# Group and particle velocities

## 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

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Anybody? 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

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 $$c^2$$.

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... 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.

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?

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 $$\hbar^2 k^2 << m^2 c^2$$).

Cheers for the help gravityandlev! Have a great day!