How to Determine Velocity in Special Relativity

[danielatha4]

How do we determine relativistic velocity if velocity is a function of distance and time and if distance and time in Special Relativity are both functions of velocity?

From the point of an outside observer, if velocity increases then time dilates. Due to the inversely proportional relation to velocity, a slower time would result in a higher velocity. Therefore would a constant acceleration not actually be a constant acceleration relative to the outside observer?

V = ∆x / t

where t = t_o * sqrt( 1 - [v/c]²)^-1

Where V equals the previous equation, etc.

 

[CompuChip]

Note that by length contraction,
∆x = ∆x_o * sqrt( 1 - [v/c]²)^-1

So when you plug that into the equation
V = ∆x / t
for one observer, it merely tells you that a similar expression, namely
V = ∆x_o / t_o,
also holds in the other frame.

So in whichever frame you measure v, both observers will get the same value. Thus they will agree unambiguously on the relative velocity between their frames (however, if they are not familiar with special relativity, they will argue about who is actually moving and who is at rest).

The v (alternatively, gamma = sqrt( 1 - [v/c]²)^-1) should be thought of as just one of the constants which specifies the system.

 

[tiny-tim]

Welcome to PF!

Hi danielatha4! Welcome to PF!

From the point of an outside observer, if velocity increases then time dilates. Due to the inversely proportional relation to velocity, a slower time would result in a higher velocity. Therefore would a constant acceleration not actually be a constant acceleration relative to the outside observer?

Yes, you're right … dv/dt constant for one inertial observer is not generally constant for others.

A constant force (as measured by any inertial observer) gives a constant rate-of-change of momentum rather than of dv/dt … good ol' Newton's second law …

so d/dt mv/√(1 - v²/c²) = constant.

This will be true for any inertial observer.

(in terms of rapidity u, and proper time s, with v = tanhu, d/dt v/√(1 - v²/c²) = d/dt sinhu = coshu du/dt = coshu du/ds ds/dt = du/ds, and changing to rapidity u' = u + w with w constant gives du'/ds = du/ds)

 

[danielatha4]

Note that by length contraction,
∆x = ∆x_o * sqrt( 1 - [v/c]²)^-1

I thought Length contraction was

∆x = ∆x_o * sqrt( 1 - [v/c]²)

which wouldn't allow the (I forgot what the variable is...) gammas? to cancel

Also, the way I understood it was that the observer would see a length contraction in the body of the moving object, not the distance it's traveling. Whereas, the moving object sees the direction it's traveling in as contracted.


And tiny-tim. If I understand you correctly you're saying that Force, or rate of change of momentum is always a constant to every inertial observer but not acceleration?

F = ma, if force is constant then with varying accelerations mass would appear different to different inertial observers?

 

[diazona]

I thought Length contraction was

∆x = ∆x_o * sqrt( 1 - [v/c]²)

which wouldn't allow the (I forgot what the variable is...) gammas? to cancel

Also, the way I understood it was that the observer would see a length contraction in the body of the moving object, not the distance it's traveling.

That would be correct, based on my understanding.

 

[vitruvianman]

How do we determine relativistic velocity if velocity is a function of distance and time and if distance and time in Special Relativity are both functions of velocity?

From the point of an outside observer, if velocity increases then time dilates. Due to the inversely proportional relation to velocity, a slower time would result in a higher velocity. Therefore would a constant acceleration not actually be a constant acceleration relative to the outside observer?

V = ∆x / t

where t = t_o * sqrt( 1 - [v/c]²)^-1

Where V equals the previous equation, etc.

(so sorry but) I don't understand what you're talking about. Firstly $$t=\gamma (t'+\frac{vx'}{c^2}$$, and "wobei" (it's german sorry again) $$t'=\gamma (t-\frac{vx}{c^2})$$ where $$\gamma$$ is the famous Lorentz constant $$\sqrt{1-\frac{v^2}{c^2}}$$. These are the famous Lorentz translation for time (not for accelerating ones) and for length, they are:

$$x'=\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}$$

and its reverse $$x=\gamma (x'-vt')$$ is also right.

That SHOULD esplain what you've asked above but if it doesn't, please let me know what you're so curious about.

 

[danielatha4]

Thanks for the reply vitruvianman. I forgot all about those equations, but what I've been trying to find out is how an outside observer would determine the velocity of a moving body.

Obviously it would be the change in distance observed divided by the time, but the change in time is given by the time dilation equation, which depends upon velocity.

Would the outside observer see a greater velocity due to the time dilation?

 

[diazona]

Hmm... having not thought about this for a couple days, let me say the following: if you are an observer, a.k.a. "outside observer," trying to determine the velocity of a body moving relative to you: you measure the position of the body at some time, and then again at some later time. So you have two sets of (time,space) coordinates. Both of these coordinates are measured in your reference frame, and thus there is no time dilation or length contraction involved.

For purposes of comparison, imagine that you make your (time,space) measurements triggered by two flashes of light coming from the moving body. (Assume the body is a strobe light or a pulsed laser.) You can measure the time interval between flashes in your reference frame, and a timer traveling with the moving body can measure the time interval between the same flashes in its reference frame, and your measurement will disagree with the timer's. That's when time dilation (and length contraction) becomes involved: measurements of the same two events in multiple, different reference frames. But when you're trying to determine a moving body's velocity with respect to yourself, there is no need to make any measurements in any reference frame other than your own. Thus, no time dilation or length contraction.

 

[vitruvianman]

Would the outside observer see a greater velocity due to the time dilation?

I think no.

Let's say you're looking at the train with a (random, doesn't matter) velocity. As you know, it's a very basic and typical relativity example. We can absolutely say that $$t'=\frac{\frac{tc^2-vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$ can be used to calculate its time dilation.

But firstly we should calculate the velocity of the train and we can calculate it how we want and the result will be right (remember there isn't any dilation for v, actually no transformation for v).

But we can say that train's "clock" is working slower than ours, and we can calculate it using the classic Lorentz transformation given above. So we may think that the train is taking the road x in a little time, so its v must be observed higher.
But we must say the length of the train got also a bit shorter, (can bu surely calculated by $x'=\gamma(x-vt)$) so the train takes the road a bit longer. I think you've missed that point.

 

[danielatha4]

I think I understand the concept that length contraction cancels the time dilation effect, but I want to see it mathematically.

If you divide X^' by t^' you should get an expression for velocity right?

$$\gamma$$(x-vt) / (tc²-vx)/c²$$\gamma$$

 

[vitruvianman]

I think I understand the concept that length contraction cancels the time dilation effect, but I want to see it mathematically.

If you divide X^' by t^' you should get an expression for velocity right?

$$\gamma$$(x-vt) / (tc²-vx)/c²$$\gamma$$

No. I didn't mean that, so, sorry I think couldn't express myself well. But I'll try once again. But firstly, I should show that [itex]\gamma(x-vt) / (tc^2-vx)/c^2\gamma[/tex] isn't right. For
$$v=\gamma(x-vt) / (tc^2-vx)/c^2\gamma$$,
i.e.:
$$v=\frac{c^2(x-vt)}{tc^2-vx}$$
and$-v^2x=c^2x$, so finally $v^2=-c^2$ which isn't correct of course.

Now return to the train example. Firstly, I'd like to say that your first question about observing a greater velocity because of time dilation was a good question but you missed a very basic relativity principle: You didn't show a reference(point). In the train example (actually in all situations of relativity), it's the most importnant point as you know.

Now we're the outside observer and the train is moving. As I said before, we can calculate its v in any sensible way. After we calculate, we have a(the road)/t(for us). After v, we can calculate its length, which is $x'=\gamma(x-vt)$, and if we want its time dilation. Now careful here; x' is observed by us, but time dilation is just calculated, can't be observed by any observer, by the train itself too. Time dilation can be seen only when the observers compare their own results about time. So the v of train, for us, isn't related to the time dilation.
Saying that "then, v may be greater only due to the length of train (x')" is also not correct. Because the train isn't accelerating here. We need only the length of the road a and t to calculate v. But if the train was accelerating, the train would be getting shorter and shorter, and so, in order to finish the road a at the same time, v must be higher. Actually, the train is already accelerating :rofl: .

Now let's get into the train. There isn't going to be anything different, because the train is moving permanently and it's the same thing to move permanently or not to move (basic relativity). Only, we would see the outside observer shorter, etc. The rest is the same film.


Now I'm awaiting another question from you, about comparison. :)

 

[danielatha4]

So the v of train, for us, isn't related to the time dilation,

I'm finding this a hard pill to swallow.

I understand you're example. I just get stuck at the part where you said

As I said before, we can calculate its v in any sensible way.

You're saying that the velocity of the train can be determined by the length of the road it travels divided by the time in our (the observer's) reference frame. I tend to think that if we view a moving object and claim that its clock is running slower (time dilation) that we must also assume that its velocity, as inversely proportional to time, must be greater. The argument's you've posed are:
1. The velocity of the moving train is dependent on the observer's time (which would make sense, I'm just questioning why)
2. that somehow, mathematically, the time dilation effect cancels

LaTeX Code: v=$$\gamma$$(x-vt) / (tc^2-vx)/c^2$$\gamma$$

I got a different answer. x' / t' is equivalent to x' * 1/t' therefore

V = $$\frac{\gamma(x-vt)}{1}$$ * $$\frac{c^2\gamma}{(tc^2-vx)}$$

In which case the gamma's would not cancel. Instead you would get a very ugly expression...

 

[pervect]

Here's the classic way of measuring velocity in SR.

1) Mark of a course of known, constant distance.

2) put two clocks on the course, one at the starting location, the other at the ending location

3) synchronize both clocks, using the Einstein convention. The frame to use is the frame in which both the starting line and the finish line are "at rest". Setting off a light pulse at the middle of the course, and making sure the two clocks read the same when the light pulse reads them is what is meant by "using the Einstein convention". Note that this notion of "synchronized clocks" is frame dependent, but the start and finish line define the frame to use.

4) record the time at which the object crosses the start line using the start-line clock. Record the time at which the object crosses the finish line using the finish-line clock. Subtract the two clock readings to get the travel time.

5) divide distance / time, this is your velocity.

I hope this answers the question.

 

[tiny-tim]

I think I understand the concept that length contraction cancels the time dilation effect, but I want to see it mathematically.

Hi danielatha4!

(have a gamma: γ )

I suspect you're trying to compare the wrong things.

A can't measure A's velocity (or rather, he can, but it's always zero ).

A can measure B's velocity, and B can measure A's velocity.

So if A says that B follows x = vt, then B says:

Well, A thinks he follows x = 0 for all t, but that's x' = γ(x - vt) = γ(0 - vt) = γvt, t' = γ(t - vx) = γ(t-v0) = γt, so x' = -vt', and A's velocity is minus v.​

 

[vitruvianman]

You're saying that the velocity of the train can be determined by the length of the road it travels divided by the time in our (the observer's) reference frame. I tend to think that if we view a moving object and claim that its clock is running slower (time dilation) that we must also assume that its velocity, as inversely proportional to time, must be greater. The argument's you've posed are:
1. The velocity of the moving train is dependent on the observer's time (which would make sense, I'm just questioning why)
2. that somehow, mathematically, the time dilation effect cancels

It's normal that you tend to think that way, it's still about a reference point that you don't show. Actually, you know what to do, but you just have to remember it (Plato's last theory)
In SR and GR, there isn't any real velocity determination. Or rather, v doesn't mean anything real or steady. The sentence "train is moving with the v 100km/h" doesn't make any sense. Or simply "the v of the train" isn't enough too. You know these basics of SR, already. We've already discussed it.
In train example, for us, there isn't any time dilation. But we can calculate the time dilation of the moving train of course. The road is at rest and because of this v_road=0 for us, so x'=x and t'=t . After we determine the length of road, you can measure the how long it took for the train to finish the road as pervect said (2,3,4).
The time dilation MAY be used (I haven't seen anything like that before but) to calculate the v of the train for the train for us.:yuck: Surely, you calculate many "for"s that way, but it's not what we're talking about.
The dilation is much interesting in GR, with accelerating frames.

I got a different answer. x' / t' is equivalent to x' * 1/t' therefore

V = $$\frac{\gamma(x-vt)}{1}$$ * $$\frac{c^2\gamma}{(tc^2-vx)}$$

In which case the gamma's would not cancel. Instead you would get a very ugly expression...

In my opinion, x'/t' doesn't mean anything. x' is the length of train for us, or train, the length of the road or us. t' is the time dilation of the train for us or the dilation of the road or us, for train.

Whatever, for v=x'(1/t') we can also say
$$v=\frac{x-vt}{\gamma ^{-1}}\frac{\gamma ^{-1}}{\frac{tc^2-vx}{c^2}}$$
In which case the gamma^-1's Would cancel. Therefore
$$tvc^2-v^2x=c^2x-vtc^2$$
and finally
$$v=\frac{x(v^2+c^2)-c^2}{2}$$
Right? lol

After all, I'm still waiting for a question about comparison of relativistic results :)

 

[danielatha4]

Thanks everyone for your answers to my questions, and I apologize if I’ve made you impatient.

Vitruvianman: I don’t know what you’re referring to when you expect me to ask you about “comparison of relativistic results.”

What pervect said about determining velocity makes much more sense to me. Thanks pervect, and it also coincides with what vitruvianman was saying. But I still have a few questions that, in my mind, are kind of bizzare.

Let me give a scenario:

Clock A and clock B are synchronized according to Einstein’s method in the “Definition of Simultaneity” section of his “On the Electrodynamics of Moving Bodies” paper. Where clock A is to the left of clock B relative to an outside observer who has an inertial reference frame. We’ll call this outside obersver Bob (for ease sake). We’ll say that the pre-determined distance between clock A and clock B is 239833966.4 meters. Now suppose that a car moves with a constant velocity, V, relative to Bob from clock A to clock B. Bob observes the time on clock A when the car passes it and observes the time on clock B when the car passes it. Bob subtracts the time on clock A from the time on clock B and sees that it took the car exactly 1 second to travel the 239833966.4 meters; therefore, Bob concludes that the car’s velocity, V, relative to him is .8c.

But clock A and clock B are at rest relative to Bob. What about a clock in the car? Let’s assume that before the car began motion it’s clock what synchronized with Bob’s via the same method as clock A and clock B. With velocity, V, relative to Bob, Bob sees that the car’s clock runs slower. And Bob concludes that the amount of time passed on the car’s clock must be equal to 1*sqrt(1-.8^2) which equals .6 second. Bob suspects that .6 of a second has passed for the car. And Bob also sees no change in the distance between the clocks. Therefore, Bob can conclude that someone sitting in the car sees themselves as going 239833966.4 meters in .6 seconds, or 1.333c. This is what BOB concludes that a passenger in the car concludes because Bob does not see the distance between the clocks contract but he DOES see the clock of the car dilate. Can this not be said?

However, we do know that if we were to step into the car the length between clock A and clock B would contract given by this equation: 239833966.4*sqrt(1-.8^2). The distance between clocks A and B from the point of view of the car is 143900379.84 meters. Now let’s suspect that a passenger of the car reads a clock on his wrist as he passes clock A and also reads it when he passes clock B and finds the difference to be exactly 1 second. This is the same amount of time as was observed by Bob for the moving car because their clocks were synchronized prior to relative motion between the two, AND the difference observed between clocks A and B (if was observed when passing them) should also read 1 second as the relative velocity of both the clocks to the car are the same; therefore, any time dilation effect on them would cancel. Therefore, the passenger of the car sees that he traveled 143900379.84 meters in 1 second, and concludes that he must be traveling .48c. And let’s say that the car passed clock A first and headed for clock B. In the inertial reference frame of the car, clock B traveled 143900379.84 meters in 1 second and was also traveling at .48c. And all this is relative to the fact that Bob sees the car traveling .8c.

Can these assumptions be made?

 

[yuiop]

Say we have rocket traveling from Earth to Mars. An observer on the Earth might say the distance from Earth to Mars is x and the time taken by the rocket is t so the velocity of the rocket relative to the Earth Mars frame is x/t. To an observer onboard the rocket the distance x is length contracted and the time according to an onboard clock is less than t by exactly the same factor, so that the rocket observer measures the velocity of the rocket relative to the Earth-Mars system as x/t too.

For example, if in some arbitary units the Earth observer measures the distance as 80 and the time as 100, then the velocity of the rocket is 0.8c. The rocket observer would measure the distance as 48 and the time as 60 and so the velocity according to him is also 0.8c. (The gamma factor for a rocket moving at 0.8c is 0.6). Simples

If the rocket observer decides to use the Earth-Mars distance as measured by an observer at rest with the Earth, but at the same time use his onboard clock he would calculate an apparent velocity of 80/60 = 1.333c but this is not velocity in the normal sense because he using a mixture of measurements taken by observers that are not at rest with respect to each other. This result of this type of calculation is called "rapidity" rather than velocity as mentioned by Tim.

In brief, you should only make velocity measurements by using rulers and clocks that are at rest with respect to you.

 

[danielatha4]

To an observer onboard the rocket the distance x is length contracted and the time according to an onboard clock is less than t by exactly the same factor, so that the rocket observer measures the velocity of the rocket relative to the Earth-Mars system as x/t too.

But the time t onboard the rocket is not dilated, as it is an inertial reference frame. The clock on Earth would appear dilated relative to the onboard passenger, and vice versa. And only the length is contracted. Am I right? Therefore, the onboard passenger should see 48 distance and 100 time, or .48c.

I understand the concept that you can't measure velocity using measurements that aren't at rest to your reference frame. I guess I'm just curious about what the observer from Earth can assume about the onboard observer's reference frame.

 

[Altabeh]

Say we have rocket traveling from Earth to Mars. An observer on the Earth might say the distance from Earth to Mars is x and the time taken by the rocket is t so the velocity of the rocket relative to the Earth Mars frame is x/t. To an observer onboard the rocket the distance x is length contracted and the time according to an onboard clock is less than t by exactly the same factor, so that the rocket observer measures the velocity of the rocket relative to the Earth-Mars system as x/t too.

For example, if in some arbitary units the Earth observer measures the distance as 80 and the time as 100, then the velocity of the rocket is 0.8c. The rocket observer would measure the distance as 48 and the time as 60 and so the velocity according to him is also 0.8c. (The gamma factor for a rocket moving at 0.8c is 0.6). Simples

The bold-faced statement is not right. From the basic laws of SR, we know that in an onboard observer's eyes (if the rocket, as assumed, is flying with a constant acceleration), everything is as normal as all measurements done on Earth by its inhabitants. It is exactly the well-known example of Einstein's "inertial elevator" whose inertia gives credit to its being equivalent to a freely falling object toward Earth: the observer located in the elevator feels he is at rest as does a freely falling person who did unexpextedly fall off of his window. So the measurements of distance x and time t of a flying rocket will be the same for all observers whether onboard or at rest relative to their own IRF. The only difference here comes along as we want to discuss what these measurements would be from the point of view of the onboard observer. Lorentz factor plays no role here as we can understand from the second basic law of SR that for all inertial observers, measurements relative to their own IRF won't involve any kind of dilation/contraction.

I understand the concept that you can't measure velocity using measurements that aren't at rest to your reference frame. I guess I'm just curious about what the observer from Earth can assume about the onboard observer's reference frame.

This case only consists of the concepts of contraction/dilation. The twins paradox comes from this setup: what can I as an observer at rest on Earth infer about the measurements of a guy in that rocket relative to HIS IRF, not to my own and vice versa?!

I think you are now able to continue to answer my question if you know it has something to do with contraction or dilation!

 

[Vanadium 50]

I don't understand why there is any confusion here. In ordinary Newtonian mechanics, observers in different frames disagree about the velocity of objects. This has nothing to do with SR.

 

[vitruvianman]

I don't understand why there is any confusion here. In ordinary Newtonian mechanics, observers in different frames disagree about the velocity of objects. This has nothing to do with SR.

absolutely...

 

[danielatha4]

Can it be said that if an object is moving .8c relative to an outside observer that the outside observer is seen moving at .8c from the inertial reference frame of the moving object?

I understand now the point of why I made this thread: How to accurately determine velocity while viewing an object moving wit ha constant velocity relative to another inertial reference frame. By understanding that you must make distance and time measurements that are stationary to your reference frame.

Now I'm just stuck on the example of the rocket ship traveling from Earth to Mars. The observer on Earth sees the rocket ship 80 distance(pre-determined) in 100 time (change in clock in the Earth system); therefore, from the Earth's reference frame the velocity of the rocket ship is .8c. But what about a passenger on the rocket ship? will it not see contracted length (48) but no time dilation (100) and therefore assume its velocity as .48c?

 

[Altabeh]

Can it be said that if an object is moving .8c relative to an outside observer that the outside observer is seen moving at .8c from the inertial reference frame of the moving object?

I understand now the point of why I made this thread: How to accurately determine velocity while viewing an object moving wit ha constant velocity relative to another inertial reference frame. By understanding that you must make distance and time measurements that are stationary to your reference frame.

Now I'm just stuck on the example of the rocket ship traveling from Earth to Mars. The observer on Earth sees the rocket ship 80 distance(pre-determined) in 100 time (change in clock in the Earth system); therefore, from the Earth's reference frame the velocity of the rocket ship is .8c. But what about a passenger on the rocket ship? will it not see contracted length (48) but no time dilation (100) and therefore assume its velocity as .48c?

I think you only have an itty-bitty problem in realizing which observer would observe the length contraction; so the following link shows you this in a very beautiful way that I doubt you'll have any problem after seeing it!

http://www.physicsclassroom.com/mmedia/specrel/lc.cfm

AB

 

[vela]

Let S be the observer's frame and S' be the rocket's frame. Use the standard set-up where the origins coincide at $$t=t'=0$$ so that the coordinates of the two frames are related by the usual Lorentz transformations.

$$x' = \gamma(x-\beta t)$$
$$t' = \gamma(t-\beta x)$$

where $$\beta$$ is the velocity of S' relative to S and $$\gamma = 1/\sqrt{1-\beta^2}$$.

Event A: The stationery observer sees the rocket passes through the $$x=0$$ at $$t=0$$ with velocity $$\beta$$. In the rocket's frame, an observer will say that this event occurred at $$x'=0$$ and $$t'=0$$.

Event B: At time $$t=T$$, the stationery observer will see the rocket is at $$x=\beta T$$. In the rocket's frame, the coordinates for this event are given by

$$x' = \gamma(x-\beta t) = \gamma(\beta T - \beta T) = 0$$
$$t' = \gamma(t-\beta x) = \gamma(T - \beta^2 T) = \gamma T (1-\beta^2) = T/\gamma$$

The rocket remains at $$x'=0$$ because the rocket isn't moving in its rest frame, but time will have elapsed between the two events. Since $$\gamma>1$$, we have $$t' < T$$, which is the usual time dilation. The observer on the rocket will say event B occurred after less time elapsed than for the stationery observer .

The observer on the rocket will say during this time interval, he traveled a distance $$d'=\beta T/\gamma$$, the distance $$\beta T$$ length-contracted. So he calculates that the rocket was moving at speed

$$\beta'=\frac{d'}{t'} = \frac{\beta T/\gamma}{T/\gamma} = \beta$$

They will both agree the rocket was moving with speed $$\beta$$ relative to S'.

 

[Altabeh]

Let S be the observer's frame and S' be the rocket's frame. Use the standard set-up where the origins coincide at $$t=t'=0$$ so that the coordinates of the two frames are related by the usual Lorentz transformations.

$$x' = \gamma(x-\beta t)$$
$$t' = \gamma(t-\beta x)$$

where $$\beta$$ is the velocity of S' relative to S and $$\gamma = 1/\sqrt{1-\beta^2}$$.

Event A: The stationery observer sees the rocket passes through the $$x=0$$ at $$t=0$$ with velocity $$\beta$$. In the rocket's frame, an observer will say that this event occurred at $$x'=0$$ and $$t'=0$$.

Event B: At time $$t=T$$, the stationery observer will see the rocket is at $$x=\beta T$$. In the rocket's frame, the coordinates for this event are given by

$$x' = \gamma(x-\beta t) = \gamma(\beta T - \beta T) = 0$$
$$t' = \gamma(t-\beta x) = \gamma(T - \beta^2 T) = \gamma T (1-\beta^2) = T/\gamma$$

The rocket remains at $$x'=0$$ because the rocket isn't moving in its rest frame, but time will have elapsed between the two events. Since $$\gamma>1$$, we have $$t' < T$$, which is the usual time dilation. The observer on the rocket will say event B occurred after less time elapsed than for the stationery observer .

The observer on the rocket will say during this time interval, he traveled a distance $$d'=\beta T/\gamma$$, the distance $$\beta T$$ length-contracted. So he calculates that the rocket was moving at speed

$$\beta'=\frac{d'}{t'} = \frac{\beta T/\gamma}{T/\gamma} = \beta$$

They will both agree the rocket was moving with speed $$\beta$$ relative to S'.

And let me sum up the whole thing for danielatha4:

Case I: The onboard observer A measures distance and time relative to his own frame or, similalry, to the frame of moving rocket the way that the observer at rest, B, does relative to his own frame. So the velocity of rocket in this case will exactly be the same, as measured by both A and B relative to their own frames.

Case II: Observer A measures distance relative to B's frame thru $$x' = \gamma(x-\beta t)$$ and time using the relation $$t' = \gamma(t-\beta x)$$. Observer B does these measurements by Newton's classical methods relative to A's frame. The vanishing of velocity of A relative to B's frame amounts to Newton's laws playing a role in A's considerations. Remember that length contraction here doesn't get involved because a device which takes advantage of Einstein's synchronization would make it possible for B to understand the real x (looking at the rocket with his own eyes would lead to a wrong measurement of distance). So as was shown by vela in the above post, the relative velocity of rocket in this case is again the same for both A and B.

Hope this'll help you.
AB

 

[danielatha4]

Let S be the observer's frame and S' be the rocket's frame. Use the standard set-up where the origins coincide at $$t=t'=0$$ so that the coordinates of the two frames are related by the usual Lorentz transformations.

$$x' = \gamma(x-\beta t)$$
$$t' = \gamma(t-\beta x)$$

where $$\beta$$ is the velocity of S' relative to S and $$\gamma = 1/\sqrt{1-\beta^2}$$.

Event A: The stationery observer sees the rocket passes through the $$x=0$$ at $$t=0$$ with velocity $$\beta$$. In the rocket's frame, an observer will say that this event occurred at $$x'=0$$ and $$t'=0$$.

Event B: At time $$t=T$$, the stationery observer will see the rocket is at $$x=\beta T$$. In the rocket's frame, the coordinates for this event are given by

$$x' = \gamma(x-\beta t) = \gamma(\beta T - \beta T) = 0$$
$$t' = \gamma(t-\beta x) = \gamma(T - \beta^2 T) = \gamma T (1-\beta^2) = T/\gamma$$

The rocket remains at $$x'=0$$ because the rocket isn't moving in its rest frame, but time will have elapsed between the two events. Since $$\gamma>1$$, we have $$t' < T$$, which is the usual time dilation. The observer on the rocket will say event B occurred after less time elapsed than for the stationery observer .

The observer on the rocket will say during this time interval, he traveled a distance $$d'=\beta T/\gamma$$, the distance $$\beta T$$ length-contracted. So he calculates that the rocket was moving at speed

$$\beta'=\frac{d'}{t'} = \frac{\beta T/\gamma}{T/\gamma} = \beta$$

They will both agree the rocket was moving with speed $$\beta$$ relative to S'.

Thank you very much vela. This cleared everything up for me. The math is what I've wanted to see this whole time and indeed the gammas cancel as originally suggested earlier in this thread. I'm just the type of guy who likes to see it.

$$\beta'=\frac{d'}{t'} = \frac{\beta T/\gamma}{T/\gamma} = \beta$$

 

[vela]

Thank you very much vela. This cleared everything up for me. The math is what I've wanted to see this whole time and indeed the gammas cancel as originally suggested earlier in this thread. I'm just the type of guy who likes to see it.

You're welcome. I totally understand how you feel. It's nice to have the math back up an intuitive explanation so you know your intuition hasn't led you astray and to help keep the details straight.