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Exactly how is causality violated by superluminal travel?

  1. Nov 15, 2009 #1
    From my amateur readings in relativity, one of the arguments against tachyons is that causality would be violated locally.

    But how?

    Let's say we have observer A and B with synchronized clocks that are separated by a reasonable distance d known to them. A sends B a photon at A's clock t0, and B sees it at his clock t0 + cd. A sends a tachyon to B at A's clock t1, and B sees it at his clock t1 + x, where x < cd. But regardless of the speed of the tachyon, x is always >= 0. How can A or B violate causality with tachyons? I.e., how can x ever be less than 0?
     
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  3. Nov 15, 2009 #2

    JesseM

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    Are you familiar with the relativity of simultaneity? For two events with a spacelike separation (i.e. a signal traveling at the speed of light or slower could not make it from one event to the other), different inertial frames will disagree on which event happened earlier and which happened later, so since there'd be a spacelike separation between the event of a tachyon signal being sent and a tachyon signal being received, some frames would say that the signal was received before it was sent (and there will be one frame that says these events were simultaneous, meaning the signal traveled infinitely fast in that frame). It works out that if you and I are sublight observers moving away from each other at relativistic speed, and I send you a tachyon signal which travels FTL in my frame but backwards in time in your frame, and then you send me a tachyon signal in reply which travels FTL in your frame but backwards in time in my frame, then I can receive your reply before I sent the original signal, which all frames will agree is a violation of causality.
     
  4. Nov 16, 2009 #3
    OK, there's the math and there's the analogies. I don't know how to apply the frame transformations to answer my question. So I was hoping to come up with an analogy that I can use to understand why the transforms work the way they do. (Or even a space-time diagram.)

    My motivation is to understand this well enough to ask, for example, can A be at rest with respect to B and still violate causality with FTL information transmission? If not, what is the minimum velocity of separation needed? And what is the analogy (or spacetime diagram) that shows this violation of causality? I'm still trying to understand why just sending a FTL signal to someone on the other side of the solar system would allow them to send me a response prior to my sending the original signal. Clearly, there's something deep I am missing in my attempt to understand the space-time of special relativity. (My mental model is: A and B start together; they synchronize clocks. A and B accelerate identically and in opposite directions and then come to a stop. At that point, they are apart, and their clocks are still synchronized. A then sends a signal to B, who then returns a signal back to A. If this signal is FTL, how does special relativity conclude that A will receive B's response prior to sending out the first signal?)
     
  5. Nov 16, 2009 #4

    JesseM

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    Well, are you familiar with the way different frames' lines of simultaneity can be depicted in a single spacetime diagram? You can see sort of a primer on this stuff here. Anyway, if you can follow that sort of thing, here's a series of illustrations showing how causality can be violated if two observers moving at different velocities each send FTL signals which are instantaneous in their own frame (i.e. they lie along their own frames' lines of simultaneity):

    http://www.theculture.org/rich/sharpblue/archives/000089.html

    I suppose it depends on the details of how the hypothetical FTL transmitter works...for example, if the transmitter always sends signals which are instantaneous in its current rest frame, then you would need two transmitters in motion relative to one another, although of course the human experimenters need not share the same rest frame as their transmitter, one could just accelerate his transmitter to a high speed relative to himself before using it.
    If they are at rest relative to one another when they send the signals (because they 'come to a stop' in the same frame before transmitting), and each perceives his own signal moving FTL rather than backwards in time in his own rest frame, then there will be no violation of causality here, since they both share the same rest frame. It's only when the first signal travels FTL in frame A but backwards in time in frame B, and the reply travels FTL in frame B but backwards in time in frame A, that you can have the situation of one observer receiving a reply before he sent the original message. And as I said, because of the way different frames define simultaneity, any signal which travels FTL in one inertial frame must automatically be traveling backwards in time in some other inertial frame; so, if you accept that the laws of physics governing FTL signaling obey the first postulate of relativity, then if it's possible for a signal to go backwards in time in one frame, this must be possible in all frames. If you want to say that FTL doesn't obey the first postulate then you can have a preferred frame where signals can travel FTL but not backwards in time, and there will be no causality violations, but of course any violation of the first postulate means the theory of relativity itself would be incorrect.
     
  6. Nov 18, 2009 #5
    OK, in the diagram at the bottom of http://www.theculture.org/rich/sharpblue/archives/000089.html, I see how causality is violated. It's very clear. (Thanks for the reference.) Let's call this F5.

    My question is as follows. In order to set up the configuration in F5, Alice, Bob, Carol, and Dave must have all been at the same space-time coordinate to start with (i.e., so they could hatch their devious plan.)

    The problem I have is, I can't see how to physically make the configuration in F5 happen assuming that ABCD were together to start. I've attached a spacetime diagram showing one attempt. In other words, I can't see how to move D ahead so he's at the same time coordinate as B (in B's frame.) Instead, in B's frame, D is in the past, and in D's frame, B is in the future.

    I'm also still trying to reconcile relativity of simultaneity with the concept of "now." Presumably the universe has some sense of absolute time, in that there is past, now, and future. Is relativity saying that "now" differs depending on your frame? But one thing at a time!

    Thanks for your responses so far!
     

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  7. Nov 18, 2009 #6

    Fredrik

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    This was discussed a couple of days ago in the quantum physics forum. Here's a quote from one of my posts in that thread:
    When I say e.g. (1,10), I mean that the coordinates that Alice assigns to that event are t=1, x=10. A spacetime diagram is drawn with time in the "up" direction. I define the "slope" of a curve to be dt/dx, so that a horizontal line in the diagram has slope 0.

    Check the other thread for some suggestions about how to avoid the paradox.
     
  8. Nov 18, 2009 #7

    JesseM

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    They could also just use radio transmissions. But I don't have a problem with assuming they all started at the same point in spacetime.
    But D's worldline is continuous, so there's some point on his worldline at every time coordinate in any given frame (at least until he dies!) Are you just talking about D's age? Why would it be important that D be the same age as B in B's frame?

    Also, just as a separate issue, you seem to have drawn your diagram a little different than diagram F5. In F5, A&B are at rest relative to each other, and C&D are at rest relative to each other; but in your diagram, once everyone has reached their final velocity, you seem to show A&C are at rest relative to each other, and B&D are at rest relative to each other. Was this intentional, or did you mean for D to be moving in B's frame as in diagram F5 when you asked the question above?
    Well, if one event is in the past light cone of another, then all frames agree on the order of the events. But for a pair of events that are spacelike-separated (neither is in the past or future light cone of the other), in relativity there is no absolute physical truth about the order of these events--some frames say one event happened first, other frames say that event happened second, and there'll be one frame that says they happened simultaneously. All these perspectives are considered equally valid in SR. Of course if you are attached to the idea of an objective universal "present" (if you subscribe to the philosophical view about time known as presentism), you are free to hold the metaphysical belief that some frame's definition of simultaneity is "metaphysically" preferred even if there is nothing physical that distinguishes this frame from any other. On the other hand, you could drop the idea of an objective "now" and just be an eternalist who says that no definition of simultaneity is metaphysically preferred. Either way, relativity is just a physical theory, so it can't finally settle such metaphysical questions, but it does say unambiguously that no definition of simultaneity is preferred in any measurable physical sense.
     
  9. Nov 18, 2009 #8
    Ugh. I mixed up the labels in my diagram. Here's the corrected one that is intended to be in the "past" of the F5 diagram. The question is, how does my diagram end up as in F5?


    Yes, I intended for C and D to be moving in A and B's frame.

    In response to Fredrik, how does Bob get to (1,10) in Alice's frame when Alice is at (0,0)? In other words, let's say that both Bob and Alice were at (-10,0). How can Bob get to (1,10) in Alice's frame (from Alice's viewpoint) when Alice is at (0,0)? I understand I may be using the word "when" incorrectly here. So what I am asking for is the spacetime diagram that starts with Bob and Alice at (-10,0) in Alice's frame, and ends up with Alice at (0,0) and Bob at (1,10).

    I suppose another way of asking this is, if Alice and Bob were together at (-10,0), and Bob is now at (1,10) in Alice's frame, isn't Alice at (0,10) in her frame?
     

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    Last edited: Nov 18, 2009
  10. Nov 18, 2009 #9

    JesseM

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    Again, they all have worldlines which go through all times, they don't exist instantaneously at a single point in both space and time! For example, if Alice is at rest at position x=0 in her frame, then if she is at that position at age 20 when t=0, then she will still be at that position at age 30 when t=10. If Bob departs from position x=0 at time t=0 at 1/10 light speed, then he will be at the position x=1 when t=10 in Alice's frame...but due to time dilation he will not have aged a full 10 years in this frame, so if he was age 20 at t=0, he will be 20 + 10*sqrt(1 - 0.1^2) = 29.95 years old at t=10 (the math is easier if we assume Bob was traveling at 0.6c, so he will be at position x=6 at t=10, and he will have aged exactly 10*sqrt(1 - 0.6^2) = 8 years).
    This doesn't make sense, if they started at position x=-10 at time t=0 in this frame, then how could Alice also be at position x=0 at time t=0 in this frame? Have you reversed the order of which coordinate is the time coordinate and which is the space coordinate?

    edit: Just noticed that Fredrik wrote "When I say e.g. (1,10), I mean that the coordinates that Alice assigns to that event are t=1, x=10"--sorry, my mistake, I was used to the convention that position is written first. In any case, my point about each observer being present at every time coordinate in a given frame still stands. And if you want Bob to be at position x=10 at time t=1, then he can't have been at position x=0 at time t=-10 as you suggest...if he was traveling at 0.1c as Fredrik suggested, then he must have been at position x=0 at a time 100 years earlier (or 100 seconds earlier if you're measuring distance in light-seconds, etc.), at time t=-99.
     
    Last edited: Nov 18, 2009
  11. Nov 18, 2009 #10
    I will proofread my messages better.

    Bob and Alice are at (0,-10). How can Bob get to (1,10) in Alice's frame when Alice is at (0,0)? (correcting my original question.)

    Ok, Fredrik's setup appears to be incorrect. The event at (1,10) has the same time (in Bob's frame) as the event at Alice's (0,9.9), not (0,0). So let's skip Fredrik for now.

    Oh. I think I understand now. You're saying that I need to look at the spacetime diagram from an "eternalist" viewpoint. Clearly, I can construct the original F5 diagram starting with ABCD at the same event and then dispersing, as long as I dispense with the concept of a "universal now."

    Wow.

    Hmm. Do all known causality violations from superluminal communication involve messages to the future in some part?
     
  12. Nov 18, 2009 #11

    JesseM

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    Fredrik's setup is correct...the slope of an observer's line of simultaneity is always just the inverse of the slope of the observer's worldline, so if if Bob's worldline has a slope of 10 in Alice's frame, then his lines of simultaneity have a slope of 1/10, so if one of Bob's lines of simultaneity passes through t=1, x=10, it will also pass through t=0, x=0. You could also verify this using the Lorentz transformation...though as I said earlier, it would be easier to have Bob's velocity be 0.6c so his gamma factor will be a nice 1.25 instead of an ugly 1.00503781525921.

    Anyway, there's no way for Bob's worldline to cross through both (t=1, x=10) and (t=0, x=-10) because that would require him to move FTL. But why did you pick that point for their original departure point? You have to pick a point that will lie on Bob's worldline if he's moving at 0.1c and he's at x=10 at time t=1...as I said in the "edit" at the end of my last post, you could have the departure point be (t=-99, x=0).
    I don't think an eternalist viewpoint is necessary--for example, you're free to imagine that Alice's definition of simultaneity is really the correct one in a metaphysical sense, and to say that anyone who's in motion relative to Alice is objectively aging more slowly. Thus if they depart at (t=-99, x=0), then when Alice is 100 years older at (t=1, x=0), this point on her worldline "really" happens at the same time as Bob being at (t=1, x=10)...but because Bob is traveling at 0.1c, his aging is slowed down, so despite the fact that 100 years have passed in "reality" he only appears to have aged by (and any clock traveling with him has ticked forward by) 100*sqrt(1 - 0.1^2) = 99.498743710662 years.
    Did you mean "messages from the future"? Messages to the future are very easy after all, any signal that takes time to get from sender to receiver would qualify! Anyway, causality violations all involve messages or causal influences that go from one event E1 to another event E2 which lies in the past light cone of E1.
     
    Last edited: Nov 18, 2009
  13. Nov 19, 2009 #12

    Fredrik

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    He doesn't. Alice's world line is the time axis, i.e. the line through all points of the form (t,0). Bob's world line (the curve that represents his motion) is the straight line through (-99,0) and (1,10). So the only event where they meet is (-99,0). But it's actually completly irrelevant if they meet or not. It's only important that Bob's world line has slope 10 at (1,10) where he receives Alice's message and immediately replies. So I could have specified that they never meet instead.

    I'm confused by the confusion this is causing. (RUTA assumed I meant the opposite too, in the other thread). I thought it was totally standard to write time first in both SR and GR. Maybe I was wrong about that.

    By the way, the reason I didn't choose a speed that gives us a nice gamma factor is that gamma doesn't enter into the argument at all.
     
  14. Nov 19, 2009 #13
    This was answered on a previous thread. I have not read all of this one so it may have already been answered here, but this is what I said the last time.

    Suppose there is an inertial frame, F, such that event A precedes event B and the two have a time seperation in F of dt and a spatial separation of dx (these are not necessarily infinatesimals). Take another inertial frame, F', such that F' is traveling at a speed of u with respect to F. The time interval between events A and B in F', call it dt', will be given by the well known lorentz transformation

    dt'=(dt-u*dx/c^2)/(1-u^2/c^2)^0.5

    Lets assume for a second that dt' is negative such that in frame F', B precedes A. The question is, in what conditions is this possible?

    If dt' < 0 then clearly that means (dt-u*dx/c^2) < 0 and thus you get the condition to be

    c*dt < (u/c)*dx

    It is important to notice that c*dt is the distance a light beam would travel between events A and B in frame F. We now have a condition for which the temporal order of two events in one reference frame can be reversed simply by switching to a different reference frame. The ways in which this can happen are that either dx > cdt, i.e. event B is not within the light-cone of event A, or if dx<cdt then (u/c)>1 in which case u>c.

    If the condition for causality to hold is states as follows, "it is not possible for an event A to be the cause of an event B if a reference frame can be found such that B precedes A in time", then it can be seen from what was given above that we have to exclude all reference frames where u>c and state the following condition "it is not possible for an event to cause another event which is outside its lightcone, i.e. where dx>cdt". Thus, information cannot travel faster than the speed of light.
     
  15. Nov 30, 2009 #14
    OK, thanks for everyone's responses. I believe I have corrected my misunderstandings.

    So, as I understand it, as long as B is at rest with respect to A, and they are both at the same time coordinate, then SR does not prohibit instantaneous communications between the two.

    Alternatively, going back to Fredrik's setup, since B is at (1,10) and A is at (0,0), there is a coordinate frame where A is in B's past. If there is such a frame, then B will find it impossible to communicate with A instantaneously. (Or the communication takes place but no paradox occurs as any attempt to create a paradox generates a large explosion...)

    The universe won't let causality be violated, but that doesn't mean you can't cut corners here and there...
     
  16. Nov 30, 2009 #15
    If the signal from A to B has a causal effect on B then the signal can not be instantaneous.

    If A could transmit a signal instantaneously to B then we could find a frame (C) where the signal from A to B goes backwards in time and B could reply to A with a normal limited to light speed signal in such a way that A receives the reply before he sent the request. The universe and SR does not allow that to happen, so instantaneous communication of (useful) information is prohibited.
     
    Last edited: Nov 30, 2009
  17. Nov 30, 2009 #16

    JesseM

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    No, it definitely does prohibit instantaneous communication. A single one-way signal would not violate causality in practice (since it wouldn't involve anyone sending a signal into their own past light cone), but any laws of physics that allow for FTL one-way signals would automatically also allow for causality violations if relativity is correct, since relativity says that anything that's possible to do in one inertial frame must also be possible to do in every other inertial frame (the first postulate of relativity), and any signal which moves instantaneously in one frame must move backwards in time in other inertial frames, so whatever laws govern the signal must allow signals to go back in time in any frame (unless relativity itself is wrong and there is a preferred frame for FTL communication).
    Why do you say "if" there is such a frame? You are always permitted to analyze any situation from the perspective of any inertial frame in SR, frames are just coordinate systems, it's not like there's an empirical question of whether they are "present" or "absent" in a given situation (you can analyze a situation from the perspective of a given frame even if no physical objects are at rest in that frame, for example).
    The issue is what the laws of physics allow, not what actually happens. Any laws of physics that would allow for the possibility of instantaneous communication would also allow for causality violation in SR, regardless of whether anyone ever takes advantage of this possibility.
     
  18. May 26, 2010 #17
    This reference is too hard for me to follow : http://www.theculture.org/rich/sharpblue/archives/000089.html

    Can somebody use a simpler analogy about why superluminal light speed would violate causality. Please no world-lines or frames, just normal English.

    Lets start with two people on distant stars signaling or shooting at each other
    and say how one could die before he is shot using superluminal speeds.

    it cannot be so hard (although it is too hard for me)

    THANK YOU.
     
  19. May 26, 2010 #18

    JesseM

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    No way to explain it without using the concept of frames. Of course you don't have to understand frames in detail, basically what you need to know is the following:

    1) A frame is just a coordinate system for assigning position and time coordinates to different events in spacetime.

    2) According to relativity, the laws of physics (including those pertaining to FTL transmissions) must work the same way in all inertial frames (not really to important to understand in detail the way 'inertial frames' are defined, the important thing is just that they are the rest frames for slower-than-light observers that aren't accelerating and thus experience no G-forces).

    3) In relativity different inertial frames define "simultaneity" differently due to the relativity of simultaneity, meaning that if you have two events A and B at different spatial locations which occured at the same time-coordinate in one inertial frame, then there are other inertial frames where A happened at an earlier time coordinate than B, and still others where B happened at an earlier time coordinate than A.

    4) If A is the event of a signal being transmitted and B is the event of the signal being received, then as long as the signal was traveling at the speed of light or slower, all inertial frames agree A happened at an earlier time coordinate than B. But if the separation between A and B is such that the signal would need to travel FTL (for example, if in the coordinates of some inertial frame A happened on Earth in 2010 while B happened 4 light-years away in 2012, so the signal moved 4 light years in 2 years in this frame), then it would always be possible to find one inertial frame where A and B happened at the same time-coordinate (so the signal traveled infinitely fast in this frame), and some other frames where the signal was actually received at an earlier time than it was transmitted (B happened at an earlier time coordinate than A).

    5) Combining #4 with #2, if it is possible in at least one frame for a signal to travel backwards in time (so it was received before it was sent), this must be possible in all frames. Thus if FTL transmitters are possible, and Alice and Bob are two slower-than-light observers moving apart and carrying FTL transmitters with them, it should be possible for Alice to send a message to Bob which moves FTL in her frame but backwards in time in his frame, and for Bob to send a reply which moves FTL in his frame but backwards in time in Alice's frame, such that Alice actually receives Bob's reply before she sent the original message, a clear causality violation in every frame.
     
    Last edited: May 26, 2010
  20. May 26, 2010 #19

    Fredrik

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    See also this post (and the comment about the typo in #138).
     
  21. Feb 10, 2012 #20
    I know this is an old post, but isn't the reason why there would be causality violations simply a function of the speed of light a constant - thereby making relativity mathematics useless in ANY case involving formulas involving v>c?
     
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