Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz contraction in other dimensions?

  1. Aug 16, 2011 #1
    Hi guys,

    this is my first post/thread, so I'd like to start with an easy one:rolleyes:

    I've searched the web and I wasn't able to find a satisfying proof of the fact that Lorentz contraction is NOT applied to the dimensions, that are not parallel to the direction of motion (e.g. perpendicularly). It's a bit annoying, because in most of the textbooks it is stated that this proof would be easy. Nevertheless it's NEVER provided!!!
    I don't need another first postulate reductio ad absurdum but rather a mathematically consistent counter-argument (if one exists...).

    Thanks a lot!
     
  2. jcsd
  3. Aug 16, 2011 #2

    Mentz114

    User Avatar
    Gold Member

    Surely it's built into the definition of the Lorentz transformation ? It's a linear transformation of a 4-vector which mixes time and space in the direction of the boost. The other directions (spatial dimensions) just don't come into it.
     
    Last edited: Aug 16, 2011
  4. Aug 17, 2011 #3
    And the reason why?

    That's what I'm looking for:wink: and if it is too fundamental and its derivation is purely axiomatic how come nobody bothers to demonstrate it?
     
  5. Aug 17, 2011 #4

    haushofer

    User Avatar
    Science Advisor

    If so, one would be able to distinguish inertial systems, violating the axiom of SR that inertial systems are physically equivalent. See for instance Moore's "Traveler's guide" for a proof.
     
  6. Aug 17, 2011 #5
    Currently I have no access to the specific textbook and if I did I'm nearly sure it would reproduce the same ab adsurdum proof that I'm already familiar with.
    As far as I can tell this is the only available approach to the matter out there. I would appreciate it if someone came up with a mathematic one or at least explained to me why this is not possible.
     
  7. Aug 17, 2011 #6

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

  8. Aug 17, 2011 #7
    Hi Trifis welcome to physicsforums. :smile:

    There are different ways to get to that result; it is necessary to assume that the speed of light is completely independent (to second order precision) of the motion of the source.
    You could start with correcting the classical calculation by Michelson-Morley here, on p.336 of
    http://en.wikisource.org/wiki/On_the_Relative_Motion_of_the_Earth_and_the_Luminiferous_Ether

    Or you could follow Einstein's simple derivation starting from (8a), here:
    http://www.bartleby.com/173/a1.html
    It follows directly from (8a)+(11a) :tongue2:

    Harald
     
    Last edited: Aug 17, 2011
  9. Aug 17, 2011 #8
    @Robphy I do understand your metersticks gedanken experiment and it is exctacly what I have in mind in order to rule out contraction perpedicular to the motion. It is designed to contradict the principle of relativity. Moreover @Andrew Mason expanded the idea and attached the concept of simultaneity (which btw arises a new question:tongue:: why simultaneity in the y-axis is preserved, considering the fact that time dilation is dimension-independent?).

    I really can't find another way to conceive this except for the above-mentioned "classical way"...

    @Harald thanks for the greeting!

    I don't need a derivation of the Lorentz transformations. What I need is to mathematically derive the equation (9) on your second link, that are always taken arbitrarily for granted!
     
    Last edited: Aug 17, 2011
  10. Aug 17, 2011 #9

    Dale

    Staff: Mentor

    I don't get what you are saying here. Eq 9 on the second link is part of the Lorentz transform. If you are taking the Lorentz transform as a given then eq 9 is a given. If not then you do in fact need a derivation of the Lorentz transform. Can you clarify? What do you want to allow as a given in order to derive 9?
     
  11. Aug 17, 2011 #10
    I allow the postulates of relativity as given and try to find out why the perpedicular dimension to the relative motion remains unaffected during the measurements of both observers. You could say that I seek how y=y' pops up all of a sudden in Lorentz formulas so as to "satisfy the postulate of the constancy of the velocity of light"

    Maybe I should after all just compromise with the idea of violating the axiom without further mathematical analysis...
     
  12. Aug 17, 2011 #11

    Dale

    Staff: Mentor

    OK, what you are looking for is a derivation of the Lorentz transform from the postulates. You might like Einstein's original derivation in section 3 of his famous OEMB paper:

    http://www.fourmilab.ch/etexts/einstein/specrel/www/

    Note that he does not assume y=y' and z=z', but he derives it. Similarly with this derivation:

    http://farside.ph.utexas.edu/teaching/em/lectures/node109.html
     
  13. Aug 17, 2011 #12
    @ DaleSpam Intresting! I hadn't noticed that when I read the paper in german. So the whole concept is based on the fact that the two systems move parallel to its other as well as the fact that the events (0,0,0,0) coincide in both of them. Moreover we can interchange the frames due to the symmetry involved. As easy and "galilean" as it sounds, am I right?

    So as long the basic transformations (=parallel motion of the systems) are considered, we can show that the contraction acts only on the direction of motion without the need of any gedanken experiments... I don't know if I'm satisfied or not :tongue:
     
  14. Aug 18, 2011 #13
    Yes indeed, and as I already told you, eq.(9) follows from eq. (8a) + (eq.11a). I told you that as Einstein didn't tell that clearly. But he did introduce it with "This may be shown in the following manner." :smile:

    It also follows directly if you plug in the PoR and length contraction in the first derivation that I referred you to; only without a change in y and z length can it work.
    Note also that simultaneity does not play a role in that two-way speed of light.

    Best,
    Harald
     
  15. Aug 18, 2011 #14
    @Harald I must admitt that I didn't read Morley's experiment description again but neither the second link was crystal clear about the consequence of (9). In fact it is also stated that (11) , the final transformation, follows from (9) and not the other way around :/

    I've grasped that in the end all comes down to the parallel motion of the axes of the frames (as mentioned above) exact like the Galilean concept. Otherwise equations like (11) would not hold! My original question seems to have a simple explanation when one considers the Lorentz transformations...
     
  16. Aug 18, 2011 #15
    If he did so then it would not be true that "This may be shown in the following manner." Just tell me where he needs (9) to derive (11)! :wink:
    As I already told you twice: (9) follows directly from (8a) and (11a). As Einstein didn't tell that clearly, that is why I told you that clearly! It's a simple proof.

    [addendum: I admit that it only gives that y'2 + z'2 = y2 + z2; but as the choice of those axes is arbitrary, evidently y'=y and z'=z.
    What happened here was that Einstein showed that the right answer (9) is conform the requirement (11), while what you wanted to see is how (9) follows from (11); which is almost as straightforward, even without changing the text].
    Yes of course, that is the definition of the frames. :smile:

    Best,
    Harald
     
    Last edited: Aug 18, 2011
  17. Aug 18, 2011 #16

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Why not start with an algebraic definition of the Lorentz group? [itex]\Lambda[/itex] is a Lorentz transformation if [itex]\Lambda^T\eta\Lambda=\eta[/itex]. We know that in 1+1 dimensions, every Lorentz transformation is of the form [tex]\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},\qquad \gamma=\frac{1}{\sqrt{1-v^2}}.[/tex] This makes [tex]\Lambda=\gamma\begin{pmatrix}1 & 0 & 0 & -v\\ 0 & \frac{1}{\gamma} & 0 & 0\\ 0 & 0 & \frac{1}{\gamma} & 0\\ -v & 0 & 0 & 1\end{pmatrix}[/tex] an obvious candidate for what we should be calling a "pure boost" in the z direction. It's easy to verify that it's a Lorentz transformation. Is there a reason why we shouldn't just take this as the definition of what we mean by a pure boost in the z direction?

    If you want simple physical arguments, how about this one? If Lorentz contraction "contracts" objects in any other direction than the direction of motion, then we could build two identical solid rings at rest in two different inertial frames and have one of them pass through the other. This violates the principle of relativity. (One of the regulars here mentioned this to me a couple of years ago, but I can't remember who it was. Mentz? Yuiop?).
     
  18. Aug 18, 2011 #17
    @harald maybe the simplest things could sometimes prove to be the hardest to be fully conceived :/ I've read the appendix twice and believe it or not I kept passing the very sentence that you told me (repeatedly) : "This may be shown in the following manner."
    I'm sorry :)

    Then I read DaleSpam's posted papers which point it out straightforward and having read that I also got what you were telling me. On the first place I was just misled by this sentence: "or (11) is a consequence of (8a) and (9)"

    @Frederik I'm already familiar with the physical proof by contradiction. I was searching for a mathematical and I found one through the original transformations (with the provided help of course)

    I like this forum :cool:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Lorentz contraction in other dimensions?
  1. Lorentz contraction (Replies: 389)

  2. Lorentz Contraction (Replies: 15)

  3. Lorentz contraction (Replies: 2)

Loading...