Acceleration is not important in the twin paradox

[Delta1212]

Frame C has no impact on the solution, all proper times are calculated in one frame, frame A. You (and Iggy) need to come with grips with this fact.

Please re-read your own post:

 

"On the other hand, the proper time accumulated by A can be calculated by a frame comoving with C, F_C as :

 

tau_A=d/v*sqrt(1-(v/c)^2)

 

(all frames agree on proper time so we might as well use F_C)"

 

Where d is the length as measured in frame A and v is A's velocity as measured in frame C.

[xyzt]

Please re-read your own post:

 

"On the other hand, the proper time accumulated by A can be calculated by a frame comoving with C, F_C as :

 

tau_A=d/v*sqrt(1-(v/c)^2)

 

(all frames agree on proper time so we might as well use F_C)"

 

Where d is the length as measured in frame A and v is A's velocity as measured in frame C.

I don't have to use the frame for C, it was just a convenient way. Did you get the line: "all frames agree on proper time so we might as well use F_C" ?

When I do the calculations, I replace [math]d[/math] with [math]d

\sqrt{1-(v/c)^2}[/math], did you notice that?

Edited May 8, 2013 by xyzt

[Iggy]

Please do so

No. I'm not going to teach you how to use the Lorentz transforms while you vacuously try to debate me. Look for attention somewhere else.

 

Frame C has no impact on the solution, all proper times are calculated in one frame, frame A. You (and Iggy) need to come with grips with this fact.

 

There is no preferred frame in relativity. If you're stuck solving this paradox in one frame then that's your problem.

[xyzt]

No. I'm not going to teach you how to use the Lorentz transforms while you vacuously try to debate me. Look for attention somewhere else.

 

 

There is no preferred frame in relativity. If you're stuck solving this paradox in one frame then that's your problem.

I don't need any lessons so it looks like you pulled the formula out of your butt and you are peddalling very fast trying to cover up that you have no derivation. Thanks for playing.

[md65536]

Please do so, there is no derivation for the second term that you are adding in.

It looks like a straight derivation from the Lorentz transformation.

It gives the correct answer.

 

Without the second term, you get [math]\tau_B/\gamma[/math] which is just the rate of A's clock relative to B's according to B [1]. With the second term you get A's clock relative to B according to A.

 

 

Refs:

[1] http://en.wikipedia.org/wiki/Lorentz_factor#Definition

Edited May 8, 2013 by md65536

[Iggy]

I don't need any lessons so it looks like you pulled the formula out of your butt and you are peddalling very fast trying to cover up that you have no derivation. Thanks for playing.

It is the time coordinate Lorentz transformation with t' set to zero and vt' substituting x'. I said this when I first used the term. If you know how to use the Lorentz transforms then you'll understand what that's all about. If not, I'm sorry. I showed that it worked with numbers and I told you where I got it. There is nothing more I'm willing to do. If anyone besides yourself asks me to do a Minkowski diagram for it I will.

 

Lorentz transformation:

 

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

 

t' = 0, x = t'v

 

[math]\tau_A = \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]

 

 

edit: Thank you, Md. We posted at the same time there.

Edited May 8, 2013 by Iggy

[xyzt]

It looks like a straight derivation from the Lorentz transformation.

It gives the correct answer.

 

Without the second term, you get [math]\tau_B/\gamma[/math] which is just the rate of A's clock relative to B's according to B [1]. With the second term you get A's clock relative to B according to A.

 

 

Refs:

[1] http://en.wikipedia.org/wiki/Lorentz_factor#Definition

This exercise is not about clock rate, it is about elapsed proper time. I have shown you how to calculate the elapsed proper time as a function of coordinate time, textbook stuff.

[Delta1212]

I don't have to use the frame for C, it was just a convenient way. Did you get the line: "all frames agree on proper time so we might as well use F_C" ?

When I do the calculations, I replace [math]d[/math] with [math]d

\sqrt{1-(v/c)^2}[/math], did you notice that?

Please show me where you did that. I honestly can't find it.

[xyzt]

Please show me where you did that. I honestly can't find it.

[math]\tau_A=\frac{d \sqrt{1-(v/c)^2}}{v}[/math] in the post where I did all the calculations, you need to go a few pages back.

Edited May 8, 2013 by xyzt

[md65536]

This exercise is not about clock rate, it is about elapsed proper time. I have shown you how to calculate the elapsed proper time as a function of coordinate time, textbook stuff.

Okay. And you've done that for the proper time C measures between an event at C that is simultaneous with AB according to A (I think) and should have found that T_C = T_A / gamma. You've also done it for the proper time C measures between an event at C that is simultaneous with AB according to F, and you found T_C = T_A. No problem there. And we've done that for the proper time C measures between BC and AC, and also for the proper time B measures between AB and BC, which are the identical calculations done in a corresponding twin paradox experiment. No problem.

 

These are all okay because they're different proper times measured between different events.

[Delta1212]

[math]\tau_A=\frac{d \sqrt{1-(v/c)^2}}{v}[/math] in the post where I did all the calculations, you need to go a few pages back.

Ok, right. You were attempting to find the proper time of A using the frame of C. To do that, you multiply the time measured in frame C by sqrt(1-(v/c)^2). Since you had the distance d and the velocity v (since C measures A traveling towards it at the same speed that A measures C traveling at). To find time you do d/v. And you wind up with the equation:

 

tau_A = d/v * sqrt(1-(v/c)^2) or (d *sqrt(1-(v/c)^2))/v, as you said. Awesome.

 

Except that in frame C, d does not equal the distance traveled by A as measured by C. It equals the distance traveled by C as measured by A. You found the time measured in frame C by dividing the distance measured in frame A by the velocity of A as measured in frame C.

 

Here, let's try this out.

 

In your example, in frame A, A starts counting from the moment B leaves, which A measures as being simultaneous with the moment that C is d distance away from A, traveling towards it at velocity v. A therefore measures the total time from the beginning of the experiment (which in frame A is also when C started moving toward A along a path of length d) as being d/v.

 

Now, you say that frame C measures A's proper time as being d/v * sqrt(1-(v/c)^2). Since proper time is agreed upon in all frames, then A's proper time as measured in frame A must be equal to A's proper time as measured in frame C. A measures it's proper time as d/v. C measures A's proper time (according to your math) as d/v *sqrt(1-(v/c)^2). Therefore:

 

d/v = d/v * sqrt(1-(v/c)^2)

 

1 = sqrt(1-(v/c)^2)

 

1 = 1-(v/c)^2

 

1 + (v/c)^2 = 1

 

(v/c)^2 = 0

 

v/c = 0

 

v = 0

 

So if d/v = d/v * sqrt(1-(v/c)^2), then for any value v, v = 0. If v = 1, then v = 0. 1 does not equal 0. Clearly something is wrong.

 

Either:

 

1. I can't do algebra, in which case please point out the mistake.

2. The proper time that A measures for frame A between the time it measures C traveling along distance d at velocity v and the time it measures C as finishing traveling along distance d at velocity v cannot be expressed as d/v and I am wrong. (In which case, please explain because I clearly have a misconception about something here).

3. The proper time of A as calculated from frame C is not d/v * sqrt(1-(v/c)^2), in which case the conclusions you draw from that are wrong.

Edited May 8, 2013 by Delta1212

[xyzt]

Ok, right. You were attempting to find the proper time of A using the frame of C. To do that, you multiply the time measured in frame C by sqrt(1-(v/c)^2). Since you had the distance d and the velocity v (since C measures A traveling towards it at the same speed that A measures C traveling at). To find time you do d/v. And you wind up with the equation:

 

tau_A = d/v * sqrt(1-(v/c)^2) or (d *sqrt(1-(v/c)^2))/v, as you said. Awesome.

 

Except that in frame C, d does not equal the distance traveled by A as measured by C. It equals the distance traveled by C as measured by A. You found the time measured in frame C by dividing the distance measured in frame A by the velocity of A as measured in frame C.

 

Here, let's try this out.

 

In your example, in frame A, A starts counting from the moment B leaves, which A measures as being simultaneous with the moment that C is d distance away from A, traveling towards it at velocity v. A therefore measures the total time from the beginning of the experiment (which in frame A is also when C started moving toward A along a path of length d) as being d/v.

 

Now, you say that frame C measures A's proper time as being d/v * sqrt(1-(v/c)^2). Since proper time is agreed upon in all frames, then A's proper time as measured in frame A must be equal to A's proper time as measured in frame C. A measures it's proper time as d/v. C measures A's proper time (according to your math) as d/v *sqrt(1-(v/c)^2). Therefore:

 

d/v = d/v * sqrt(1-(v/c)^2)

 

1 = sqrt(1-(v/c)^2)

 

1 = 1-(v/c)^2

 

1 + (v/c)^2 = 1

 

(v/c)^2 = 0

 

v/c = 0

 

v = 0

 

So if d/v = d/v * sqrt(1-(v/c)^2), then for any value v, v = 0. If v = 1, then v = 0. 1 does not equal 0. Clearly something is wrong.

 

Either:

 

1. I can't do algebra, in which case please point out the mistake.

2. The proper time that A measures for frame A between the time it measures C traveling along distance d at velocity v and the time it measures C as finishing traveling along distance d at velocity v cannot be expressed as d/v and I am wrong. (In which case, please explain because I clearly have a misconception about something here).

3. The proper time of A as calculated from frame C is not d/v * sqrt(1-(v/c)^2), in which case the conclusions you draw from that are wrong.

The calculation I gave you is correct, here is your obvious mistake, right off the bat:

 

"A measures it's proper time as d/v."

 

No , he doesn't.

Edited May 8, 2013 by xyzt

[Delta1212]

Let me take this one step at a time and see if I can find a fault in my understanding.

 

A measures C traveling a distance d at velocity v. A measures the time that this takes as being d/v. Is this, at least, correct?

Edited May 8, 2013 by Delta1212

[xyzt]

Let me take this one step at a time and see if I can find a fault in my understanding.

 

A measures C traveling a distance d at velocity v. A measures the time that this takes as being d/v. Is this, at least, correct?

This is exactly as in the muon exercise: the muons ( C ) are located at altitude d , as measured by the Earh (A) and they move at speed v (wrt the Earth as well). It takes the muons [math]\frac{d \sqrt{1-(v/c)^2}}{v}[/math] to hit the Earth (A).

Where you make the mistake is that, indeed, A measures the coordinate time [math]T=\frac{d}{v}[/math] for C to reach him (A). But what A wants, is the proper time, [math]\tau_A=T\sqrt{1-(v/c)^2}[/math]

Edited May 8, 2013 by xyzt

[Delta1212]

This is exactly as in the muon exercise: the muons ( C ) are located at altitude d , as measured by the Earh (A) and they move at speed v (wrt the Earth as well). It takes the muons [math]\frac{d \sqrt{1-(v/c)^2}}{v}[/math] to hit the Earth (A).

If that is the time the muon takes as measured by the Earth, what would the proper time of the muon be?

[Iggy]

This seems to make no sense: according to "Delta" AC is the event of C

passing through A and BC is the event of B passing through C so

[math]\tau_B[/math] seems to have no sense for the problem.

Delta is correct. AC (what I said) is the same as AB. A and B start the experiment collocated.

I'm sorry, I did mislabel that. AC is certainly not the same as AB. Total brain lock. Force of contrition, I'll make that diagram. Give you something else to muddle...

 

It'll take a few...

[md65536]

The calculation I gave you is correct, here is your obvious mistake, right off the bat:

 

"A measures it's proper time as d/v."

 

No , he doesn't.

I thought that d was the distance that C travels, according to A.

In post #164 you use d "as measured by the Earth (A)".

 

If C is moving at v relative to A, it will close the distance d in a time of d/v, ie. the proper time that elapses at A while (according to A) C closes that distance. (Edited for clarity)

 

Using an example with numbers would clarify what you mean (maybe let you see where your equations are not working out)...

Edited May 8, 2013 by md65536

[xyzt]

If that is the time the muon takes as measured by the Earth, what would the proper time of the muon be?

Proper time is frame invariant, all frames agree on proper time, please stop asking these type of questions, it shows that you aren't learning.

 

I thought that d was the distance that C travels, according to A.

In post #164 you use d "as measured by the Earth (A)".

 

If C is moving at v relative to A, it will close the distance d in a time of d/v, ie. the proper time that A measures to close that distance.

 

Using an example with numbers would clarify what you mean...

No, it won't, you (and Delta) need to learn how to use symbolic calculations.

[md65536]

No, it won't, you (and Delta) need to learn how to use symbolic calculations.

Equations are meaningless if improperly applied.

[Iggy]

Ok, if you're unfamiliar with spacetime diagrams then I'm sure this is just gonna look like a gigantic mess, but I'll do my best to be as descriptive as possible:

I adapted a diagram I had already made for Michel. The velocity is v = 0.6. If I used the velocity we've been using 13/15c the angles would be too small to work well.



The green and black arrowed line marked earth is earth's world line. The line marked [math]X_A[/math] is earth's spatial axis. The green dotted lines and green numbers are earth's coordinate system. The bold red line marked [math]\tau_B[/math] is B's world line. It bumps into C at the event marked BC. The bold red line marked [math]X_B[/math] is B's spatial axis. The red dotted lines and red numbers are B's coordinate system.

We are solving A's proper time (earth's proper time) for half of the experiment from B's perspective. Half the experiment is from event AB to event H. So, we know B's proper time and we know A's velocity relative to B, and that's it. That is all we know, and we cannot assume that B moves while A is stationary. That is cheating. We are solving from B's perspective. B's coordinate system.

We know the length of the red line from the event AB to BC in the red coordinate system. It is 5 years. Using the time dilation equation we can solve the length of the black line from the event AB to the event P in the green coordinate system.

[math]\tau_A = \tau_B \sqrt{1-v^2}[/math]

[math]\tau_A = 5 \sqrt{1-0.6^2}[/math]

[math]\tau_A = 4[/math]

AB -> P in the green coordinate system should be 4 years. The diagram verifies this.

The red line that joins P and BC is the line of simultaneity for B when B bumps into C. The green line that joins H and BC is the line of simultaneity for A when B bumps into C. The length of P -> H has yet to be accounted for and that's what I used the following time Lorentz transformation for:

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

It will find the length of the black line P2 -> H2 in the green coordinate system if you set t'=0 and x' = 3. We are looking for the black line P -> H which is the same length as P2 -> H2. Setting t'=0 and [math]x' = \tau_B \cdot v[/math] (which always equals the correct x') we get:

[math]\tau_A = \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]

[math]\tau_A = 5 \frac{0.6^2}{\sqrt{1-0.6^2}}[/math]

[math]\tau_A = 2.25[/math]

The black line P -> H should therefore be 2.25 years in the green coordinate system. Checking the diagram verifies.

Add the two together

[math]\tau_A = 4 + 2.25[/math]

[math]\tau_A = 6.25[/math]

and one gets the proper time of A for half the trip. In other words, the length of the black line AB -> H is 6.25 years in the green coordinate system.

That is why I summed those two terms when solving from B's perspective. One finds the length of AB -> P and the other finds the length of PH. If one is familiar enough with the Lorentz transforms they should be able to do the above in their head (or with a small sketch) and come up with the equations needed.

enough said, I think.

---

d/v = d/v * sqrt(1-(v/c)^2)

1 = sqrt(1-(v/c)^2)

1 = 1-(v/c)^2

1 + (v/c)^2 = 1

(v/c)^2 = 0

v/c = 0

v = 0

 

Your algebra is correct.

 

At this point:

d/v = d/v * sqrt(1-(v/c)^2)

1 = sqrt(1-(v/c)^2)

You assumed that d/v on each side of the equation is the same. If that is the case then the velocity would be zero. d/v is time, so..

t' = t * sqrt(1-(v/c)^2)

if the equation you wrote reduces to the time dilation equation then it should be intuitive that the time on the left side isn't the same as the time on the right (if the velocity isn't zero). The one on the right, in this case, is bigger. The equation you wrote should be:

d1/v = d2/v * sqrt(1-(v/c)^2)

where d1 is the distance from the perspective of the moving particle and d2 is the distance from rest.

Edited May 8, 2013 by Iggy

[michel123456]

Fascinating thread. going too fast for the time i can spend here.

From post #156 on page 8

It is the time coordinate Lorentz transformation with t' set to zero and vt' substituting x'. I said this when I first used the term. If you know how to use the Lorentz transforms then you'll understand what that's all about. If not, I'm sorry. I showed that it worked with numbers and I told you where I got it. There is nothing more I'm willing to do. If anyone besides yourself asks me to do a Minkowski diagram for it I will.

Lorentz transformation:

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

t' = 0, x = t'v

[math]\tau_A = \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]


edit: Thank you, Md. We posted at the same time there.

If t'=0 then I suppose x=t'v=0 also.

Edited May 8, 2013 by michel123456

[Delta1212]

Proper time is frame invariant, all frames agree on proper time, please stop asking these type of questions, it shows that you aren't learning.

Proper time is the time read by a clock between two events that the clock is present at. Everyone agrees on what a clock measures in its own frame. This is why proper time is frame invariant.

 

If the Earth measures a muon traveling to Earth from distance d (as measured by Earth) at velocity v (as measured by the Earth), then d/v * sqrt(1-(v/c)^2) will give you the proper time of the muon as it travels that path in the frame of the muon. But d/v will be the time that the Earth measures the muon taking in Earth's frame. It will, however, agree that the muon experiences d/v *sqrt(1-(v/c)^2) time during that period because that is the muon's proper time, which is agreed upon in all frames.

 

Proper time being frame invariant does not mean that every frame measures the same time between two events. It just means they all agree on what each other measures.

[xyzt]

Ok, if you're unfamiliar with spacetime diagrams then I'm sure this is just gonna look like a gigantic mess, but I'll do my best to be as descriptive as possible:

 

I adapted a diagram I had already made for Michel. The velocity is v = 0.6. If I used the velocity we've been using 13/15c the angles would be too small to work well.

 

 

The green and black arrowed line marked earth is earth's world line. The line marked [math]X_A[/math] is earth's spatial axis. The green dotted lines and green numbers are earth's coordinate system. The bold red line marked [math]\tau_B[/math] is B's world line. It bumps into C at the event marked BC. The bold red line marked [math]X_B[/math] is B's spatial axis. The red dotted lines and red numbers are B's coordinate system.

 

We are solving A's proper time (earth's proper time) for half of the experiment from B's perspective. Half the experiment is from event AB to event H. So, we know B's proper time and we know A's velocity relative to B, and that's it. That is all we know, and we cannot assume that B moves while A is stationary. That is cheating. We are solving from B's perspective. B's coordinate system.

 

We know the length of the red line from the event AB to BC in the red coordinate system. It is 5 years. Using the time dilation equation we can solve the length of the black line from the event AB to the event P in the green coordinate system.

 

[math]\tau_A = \tau_B \sqrt{1-v^2}[/math]

 

[math]\tau_A = 5 \sqrt{1-0.6^2}[/math]

 

[math]\tau_A = 4[/math]

 

AB -> P in the green coordinate system should be 4 years. The diagram verifies this.

 

The red line that joins P and BC is the line of simultaneity for B when B bumps into C. The green line that joins H and BC is the line of simultaneity for A when B bumps into C. The length of P -> H has yet to be accounted for and that's what I used the following time Lorentz transformation for:

 

[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]

 

It will find the length of the black line P2 -> H2 in the green coordinate system if you set t'=0 and x' = 3. We are looking for the black line P -> H which is the same length as P2 -> H2. Setting t'=0 and [math]x' = \tau_B \cdot v[/math] (which always equals the correct x') we get:

 

[math]\tau_A = \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math]

 

[math]\tau_A = 5 \frac{0.6^2}{\sqrt{1-0.6^2}}[/math]

 

[math]\tau_A = 2.25[/math]

 

The black line P -> H should therefore be 2.25 years in the green coordinate system. Checking the diagram verifies.

 

Add the two together

 

[math]\tau_A = 4 + 2.25[/math]

 

[math]\tau_A = 6.25[/math]

 

and one gets the proper time of A for half the trip. In other words, the length of the black line AB -> H is 6.25 years in the green coordinate system.

 

That is why I summed those two terms when solving from B's perspective. One finds the length of AB -> P and the other finds the length of PH. If one is familiar enough with the Lorentz transforms they should be able to do the above in their head (or with a small sketch) and come up with the equations needed.

 

enough said, I think.

What you did is to work out the trivial case where C's speed wrt A is equal to B's speed wrt A. This is the particular, trivial case I described in my earler post that makes C irrelevant. It is equivalent with the case where B makes an instantaneous, infinite acceleration turnaround at the midpoint. See here, for example.

I have shown what happens when [math]v_B \ne v_C[/math] already.

 

Proper time is the time read by a clock between two events that the clock is present at. Everyone agrees on what a clock measures in its own frame. This is why proper time is frame invariant.

 

If the Earth measures a muon traveling to Earth from distance d (as measured by Earth) at velocity v (as measured by the Earth), then d/v * sqrt(1-(v/c)^2) will give you the proper time of the muon as it travels that path in the frame of the muon. But d/v will be the time that the Earth measures the muon taking in Earth's frame. It will, however, agree that the muon experiences d/v *sqrt(1-(v/c)^2) time during that period because that is the muon's proper time, which is agreed upon in all frames.

 

Proper time being frame invariant does not mean that every frame measures the same time between two events. It just means they all agree on what each other measures.

The above shows why your earlier attempt to construct the equation [math]\frac{d}{v}=\frac{d}{v}*\sqrt{1-v^2}[/math] and arrive to the conclusion [math]v=0[/math] was wrong. You were trying to equate C's proper time with A's cordinate time.

Edited May 8, 2013 by xyzt

[Delta1212]

Ok, if you're unfamiliar with spacetime diagrams then I'm sure this is just gonna look like a gigantic mess, but I'll do my best to be as descriptive as possible:I adapted a diagram I had already made for Michel. The velocity is v = 0.6. If I used the velocity we've been using 13/15c the angles would be too small to work well.The green and black arrowed line marked earth is earth's world line. The line marked [math]X_A[/math] is earth's spatial axis. The green dotted lines and green numbers are earth's coordinate system. The bold red line marked [math]\tau_B[/math] is B's world line. It bumps into C at the event marked BC. The bold red line marked [math]X_B[/math] is B's spatial axis. The red dotted lines and red numbers are B's coordinate system.We are solving A's proper time (earth's proper time) for half of the experiment from B's perspective. Half the experiment is from event AB to event H. So, we know B's proper time and we know A's velocity relative to B, and that's it. That is all we know, and we cannot assume that B moves while A is stationary. That is cheating. We are solving from B's perspective. B's coordinate system.We know the length of the red line from the event AB to BC in the red coordinate system. It is 5 years. Using the time dilation equation we can solve the length of the black line from the event AB to the event P in the green coordinate system.[math]\tau_A = \tau_B \sqrt{1-v^2}[/math][math]\tau_A = 5 \sqrt{1-0.6^2}[/math][math]\tau_A = 4[/math]AB -> P in the green coordinate system should be 4 years. The diagram verifies this.The red line that joins P and BC is the line of simultaneity for B when B bumps into C. The green line that joins H and BC is the line of simultaneity for A when B bumps into C. The length of P -> H has yet to be accounted for and that's what I used the following time Lorentz transformation for:[math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math]It will find the length of the black line P2 -> H2 in the green coordinate system if you set t'=0 and x' = 3. We are looking for the black line P -> H which is the same length as P2 -> H2. Setting t'=0 and [math]x' = \tau_B \cdot v[/math] (which always equals the correct x') we get:[math]\tau_A = \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math][math]\tau_A = 5 \frac{0.6^2}{\sqrt{1-0.6^2}}[/math][math]\tau_A = 2.25[/math]The black line P -> H should therefore be 2.25 years in the green coordinate system. Checking the diagram verifies.Add the two together[math]\tau_A = 4 + 2.25[/math][math]\tau_A = 6.25[/math]and one gets the proper time of A for half the trip. In other words, the length of the black line AB -> H is 6.25 years in the green coordinate system.That is why I summed those two terms when solving from B's perspective. One finds the length of AB -> P and the other finds the length of PH. If one is familiar enough with the Lorentz transforms they should be able to do the above in their head (or with a small sketch) and come up with the equations needed.enough said, I think.

---

Your algebra is correct. At this point:

d/v = d/v * sqrt(1-(v/c)^2)1 = sqrt(1-(v/c)^2)

You assumed that d/v on each side of the equation is the same. If that is the case then the velocity would be zero. d/v is time, so..t' = t * sqrt(1-(v/c)^2)if the equation you wrote reduces to the time dilation equation then it should be intuitive that the time on the left side isn't the same as the time on the right (if the velocity isn't zero). The one on the right, in this case, is bigger. The equation you wrote should be:d1/v = d2/v * sqrt(1-(v/c)^2)where d1 is the distance from the perspective of the moving particle and d2 is the distance from rest.

The fact that it should be d1 and d2 is actually the entire basis of my argument.

 

If you go back and look at xyzt's post that I'm taking this from, d and v aren't generic variables for distance and velocity. He specifically defines them as d being the distance travelled by C as measured by A and v as the velocity of C as measured by A.

 

He then does tau_A = d/v *sqrt(1-(v/c)^2) from frame C. d here should be d(2). However, he takes that expression and subtracts it from

 

d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)+d/v*sqrt(1-(v/c)^2) which defines the proper time experienced by B on the first leg with C on the second leg, where all of the d's are d(1) as if d(1) and d(2) are the same variable. He then uses the result to explain why the whole thing falls apart, when he was actually using a single variable to define to separate distances, which resulted in him subtracting the proper time experienced by C over the course of the experiment while thinking he was subtracting the proper time of A over the course of the experiment.

 

When I used the same variables that he did, I derived that contradiction because it's wrong.

 

Please go back and check his math to see whether I'm way off base here. It's in post #73.

 

What you did is to work out the trivial case where C's speed wrt A is equal to B's speed wrt A. This is the particular, trivial case I described in my earler post that makes C irrelevant. It is equivalent with the case where B makes an instantaneous, infinite acceleration turnaround at the midpoint. See here, for example.I have shown what happens when [math]v_B \ne v_C[/math] already.

The above shows why your earlier attempt to construct the equation [math]\frac{d}{v}=\frac{d}{v}*\sqrt{1-v^2}[/math] and arrive to the conclusion [math]v=0[/math] was wrong. You were trying to equate C's proper time with A's cordinate time.

I wasn't trying to do that. I was pointing out that you did do that.

[xyzt]

The fact that it should be d1 and d2 is actually the entire basis of my argument.

 

If you go back and look at xyzt's post that I'm taking this from, d and v aren't generic variables for distance and velocity. He specifically defines them as d being the distance travelled by C as measured by A and v as the velocity of C as measured by A.

 

He then does tau_A = d/v *sqrt(1-(v/c)^2) from frame C. d here should be d(2). However, he takes that expression and subtracts it from

 

d/(u+v)(sqrt(1-(u/c)^2-sqrt(1-(v/c)^2)+d/v*sqrt(1-(v/c)^2) which defines the proper time experienced by B on the first leg with C on the second leg, where all of the d's are d(1) as if d(1) and d(2) are the same variable. He then uses the result to explain why the whole thing falls apart, when he was actually using a single variable to define to separate distances, which resulted in him subtracting the proper time experienced by C over the course of the experiment while thinking he was subtracting the proper time of A over the course of the experiment.

 

When I used the same variables that he did, I derived that contradiction because it's wrong.

 

Please go back and check his math to see whether I'm way off base here. It's in post #73.

 

I wasn't trying to do that. I was pointing out that you did do that.

You don't know what you are doing, that much is pretty clear.