Group and particle velocities

[phyzmatix]

Homework Statement

In relativistic wave mechanics the dispersion relation for an electron of velocity $$v=\frac{\hbar k}{m}$$ is given by $$\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}$$ where c is the velocity of light, m is the electron mass (considered constant at a given velocity) $$\hbar=\frac{h}{2\pi}$$ and h is Planck's constant.

Show that the product of the group and particle velocities is $$c^2$$

Homework Equations

$$v_g=\frac{d\omega}{dk}$$

The Attempt at a Solution

From the dispersion relation I got

$$\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}$$
$$\omega = c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}$$

so that

$$v_g=\frac{d\omega}{dk}$$

$$v_g=\frac{d}{dk}(c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}})$$

$$v_g=\frac{ck}{\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}}$$

But this answer, multiplied with the particle velocity will obviously not give c^2. What am I missing?

Thanks!
phyz

 

[phyzmatix]

Anybody?

 

[Sebbie]

you got the same problem as me. I did notice that the particle velosity was the velosity about equilibrim position not through the medium (cant remember the page no. Waves and Vibrations). but I didnt get much further. also try wikipedia

 

[gravityandlev]

Try using implicit differentiation on your dispersion relation:

$$\frac{2 \omega}{c^2} d\omega = 2 k \dk \Rightarrow \frac{d\omega} {dk} \frac{\omega}{k} = c^2$$

$$d\omega/dk$$ is the group velocity and $$\omega/k$$ is the phase velocity. The product of the two is $$c^2$$.

 

[phyzmatix]

Ye gods! I have a bite!

Thank you kindly for the reply gravityandlev, but I'm afraid you'll have to dumb it down for me a bit please. I'm not sure I follow...

 

[gravityandlev]

Sorry. I gave a quick, unhelpful reply, and it had a typo in it to boot.

I meant that you could start with your relation between frequency $$\omega$$ and wavevector $$k$$ (which we usually call the "dispersion relation"):
$$\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}$$

and do an implicit differentiation (http://en.wikipedia.org/wiki/Implicit_differentiation#Implicit_differentiation). That's where you take the derivative of all terms containing $$\omega$$ with respect to $$\omega$$ and all terms containing $$k$$ with respect to $$k$$.

That way $$\omega^2$$ becomes $$2 \omega d\omega$$ and $$k^2$$ becomes $$2 k dk$$. The constant term $$\frac{m^2 c^2}{\hbar ^2}$$ does not contribute to the derivative.

So implicit differentiation of your dispersion relation gives
$$\frac{2 \omega}{c^2} d\omega = 2 k dk$$

and you can rearrange to get
$$\Rightarrow \frac{d\omega} {dk} \frac{\omega}{k} = c^2 .$$

The quantity $$d\omega/dk$$ is your group velocity. The term $$\omega/k$$ is called the "phase velocity". It is the velocity at which a single wave of frequency $$\omega$$ and wave vector $$k$$ would propagate. So in this case you interpret it as the particle velocity.

 

[phyzmatix]

Thank you so much for your help! I would never have got this from my textbook alone...

Two last questions though if I may:

1. Does this mean that the part "...an electron of velocity $$v=\frac{\hbar k}{m}$$" really contributes nothing to the question?

2. How would the particle velocity as derived by your method be related to this given electron velocity?

 

[gravityandlev]

The phrasing of the question actually seems a little wrong to me. Generally what we call the "group velocity" actually IS the velocity a particle would move at.

The expression $$v = \hbar k/m$$ is only true in the non-relativistic limit (take your expression above for $$v_g$$ and consider the limit $$\hbar^2 k^2 << m^2 c^2$$).

 

[phyzmatix]

Cheers for the help gravityandlev! Have a great day!