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

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Discussion Overview

The discussion revolves around the concept of what contributes to the force measured by a scale when a particle with momentum is placed on it. Participants explore the implications of measuring momentum and energy in different frames of reference, particularly in the context of special relativity.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants suggest that in the rest frame of a particle, the energy component of momentum is what is measured by the scale.
  • Others propose that the momentum being non-zero complicates the situation, implying that the particle may not remain stationary on the scale.
  • One participant introduces the idea of gravitational mass being defined under certain conditions, linking it to the equivalence principle and the forces acting in the lab frame.
  • Another viewpoint is that neither energy nor momentum directly pushes on the scale; rather, the scale measures the force exerted by the particle.
  • Participants discuss the relationship between force, acceleration, and the measurements taken in different frames, raising questions about how these quantities interact.
  • Some express confusion regarding the equations presented in linked resources, indicating a need for clearer explanations and connections to the discussion.

Areas of Agreement / Disagreement

There is no consensus on what specifically contributes to the force measured by the scale, with multiple competing views and interpretations of the relationship between energy, momentum, and force. The discussion remains unresolved.

Contextual Notes

Participants express uncertainty regarding the definitions and implications of the terms used, particularly in the context of special relativity. There are mentions of assumptions that may not be fully articulated, and the mathematical steps involved are not universally agreed upon.

Phrak
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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?
 
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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].
 
What if the momentum is non-zero?
 
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Then it won't be in the palm of your hand for very long.
 
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?
 
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I have some notes which might help:

http://www.physics.thetangentbundle.net/wiki/Special_relativity/force
http://www.physics.thetangentbundle.net/wiki/Special_relativity/motion_under_constant_force

The "3-force" is what the experimenter applies to the object. The measurement on the scale, if you will.
 
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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.
 
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?
 
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.
 
  • #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.
 
  • #11
The first link assumes basic familiarity with SR, but I've tried to make it more self contained. Thanks.
 
  • #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.
 
  • #13
Phrak said:
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?
Since you are already starting with the http://en.wikipedia.org/wiki/Four-momentum" is simply dp/dtau.
 
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