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Deriving Lorentz transformations

  1. Dec 17, 2015 #1
    Why is relative speed taken to be symmetrical i.e. speed of one frame of reference from a second frame is equal to that of the second frame of frame refrence from the first frame
     
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  3. Dec 17, 2015 #2

    PeroK

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    Which one do you think should be the greater?
     
  4. Dec 17, 2015 #3
    Well I think that if I observe an object moving with velocity v then how can I know how is the object observing me.
    Also if the speeds are equal then can I say that in my refrence frame
    X=vt
    And in his frame X'=vt'
    Please forgive me if I m mistaken
    I m not so good at physics...
     
  5. Dec 17, 2015 #4

    PeroK

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    That wasn't the question. The question is:

    If you observe an object moving with velocity ##v## does that object measure your velocity as ##> v## or ##< v##? (Assuming it's not ##= v##)
     
  6. Dec 17, 2015 #5
    I don't have any reason to prefer >v or <v
     
  7. Dec 17, 2015 #6

    PeroK

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    Then it must be ##=v##.

    Let's say you see an object moving towards you at speed ##v## and it sees you moving towards it at speed ##v_1 > v##. Then, unless there is a lack of symmetry in the universe, you must see it moving towards you at ##v_2 > v_1##. That can't be. You have the same argument if ##v_1 < v##

    The symmetry of the universe implies equality of relative velocities.
     
  8. Dec 17, 2015 #7

    Dale

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    Yes, exactly. This is called a symmetry principle.
     
  9. Dec 17, 2015 #8

    bcrowell

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    I have a discussion of this in my SR book http://lightandmatter.com/sr/ , section 1.4, example 11, "Observers agree on their relative speeds."
     
  10. Dec 17, 2015 #9
    The book seems excellent. I am glad to have found about that example, it is the first time that I see, explicitly stated, that the question is not completely obvious or trivial. A teacher of mine solved it by printing a paper and fliping it the other side to show the symmetry. IMHO, that is not completely trivial and the matter is a little more subtle than that.
     
  11. Dec 17, 2015 #10
    Let velocity of IFR 2 measured in IFR 1 be $$V_{12}$$.
    Let velocity of IFR 1 measured in IFR 2 be $$V_{21}$$.
    Let X be axis along to the velocity. Changing the direction of X, V_12 also changes its sign.
    Chanbing the direction of X, IFR 1 and IFR 2 exchange in the sense that one is along X and the other is reverse.
    So we may have a reason to say $$V_{12}=-V_{21}$$.
     
  12. Dec 17, 2015 #11

    strangerep

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    Afaik, one only needs the following assumptions:

    1) The usual relativity principle,

    2) An assumption of 3D spatial isotropy,

    3) An assumption that the axes transverse to the boost direction remain unchanged (as usual when deriving Lorentz transformations), [Edit: meaning that the rotational freedom allowed by spatial isotropy is exploited to align the transverse axes of the original and transformed frames.]

    4) An assumption that velocity boost transformations form a 1-parameter Lie semigroup (possibly a group).

    From (1)-(4) one can derive that, for a boost transformation with parameter ##v##, the inverse transformation must have parameter ##-v##. Additional hand waving intuition is not necessary.

    [Edit: I modified this post slightly in an attempt to soften my previous tone.]
     
    Last edited: Dec 17, 2015
  13. Dec 17, 2015 #12

    bcrowell

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    This isn't a necessary assumption. This fact can be derived.

    By assuming a Lie group you are assuming what you claim to prove.To justify this assumption, you will find yourself having to reason about physics. Physical reasoning does not equate to "hand waving intuition."

    [Mentors note: This post has been edited as part of some overall thread moderation]
     
    Last edited by a moderator: Dec 17, 2015
  14. Dec 17, 2015 #13

    strangerep

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    I suspect we are talking at crossed purposes on this point. Please see my edit in post #11. If that doesn't put the discussion back on track, then please explain how this is "derived". (I did look in your book, but maybe I wasn't looking in the right place.)

    I do not see this. What do you think I "claim to prove".

    Sure.
     
  15. Dec 18, 2015 #14

    bcrowell

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    @strangerep: Sorry for my grumpy tone in my previous post.

    I see. From your original version, I thought you were talking about an assumption that there is no transverse Lorentz contraction. From your edited version, I can see that you just mean there is no rotation between the two frames.

    Maybe I'm misunderstanding this as well. Is the complete argument written up somewhere?

    In #4, you only assume a semigroup, but then at the end you refer to "the inverse transformation." Are you claiming that based on assumptions 1-4 you can prove that it's a group, and not just a semigroup?

    My general comments on the outline you've presented are:

    (1) I don't see the point of doing this in 3+1 dimensions rather than 1+1. Any successful argument is going to require an assumption of spatial isotropy, but that assumption can be expressed in one spatial dimension, where it's equivalent to parity symmetry.

    (2) Proving some group-theoretical facts doesn't establish anything about the physics unless you provide some "glue" between the math and the physics, which you haven't done. The mathematical assumptions require physical justification, and the mathematical results require physical interpretation. Using fancy math may give the superficial impression of rigor, but fancy math without the glue is actually less rigorous than simple math with the glue. In any case, all you seem to be claiming to prove is that "for a boost transformation with parameter v, the inverse transformation must have parameter −v." This is easy to prove in even the simplest treatment of SR using high school algebra. The gee-whiz stuff about Lie semigroups comes off to me as obscurantism.
     
    Last edited: Dec 18, 2015
  16. Dec 18, 2015 #15

    strangerep

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    Thank you. Actually, I see now that (the original version of) my post could indeed be perceived as supercilious. I take your remarks on board for the future.

    My writeup of the complete argument is still not in final form (and no one else has yet proofread the draft). Probably, I should keep my mouth shut until this occurs, but I don't always have the necessary self control.

    I believe so, yes. (Strictly speaking, there's a couple of other inputs: i.e., that the original and transformed origins coincide, and some physical input to establish that the effect of the transformations does indeed correspond to a natural interpretation of what "velocity boost" means in terms of coordinates.)

    I wanted to investigate the most general case, to be sure I wasn't accidentally overlooking something.

    I agree with all of the above. One of the reasons I haven't finished my writeup is that it's challenging to do this well, and without subtly introducing extra assumptions.

    Well, I'm trying to write a treatise covering the most general (fractional-linear) case, with as few assumptions as possible. I decided to be more careful about the semigroup stuff when I investigated time displacement transformations in the same framework. The advanced investigations along related lines that I'm aware of always seem to assume time reversal symmetry. (E.g., Bacry & Levy-LeBlond, "Possible Kinematics", JMP vol 9, 1968, p1605.) Relaxing that assumption, one must investigate the physical domain (in velocity phase space) on which all the transformations are well-defined, and ask on what domain the inverse transformations are also well-defined. I found some interesting consequences (beyond the scope of this thread), and this motivated me to relax the group assumption for other transformations such as boosts and spatial displacements, and investigate how much could be derived starting from only a semigroup assumption. Whether this is perceived as "obscurantism" depends on one's interests, I suppose.
     
    Last edited: Dec 18, 2015
  17. Dec 22, 2015 #16

    Erland

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    Symmetry, spatial isotropy, how are these principles formulated in a mathematically stringent way?

    Early in this thread, it was claimed that the relative speeds ##v_1## of Frame 1 w.r.t. Frame 2 and ##v_2## of Frame 2 w.r.t. Frame 1 must be equal because of "symmetry". But then, suppose an object is at rest w.r.t Frame 1. Then, it has speed ##v_1>0## w.r.t Frame 2. Why doesn't this violate symmetry?

    Also, I am reading bcrowells interesting book and looked up the example he mentioned, but I don't understand how "spatial isotropy" is applied in that case.
     
  18. Dec 22, 2015 #17

    strangerep

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    Let's take spatial isotropy as an example. Heuristically, this means there is no distinguished (or "preferred") direction in 3-space. In mathematically more precise terms, I'd express it as follows:

    A theory is spatially isotropic iff every equation of the theory remains invariant when all quantities in the equation are substituted by their (respective) rotated counterparts. To understand this in more detail, let's consider an equation in a theory, written as $$F(t,{\bf x}, {\bf v}, ...) ~=~ 0 ~~~~~~ (1) ~,$$ where the bold symbols denote 3-vectors. Now consider an arbitrary 3D rotation matrix ##\bf M##, and substitute all 3-vectors in (1) by their rotated counterparts, i.e., ##{\bf Mx}, {\bf Mv}##, etc. So we have $$F(t,{\bf Mx}, {\bf Mv}, ...) ~=~ 0 ~~~~~~ (2) ~.$$ If (2) is equivalent to (1), i.e., if by purely algebraic manipulations we can make (2) look exactly like (1), then (1) is called a spatially isotropic equation. If every equation in the theory has this property, then the theory has the property of spatially isotropy.

    A similar idea applies to other symmetries: if we can take any equation in the theory, substitute the various quantities therein by their transformed counterparts, and by purely algebraic manipulations make the new equation look exactly like the original, then the theory has that symmetry.

    Sometimes it's obvious that an equation is spatially isotropic, e.g., if it contains only scalar expressions explicitly, like (say) ##{\bf x \cdot \bf p}##.

    BTW, although I've only explained the case with 3-vectors explicitly, the same idea applies if there are higher rank tensors (or spinors) in the theory -- one simply has to use the correct rotation transformation formula for each type of quantity appearing.
     
    Last edited: Dec 22, 2015
  19. Dec 22, 2015 #18

    strangerep

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    By introducing "an object" (which I'll denote by ##O##), you must now consider 4 relative velocities: ##v_{12}##, ##v_{21}##, ##v_{O1}##, ##v_{O2}##. (Here, my notation ##v_{AB}## denotes velocity of "A" relative to "B". More precisely, ##v_{12}## denotes the velocity of the origin in Frame 1 relative to the origin of Frame 2, where we assume the origins coincide momentarily.)

    We have ##v_{21} = -v_{12}## by spatial isotropy (or rather, parity reversal in the 1+1D case), but that (by itself) doesn't say anything about the speed of O relative to either frame. So even if ##v_{O1} = 0##, it doesn't necessarily mean ##v_{O2} = 0##, since ##v_{12} \ne 0##.
     
    Last edited: Dec 22, 2015
  20. Dec 23, 2015 #19

    Erland

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    Very good, but this then leads to the question: Which are the equations in the theory to which this applies? This should be specified in a stringent exposition.

    Also I wonder, how can this be used to prove that, for example, distances in directions perpendicular to the direction of motion between the frames are not changed by the transformation? (In this case, only rotations fixing the direction of motion should be used above.)

    And how is it used to prove that ##v_{12}=-v_{21}##?
     
  21. Dec 23, 2015 #20

    Dale

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    The presence of the object breaks the symmetry. This is not because of asymmetry in the laws, but asymmetry in the boundary conditions.
     
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