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Mass in Relativity

by AlbertEinstein
Tags: mass, relativity
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jtbell
#19
Nov25-06, 09:22 AM
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Maybe it will help in this case to think of the momentum in terms of the invariant (rest) mass instead of the relativistic mass:

[tex]p = \frac{m_0 v} {\sqrt{1 - v^2 / c^2}}[/tex]

Then we have

[tex]\Delta x \Delta p = m_0 \Delta x \Delta \left( \frac{v} {\sqrt{1 - v^2 / c^2}} \right)[/tex]

and we can express that complicated [itex]\Delta[/itex] on the right in terms of [itex]\Delta v[/itex]. I'm in a bit of a hurry so I won't do that last step right now.
Meir Achuz
#20
Nov25-06, 04:21 PM
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Quote Quote by AlbertEinstein View Post
But my question reamains unanswered! As in post #8
"By that I suppose that mass is a scalar qty and hence invariant. But the relativistic mass is just due to relative motion between two observers. But then how mass is defined in SR and QM and GR ? Is the definition of mass taken the same in all these fields? Please explain. Principle of equivalence is applied in GR while not in SR and QM why?"

At least please answer my questions.
After reading your posts, I resent your choice of name.
I bet you aren't really who you say you are.
rbj
#21
Nov25-06, 07:46 PM
P: 2,251
Quote Quote by jtbell View Post
Maybe it will help in this case to think of the momentum in terms of the invariant (rest) mass instead of the relativistic mass:

[tex]p = \frac{m_0 v} {\sqrt{1 - v^2 / c^2}}[/tex]

Then we have

[tex]\Delta x \Delta p = m_0 \Delta x \Delta \left( \frac{v} {\sqrt{1 - v^2 / c^2}} \right)[/tex]

and we can express that complicated [itex]\Delta[/itex] on the right in terms of [itex]\Delta v[/itex]. I'm in a bit of a hurry so I won't do that last step right now.
i'll do it.

[tex] \frac{d}{dv} \left( \frac{v} {\sqrt{1 - \frac{v^2}{c^2}}} \right) = \frac{d}{dv} \left( v \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \right)[/tex]
[tex] = -\frac{1}{2} v \left(1 - \frac{v^2}{c^2} \right)^{-3/2} \frac{-2 v}{c^2} + \left(1 - \frac{v^2}{c^2} \right)^{-1/2}[/tex]
[tex] = \frac{v^2}{c^2} \left(1 - \frac{v^2}{c^2} \right)^{-3/2} + \left(1 - \frac{v^2}{c^2} \right)^{-1/2}[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \frac{v^2}{c^2}\left(1 - \frac{v^2}{c^2} \right)^{-1} + 1 \right)[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \left(\frac{c^2}{v^2} - 1 \right)^{-1} + 1 \right)[/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \frac{1+\left(\frac{c^2}{v^2} - 1 \right)}{\frac{c^2}{v^2} - 1} [/tex]
[tex] = \left(1 - \frac{v^2}{c^2} \right)^{-1/2} \left( \frac{1 }{1 - \frac{v^2}{c^2} } \right)[/tex]
[tex] =\left(1 - \frac{v^2}{c^2} \right)^{-3/2}[/tex]


so

[tex] \Delta x \Delta p = \left(1 - \frac{v^2}{c^2} \right)^{-3/2} \Delta x \Delta (m_0 v) [/tex]

looks sorta ugly. dunno if it's right.
pervect
#22
Nov25-06, 08:17 PM
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Quote Quote by AlbertEinstein View Post
Ok then. what should be the correct answer to my question posted -
"Can we mix SR and uncertainty principle? for ex. from uncertainty principle we have [tex]\Delta x \Delta p \geq h/4\pi [/tex]Books then write [tex]m \Delta x \Delta v \geq h/4\pi [/tex]
with an assumption that m can be measured accurately.
However form SR we know that mass depends on velocity; with mass increasing with velocity. Now if we can't measure v exactly how can we measure m exactly?"
I am bit confused with all these discussions. Please help me out.
I'm not quite sure what your question is. Standard undergraduate quantum mechanics (the Schrodinger wave equation, for instance) is not compatible with special relativity, i.e. its not Lorentz invariant.

To get compatibility with SR, you either need to do quantum field theory rather than quantum mechanics (the most useful and most accurate choice but also the most demanding), or you need to do things like replace the Schrodinger equation with the Klein-Gordon equation (and if you have particles with spin, you need to do even more than that, the Klein-Gordon equation is the QM formulation of a spin 0 particle.)

Is your problem purely quantum mechanical, or are you also having problems with the classical aspects of mass? If so, you might want to stick to the classical issues and forget about QM for the time being.

It seems to me from reading your posts that you are "stuck" on the Newtonian idea that momentum = mass * velocity. All you really need to do is to unlearn this, then learn how SR defines momentum (mass * velocity * gamma, where mass is invariant mass), and how quantum mechanics defines momentum - as an operator - (which is a much trickier topic) - you might want to start with

http://en.wikipedia.org/wiki/Momentu...ntum_mechanics

And if you want to tackle QFT, it's got an even different notion of momentum. (Actually, I suppose it might be better to say it's got a different notion of particles. The wikipedia article on QFT explains it reasonably well.)

You also might find it helpful to take a side-trip down classical mechanics into the Lagrangian formulation, where you'll learn that momentum is the partial derivative of the Lagrangian with respect to the velocity. This leads to the Hamiltonian formulation of classical mechanics, which is where more serious treatments of quantum mechanics begin.
AlbertEinstein
#23
Nov26-06, 07:00 AM
P: 113
Originally Posted by Pervect
Is your problem purely quantum mechanical, or are you also having problems with the classical aspects of mass? If so, you might want to stick to the classical issues and forget about QM for the time being.

It seems to me from reading your posts that you are "stuck" on the Newtonian idea that momentum = mass * velocity. All you really need to do is to unlearn this, then learn how SR defines momentum (mass * velocity * gamma, where mass is invariant mass), and how quantum mechanics defines momentum - as an operator - (which is a much trickier topic) - you might want to start with
This is exactly what I wanted to know that how momentum is exactly defined in QM.Thanks guys I suppose I have got an answer to the question.
pervect
#24
Nov26-06, 01:06 PM
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Quantum mechanics is weird (IMO) - a lot weirder than relativity.

But it turns out that if you've got a wavefunction [itex]\Psi(x,t)[/itex], the momentum can be represented by a vector

[tex]
[px,py,pz] = -i \hbar \left[\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right] \Psi
[/tex]

which is the momentum operator that I was talking about. See for instance

http://scienceworld.wolfram.com/phys...mOperator.html

This defintion of momentum doesn't involve velocity at all (which as you point out isn't really defined in quantum mechanics)- just partial derivatives of the wave function. There might be someone in the QM forums who can explain in detail why this works, but it's not me. I just use the results when I need to. I'll add that

http://www.missioncollege.org/depts/...chrodinger.htm

does a somewhat reasonable job of motivating this result (they don't talk about momentum operators specifically but are mainly interested in the Schrodinger equation) starting from the de-Broglie relationship.


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