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Group and particle velocities

  • Thread starter phyzmatix
  • Start date
  • #1

Homework Statement

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

Show that the product of the group and particle velocities is [tex]c^2[/tex]

Homework Equations


The Attempt at a Solution

From the dispersion relation I got

[tex]\frac{\omega ^2}{c^2}=k^2 + \frac{m^2 c^2}{\hbar ^2}[/tex]
[tex]\omega = c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}[/tex]

so that


[tex]v_g=\frac{d}{dk}(c\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}})[/tex]

[tex]v_g=\frac{ck}{\sqrt{k^2 + \frac{m^2c^2}{\hbar ^2}}}[/tex]

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


Answers and Replies

  • #2
Anybody? :smile:
  • #3
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
  • #4
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

[tex] d\omega/dk[/tex] is the group velocity and [tex] \omega/k [/tex] is the phase velocity. The product of the two is [tex] c^2 [/tex].
  • #5
Ye gods! I have a bite! :biggrin:

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... :confused:
  • #6
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 [tex] \omega [/tex] and wavevector [tex] k [/tex] (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 [tex] \omega [/tex] with respect to [tex] \omega [/tex] and all terms containing [tex] k [/tex] with respect to [tex] k [/tex].

That way [tex] \omega^2 [/tex] becomes [tex] 2 \omega d\omega [/tex] and [tex] k^2 [/tex] becomes [tex] 2 k dk [/tex]. The constant term [tex] \frac{m^2 c^2}{\hbar ^2} [/tex] 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 [tex] d\omega/dk [/tex] is your group velocity. The term [tex] \omega/k [/tex] is called the "phase velocity". It is the velocity at which a single wave of frequency [tex] \omega [/tex] and wave vector [tex] k [/tex] would propagate. So in this case you interpret it as the particle velocity.
  • #7
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 [tex]v=\frac{\hbar k}{m}[/tex]" really contributes nothing to the question?

2. How would the particle velocity as derived by your method be related to this given electron velocity?
  • #8
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 [tex] v = \hbar k/m [/tex] is only true in the non-relativistic limit (take your expression above for [tex] v_g [/tex] and consider the limit [tex] \hbar^2 k^2 << m^2 c^2 [/tex]).
  • #9
Cheers for the help gravityandlev! Have a great day!

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