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Hooke's law and SR

  1. Sep 3, 2007 #1
    A spring pushed or pulled with a force F is elongated by x according to Hooke's law:

    F = -k . x

    where k is a constant of the spring.

    Does the k constant changes when seen from a reference frame at rest ?
    The spring is moving.

    This comes from "https://www.physicsforums.com/showthread.php?t=182175" where I didnt get any answer.
    Thank you.
  2. jcsd
  3. Sep 3, 2007 #2
    Hooke's law is a rough approximation which does not apply to all springs. Relativistic effects are tiny compared to the error in Hooke's law.
  4. Sep 3, 2007 #3
    This is not the question.
    The question is: are there any relativistic effects ?
    I dont think they are tiny al 0.999 c.
  5. Sep 3, 2007 #4


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    I would rather consider that the spring has some spring constant k in the rest frame, and ask what it looks like in the moving frame.

    The simplest case is where the spring is perpendicular to the direction of motion of moving frame, i.e. the direction of the boost.

    We can analyze the problem conveniently but possibly not in the most elementary terms by considering how the stress-energy tensor of the spring transforms. See for example Rindler, "Intro to SR", 2nd ed, pg 133 for the formula for the transformation of the stress-energy tensor (which I am refeing to while typing this up).

    First of all, we must realize that [itex]T^{11}[/itex] is the quantity of primary interest, the force / unit area. See for instance

    http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec12.html [Broken]

    for why [itex]T^{11}[/itex] is the component of the stress energy tensor that is equal to the force / unit area, assuming that [itex]x^{1}[/itex] points in the direction along the spring.

    In the case where the spring is perpendicular to the boost, [itex]T^{11}[/itex] remains the same.

    However, the area is multiplied by a factor of [itex]1/\gamma[/itex], where [itex]\gamma[/itex] is a number greater than one given by the usual formula

    [tex]\gamma = \frac{1}{\sqrt{1-\frac{v}{c}^2}}[/tex]

    The length of the spring remains unchanged.

    So we see that the spring constant, which is force / length, goes down by a factor of gamma. This is sufficient to show that the spring constant does change when you change reference frames, i.e we have for the perpendicular case:

    [tex]k' = \frac{k}{\gamma}[/tex]

    This may be enough for your purposes, but let us also consider the case where the spring is parallel to the boost.

    The case where the spring is parallel to the direction is much more complicated, and we may not have enough information to solve the problem unless we know the density of the material that the spring is made of.

    The stress energy tensor transforms as:

    [tex]T^{'11}= \gamma^2 \left( T^{11}+\rho v^2 \right)[/tex]

    We introduce here [itex]\rho[/itex] aas the proper density (mass per unit volume) of the spring.

    The area of the spring remains unchanged.

    The length of the spring is shorter due to Lorentz contraction, by a factor of [itex]1/\gamma[/itex]. Let L be the proper length of the spring, and A be the proper cross-sectional area of the spring. Then, if I'm doing all this math correctly, when we put this altogether we get

    [tex]k' = \gamma^3 \left( \frac{T^{11} A}{L} + \frac{\rho A v^2}{L} \right) = \gamma^3 \frac{T^{11} A}{L} \left( 1 + \frac{\rho v^2}{T^{11}} \right) [/tex]

    and since k = [itex]T^{11} A / L [/itex]

    [tex]k' = \gamma^3 k \left( 1 + \frac{\rho v^2}{T^{11}} \right) [/tex]

    This is complicated enough and non-intuitive enough that I may have made an error somewhere and the math should be double checked. (I'm fairly confident in the formulas I gave for how the stress-energy tensor transforms, the remaining steps are what needs to be checked in detail). The previous simpler case was good enough to show that k does not remain constant, which answers the original question.
    Last edited by a moderator: May 3, 2017
  6. Sep 3, 2007 #5
    Hooke's law

    Have a look at

    Covariant formulation of Hooke's law
    Ø. Grøn
    Am. J. Phys. 49, 28 (1981) Full Text: [ PDF (247 kB)]
  7. Sep 4, 2007 #6


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    There's at least one meta- point I should probably make. I'm assuming that we have a static spring that we know (at least approximately) obeys Hooke's law in the laboratory, which is a reasonable assumption. I'm then computing the static behavior of the spring as seen by a moving frame.

    It would be a mistake to apply this analysis to a dynamic situation (such as for instance a spring/mass clock) without more refinement. Specifically, one is likely to run into simultaneity issues, in the Newtonian model the force is present on both ends of the spring at the same time, in the relativistic case one knows in advance that this can't be correct.

    If you want to analyze a mechanical clock, you need a more sophisticated analysis than the one I've done. I would suggest using a continuous Lagrangian approach, but it should also be possible to use an approach similar to Greg Egan's


    (also )

    and there's some relevant material buried in the huge thread

    specifically, I'd suggest:
    Relatistic elastodynamics, Wernig Piilchler
    http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Agr-qc%2F0403073 [Broken]

    Finding the proper volume element to integrate the Lagrangian density over can be very tricky! The best approach I found was that any small volume of the elastic material that has some volume dV in its proper frame would be stretched by the "stretch factor" s (discussed in some of the above links), then multiplied by a factor of 1/gamma due to its velocity (if it's vibrating, it's moving, and this effect is important to model to get the right answer).

    The basic Lagrangian density that describes an idealized spring is fairly simple, though - it's just -sqrt(rho_0), where rho_0 is the density of the material in its rest frame. The Lagrangian transforms as a scalar quantity which is different from how the actual density transforms (which transforms as one component of a rank 2 tensor).

    One can realize in advance that an isotropic material, seen from a relativistic moving frame, is no longer going to behave isotropically from the Lorentz transforms.
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  8. Sep 4, 2007 #7
    am i mistaken in simply saying that the oscillations will be slowed down when viewed from another inertial frame via time dilation?

    i seem to recall deriving the time dilation effect by considering light bouncing between two mirrors (i.e. a light clock) as seen from a rest frame and an non-rest inertial frame..couldn't this same HO, given it's periodic motion, be seen as a clock in this case? in which case the result is time dilation slowing the period of the HO.
  9. Sep 4, 2007 #8


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    The oscillations should wind up being "slowed down", though I haven't done the calculations, that's what I would expect.

    Furthermore, the length of the oscillating gadget will change depending on how you rotate it, so the amplitude of the oscillations will also change.

    Doing this with light, as in the light clock, is a standard exercise. Doing it with matter gets a lot more involved as you can see if you read some of the links :-(.
    Last edited: Sep 4, 2007
  10. Sep 4, 2007 #9
    Thanks to all, especially to pervect.
    The answers, and the links, are far away the scope of my knowledge, although they will be helpful for others.

    What does HO mean ?

    1 - Its a wrong approach to say that the period T of any oscillator must be slowed down by the same factor, when viewed from a frame of reference at rest ?

    2 - Its a wrong approach to say that any transversal movement ( relative to the boost ) must be viewed the same when viewed from a frame of reference at rest or from a frame of reference moving ( in the direction of boost )?

    Cant we use 1 and 2 the same way we use the conservation of energy or momentum in order to resolve any problem ?

    A short answer is enough.
  11. Sep 4, 2007 #10


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    No, it's not wrong to say that the period of any oscillator must be slowed down by the same factor. In fact, if it were not, we'd have a way of distinguishing "absolute motion", which we already know is not possible.

    Verifying or demonstrating this gets very messy, though. But if I got any other answer, I'd know that I had made an error somewhere in the calculation.

    As far as viewing transversal movement, assume we have an unboosted frame S with coordinates (t,x,y) and a boosted frame S', with coordinates (t', x',y'). Let the boost be in the x direction. We can then write y' =y via the Lorentz transforms. But we can't write t'=t.

    "Viewed the same" is a bit ambiguous given that y'=y, but t' is not equal to t.
  12. Sep 5, 2007 #11
    spring constant transformation

    We can analyze the problem conveniently but possibly not in the most elementary terms by considering how the stress-energy tensor of the spring transforms
    I think that would be the case if we express the elasticity law as a function of Young's modulus and the surface of the transversal section.
    Why not work with the transformation of force and lengths?
    Thanks for your answer
    Last edited by a moderator: May 3, 2017
  13. Sep 5, 2007 #12
    spring constant transformation

    As I see you consider that the spring is located along the OX axis. Let I(0) its rest frame which moves with speed u relative to u and with u' relative to I', I' moving with V relative to I all in the positive direction of the overlapped axes,
    u=(u'+V)/(1+u'V/cc) (1)
    We have
    F(0)=k(0)x(0) (2)
    in I(0), all proper values.
    In I we have
    F=kx (3)
    whereas in I' we have
    F'=k'x'. (4)
    Invariance of the OX compoents requires
    F(0)=F=F'. (5)
    Combining (2) and (3) and taking into account (4) we have
    The problem is to find out a physical meaning for k'u'. dE'=k'x'dx'....
    I come on the Forum for learning from others and so please leat me know where I am wrong. Thanks
  14. Sep 5, 2007 #13
    These are the kind of answers that I understand. Thanks

    But, I will try to define "Viewed the same". You can view a video at 2 times the velocity or 2 times slower, or even sometimes quickier or slower in a continuous change of velocity, but its always the same video. This will apply only to the movements on the axis transversal to the axis of motion.

    I expect you are not referring to me. If this is the case you are completely wrong.

    But reading your post is very difficult to understand anything. I dont know what you want to prove, or show me.
  15. Sep 7, 2007 #14
    k transformation

    I consider that k when the spring is located along the X axis transforms as

    with k in I and k' in I'. Where am I wrong?
  16. Sep 7, 2007 #15
    To bernhard.rothenstein:
    I read your last post many times and I dont understand a bit.
    I dont know if you are talking about 2 or 3 frames of reference.

    This is what Im talking about:
    A spring located along the X axis moves along the Y axis with velocity v respect to a frame of reference ( "at rest" ).

    The spring, at rest, has a k constant.
    The moving spring, as seen from the frame at rest has a k' constant.

    The relation between k and k' must be:

    k' = k / sqrt(1-VV/cc) ( 0 )

    Why ? Because a clock made up of a mass M and a spring whit constant k must oscillate with a period T when the clock is at rest and T' when the clock is moving and seen by a frame of reference at rest, such that:

    T' = T / sqrt(1-VV/cc) ( 1 ) ( time dilation )

    M and M' are related by:

    M' = M . sqrt(1-VV/cc) ( 2 )

    The period T depends upon M and k:

    sqrt (k/M) ( 3 )

    From ( 1 ), ( 2 ) and ( 3 ) we obtain k' = k / sqrt(1-VV/cc) ( 0 )
  17. Sep 7, 2007 #16


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    If the spring at rest has a spring constant of k, the moving spring should have a weaker spring constant. This is the result I got earlier, and it's simple enough that I don't think there is any issue here.

    In terms of the stress-energy tensor, the tension/unit area in the spring shouldn't change, but the area is lower, hence the force is lower. (There are other routes to the same conclusion as well).

    But that means that [itex]k' = \sqrt{1-\beta^2} k[/itex], where[itex]\beta[/itex] = v/c, which is different from your result, unless I'm misinterpreting some variable name.

    I.e the spring constant gets divided by gamma, your result multiplies it by gamma (if I'm reading it correctly?).

    I'll defer the other case until I'm sure we agree about this one.

    A slightly more complex issue is the issue of mass. The spring is weaker by a factor of gamma, but to get the proper period, the "mass: must also be heavier by a factor of gamma. But I believe that the "mass" here is the so-called transverse mass, not the relativistic mass or the invariant mass. I.e. the Newtonian derivation assumes

    F = ma = kx

    but the mass m in F = m a is really the "transverse mass", see for instance


    in the appendix "What is the relativistic version of F=ma".

    Why do I say the mass must be heavier? Because the newtonian oscillation frequency is proportional to sqrt(k/m), so k being smaller by a factor of gamma doesn't give the proper behavior alone, m must also scale.

    Note this means that acceleration decreases by a factor of gamma^2, this is right because acceleration is meters/second^2.

    Much of the problem is in using Newtonian ideas to solve a relativistic problem. But you need to do something like the relativistic Lagrangian approach I'm writing about in another thread to really handle the dynamics correctly, IMO. Unfortunately, that's a bit advanced for this thread :-(.
    Last edited: Sep 7, 2007
  18. Sep 7, 2007 #17
    k transformation

    As I understand from the way in which you state the problem, the spring is at rest in I' and moves relative to I with speed u. Then I' is the rest frame and I is the reference frame relative to which it moves with speed u in the positive direction of the overlapped axes I and I' being in the standard arrangement. The physical quantities measured in I' are presented with an index (0). We have in I'
    F(0)=k(0)x(0) (1)
    F=kx (2)
    neglecting the sign. The invariance of the O'X' requires that
    F=F(0) (3)
    and so, taking into account the length contraction we obtain
    k=k(0)x(0)/x=k(0)/sqrt(1-uu/cc). (4)
    Consider a third inertial reference frame K in which the spring is at rest. K moves with speed U relative to I and with U' relative to I, I' moving with V relative to I.
    In order to be able to continue, please restate the problem mentioning clearly the way in which the spring moves.
    If I am "completly wrong" please tell me where using "soft words and hard arguments".
  19. Sep 7, 2007 #18
    k transformation

  20. Sep 8, 2007 #19
    To bernhard.rothenstein:
    I give it up. But:

    No, I is the system of reference at rest and M, k, T are the constants at rest.
    The spring moves with I'.
    I' moves with speed u relative to I and M', k' and T' are the constants when seeing from I.

    There is no length contraction, the spring is 90º to the direction of movement

    a mistake
    M' = M / sqrt(1-VV/cc) because mass increases with speed.
  21. Sep 8, 2007 #20
    Thanks. Was all you say there, comprised in your initial thread? Did you mention there the oscillations of the spring?
    I have considered the spring along the OX axis and moving along it. I have also considered the rest frame of the spring which moves with u and u' relative yo I and I' respectively involving the invariance of the OX component of the force. I have derived transformation equation for k in I and K(0) in the rfest frame and between k' in I'and k(0). Eliminating k(0) I have obtained the transformation between k and k'.
    will have a look at that new facet of your problem which has large pedagogical power.
  22. Jan 3, 2011 #21
    Is there any link which gives some direct information of elastodynamics??
    Really interested to study that
  23. Jan 4, 2011 #22


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  24. Jan 5, 2011 #23
    THANKS, Pervect i visite the link you provided that was very informative, actually the situation is i want to teach elasto dynamics to BS students volentierly so that i can learn about elastodynamics ,, i need is lecture notes or book that can give me full material of my course from basic to advance level.....

    thanks again
  25. Jan 5, 2011 #24
    Here is a quick analysis using simpler methods to the earlier ones in this thread.

    The static force in spring (actually any force) is know to transform as:

    [tex]f_{\perp} = f_0 \gamma^{-1} [/tex]

    where [tex]f_{\perp} ' [/tex] is the force perpendicular to the relative motion of the spring and [tex] f_0 [/tex] is the proper force in the rest frame of the spring.

    For the parallel case:

    [tex]f_{\parallel}' = f_0 [/tex]

    The spring constant in the rest frame of the spring with length [itex]L_0[/itex] is:

    [tex]k_0 = -f_o /L_0 [/tex]

    Under transformation and allowing for length contraction we get:

    [tex]k_{\perp} ' = -\frac{f_{\perp} '}{L_{\perp} '} = -\frac{(f \gamma^{-1})}{L} = k_0 \gamma^{-1}[/tex]

    (which agrees with the result pervect got for the transverse case) and:

    [tex]k_{\parallel} ' = -\frac{f_{\parallel} '}{L_{\parallel} '} = -\frac{f }{(L \gamma^{-1})} = -\frac{f \gamma}{L} = k_0 \gamma[/tex]

    (This last result agrees with a result obtained by alvaros in post #15 using his oscillating method https://www.physicsforums.com/showpost.php?p=1421275&postcount=15)

    It is interesting to see how this fits in with Young's modulus (Y) that pertains to the elastic properties of the material the spring is made of.
    This is defined in the rest frame as:

    [tex]Y_{0} = \frac{k_0 L_0}{A_0} [/tex]

    where [itex]A_0[/itex] is the cross sectional area of the spring.

    Under transformation for the transverse case we get:

    [tex]Y_{\perp} ' = \frac{k_{\perp} ' L_{\perp} ' }{A_{\perp} '} = \frac{(k_0 \gamma^{-1}) (L_0)}{(A_0 \gamma^{-1})} = Y_0 [/tex]

    and for the parallel case we get:

    [tex]Y_{\parallel} ' = \frac{k_{\parallel} ' L_{\parallel} ' }{A_{\parallel} '} = \frac{(k_0 \gamma) (L_0 \gamma^{-1})}{A_0} = Y_0 [/tex]

    so it appears Young's modulus is invariant under transformation and independent of orientation to the motion.

    Put another way, IF Young's modulus is invariant and a scalar, THEN [tex]k_{\parallel} ' = k_0 \gamma[/tex]

    Does that seem about right?
    Last edited: Jan 5, 2011
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