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

1. May 7, 2008

Phrak

You hold in the palm of your hand a classical particle with momentum

$$P = (E,cp_{x},cp_{y}, cp_{up})$$ where $$cp_{up}=0$$ .

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

$$P = (E,cp_{x},cp_{y}, cp_{up})$$

is pushing down on the scale?

2. May 7, 2008

lbrits

Usually you weigh particles in their rest frame, otherwise they move around too much =) In that case, you are measuring the first component of $$P$$.

You might consider the following: A gas of such non-interacting particles, in a box. The average 4-momentum of the fluid is then $$\left\langle P \right\rangle = ( \left\langle E \right\rangle, 0, 0, 0)$$, so the mass if the gas is going to be $$\left\langle E\right\rangle$$.

3. May 8, 2008

Phrak

What if the momentum is non-zero?

Last edited: May 8, 2008
4. May 8, 2008

lbrits

Then it won't be in the palm of your hand for very long.

5. May 8, 2008

Phrak

OK, so backing up a little, in everyday life, where things sit still on a scale.

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

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
6. May 8, 2008

lbrits

I have some notes which might help:

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

The "3-force" is what the experimenter applies to the object. The measurement on the scale, if you will.

Last edited by a moderator: May 3, 2017
7. May 8, 2008

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.

8. May 8, 2008

Phrak

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?

9. May 8, 2008

lbrits

A quick perusal of the links I gave would show that the answer to that question is
$$\vec{f} = \frac{1}{\gamma} \frac{d^2 \vec{x} }{d\tau^2} = \frac{1}{\gamma} \frac{d}{d\tau} \left( \gamma \frac{d\vec{x}}{dt} \right) = \frac{d}{dt} \left( \gamma \frac{d\vec{x}}{dt} \right)$$

Cmon people... no need for spoon feeding.

10. May 8, 2008

Phrak

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. May 8, 2008

lbrits

The first link assumes basic familiarity with SR, but I've tried to make it more self contained. Thanks.

12. May 8, 2008

Phrak

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. May 9, 2008

Staff: Mentor

Since you are already starting with the http://en.wikipedia.org/wiki/Four-momentum" [Broken] is simply dp/dtau.

Last edited by a moderator: May 3, 2017