Inertia and momentum, the modern view.

In summary, momentum is proportional to the spacelike rate of change of phase of a quantum mechanical amplitude. Energy is proportional to the timelike rate of change of the phase. The factor of proportionality is Planck's Constant h. We represent a moving object as a superpostion of waves, a wave train or wavegroup. The probability that we will find the object represented by the wavegroup, e.g. an electron, is the absolute square of its amplitude, so the object classically moves at the group velocity. The group velocity is: Vg=pC2/E=kC2/ω. Inertia is m=E/C2 classically or i=&omega
  • #1
Tyger
398
0
The classical quantity momentum is proportional to the spacelike rate of change of phase of a quantum mechanical amplitude. The classical quantity energy is proportional to the timelike rate of change of the phase. The factor of proportionality is Planck's Constant h. We represent a moving object as a superpostion of waves, a wave train or wavegroup. The probability that we will find the object represented by the wavegroup, e.g. an electron, is the absolute square of its amplitude, so the object classically moves at the group velocity. The group velocity is:

Vg=pC2/E=kC2/ω

where p=momentum, E=rest energy, k=wavenumber and ω=frequency and C the speed of light,

so that inertia is m=E/C2 classically or

i= ω/C2 quantum mechanically

As long as we don't change the rest energy or total momentum (wavenumber or total frequency) of a wavegroup it will continue to propagate in a straight, unaccelerated line. The property of inertia arises from the combination of quantum mechanics and special relativity, no deep mystery there.

And that's the modern view.
 
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  • #2
Tyger,
sorry this doesn't make sense to me. How can a wavegroup have just one value of &omega and just one value of k?
 
  • #3
Originally posted by arcnets
Tyger,
sorry this doesn't make sense to me. How can a wavegroup have just one value of &omega and just one value of k?

A wavegroup is a superpostion of several waves with the same value of rest energy (rest frequency) but slightly different values of momentum (wavenumber). The kinetic energy is p2C2/2E, so it differs slightly for each wave component, this causes the "bump" in the interference pattern to move with time, and that is why the object moves. The formula for the group velocity is an average of the different values.

You can find a more complete description in just about any book on quantum mechanics.
 
  • #4
But Tyger, you give a formula which links energy to momentum:
Originally posted by Tyger
Vg=pC2/E
So, how can we have different values of momentum p while energy E is constant?
 
  • #5


Vg=pC2/E

Originally posted by arcnets
But Tyger, you give a formula which links energy to momentum: So, how can we have different values of momentum p while energy E is constant?

That formula gives the group velocity for a wave group with an average momentum p and rest energy E. The individual momenta and kinetic energy for each wave in the group will be different.

I was trying to present the basic idea involved without writing a book on the subject, so some of the details are missing.
 
  • #6


Originally posted by Tyger
The individual momenta and kinetic energy for each wave in the group will be different.

Yes, OK. Now I think I understand. Up to now, I only knew the formula
vg = d(omega)/dk.
Now, since
omega(k) = hbar * E(k)
= hbar * c * ((m0 c)^2 + (hbar k)^2)^(1/2),
we get
vg = hbar * c * 1/2 * ((m0 c)^2 + (hbar k)^2)^(-1/2) * 2 hbar k * hbar
= hbar * c * k ((m0 c)^2 + (hbar k)^2)^(-1/2)
= pc^2/E

So the wavegroup will travel with this velocity if the expectation value (average) of momentum is p. Like you say. And that's indeed the classical particle velocity. OK.

But, since different omegas are involved, the wavegroup will of course disperse very quickly, and the spatial distribution will soon be 'smeared out' over a large area. Well, you can state that it represents a 'moving object'. But not a 'moving particle'. There are no trajectories of particles in QM.
 
  • #7
err, sorry if that's stupid question, but is there in this explanation anything that does not depend on axiom of inertia? Like C is const, h is const, frequency doesn't change on its own, uniformity of time, group velocity...?
 

1. What is the difference between inertia and momentum?

Inertia is the resistance of an object to change its state of motion, while momentum is the quantity of motion an object has.

2. How is inertia related to an object's mass?

Inertia is directly proportional to an object's mass. The greater the mass, the greater the inertia.

3. How does Newton's first law of motion relate to inertia?

Newton's first law of motion states that an object will remain at rest or in motion with constant velocity unless acted upon by an external force. This is essentially describing the concept of inertia.

4. Can inertia and momentum be changed?

Inertia and momentum can be changed by applying an external force to an object. This can cause the object's motion to change, thus altering its inertia and momentum.

5. How does the modern view of inertia and momentum differ from the classical view?

In the modern view, inertia and momentum are considered to be properties of matter, while in the classical view they were seen as separate entities. Additionally, the modern view takes into account the effects of relativity and quantum mechanics on inertia and momentum, whereas the classical view did not.

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