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B Does SR length contraction produce compressive forces?

  1. Jul 13, 2017 #1

    ibkev

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    I'd like to get some conceptual clarification on whether or not length contraction is only apparent to an observer or if it's physically real in the sense that the object under contraction would experience compressive forces?

    My thinking on this has been along the lines of: I know that time dilation is objectively real given atomic clock experiments, etc. so the associated length contraction must also be objectively real and I'm thinking that means compressive forces would result. Possibly these forces could be large enough to be destructive. On the other hand, I try to imagine the perspective of the astronaut in their almost-light-speed-spacecraft and I can't convince myself that the length contraction is real from their perspective.

    Any clarification would be helpful. I indicated a basic level for this question because I'm learning this in my spare time and only just getting started on the actual math.

    Thanks!
     
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  3. Jul 13, 2017 #2

    PeterDonis

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    It's only apparent. More precisely: if you measure the length of an object moving relative to you, you will get a result that is shorter than what you would get if you were at rest relative to the object. This is just "apparent" and has no effect on the object itself.
     
  4. Jul 13, 2017 #3

    ibkev

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    Many thanks Peter! I will keep that in mind as I read more and maybe have more questions later :)
     
  5. Jul 14, 2017 #4

    Ibix

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    It might be worth pointing out that one way to understand relativity (probably the easiest one) regards objects as four dimensional - three dpatial dimensions and an extent in time. What you think of as a 3d object is actually a 3d cross-section of the 4d object.

    It turns out that observers in relative motion don't share a notion of simultaneity. In our 4d world, this means that they don't agree on the cross-section that they will call the 3d object. So length contraction is related to the fact that a sausage sliced perpendicular to its length has a circular cross-section, but sliced at an angle has an elliptical cross-section. It's still the same sausage. It hasn't been squashed. You're just looking at a slightly different part of it.

    The maths in relativity is a bit different (the Euclidean example of the sausage exhibits length dilation, not length contraction as in relativity). But the basic reasoning is the same.
     
  6. Jul 14, 2017 #5

    ibkev

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    That's a really cool way to think about it - thanks.

    Sausages will now forever remind me of Douglas Adam's mice ("the protrusion into our dimension of vast hyperintelligent pandimensional beings!)

    Ha!
     
  7. Jul 14, 2017 #6

    pervect

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    My first reaction was to say no, but might second reaction is not the same. Unfortunately my answer at this point is A level. If we reach a consensus on the correct answer, it will be worthwhile trying to write a simpler answer.

    The key point is what one might mean by "compressive forces". What immediately comes to mind is that compressive forces would be the pressure part of the stress-energy tensor ##T^{ab}##.

    By this definition, the somewhat surprising answer is that the object under motion would experience compressive forces, i.e. pressure. The stress-energy tensor of an unstressed rod at rest, using geometric units in which c=1, would be:

    $$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

    The pressure term here would be ##T^{11}## which is zero. A presureless rod at rest has zero pressure.

    Next we do a Lorentz boost: ##\tilde{T}^{cd} = \Lambda^c{}_a \Lambda^d{}_b T^{ab}## with

    $$\Lambda^i{}_j = \begin{bmatrix} \gamma & \beta \, \gamma & 0 & 0 \\ \beta \, \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

    The term of interest here is now ##\tilde{T}^{11}## , the only non-zero term of this will be ##\Lambda^1{}_0 \Lambda^1{}_0 T^{00}##. Which will be ## \beta^2 \gamma^2 \rho## by my calculations. This is non-zero, so the answer with this interpretation of "compressive forces" would be yes. And the only rigorous relativistic defintion of "compressive forces" that comes to mind is the appropriate components of the stress-energy tensor.

    One point worth further exploring would be the differences between the physicists conception of the stress-energy tensor and the engineer's concept of the stress tensor. I seem to recall reading something about a different aproach with convective terms, but the details elude me.
     
    Last edited: Jul 14, 2017
  8. Jul 14, 2017 #7

    PeterDonis

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    No, that's compressive stress-energy. Compressive forces would be something that produces nonzero proper acceleration of at least some part of the object. Lorentz transformation does not affect that, so if the object is moving inertially (zero proper acceleration) in one frame, it will be moving inertially (zero proper acceleration) in all frames. So no compressive forces can be associated with length contraction.
     
  9. Jul 14, 2017 #8

    Vanadium 50

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    There are no compressive forces. We don't feel them now, and in certain frames we are moving at almost the speed of light.
     
  10. Jul 14, 2017 #9

    pervect

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    What mathematical entity models the "compressive force"? Is it a tensor, or part of a tensor?
     
  11. Jul 14, 2017 #10

    PeterDonis

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    The path curvature of the worldline (times the rest mass of the object, if you want to use force units instead of acceleration units).
     
  12. Jul 14, 2017 #11

    ibkev

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    I've never heard of compressive stress-energy before. Is it easy to explain? Otherwise, where could I find an intro?
     
  13. Jul 14, 2017 #12

    PeterDonis

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    I don't know if it's a standard term; I just meant by it the "stress" components (i.e., the "space-space" components) of the stress-energy tensor. Those are the ones that capture stress-energy due to pressure and other stresses (as opposed to stress-energy due to energy and momentum density).
     
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