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I Are the Lorentz transformation formulas on wikipedia correct?

  1. May 8, 2016 #1
    They seem to defy the most fundamental principle of SR. The first postulate/equivalence principle.

    According to wikipedia, we get

    Lorentz boost (x direction)
    d3140dec18ab1103cc4aebe3fe182f98.png

    and slightly different formulas for the inverse Lorentz boost

    017ab51d8e47556d5c9f739ae48e8e29.png

    "This "trick" of simply reversing the direction of relative velocity while preserving its magnitude..." - Wikipedia *snap*

    You cannot reverse the direction of relative velocity. The whole point is that it is *relative*

    Taking two observers A and A' which move at v relative to each other, with A measuring an even e1(x1,t1) and A' measuring e1'(x2,t2) then it SHOULD follow by the equivalence principle,
    that if A' was to measure an event e2 with the same values of e1, hence e2'(x1,t1) then A has to necessarily measure this even happening with the SAME values as e1', hence e2(x2,t2).

    Only then can we talk about this being equivalent.

    But this is NOT the case, as according to wikipedia we have to use different formulas when transforming events from A to A' compared to transforming events from A' to A

    x=γ(x'+vt') should be x=γ(x'-vt') instead OR the former for both cases.

    What changes is not the velocity sign, since the velocity is relative.
    What changes are the signs of the events' coordinates depending on how the observers are facing in relation to each other.

    Of course, i might be wrong, but my brain is starting to wobble, so i will have to stop here before serious damage. I might pick up on it later.
     
  2. jcsd
  3. May 8, 2016 #2

    Orodruin

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    No, you are wrong. If A' moves with velocity ##v## relative to A, then A moves with velocity ##-v## with respect to A', hence the sign in the inverse Lorentz transformation. You can also easily check that this is the case by inverting the transformation.

    Note that this is the Lorentz transformation in standard configuration, i.e., when the x and x' axes are pointing in the same direction, which is usually the case by convention. In general you can write down Lorentz transformations that boost along arbitrary axes and include rotations.
     
  4. May 8, 2016 #3
    But then they wouldn't be equivalent. You are probably right, but it certainly does not click with me.
     
  5. May 8, 2016 #4

    PeroK

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    The Lorentz Transformation has an assumption that the origins of the coordinates coincide and the axes point in the same physical direction.

    If the velocity were the same in each frame, then the x-axes would be pointing in opposite directions. The Lorentz Transformation would then be more complicated to take account of the x-axis being in the opposite direction from the x'-axis.

    Although, then it would be exactly the same transformation for each.

    It might be an interesting exercise to work this out for yourself!
     
  6. May 8, 2016 #5

    Orodruin

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    I think you have a misconception about what the equivalence principle states. It states that physical laws will take the same form in all inertial systems. Both the Lorentz transform in standard configuration and its inverse preserve the speed of light.
     
  7. May 8, 2016 #6

    Orodruin

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    Also, the change in sign is not something peculiar to Lorentz transformations. It is already present when inverting the classical Galilean transformations.
     
  8. May 8, 2016 #7

    Ibix

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    It's a fairly straightforward requirement of the equivalence principle that the forward and reverse transforms should be identical except for the sign on the velocity. After all, if we both agree which way forwards is, and I think you are moving forwards you must think I'm moving backwards. Thus my velocity term and yours must have opposite signs.

    Note that this discussion is about the transform from some frame S to S' and its inverse. The transforms aren't different except in the sense that one is translating to a frame going in the +x direction at speed v and the other to a frame going in the -x direction at speed v.
     
  9. May 8, 2016 #8
    I don't think so.

    If we were both to meet in empty space, flying by each other and facing each other, we would not be able to tell who is the moving one and who is standing still. At some point we would have to agree that none of us is special, hence we are equivalent.

    From my perspective, you are coming from the front, pass me, and then you move backwards away of me.
    From your perspective, i am doing pretty much the same.

    So why would i assume that if i see an event happening at x1,t1 from my perspective and this event was observed by you at x2,t2 i could compute using the Lorentz transformation formulas, that the other way around would be different?

    To put it differently. If this was some virtual reality with virtual astronauts and i was able to place you into any of the two astronauts as they are passing by, and given the astronauts were also inside exactly the same spaceships with the same events happening, you would not be able to tell which of the two astronauts you are watching at any given point.

    You would have a very hard time to decide on if to use the "standard" or the inverse Lorentz transformation formulas if i was to ask you how the other astronaut would see a given event, you see happening at x1,t1. But you shouldn't have to be puzzled and i will show (hopefully) that this can be resolved without the astronauts having to agree prior to who is the one moving forward and who is the one moving backwards.
     
  10. May 8, 2016 #9

    PeterDonis

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    No, but we can tell that your direction, relative to me, must be opposite to my direction, relative to you. And that is what the sign change from the Lorentz transformation to its inverse is representing.

    Only if "pretty much the same" takes into account that, since we are facing each other (at least, you appear to assume that we are), "front" and "back" relative to me are opposite directions from "front" and "back" relative to you.

    Because of the opposite directions. From the perspective of a faraway observer watching both of us, if our coordinate axes both point in the same directions--which is what the standard Lorentz boost formulas we are using assume, they assume no spatial rotation, just a change in velocity--then the signs of our velocities, relative to each other, must be opposite.

    What you are imagining is that you are using coordinates in which the direction "towards me" is the positive ##x## direction, but I am using coordinates in which the direction "towards you" is the positive ##x## direction. But these are opposite directions, so a simple Lorentz transformation like the ones we are discussing cannot transform between our coordinates in this case. You would also have to include a spatial rotation by 180 degrees, so that the direction of the ##x## axis flips. But that would require different (and more complicated) formulas for the transformations; the ones given in this thread won't work. It's much simpler to have our coordinate axes point in the same direction and let the sign change in the transformation vs. its inverse take care of things.
     
  11. May 8, 2016 #10

    Ibix

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    Then we are disagreeing on which way is forwards, and we can't use this form of the Lorentz transforms. You can derive a version of them for this setup where my +x direction is your -x direction (just flip the signs on x' and introduce a v') but it seems unnecessarily complicated.

    I think you are not thinking clearly about the physics here. You are correct that there is no global sense in which either of us is moving. However one of us is always moving and setting up an experiment to show that in my rest frame you are moving in the opposite direction from the way I am moving in yours is fairly trivial.

    Edit: beaten to it by Peter, I see.
     
  12. May 8, 2016 #11

    Orodruin

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    I think the OP is assigning too much importance to what is the "forward" direction. This is a matter of definition for both observers, but is well defined for the Lorentz transformation in standard configuration. You can also define Lorentz transformations with the coordinate axes pointing in arbitrary directions for either observer, but it will no longer be in standard configuration.
     
  13. May 8, 2016 #12

    ogg

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    Generally, coordinates should be distinguished by the f.o.r. they are relative to (if they are relative coordinates). "Taking two observers A and A' which move at v relative to each other, with A measuring an even[t] e1(x1,t1) and A' measuring e1'(x2,t2) then..."
    The event E1 which is measured by both A and A' is better signified by e1(x1,t1) and e1'(x1', t1') {or, equivalently, e1'(x'1,t'1) or even e1(x1',t1')}. Unless the frames of reference are identical, the coordinates will not be identical, but they're both describing the SAME event, E1, which must, obviously, not change with a change in coordinates. Writing the way you did suggests that x1 & x2 are different but IN the same frame of reference (coordinates) which is just plain wrong. The event is invariant, but the values of the coordinates (given different f.o.r.s) can not be.) As far as the written equations, if for A the closing velocity is +v then for A' the closing velocity (IF you assume they share the same x direction!) is -v and using those values in the equations will give identical results. But note the problem (ambiguity) that is injected by not clearly specifying the difference between the velocity in a f.o.r. (which should be v and v' respectively) and its VALUE in those frames (+v, and -v, resp.). if v' = -v, then v = -v'. Also of note is that the word "coordinates" may designate the spatial coordinates, the space-time coordinates, or the more abstract coordinates of phase space (eg. 6N elements for N objects; 3 for xyz and 3 for vx,vy,vz for each object) - you need to get on the same page as the author using the terminology. (Also note that time isn't included as a "coordinate" in the (elementary) classical treatment of dynamical systems)...
     
  14. May 8, 2016 #13

    Dale

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    This isn't the standard for equivalence. Two frames are equivalent if the laws of physics are the same in them.
     
  15. May 8, 2016 #14

    PeroK

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    No, because the two astronauts have to agree on a common set of axes. If they agree that the positive x-axis is in the forward direction for astronaut A, then he will use the "normal" transformation and astronaut B will use the inverse transfomation. And, if they agree that the positive x-axis is in the forward direction for astronaut B, then he will use the normal transformation and the astronaut B will use the inverse.

    And, if they agree to use different coordinate systems, then they will both have to apply an additional transformation on top of the Lorentz. Just as they would if they were at rest with respect to each other.

    The Lorentz transformation (it often neglected to say) assumes this common agreement of the origin and orientation of the axes. In particular, ##x = 0, t = 0## must represent the same spacetime event as ##x' = 0, t' = 0## AND the ##x, y, z## and ##x', y', z'## axes must coincide.
     
  16. May 8, 2016 #15
    What i am questioning is if them having to "meet"/communicate and agree on anything is at all necessary, which is why i used a scenario which looks exactly identical, no matter whose astronaut's perspective you observe it from.
     
  17. May 8, 2016 #16

    PeroK

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    It's clear that if someone (anyone) is trying to transform between two coordinate systems, then they need to know them both.
     
  18. May 8, 2016 #17

    Dale

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    Coordinate systems are not objects in nature waiting to be found. They are mathematical conventions used by physicists to make it easier to analyze a situation. As PeroK succinctly put it, if you want to transform between two conventions then obviously you must know both conventions.

    The Wikipedia article carefully explains the convention used. Under that convention the transformation equations are correct. There are certainly other conventions you could adopt and then other transformation equations would be required.

    I hope that is clear to you now.
     
  19. May 8, 2016 #18
    The only reason the velocity of an object is deemed negative is because it's moving in a direction of decreasing position. So if your x-axis points away from you, with the values of x increasing in the direction you are facing, then in the scenario you describe you will conclude that my velocity is negative. Likewise I will conclude that your velocity is negative. On the other hand, if I were instead moving in the opposite direction you would conclude that my velocity is positive and I will conclude that your velocity is positive.

    Nothing about this allows either of us to conclude that we the one who's at rest and it's the other who's moving, so nothing about this violates the Principle of Relativity. Each of us is at rest in an inertial frame of reference, and those two frames of reference are equivalent to each other!
     
  20. May 12, 2016 #19
    Correct, since this is a symmetrical scenario, both astronauts facing each other (or both having their backs turned towards each other) what one measures, the other will measure as well. Now if one of them was to turn, hence one watching the other astronaut's back, while the other astronaut was "looking away", the scenario would not be symmetric any more. The astronaut who was looking away, having his back turned, would consider himself at rest, with someone approaching from his back(negative x axis) and then moving away from him. It would be a matter of switching the signs of the x axes on the diagram.
    But that is not the point. I used the symmetric scenario to make the point clearer.
    More important, using this symmetric scenario is something that can be used as a standard, which requires no communication at all once declared as such.

    Since none of the astronauts can tell who is the moving one, and the scenario is symmetric, several statements can be made by logic alone IMO.

    For example, what you said about velocities. If one astronaut sees the other approaching from the front, hence negative velocity, the other will see the same from his perspective. Hence, also negative velocity.

    If astronaut A measures a ruler at length L but astronaut A' measures that ruler which is moving at Vrel to be shorter by a factor 1/γ, hence L' = L*(1/γ) then a ruler which A' measures to be of the length L2' = L will be measured by astronaut A to be shorter by the same factor 1/γ. Hence L2 = L2' * (1/γ) = L * (1/γ) = L'
    In other words. If A measures a non-moving ruler that at 10m from his perspective and A' measures this ruler to be 8m from his perspective, then the other way around HAS to be also true. If A' measures a non-moving ruler at 10m from his perspective then A would measure this ruler at 8m from his perspective.

    If A measures moving clocks which are at rest relative to A' to be "ticking slower" by some factor 1/γ, then A' will measure moving clocks at rest relative to A to be "ticking slower" by the same factor 1/γ.

    Continuing this line of reasoning, it follows (unless my mind plays tricks on me) that

    If A measures an event e1 at x=5 and t=10 and A' measures this event to be "happening" at let's say e1' at x'=3 t'=11 (no calculations made, just fictive numbers to make the point), then it would follow that if A' was to measure an event e2' at x'=5 and t'=10, A would measure this event to be happening at e2 with coordinates x=3 t=11.
    For this to be the case, we could not possibly use the two _different_ formulas we get on wikipedia. As we would use the Lorentz transformation formula on e1 but we would have to use the inverse Lorentz transformation formula on e2'.

    What would also have to be correct, if i am not mistaken, would be for the formula i am looking for to have the property of mapping e1' x'=3 t'=11 right back to e1 x=5 t=10. The formula would be the same going from e1 to e1' and back from e1' to e1.

    The astronatus would not have to communicate anything at all in order to get their values right and decide who is A and who is A'. And neither would have any of the other observers potential moving relative to A and A'.

    I think the correct term for the formula i am looking for would be an involution.

    as a simple example:
    x' = 1/x x=3 we would get x'= 1/3 and with x= 1/x' we would get back to x=3. The Lorentz transformation formulas for the symmetric case of the two astronauts approaching each other and facing each other or having both their backs turned against each other should be involutions.
     
    Last edited: May 12, 2016
  21. May 12, 2016 #20

    Ibix

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    You can use that transformation if you wish. No one will stop you. What will you do when there are three astronauts moving at different velocities? Who will face who? Or will you rotate the coordinate system of the middle astronaut when transforming to the frame of the astronaut behimd him but facing him?
     
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