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Relativity of spring constant

  1. Jan 15, 2010 #1
    Relativity of spring "constant"

    Let equal-magnitude, oppositely signed charges be at rest at the Origin and on the y-axis of IRF K. They are held apart by a compressed spring. The force exerted by the spring on either charge is equal and oppositely directed to the electrostatic force.

    Viewed from IRF K’, which moves in the positive x-direction of K at speed v, the charges and the spring move in the –x’ direction at common speed v. But according to the general field transformations the Lorentz force on either charge in K’ is less than it is in K by a factor (1-v^2/c^2)^(1/2). Is the spring constant actually a function of the spring’s motion relative to an IRF? And if so, what is the general rule for arbitrary spring orientations in frame K’?
     
  2. jcsd
  3. Jan 15, 2010 #2
    Re: Relativity of spring "constant"

    If I can understand you correctly, as soon as the spring is released, the charges start to move along the y-axis of IRF K. But the point is that in order to have a relativistic motion, the trajectory of motion in both K and K' must in the the same direction (as is assumed here, the direction of x-axis), while in this example, K' moves in the positive x-direction of K, so it is impossible for an observer in K' to measure any relativistic thing in K.

    AB
     
  4. Jan 15, 2010 #3

    bcrowell

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    Re: Relativity of spring "constant"

    This may be helpful: Grøn, Covariant formulation of Hooke's law, Am. J. Phys. 49, 28-30 ( 1981 )
     
  5. Jan 15, 2010 #4
    Re: Relativity of spring "constant"

    It is not gonna be helpful that much, since its framework is surely GR and in the SR, unfortunately the matter tensor vanishes so no contribution of the spring constant exists anymore. The same topic can be found in a very better and glib language http://arxiv.org/pdf/gr-qc/0005099".

    AB
     
    Last edited by a moderator: Apr 24, 2017
  6. Jan 15, 2010 #5
  7. Jan 16, 2010 #6
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