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Is it true that in general relativity length does not exist?

  1. Oct 30, 2012 #1
    I think I read once that length does not exist in general relativity, though I'm not sure if I got that hopelessly muddled or if there are subtleties that I am missing. I have a vague understanding of length contraction, but I thought that what I was reading said that there simply is no length in general relativity, it is not a meaningful concept.
     
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  3. Oct 30, 2012 #2
    Someone correct me if I'm missing something as well, as I'm mostly familiar with SR, but I believe lengths certainly do exist in GR as well. What you may be thinking of is the fact that there isn't one "right" length i.e. no particular reference frame is any better than the others. In this sense, length is variable, although there is what's called "proper length." In any reference frame, proper length L is:

    [itex] L = \sqrt{\Delta x^{2} + \Delta y^{2} + \Delta z^{2} - (c \Delta t)^{2}} [/itex]

    Proper length is defined as the invariant interval of a space-like path between two events. Two events are space-like separated if there exists a reference frame in which the two events can be observed to occur at the same time, but no reference frame such that they are observed to be located the same point in space.** This would be the case for defining the length between two spatially separated points like the endpoints of an object, since we can take proper length to be distance between simultaneous "events" at the ends of the object, in the object's rest frame. That is, the proper length of an object is the length of the object as it would be measured in the object's own rest frame (though technically this length is not any more "correct" than any other possible lengths). This invariant quantity could be inferred from any reference frame by 'boosting' to the object's rest frame. I should note that this formula assumes a flat spacetime, and it looks much more complicated in GR as it accommodates curved spacetime.

    **If you think of it in terms of length contraction, we're just saying that there is no reference frame such that the length contracts to 0.

    I also just had a thought that you might be muddling length contraction in GR with something you read about the plank length. Just wikipedia "plank length" for more details.
     
    Last edited: Oct 30, 2012
  4. Oct 30, 2012 #3

    Nugatory

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    Bossman, you left out a fairly important minus sign. It's:
    [itex] L = \sqrt{\Delta x^{2} + \Delta y^{2} + \Delta z^{2} - (c \Delta t)^{2}} [/itex]
    Or if you're using the other sign convention:
    [itex] L = \sqrt{(c \Delta t)^{2} - \Delta x^{2} - \Delta y^{2} - \Delta z^{2}} [/itex]

    Either way, the sign of the time and the space coordinates must be different, so that the path followed by a flash of light has length zero.
     
  5. Oct 30, 2012 #4
    Woops! My mistake, corrected now. Thanks for pointing that out.
     
  6. Oct 30, 2012 #5

    PeterDonis

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    Unless you can give a specific reference, I don't think you're going to get much clarification, except for the obvious point that "length" *is* a meaningful concept in GR.
     
  7. Oct 30, 2012 #6

    PeterDonis

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    This formula is correct (at least, it is now since you adopted Nugatory's correction) for special relativity, where spacetime is always flat. It is not correct in GR because spacetime can be curved. (Technically, it can still be used in GR in a local inertial frame, but it certainly can't be used in *any* frame.)
     
  8. Oct 30, 2012 #7

    PAllen

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    I think it's fair to say that length and distance have major additional difficulties in GR compered to SR. In SR, there are well known issues with rigidity for non-inertial bodies. Since, in GR, there are no global inertial frames, this translates to difficulties for what is meant by rigidity for an extended object in a region with substantial spacetime curvature. Then, without a clear concept of rigidity, it becomes hard to talk about the length of an extended object in such a context (length here referring to rest length - length measured in rigid object's rest frame ; in GR the latter does not exist for an extended object).

    For large distances in GR, the problems are less acute, but still present. The issue is lack of a preferred way to define simultaneity over great distances (locally, you can say that a relevant local inertial frame gives you a preferred slicing). Thus, it is not that you can't compute distances, but that, in the general case, you don't know which of an infinite number of answers to prefer.
     
  9. Oct 30, 2012 #8

    PeterDonis

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    Agreed, but none of this implies that "length" is not a meaningful concept. It just means that, in any particular case, there are multiple "lengths" that could be applicable.
     
  10. Oct 30, 2012 #9
    I did mention that the formula is altered in GR to accomodate curved spacetime. I didn't bother to copy down the formula because I (let alone the OP) don't have a firm grasp on it, and I'm under the impression that the qualitative idea of proper length is the same regardless. Is it somehow not?
     
  11. Oct 30, 2012 #10

    PeterDonis

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    Ah, sorry, I missed that in your post.

    There isn't actually a single formula for curved spacetime; it depends on *which* curved spacetime you are dealing with. But you're right that the qualitative idea is the same: the "length" between two nearby events is [itex]ds = \sqrt{ \sum_{a, b} g_{ab} dx^a dx^b }[/itex], where [itex]g_{ab}[/itex] is the metric and [itex]dx^a[/itex] and [itex]dx^b[/itex] are coordinate differentials (like dx, dy, dz). In the flat spacetime case, we have [itex]g_{00} = -1[/itex] and [itex]g_{11} = g_{22} = g_{33} = 1[/itex], and [itex]dx^0 = dt[/itex], [itex]dx^1 = dx[/itex], [itex]dx^2 = dy[/itex], [itex]dx^3 = dz[/itex], which gives the formula you wrote down. The specific coefficients in the matrix [itex]g_{ab}[/itex] will depend on the particular spacetime.
     
  12. Oct 30, 2012 #11
    Ahh, that's very interesting. Thank you for that explanation. I'm very comfortable with SR but have yet to deal with any GR in my undergrad courses, do you by chance have a preferred book or other source that I might try and do a bit of self-teaching from?
     
  13. Oct 30, 2012 #12

    PeterDonis

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  14. Oct 30, 2012 #13

    Nugatory

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  15. Oct 30, 2012 #14

    Matterwave

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    I just want to mention that the metric "g" actually also depends on the coordinate system and, more importantly, the basis vectors you use. One could always adopt a tetrad field as our basis vectors so that "g" has the form it takes in SR always...but the formulas we must us to calculate our other tensors must all change (which is why most GR courses just stick with coordinate bases).
     
  16. Oct 30, 2012 #15

    PAllen

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    This is interesting. Wouldn't it not just be tensor calculation that changes? Wouldn't you also have have to do something different to compute integral invariants (distances, time, volume, etc)? If you used the coordinate formula with the Minkowski metric, wouldn't you get the wrong anser? Or is it that you don't compute integral invariants at all with tetrad fields - you only compute tensor fields using them?
     
  17. Oct 30, 2012 #16

    Matterwave

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    Tetrad fields are not coordinate bases, so you don't have one forms which are just dx, dy, dz, etc., or vectors which are d/dx, d/dy, d/dz, etc. You have basis one forms and basis vectors e_x, e_y, and e_z, which maybe expressed as some functions of d/dx, d/dy, and d/dz, etc. So, you don't have ds^2=gdxdx or some such because the dx's are no longer your basis vectors. You have ds^2=sum_ij(e_i dot e_j)=sum_i(e_i)^2.

    EDIT: Sorry, my memory is getting hazy on this...I'll have to think about this a little more...What I wrote above (the formula for ds) might be wrong.

    Probably, it's better to just look at the wikipedia...http://en.wikipedia.org/wiki/Frame_fields_in_general_relativity
     
    Last edited: Oct 30, 2012
  18. Oct 30, 2012 #17

    pervect

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    Length becomes observer dependent in Special relativity (Lorentz contraction means that a moving observer measures an object to have a different length than a stationary one).

    Length is pretty much the same in GR as it is in SR, except that the concept of "an observer" becomes trickier. In SR, you can specify "an observer" by picking an inertial frame. In GR, to define distance, you need to pick a notion of "simultaneity" - or a class of observers For instance, observers who are isotropic (moving with the Hubble flow) will have a different "now", and hence a different notion of distance, than fermi observerers). The later sort of distance is the sort that one would measure with a ruler, but it's little used in cosmology, the sort you'll see reported in most papers is the first sort (which is much easier to compute and has become semi-standard).
     
  19. Oct 30, 2012 #18

    PeterDonis

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    This is true, and it's good to mention it. If we're going to talk about the complications, we should talk about all of them. :wink:

    Strictly speaking, this is only true in a small local patch of spacetime around a given event. You can't construct a global "reference frame" this way like you can in flat spacetime.
     
  20. Oct 30, 2012 #19

    PeterDonis

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    I think this is the right way to look at it; a "tetrad field" is, strictly speaking, a local object only. You can, of course, define a frame field over an entire spacetime region (for example, the frame field of static observers in Schwarzschild spacetime, where the basis vectors vary with r), but I don't think you can compute integral invariants using it because there isn't a consistent integration measure. You need a single chart that covers the entire region. (I would welcome comment from experts on this, though.)
     
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