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Lorentz transformations in non-inertial frames

  1. Sep 25, 2012 #1
    Are Lorentz transformations only work between inertial frames? if so, is there a simple counter-example e.g. for them not to work?
     
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  3. Sep 25, 2012 #2

    PAllen

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    v is a constant across all of space and time in the transform, so that trivially rules it out as normally understood. If you propose that you can use the Lorentz transform of a momentarily comoving inertial frame at each event of a non-inertial observer, you run into the problem that you assign two sets of coordinates to points of a region. Specifically, if you assume an observer moving in the +x direction, accelerating in the -x direction, then in a region on +x side of the point where direction of motion changes, you assign conflicting coordinates to the same event.
     
  4. Sep 26, 2012 #3

    bcrowell

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    The notion of an accelerated frame of reference has sort of a 1920 feel to it. Einstein originally described GR as a theory in which accelerated frames were as valid as inertial frames. This is no longer the way relativists think about GR. You will also see statements in old textbooks that SR only deals with inertial frames, not accelerated ones, but relativists today don't agree. They describe the distinction between SR and GR as being one between flat spacetime and curved spacetime.

    Because of these issues, it is probably going to be hard to come up with an answer to your question that satisfies everyone unless you are very, very explicit about what you mean.

    If local frames A and B coincide at event O, then the transformation between them, in a sufficiently small neighborhood of O, is purely a Lorentz transformation. Their proper accelerations, if any, are irrelevant.

    If you mean these frames to be non-local, then you have a problem, because GR doesn't have global frames of reference. There are various ways of defining an accelerated frame (or, equivalently, a spacetime representing a uniform gravitational field), but none of them are completely satisfactory: http://www.lightandmatter.com/html_books/genrel/ch07/ch07.html#Section7.4 [Broken]
     
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  5. Sep 26, 2012 #4

    Demystifier

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  6. Sep 26, 2012 #5
    That's not a problem at all. The CADO reference frame specifies the current local time at, and the current distance to, any event in (assumed flat) spacetime, from the perspective of an arbitrarily accelerating observer at any instant in her life. That's ALL that is required of the CADO reference frame: the CADO frame is NOT a GR chart, and has no need to BE a GR chart. See, in particular, Section 9 of this webpage:

    https://sites.google.com/site/cadoequation/cado-reference-frame
     
  7. Sep 26, 2012 #6

    bcrowell

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    From section 7 of the paper:

    In other words, *locally*, there is no need for anything but Lorentz transformations, and any relative acceleration is irrelevant.
     
  8. Sep 26, 2012 #7

    PAllen

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    The OP asked about transformation between non-inertial frames. An inertial frame in SR covers all of space time with one set of labels, and the Lorentz transform converts the labels of all events as calculated in one frame to those as calculated in another frame. I believe the OP assumed you could construct non-inertial frames in the same sense, and transform between them. Then, one problem you do run into with the simplest approach based on Lorentz frames is that you cannot cover all of spacetime. If you accept this limitation, then a further consequence is that the transform between two such non-inertial frames (interpreted as Fermi-Normal coordinates based on two different non-inertial observers) that cover a common spacetime region is not the Lorentz transform.

    CADO is irrelevant to this as it does not construct a non-inertial frame at all. Instead it is one technique for calculating observables and certain unobservable interpretations for a non-inertial observer, using a different frame at each event on the observer's world line. A non-inertial frame would be expected to represent world lines of various objects in the non-inertial frame. CADO does not attempt this at all, so it is not an answer to the OP question.
     
  9. Sep 26, 2012 #8
    Let me clarify my question. I have two frames that are moving with the accelerations against an inertial frame, therefore one frame moves with a constant speed against another. Can I use Lorentz transformations between these two frames?
     
  10. Sep 26, 2012 #9

    PAllen

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    My answer does relate to this. Are you presuming there is a well defined frame (extending in space and time) for each of these accelerating observers? If so, there is a problem. You have to define what you mean by such a frame before you can even ask what the transform is between them. That already presents a (not insurmountable) problem, as I described.

    If instead of asking about transforming between non-inertial frames, you ask: can each observer compute observations using a momentarily co-moving inertial frame, the answer is definitely yes.
     
  11. Sep 26, 2012 #10
    That is EXACTLY what the CADO reference frame does.
     
  12. Sep 26, 2012 #11

    PAllen

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    No, it does not. A frame as defined by everyone except this author, must assign one label to one event, not 2 or more labels to the same event.

    Please be aware that the views in this website are not accepted as valid by most relativity experts. I would go so far as to say about this website:

    - what's true is not new
    - What's new is not true

    What's not true:

    - an alleged frame that multiply maps events (at best, this is definition at odds with normal usage, that almost no one finds useful)
    - the claim that distant simultaneity has an objective definition such that there is only one right answer. This is the biggest error.
     
    Last edited: Sep 26, 2012
  13. Sep 26, 2012 #12

    PAllen

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    Let's get further at why CADO is not a reference frame at all in the sense it is normally understood, and in the sense the OP asked.

    Let us imagine two CADO reference frames for uniformly accelerating observer A and for uniformly accelerating observer B, as the OP apparently desires. Assume further that the accelerations are the same magnitude and direction, but with significant separation. Assume each is initially moving in the +x direction of some inertial frame, and that acceleration is in the -x direction so they soon reverse direction. Let us assume B located in the +x direction compared to A in this same inertial frame.

    Then, in the CADO frame for A, there are events on B's world line that have two labels. In B's CADO frame, all events on B's world line have only one label (x,t)= (0,t). So what is the transform from the CADO A frame to the CADO B frame? Not only isn't it Lorentz, but it isn't a transform at all, because it must map two labels into one label.

    [Edit: For completeness, I should mention application of a technique used in GR to this situation. Instead of talking about a CADO frame, you could set up multiple Fermi-Normal charts, such that each was a valid chart of part of spacetime. Then you have a mapping between overlapping parts of the charts. This gives an atlas of Fermi-Normal charts together covering all of spacetime, that you could call the 'point of view' of an accelerated observer. Then, for two accelerating observers, you would be able to define a series mappings between the charts of one observer's atlas, and the charts of the other observer's atlas. This obviously doesn't resemble the Lorentz transform between inertial frames, and I've never heard of anyone trying to execute such a technique in SR. The reality is that sane physicists analyze observable for accelerated observers using any convenient inertial frame.]
     
    Last edited: Sep 26, 2012
  14. Sep 26, 2012 #13
    I dont think your right about no experts believing it. I saw a NOVA show not too long ago where the guy told about someone riding a bike around in a circle at a far away place, and how he says time here is moving lots of centuries back and forth. That sounds like the same thing CADO says.
     
  15. Sep 26, 2012 #14

    PAllen

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    Pop sci show is not expert opinion. Given a number of ways of looking at something, you present the most 'amazing' one in a pop sci show. Assuming the presenter was a physicist, if you talk to them off the show they would agree that this is only one possible notion of simultaneity.

    Also, no one in the field attempts to define a frame the way he does. What they do instead is subtract the multiply mapped region, and call the result Fermi-Normal coordinates (now a well defined coordinate chart of part of spacetime), one of several possible approaches, all very ancient (before the author of this website was born).
     
  16. Sep 26, 2012 #15

    pervect

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    I concur with PAllen.
     
  17. Sep 26, 2012 #16

    pervect

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    If you take the result that MTW calcluates for "the frame of reference of an accelerated observer", right after reading their cautions about this phrase regarding uniquness and existence, you'll find that Lorentz transforms do not work globally in this "frame".

    This should serve as the counterexample you asked for. Read the section on the "accelerating observer" (assuming you track down the reference).

    You'll also find that their "frame" doesn't cover all of space-time, and see some of the compelling reasons why it can NOT cover all of space-time. (This relates to the existence part of the caution).

    If you accept that Fermi Normal coordinates are an excellent and more rigorously defined substitute for "inertial frame of reference", then you can say that the Lorentz transform does not work globally in Fermi Normal coordinates for an accelerated observer. Demystifer's paper should cover this I think. In other words, if you have Fermi normal coordinates t, x, y,z you can't simply say x' = x - vt, t' = t - vx/c^2 as a global relationship to define new Fermi normal coordinates for a differently moving observer..


    Perhaps Demystifier can also quote some papers about the issues and justifications for considering Fermi Normal coordinates to be a substitute for the concept of "an inertial frame of reference". The only thing that comes to my mind is http://arxiv.org/abs/0901.4465 and the references it cites (many of which are behind paywalls, and which I haven't read)




    You should note that Fermi Normal coordinates don't cover all of space-time, for instance that the coordinates that MTW proposes as "the" coordinate system of an accelerated observer don't cover all of space-time, and are essentially Fermi Normal coordinates.

    However the Lorentz transform will work quite well locally, even though it doesn't work normally, given that the proper basis vectors have been chosen. This gets into Ben's point, about how we now consider space-time to be a manifold, and when we talk about "frames", we're usually talking about the tangent space to the manifold, not the manifold itself.
     
    Last edited: Sep 26, 2012
  18. Sep 27, 2012 #17

    Demystifier

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    You cannot, but let me be more precise. Let S1 and S2 be the proper coordinates of two non-inertial observers, such that they move with a constant non-zero velocity with respect to each other, as seen either* by an inertial or an accelerated observer. Then the coordinate transformation from S1 to S2 is not a Lorentz transformation.

    *These two possibilities are not equivalent.
     
  19. Sep 27, 2012 #18

    Dale

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    Me too (or three).
     
  20. Sep 27, 2012 #19

    bcrowell

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    I agree with PAllen's analysis of CADO in #12. Unfortunately this thread's signal-to-noise ratio is going to pot because of all the shouting by GrammawSally, who I assume is actually a sock puppet for Mike Fontenot, promoting his crackpot theory. In July 2011, the mentors had a discussion about CADO, with the result described here: https://www.physicsforums.com/showpost.php?p=3396390&postcount=21 Since I'm no longer active as a mentor, I don't know whether the mentors' lack of action on this thread is due to their having changed their minds since last year, or whether they just have been too busy to do anything about it.

    Getting back to non-crackpot science, I agree with PAllen's analysis in #16.
     
    Last edited by a moderator: Sep 29, 2012
  21. Sep 27, 2012 #20

    PAllen

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    You mean Pervect's #16, I assume. I gave a brief statement of the same result way back in #7, first paragraph.
     
    Last edited by a moderator: Sep 29, 2012
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