matt@oddjob.UUCP (Matt Crawford) (07/18/85)
In article <851@oddjob.UUCP> I give an example of how faster
than light signalling can lead to causality violations. As
an exercise for the readers of this newsgroup who think they
understand special relativity, I pose the following question.
I provide the answer at the end of this article.
Suppose that at time t=0 person A emits a signal with velocity
u > c in A's own frame of reference. This signal is received
by B who at that instant is at a distance d from A (as measured
by A) and is moving away from A at speed v, with c^2/u < v < c.
B immediately replies by sending back a signal at speed u in
B's own reference frame.
At what time does the reply signal from B reach A?
Answer: In A's frame, the reply arrives at time
t = d/u - d*(uv/c^2 - 1)/(u-v).
Under the assumption c^2/u < v < c, this is negative.
_____________________________________________________
Matt University crawford@anl-mcs.arpa
Crawford of Chicago ihnp4!oddjob!mattjs2j@mhuxt.UUCP (sonntag) (07/18/85)
> In article <851@oddjob.UUCP> I give an example of how faster > than light signalling can lead to causality violations. As > an exercise for the readers of this newsgroup who think they > understand special relativity, I pose the following question. > I provide the answer at the end of this article. Can those of us who *don't* think they understand special relativity play too? Especially since we're the ones who don't understand how FTL signalling can cause causality violations? > > Suppose that at time t=0 person A emits a signal with velocity > u > c in A's own frame of reference. This signal is received > by B who at that instant is at a distance d from A (as measured > by A) and is moving away from A at speed v, with c^2/u < v < c. > B immediately replies by sending back a signal at speed u in > B's own reference frame. > > At what time does the reply signal from B reach A? > Well, the answer I got (which evidently was wrong) was that the time for the first signal to get to B would be t1=d/(u-v), and that the time for the second signal to get back to A would be t2=du/(u-v)^2. Both of these quantities are positive for u<v. > Answer: In A's frame, the reply arrives at time > > t = d/u - d*(uv/c^2 - 1)/(u-v). > > Under the assumption c^2/u < v < c, this is negative. I can see that. What I can't see is how you went about deriving this. I would appreciate seeing that, since *if I could be convinced about the correctness of the above formula*, I would then understand how FTL signalling violates causality > Matt University crawford@anl-mcs.arpa > Crawford of Chicago ihnp4!oddjob!matt -- Jeff Sonntag ihnp4!mhuxt!js2j "Well I've been burned before, and I know the score, so you won't hear me complain. Are you willing to risk it all, or is your love in vain?"-Dylan
mcgeer%ucbkim%Berkeley@sri-unix.ARPA (07/19/85)
From: Rick McGeer (on an aaa-60-s) <mcgeer%ucbkim@Berkeley> Unfortunately, your example required that A measure the distance between B and A simultaneous with B's receipt of A's original message. This is impossible, since simultaneity is a concept with no meaning. Rick.
throopw@rtp47.UUCP (Wayne Throop) (07/22/85)
I'd like to be able to reference a text for those who don't see how FTL implies time-travel (or at least signaling into the past). However, I don't know of such a text. What I know of the subject, I found out in a physics class word-of-mouth from the teacher so to speak (as the result of a wise-guy in the class posing special relativity "paradoxes", and the teacher eventually spending an entire class explaining things). SO: does anybody else know of a text readable by laycritters that covers this ground? I have read the one by C.P. Steinmetz, and have heard of the one by Einstein himself, but neither of these seem to address how (enter doubletalk mode) spacelike worldlines imply time-travel (end doubletalk mode). In any event, let me add a few long-winded points to the discussion. > > Suppose that at time t=0 person A emits a signal with velocity > > u > c in A's own frame of reference. This signal is received > > by B who at that instant is at a distance d from A (as measured > > by A) and is moving away from A at speed v, with c^2/u < v < c. > > B immediately replies by sending back a signal at speed u in > > B's own reference frame. > > > > At what time does the reply signal from B reach A? > > Well, the answer I got (which evidently was wrong) was that the time > for the first signal to get to B would be t1=d/(u-v), and that the time > for the second signal to get back to A would be t2=du/(u-v)^2. Both of > these quantities are positive for u<v. Since we assume u>c and v<c, u<v cannot be the case. Second, you seem to be adding quantities from two different frames, which is a no-no. Third, I'm not sure how you derive that second formula. > > Answer: In A's frame, the reply arrives at time > > > > t = d/u - d*(uv/c^2 - 1)/(u-v). > > > > Under the assumption c^2/u < v < c, this is negative. > > I can see that. What I can't see is how you went about deriving this. > I would appreciate seeing that, since *if I could be convinced about the > correctness of the above formula*, I would then understand how FTL signalling > violates causality Now these formulas are more familiar, just a simple d/u going out, and a Lorentz transform on the velocity coming back (so that it is velocity in A's frame rather than in B's). "Look" more familiar. I still don't quite follow the exact transform on the second term there (the negative one). Can anybody clarify this for me? (Matt?) The most convincing (to me) method of demonstrating the equivalence of FTL with time-travel is with space-time diagrams and assuming "u" infinite (that is, "instant messages"). I have several times tried to make these diagrams using character graphics so that they could be posted to the net, but they just don't look good, and are confusing as a result. Let me try this anyhow, and y'all can let me know if it is of any use. Consider two world lines, A and B (in stars), the three events x, y and z, and the "lines of simultenaity" drawn in A's frame from x and in B's frame from z (in dots). (It so happens that B crosses A at z.) | . . * | . . * | . . * | . . * ^ | . . * | | A *******************x****z******************************* space | .* . | * . . | * . . | * .. | * y | B * .. | . . | . . +--------------------------------------------------------------- time --> Now then, the vertical "line of simultaneity" is with respect to A's world-line, and the "tilted" one is with respect to B's. The exact angles and so on can be worked out (see below), but this diagram is just to give you the idea. Note that A considers events x and y to be simultaneous, while B considers events y and z simultaneous. Thus, if somebody in B's frame sends an instant message from z, it can go to y. Now, if somebody at y *in A's frame* sends an instant message from y, it can go to x. Thus a message relay can send messages from a "later" point in a timeline to an "earlier" point in that timeline. Note also that if a message can be sent by any ammount faster than light in A's frame, there exists a frame such that that message is seen as an "instant message". If there is a "preferred frame" in which instant messages are possible (and they are possible in no other frames), then this problem can be avoided. This may be so, but no such preferred frames are known, nor is there (strong) reason to suppose that they exist. Note that a relay is not required for time-travel to be involved. The message in B's frame from z to y "looks like" it is traveling backwards in time when somebody in A's frame looks at it. The only reason a relay is needed is to loop back on a single world-line. Now, about the "lines of simultenaity". We can find two events that B will consider simultaneous by following B's world-line forward x A-seconds (or x/(1-v^2/c^2)^-2 B-seconds) forward from event e.1, then follow A's world line back x/(1-v^2/c^2)^-2 B-seconds (or (x/(1-v^2/c^2)^-2) / (1-v^2/c^2)^-2 A-seconds) to find event e.2. B will consider e.1 and e.2 simultaneous (and all events on the line of which e.1 and e.2 are members). (I think this double-transform is how we get the second term in Matt's answer, above, but I haven't worked it all out to be sure of this). I suggest actually drawing several Ls of S for various values of v, and following the consequences of sending messages along any space-like line. It turns out that *somebody* will see a message on *any* space-like line as going "pastward". I hope this helps rather than hinders. (I also hope I haven't grossly distorted the concepts involved... this is, after all, remembered from a class 10 years ago.) > > Matt University crawford@anl-mcs.arpa > > Crawford of Chicago ihnp4!oddjob!matt > -- > Jeff Sonntag > ihnp4!mhuxt!js2j -- Wayne Throop at Data General, RTP, NC <the-known-world>!mcnc!rti-sel!rtp47!throopw
throopw@rtp47.UUCP (Wayne Throop) (07/22/85)
Evil phychic emanations from space aliens caused me to post the expression "(blah-blah)^-2", when I meant "(blah-blah)^.5". I have taken to wearing an aluminum cap, so I'm OK now. -- Wayne Throop at Data General, RTP, NC <the-known-world>!mcnc!rti-sel!rtp47!throopw
matt@oddjob.UUCP (Matt Crawford) (07/29/85)
I have received a lot of mail and seen a few followups which indicate to me that posting a solution would be appreciated. Several people showed that they could write equations without understanding the con- cepts of relativity, while one person showed that he could understand the concepts without being able to do all the arithmetic. (A gold star to you, W.T.) I will not attempt to draw figures on the CRT but I will try to make the reasoning clear. (I'll even leave in the c's!) The question: Suppose that at time t=0 person A emits a signal with velocity u > c in A's own frame of reference. This signal is received by B who at that instant is at a distance d from A (as measured by A) and is moving away from A at speed v, with c^2/u < v < c. B immediately replies by sending back a signal at speed u in B's own reference frame. At what time does the reply signal from B reach A? The solution: (gee, this is like being a TA again!) Outline: Find where the signal is received by B in A's coordinates. Transform this to B's coordinates. Then trace the path of the reply in B's coordinates and transform the result to A's coordinate system. Let the origins of both A's and B's coordinates be at the event of A sending the initial signal. Points in spacetime will be denoted as P = [t, x] in A's coordinates or P' = [t', x'] in B's. (One spatial dimension is enough for this problem, so I won't write the y's and z's.) The quantity usually denoted by gamma will be G = (1-v^2/c^2)^(-1/2). The Lorentz transformation between frames is given by: x' = G (x-vt) x = G (x' + vt') t' = G (t-xv/c^2) t = G (t' + vx'/c^2) The reception by B of A's signal will occur at point P = [ d/u, d ], In B's frame this is P' = [ Gd/u-Gdv/c^2, Gd-Gdv/u ]. (You can already see the effects of the faster-than light velocity u here. In B's coordinates the signal was emitted at time t' = 0 but was received at an earlier time. The time component of P' is negative.) The reply signal from B will be emitted at P' in B's frame and will travel to a point R' (to be determined) at speed u relative to B. The distance travelled by the reply signal, in B's system, will be some length r'. The reception of the reply is at: R' = P' + [ r'/u, -r' ] = [ Gd/u - Gdv/c^2 + r'/u, Gd - Gdv/u - r' ]. Transforming back to A's frame and simplifying gives: R = [ d/u - Gr'(v/c^2-1/u), d - Gr'(1-v/u) ], but r' is still unknown. Because A's position is at x = 0 for all times and R is the event of the reception of the reply by A, set the x-component of R equal to zero to solve for r'. This yields r' = d / G(1-v/u) = du / G(u-v) Substitute this into the t-component of R and find: R = [ d/u - d(vu/c^2 - 1)/(u-v), 0 ]. The condition uv > c^2 in my original posting is not strong enough to ensure that the time component of R is negative. The slightly stronger condition uv > (1 + (1-v^2/c^2)^(1/2)) c^2 is needed to make the reception of the reply precede the sending of the original signal. If the relative speed of A and B of v = 0.5c, then the condition is u > (2 + 3^(1/2))c. (Note that u > 2c^2/v will always suffice.) To anyone who wants to learn about relativity I enthusiastically recommend the book _Spacetime_Physics_, by Taylor and Wheeler [W.H. Freeman & Co., San Francisco]. The math needed does not go beyond the v = dx/dt level, yet the content is sufficient for first- year physics students at the best departments in the country. The book is accurately described on the cover as "A Brief, Readable Exposition of Modern RELATIVITY THEORY Illustrated and Amplified by a Wealth of PROBLEMS, PUZZLES, and PARADOXES and their Detailed Solutions." _____________________________________________________ Matt University crawford@anl-mcs.arpa Crawford of Chicago ihnp4!oddjob!matt
Purtill@MIT-MULTICS.ARPA (07/29/85)
From: Mark Purtill <Purtill@MIT-MULTICS.ARPA> <Fnord> > t = d/u - d*(uv/c^2 - 1)/(u-v). > >Under the assumption c^2/u < v < c, this is negative. I think you're wrong. Try u = 3c, v = .5c. (I assume d > 0). You get: t = (d/c)*( 1/3 - ( ((3/2)-1)/(3-.5) ) ) = (d/c)*(1/3 - 1/5) = 2d/15c > 0. The correct condition is: u > v > 2u/(1+(u/c)^2) > 0. This implies u > c, but note that v can be >= c. Now, when u >> c, we can ignore the 1, and get v > 2u/(u/c)^2, i.e., v > 2 c^2/u, which is sort of like what you had. Mark ^.-.^ Purtill at MIT-MULTICS.ARPA **Insert favorite disclaimer here** ((")) 2-032 MIT Cambrige MA 02139