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(E,p)(E,-p) What pushes on a scale?

  1. May 7, 2008 #1
    You hold in the palm of your hand a classical particle with momentum

    [tex]P = (E,cp_{x},cp_{y}, cp_{up})[/tex] where [tex]cp_{up}=0[/tex] .

    Placing it gently on a scale you ask, what part of

    [tex]P = (E,cp_{x},cp_{y}, cp_{up})[/tex]

    is pushing down on the scale?
     
  2. jcsd
  3. May 7, 2008 #2
    Usually you weigh particles in their rest frame, otherwise they move around too much =) In that case, you are measuring the first component of [tex]P[/tex].

    You might consider the following: A gas of such non-interacting particles, in a box. The average 4-momentum of the fluid is then [tex]\left\langle P \right\rangle = ( \left\langle E \right\rangle, 0, 0, 0)[/tex], so the mass if the gas is going to be [tex]\left\langle E\right\rangle[/tex].
     
  4. May 8, 2008 #3
    What if the momentum is non-zero?
     
    Last edited: May 8, 2008
  5. May 8, 2008 #4
    Then it won't be in the palm of your hand for very long.
     
  6. May 8, 2008 #5
    OK, so backing up a little, in everyday life, where things sit still on a scale.

    [tex]\langle P \rangle ^2 = m^2 = E^2/c^4 - p^2/c^2[/tex], defining m after some units adjustment, and obtaining your equation when [tex]p=0[/tex] .

    Under restricted conditions, m is what could be called gravitational mass.

    But I think you misunderstand me; I really don't know what the general solution is.

    I can guess halfway through it. Using the equivalence principle, the problem is equivalent to asking what resists acceleration a_z, by a force f_z, in the lab frame. So I might guess that it involves the 4-vectors of force and acceleration, whatever they are. Do you think so?
     
    Last edited: May 8, 2008
  7. May 8, 2008 #6
  8. May 8, 2008 #7

    Dale

    Staff: Mentor

    I would say that neither E nor p push on the scale. Instead, the scale pushes on the particle and measures the force with which it pushes.
     
  9. May 8, 2008 #8
    You've got a point Dale. If the scale acts with Fz in the lab frame, what dv/dt, in either frame, does the particle have?
     
  10. May 8, 2008 #9
    A quick perusal of the links I gave would show that the answer to that question is
    [tex]\vec{f} = \frac{1}{\gamma} \frac{d^2 \vec{x} }{d\tau^2} = \frac{1}{\gamma} \frac{d}{d\tau} \left( \gamma \frac{d\vec{x}}{dt} \right) = \frac{d}{dt} \left( \gamma \frac{d\vec{x}}{dt} \right)[/tex]

    Cmon people... no need for spoon feeding.
     
  11. May 8, 2008 #10
    lol, lbrits. I need spoon feeding. To be honest, the equations in those links were pretty haphazardly done, but for review purposes. The deductive chain was obscure.
     
  12. May 8, 2008 #11
    The first link assumes basic familiarity with SR, but I've tried to make it more self contained. Thanks.
     
  13. May 8, 2008 #12
    Sorry, I didn't know you wrote it. I hope you can take it as constructive criticism from someone who hasn't delt with SR in a while, and didn't devote a great deal of time reading it. To be fair, I've been involved in another persuit that takes up what little brain power my wee brain can handle.
     
  14. May 9, 2008 #13

    Dale

    Staff: Mentor

    Since you are already starting with the four-momentum it is pretty easy. The four-force is simply dp/dtau.
     
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