Group velocity and phase velocity of a matter wave

[andrepd]

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 \omega^{2}=k^{2}c^{2}+m^{2}c^{4}/\hbar^{2} 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.

 

[Bill_K]

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.

 

[andrepd]

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 = \sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}} to vf=c^2/v.

 

[Bill_K]

If I'm not mistaken

You're mistaken.

 

[andrepd]

You're mistaken.

Care to explain then? dω/dk=d/dk(\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

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

EDIT: This, however, presumes previous knowledge that vf=c^2/v. Hence my question.

 

[Bill_K]

Care to explain then? =\frac{k}{\omega}c^{2}=\frac{v}{c^{2}}*c^2=v
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 + m²c⁴/ħ², and its derivative is 2 k c². That gives what you want immediately.

 

[andrepd]

This yields the the same result for the *group* velocity. However, I am having trouble with the derivation of the *phase* velocity, \omega/k, which, in the context of matter waves, is not equal to v, but rather to c^2/v.

 

[Bill_K]

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.

 

[andrepd]

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 \omega/k=\sqrt{\frac{k^{2}c^{2}+m^{2}c^{4}/\hbar^{2}}{k^{2}}}=c^{2}/v

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-v^{2}/c^{2})\frac{c^{4}}{v^{2}}}=\sqrt{c^{2}+\frac{c^{4}}{v^{2}}-c^{2}}=c^{2}/v

That was it.

 

[Bill_K]

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.

 

[andrepd]

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. v_p = 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,