Iggy

How so? A lot has happened in the thread since I was here last. Oh, I do like that better.

I was proving that the velocity of B and C could be anything shy of c and you would get the same results as the traditional twin paradox. The easiest way to do that was to solve from the perspective of B and C. Solving from their perspective is more complicated. That is probably why you couldn't do it. I am truly sorry about that, but it's no reason for you to be belligerent.

I found the proper time of A given the proper time of B assuming B is at rest. You just found the proper time of B given A. That is not "the exact same results". It isn't even the same problem. I'll say again, I know you can find the proper time of the traveling twin. That's all you've done in this thread, and we're all very aware of how to do it.

Good argument then. This: [math]\tau_C = \frac{d}{v}\sqrt{1-v^2}[/math] would give the right answer for [math]\tau_C[/math] if d were measured by A. v, by the way, is measured the same by A and C. They get the same value for that. The above equation assumes that C is moving and A is at rest. C's clock runs slow, in other words. That wouldn't work for D or D2 (I know I implied that it might in my last post but when talking about specific events, it won't). The equation would be: [math]\tau_A = \frac{D_C}{v \sqrt{1-v^2}}[/math] That equation would work. It, again, assumes that C is moving and A is at rest. I don't quite follow that. We know the experiment works. xyzt has admitted as much, just saying that the acceleration is hidden. If a certain method doesn't work then it is the method.

No, t'=0 doesn't necessitate that x' is zero. x' can be any value in the lorentz transforms regardless of the value of t'. The math I did proves the general case. The diagram I did only illustrates a specific case for elucidation. If you want C's speed to be different then simply rotate the blue line. Nothing else has to change.

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.

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...

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.

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 just explained the derivation of both terms. They are added together because we are looking for the proper time of A along half of his trip and they each find part of the length of that line. The time dilation formula finds the first segment and the shift of simultaneity finds the second. If I didn't think you were trying to debate me I would make a minkowski diagram and show you visually what is happening. You think I don't know this? I solved the proper time of A from the frames of B and C. You haven't done that. All you did was find their proper time from A's frame. That is just A's time multiplied by the gamma factor. That's all you've done. If you want to disagree with what I'm doing then at least have a different method of doing it. Or show something that I've done wrong. I already showed with numbers that the equation you're bickering about works. Once again you're just objecting because somebody is doing something that you haven't considered or that you are unable to do. I'm not going to encourage that argument any more.

I used the standard time dilation formula which is derived from the Lorentz transforms at the link I gave you: hyperphysics time dilation I'm not going to copy it over. The other term I used is, like I said, a straightforward application of the Lorentz transforms to account for the relativity of simultaneity. The t formula for switching coordinate systems is, [math]t = \frac{t' + vx'}{\sqrt{1-v^2}}[/math] You can get it here. All I did was set t'=0 (imagining the origin of the coordinate system be event BC). And, like I said when I gave the term, I subbed x' = vt', since that locates A's world line and reveals the deficit of proper time from the shift of simultaneity along that world line. [math]\tau_B = \frac{v^2}{\sqrt{1-v^2}}[/math] I've shown that the equation works, and now explained how I got it. If you have a different way of switching coordinate systems then that's fine, I'd love to see it, but I'm not going to debate the efficacy of something as common as the Lorentz transformations.

Like I said when I first used it, it's a straightforward application of the Lorentz transformations. I can give a thought experiment explaining if you're interested, but I think you're more trying to debate me. Would anyone like to fix some of the (-1)'s xyzt is throwing around? I fixed what I could, but can't get them all if you know what I mean.

That is correct. We can use Md's velocity of v = 13/15c (.8666c) and 4 total years of proper time. Half of that time B travels away until bumping into C. That means A has 2 years of proper time between AC and BC. To calculate the proper time of B it is like you say, [math]\tau_A = \tau_B \frac{1}{\sqrt{1-v^2}}[/math] rearrange [math]\tau_B = \tau_A \sqrt{1-v^2}[/math] [math]\tau_B = (2) \sqrt{1-(0.8666)^2 } = 2 \cdot 0.5 = 1[/math] A therefore calculates that the proper time of B is 1 year. We can also calculate the proper time of A during half of the trip using the perspective and proper time of B. This is the equation that Xyzt disagrees with, but let's see if it works, [math]\tau_B = \tau_A \frac{1}{\sqrt{1 - v^2 }} - \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math] rearrange, [math]\tau_A = \tau_B \sqrt{1 - v^2 } + \tau_B \frac{v^2}{\sqrt{1-v^2}}[/math] [math]\tau_A = 1 \sqrt{1 - (0.8666)^2 } + 1 \frac{(0.8666)^2}{\sqrt{1-( \mbox{0.8666})^2 }}[/math] [math]\tau_A = 1 (0.5) + 1 (1.5) = 2[/math] So, B, from his frame, calculates that A measures a proper time of 2 years over the span of half of his trip. And this, we know, is correct. Calculating things in the coordinate system of A is hardly a good way of disagreeing with my method of calculating things in the coordinate system of B and C, is it? If you disagree with my method of calculating things from B and C's perspective then I welcome you to do it differently. It would help to use numbers like I have done above.

Delta is correct. AC (what I said) is the same as AB. A and B start the experiment collocated. One must account for the relativity of simultaneity when an event is considered in two different frames. In A's frame, BC is simultaneous with one event along A's world line, but it is simultaneous with two different events along A's world line in the frames of B and C. If one doesn't account for this when calculating the proper time from 3 different perspectives then one will most assuredly get the wrong answer. The simplification to inertial motion is explained here. The standard time dilation equation is given here.

That is not correct. It is true for any speed (different or the same) that B and C might have relative to A so long as it is less than c. You can easily prove this by figuring the proper time of A from the perspective of B and C, adding them, and seeing if it is the same as A would figure from his perspective. If B measures [math]\tau_B[/math] between AC and BC then he will calculate [math]\tau_A[/math] between those events as follows, [math]\tau_A = \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}}[/math] The second term accounts for the relativity of simultaneity (since A is not at BC) and is taken from the lorentz transforms where tv has substituted the distance, x, and this of course assumes light units. If C measures [math]\tau_C[/math] between BC and CA then he will calculate [math]\tau_A[/math] between those events as follows, [math]\tau_A = \tau_C \sqrt{1 - {v_C}^2 } + \tau_C \frac{{v_C}^2}{\sqrt{1-{v_C}^2}}[/math] The question is then, if we add these together do we get the correct equation from A's perspective. Let's see... [math]\tau_A \stackrel{?}{=} \tau_B \sqrt{1 - {v_B}^2 } + \tau_B \frac{{v_B}^2}{\sqrt{1-{v_B}^2}} + \tau_C \sqrt{1 - {v_C}^2 } + \tau_C \frac{{v_C}^2}{\sqrt{1-{v_C}^2}}[/math] Pull [math]\tau_B[/math] and [math]\tau_C[/math] out, [math]\tau_A \stackrel{?}{=} \tau_B \left( \sqrt{1 - {v_B}^2 } + \frac{{v_B}^2}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \sqrt{1 - {v_C}^2 } + \frac{{v_C}^2}{\sqrt{1-{v_C}^2}} \right)[/math] Common denominator, [math]\tau_A \stackrel{?}{=} \tau_B \left( \frac{{v_B}^2 + 1 - {v_B}^2}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \frac{{v_C}^2 + 1 - {v_C}^2}{\sqrt{1-{v_C}^2}} \right)[/math] add, [math]\tau_A \stackrel{?}{=} \tau_B \left( \frac{1}{\sqrt{1-{v_B}^2}} \right) + \tau_C \left( \frac{1}{\sqrt{1-{v_C}^2}} \right)[/math] Now that is indeed the proper equation from A's perspective. It is therefore true for any velocity of B and C. The proper times of B and C, each multiplied by the gamma factor equals the proper time of A, just like your good old fashion twin paradox if the two legs of the trip had a different velocity.

Please retract this claim: while you're at it, retract this one: Md clearly has a good understanding of relativity, and you keep insulting him for saying things that are perfectly true and consistent with SR. Stop hijacking this thread, indeed.

I understand that you only want to discuss proper times. I also understand that Md mentioned something that wasn't a proper time, but that was nevertheless 100% correct. I also understand that you misunderstood him, and thought that he was giving a proper time, so you took the occasion to belligerently insult his knowledge of relativity telling him how wrong he was. It was the second time he said something perfectly reasonable and true only for you to dismissively insult the claim as wrong, because it wasn't something that you were considering. See, Xyzt, Everyone here understands what you are perfectly.

We've been over that. Everyone knows that proper time is invariant. They aren't proper times. The things in bold are *not* proper times: According to A, A ages 4 years, while C ages 2 years. According to C, A ages 4 years, while C ages 8 years.. It is worded as clearly as could be, and is the most common of language when discussing relativity. According to C, A ages 4 years while C ages 8 years, obviously. You disagree? You think A ages some different amount while C ages 8 years according to C? In C's frame of reference, how long exactly do you think A ages while *in C's frame of reference* C ages 8 years? Give me a number. Everything is in C's frame of reference. That's what "according to C" means. In that frame: A approaches C. A meets up with C. How much younger would he calculate A to be when he was 8 years younger. Give me a number. Don't change the problem. Don't answer a different question. Give me a number.

Ok, X. If anyone discusses a frame of reference, or anything else, that you aren't talking about then they are wrong. Even if they are correct, they are wrong. Ok.

I'm terribly sorry, but your post: belligerently addresses Md's calculation, and calls it wrong when it is plainly correct. If you solved an unrelated problem and got a different answer then that's fine, but that isn't what you're talking about above. If it doesn't matter then I suggest you go back and edit your post.

Md never mixed frames in the post referenced. He discussed the perspective of A and the perspective of C and got the correct answer for both. You didn't understand him. That is no reason to be belligerent.

No, the expression is as follows: According to A, A ages 4 years, while C ages 2 years. According to C, A ages 4 years, while C ages 8 years.. The only reference to proper time here is the age of A according to A and the age of C according to C. A is not talking about the proper time of C and C is not talking abut the proper time of A. I'm sorry you misunderstood that, but Md did clarify and the language used in the indented section above is typical when discussing relativity problems.

Am I to understand that you think this refers to one frame?... You do understand that "according to A" is a different frame from "according to C", and that their simultaneity is relative? I think this may be the source of the problem.

Would it make more sense to use 13/15c, to directly compare the numbers with the original problem? edit: for the velocity I mean, and where [math]\tau_A[/math] is 4 yrs

He is right. Post 88 is correct, and post 90 correctly explains the basics for why it is right. The statement in quotes: is 100% accurate. The invariance of proper time doesn't conflict with it, and the relativity of simultaneity is the thing that makes it correct.