# Group velocity and phase velocity of a matter wave

1. Jun 20, 2013

### 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.

2. Jun 20, 2013

### 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.

3. Jun 20, 2013

### 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.

4. Jun 20, 2013

### Bill_K

You're mistaken.

5. Jun 20, 2013

### andrepd

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.

6. Jun 20, 2013

### Bill_K

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. Jun 20, 2013

### 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.

8. Jun 20, 2013

### 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.

9. Jun 21, 2013

### 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.

10. Jun 21, 2013

### Bill_K

You've said repeatedly that it wasn't.
Good. Now try figuring out the easier approach in #2.

11. Jun 21, 2013

### andrepd

You. are. wrong. I don't know if you still haven't understood what I needed, but you nevertheless made several mistakes.

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.

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

I did not. I assumed vg = ω/k = c2/v.

Correct

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

I searched the thread, and can find no statement of the sort. I find this: