the old fashioned Galilean concepts (separation and closing velocities which are always betwwen -2c and + 2c - which I didn't know until you folks pointed it out) versus the relative velocities as explained in SR even by Einstein. Thanks to you I have it now but if someone could corroborate my calculations from above that would "ice" it.

"separation/closing" speed are not specific to Galilean kinematics. They are equally applicable to SR. As I explained, the resultant composition of speeds is a function of the choice of reference frames. See post #25.

 

[stevmg]

Let me clarify, once and for all:

1. If u and v are specified wrt a frame C, then the relative speed of B wrt A as measured in frame C is v-u

2. If v is specified in frame of particle A then the relative speed of B wrt A is...v.

3. If u and v are specified wrt a frame C then the relative speed of B wrt A as measured wrt A is (u-v)/(1-uv/c^2)

The above is a description that is devoid of any naming convention, it is based solely on the choice of reference frames, as it should be.
Like I said, you need to be precise in terms of defining the frames of reference.

I do understand your explanation. What I meant was that once we hit velocities that required SR I was under the impression, WHICH WAS WRONG, that there always was a "relativistic" adjustment. Hence, with the explanations from you, matheinstei and kev, I see the appropriate usage of "separation/closing speeds" terminology even in SR.

You folks are the back-and-forth sounding board for my more elemental questions. If there is another site which I can do this without tying up this forum I would then be happy to go there but I have not found it. Certainly Yahoo Answers is not a good source as there is no place for give and take or interactive discussion.

 

[Doc Al]

You win, jtbell, you caught me.

1) If reality means from S (where the Earth is the frame of reference (FR), then point B and A are 360,000 km apart. Let us say the path BEA is 360,000 km.

After 1 second passes according to Earth clocks, the Earth observers will say that B and A are 360,000 km apart. Let's call this scenario #1. (Note that B frame observers will disagree that the Earth observers are measuring the positions of A and B at the same time.)

2) If reality means from S' (where B is the origin of the new FR) the path BEA is only 264,705.88 km (again, I'm using the exact 300k km/sec).

After 1 second passes according to the B frame clocks, the B frame observers will say that B and A are 264,705.88 km apart. Note that this is a completely different situation than scenario #1. (B and A are not in the same positions as they were when measurements were made in scenario #1.) Let's call this scenario #2.

3) In S' path BE should be not 180,000 km but 144,000 km. (v = 0.6c, gamma = 0.8, L' = gamma*L

4) Again, in S' path EA should be 264,705.88 - 144,000 = 120,705.88 km. The reason for EA being shorter than BE is that EA is moving more quickly relative to B than is BE.

To talk meaningfully about distances between A, E, and B, you must specify who is doing the measuring and what scenario you are interested in.

Things get complicated. In addition to clocks running slow, and distances shrinking, frames in relative motion will disagree about whether separated events took place at the same time or not. (The 'relativity of simultaneity'.)

 

[stevmg]

After 1 second passes according to Earth clocks, the Earth observers will say that B and A are 360,000 km apart. Let's call this scenario #1. (Note that B frame observers will disagree that the Earth observers are measuring the positions of A and B at the same time.)


After 1 second passes according to the B frame clocks, the B frame observers will say that B and A are 264,705.88 km apart. Note that this is a completely different situation than scenario #1. (B and A are not in the same positions as they were when measurements were made in scenario #1.) Let's call this scenario #2.


To talk meaningfully about distances between A, E, and B, you must specify who is doing the measuring and what scenario you are interested in.

Things get complicated. In addition to clocks running slow, and distances shrinking, frames in relative motion will disagree about whether separated events took place at the same time or not. (The 'relativity of simultaneity'.)

Appreciate the response, 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? 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.

- 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. 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.)

- 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'?

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

- 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)?

 

[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