Group velocity and phase velocity of a matter wave

In summary: I've figured it now. For the record:\sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}}=\sqrt{c^{2}+\frac{m^{2}c^{4}}{k^{2}\hbar^{2}}}=\sqrt{c^{2}+\frac{m^{2}c^{4}}{\gamma^{2}m^{2}v^{2}}}=\sqrt{c^{2}+(
  • #1
andrepd
10
0
Hi. Today I sat my final first year Modern Physics exam. It went very well, however I got stuck in one question. It asked (i) to prove the following relation for the matter wave [itex]\omega^{2}=k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}[/itex] and (ii) to obtain the group velocity and phase velocity of a matter wave from that relation. (i) was easy, so was obtaining the group velocity, but I got stuck on obtaining c^2/v for the phase velocity. If someone could walk me thorough that deduction, I would appreciate it. Thanks.
 
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  • #2
The phase velocity is v = ω/k. The group velocity is g = dω/dk. Differentiate the relationship you have, and you get 2 ω dω/dk = 2k c2, which immediately gives you vg = c2.
 
  • #3
If I'm not mistaken, differentiating our expression for ω(k) yields vg=(k/ω)c^2, which, if we know beforehand that the phase velocity equals c^2/v, can express as vg=v. However, I'm unsure how we get from vf = ω/k = [itex]\sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}}[/itex] to vf=c^2/v.
 
  • #4
andrepd said:
If I'm not mistaken
You're mistaken.
 
  • #5
Bill_K said:
You're mistaken.

Care to explain then? dω/dk=d/dk[itex](\sqrt{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}})=1/2*(k^{2}c^{2}+m^{2}c^{4}/\hbar^{2})^{-1/2}*2kc^{2}=\frac{k}{\omega}c^{2}=\frac{v}{c^{2}}*c^2=v[/itex]

This was my thought process. Care to point my mistake?

EDIT: This, however, presumes previous knowledge that vf=c^2/v. Hence my question.
 
  • #6
andrepd said:
Care to explain then? [itex]=\frac{k}{\omega}c^{2}=\frac{v}{c^{2}}*c^2=v[/itex]
This was my thought process. Care to point my mistake?.
In the last step you've used v = k/ω, whereas actually v = ω/k.

It's easier anyway to do it the way I showed. Do NOT take the square root. Simply differentiate the given relationship as it stands. The LHS is ω2, and its derivative is 2 ω dω/dk. The RHS is k2c2 + m2c42, and its derivative is 2 k c2. That gives what you want immediately.
 
  • #7
This yields the the same result for the *group* velocity. However, I am having trouble with the derivation of the *phase* velocity, [itex]\omega/k[/itex], which, in the context of matter waves, is not equal to v, but rather to c^2/v.
 
  • #8
The phase velocity is always ω/k, andrepd, I don't care what the context. It's the group velocity that may differ. This is a general fact for wave motion of any kind: sound waves, water waves, gravity waves, matter waves or what have you. The phase velocity is the velocity at which a wave crest travels, and the expression for it just comes from taking exp(i(kx - ωt)) and rewriting it as exp(ik(x - vt)), from which v = ω/k. If you're still not clear, I suggest you read the Wikipedia page.
 
  • #9
I know that, however, that is now what I was looking for. I know that the phase velocity equals ω/k. Again, I wanted to know how I can prove this relation [itex]\omega/k=\sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}}=c^{2}/v[/itex]

I've figured it now. For the record:

[tex]\sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}}=\sqrt{c^{2}+\frac{m^{2}c^{4}}{k^{2}\hbar^{2}}}=\sqrt{c^{2}+\frac{m^{2}c^{4}}{\gamma^{2}m^{2}v^{2}}}=\sqrt{c^{2}+(1-v^{2}/c^{2})\frac{c^{4}}{v^{2}}}=\sqrt{c^{2}+\frac{c^{4}}{v^{2}}-c^{2}}=c^{2}/v[/tex]

That was it.
 
  • #10
I know that the phase velocity equals ω/k
You've said repeatedly that it wasn't.
I've figured it now.
Good. Now try figuring out the easier approach in #2.
 
  • #11
You. are. wrong. I don't know if you still haven't understood what I needed, but you nevertheless made several mistakes.

"vg = c2"
It's not, according to the very wikipedia article you linked me to, vg = c2/v, where vg is the group velocity, vp is the phase velocity, and v is the particle velocity.

"You're mistaken."
I most certainly wasn't, seeing as your next post contained the exact same differentiation I made, calculated in a slightly different way.

"In the last step you've used v = k/ω, whereas actually v = ω/k."
I did not. I assumed vg = ω/k = c2/v.

"The phase velocity is always ω/k"
Correct

"v = ω/k"
Incorrect. the phase velocity is ω/k, not the particle velocity. Once again. vp = ω/k = c2/v

"You've said repeatedly that [phase velocity] wasn't [equal to] ω/k."
I searched the thread, and can find no statement of the sort. I find this:
derivation of the *phase* velocity, ω/k,
 

1. What is the difference between group velocity and phase velocity of a matter wave?

The group velocity of a matter wave refers to the speed at which the overall shape or envelope of the wave travels. The phase velocity, on the other hand, refers to the speed at which the individual crests or troughs of the wave travel. In some cases, these two velocities may be equal, but in others, they can differ significantly.

2. How are group velocity and phase velocity related?

The group velocity and phase velocity are related by the dispersion relation, which describes the relationship between the frequency and wavelength of a wave. The group velocity is equal to the derivative of the phase velocity with respect to the wave number.

3. Can the group velocity of a matter wave be greater than the speed of light?

Yes, the group velocity of a matter wave can exceed the speed of light in certain circumstances. This is known as superluminal group velocity and is possible due to quantum effects. However, the phase velocity of a matter wave can never exceed the speed of light.

4. How do the group velocity and phase velocity of matter waves differ from those of electromagnetic waves?

The main difference between the group and phase velocities of matter waves and electromagnetic waves is that the former are dependent on the mass and energy of the particles, while the latter are determined by the properties of the medium through which they are traveling. Additionally, matter waves can have a group velocity greater than the speed of light, while this is not possible for electromagnetic waves.

5. What are the practical applications of understanding group and phase velocity of matter waves?

Understanding the group and phase velocity of matter waves is crucial for fields such as quantum mechanics, solid-state physics, and particle physics. It allows scientists to study and manipulate the behavior of particles at the quantum level, leading to advancements in technologies such as transistors, lasers, and quantum computing. Additionally, studying matter wave velocities can provide insights into the behavior of fundamental particles and the laws of physics governing their behavior.

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