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[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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- Thread starter Phrak
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- #1

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[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

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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].

- #3

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What if the momentum is non-zero?

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- #4

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

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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?

[tex]\langle P \rangle ^2 = m^2 = E^2/c^4 - p^2/c^2[/tex], defining

Under restricted conditions,

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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- #6

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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.

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.

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- #9

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[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.

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- #12

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

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