Understanding the Group and Particle Velocities in Relativistic Wave Mechanics

[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:

<br /> \frac{2 \omega}{c^2} d\omega = 2 k \dk<br /> \Rightarrow \frac{d\omega} {dk} \frac{\omega}{k} = c^2<br />

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"):
<br /> \frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}<br />

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
<br /> \frac{2 \omega}{c^2} d\omega = 2 k dk<br />

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

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 &lt;&lt; m^2 c^2).

 

[phyzmatix]

Cheers for the help gravityandlev! Have a great day!