
#19
Nov2506, 09:22 AM

Mentor
P: 11,216

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. 



#20
Nov2506, 04:21 PM

Sci Advisor
HW Helper
P: 1,930

I bet you aren't really who you say you are. 



#21
Nov2506, 07:46 PM

P: 2,265

[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. 



#22
Nov2506, 08:17 PM

Emeritus
Sci Advisor
P: 7,433

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 KleinGordon equation (and if you have particles with spin, you need to do even more than that, the KleinGordon 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 sidetrip 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. 



#23
Nov2606, 07:00 AM

P: 113





#24
Nov2606, 01:06 PM

Emeritus
Sci Advisor
P: 7,433

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 deBroglie relationship. 


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