# Relativistic de Broglie Wavelength

1. Mar 25, 2005

### metrictensor

Is there a relativistic formulation of the de Broglie wave length equation?

2. Mar 25, 2005

### dextercioby

What do you mean...?U mean if one can use in

$$\lambda_{pilot wave}=\frac{h}{p_{particle}}$$

the relativistic expression of momentum

$$p=\gamma m v$$

,or something totally different...?

Daniel.

3. Mar 25, 2005

### Staff: Mentor

It's the same basic equation as the non-relativistic one:

$$\lambda = \frac {h}{p}$$

Simply use the relativistic momentum

$$p = \frac {mv} {\sqrt {1 - v^2 / c^2}}$$

instead of the non-relativistic $p = mv$.

[added] Argh, I've got to learn to type faster!

4. Mar 25, 2005

### metrictensor

if you use the relativistic invariant

$$E^{2}=p^{2}c^{2}+m^{2}c^{4}$$

Are there equations representing the wave form of energy and momentum that satisfy this equation for a massive particle.

5. Mar 25, 2005

### dextercioby

De Broglie's relations

$$E_{particle}=\hbar \omega_{pilotwave}$$ (1)

$$\vec{p}_{particle}=\hbar \vec{k}_{pilotwave}$$ (2)

allow us to pass from

$$e^{\frac{1}{i\hbar}p^{\mu}x_{\mu}}$$ (3) (typical for a particle,see QM)

to

$$e^{-ik^{\mu}x_{\mu}$$ (4) (typical for a wave)

and therefore describing particles and de Broglie waves unitarily.

Daniel.

6. Mar 26, 2005

### metrictensor

If we define relativistic momentum as $$\gamma mv$$ and apply that to the relativistic energy-momentum invariant equation we get $$E^2=h^2f^2 \bigg(\frac{c^2}{v^2}\bigg)$$. When $$v$$ goes to 0 this should reduce to the rest energy but we get an infinity here.

7. Mar 26, 2005

### metrictensor

If we define relativistic momentum as

$$\gamma mv=\frac{h}{\lambda}$$

and apply that to the relativistic energy-momentum invariant equation we get

$$E^2=h^2f^2 \bigg(\frac{c^2}{v^2}\bigg) + m^2c^4$$.

When v -> 0 this should reduce to the rest energy but we get an infinity in the second term.

8. Mar 26, 2005

### Staff: Mentor

You got that equation using $v = f \lambda$, right? In that equation, $v$ is the phase velocity of a wave, $v_{phase}$, which is not equal to the particle velocity, $v_{particle}$.

$$v_{phase} = \frac {\omega}{k} = \frac {E}{p} = \frac {\gamma mc^2} {\gamma mv_{particle}} = \frac {c}{v_{particle}}$$

So $v_{phase}$ goes to infinity as $v_{particle}$ goes to zero, and your equation above goes to $E = m c^2$ as expected.

To get a realistic wave function for a particle, we have to add (actually, integrate) a bunch of waves with different frequencies and wavelengths to form a wave packet. The group velocity of the packet is

$$v_{group} = \frac {d \omega} {dk}$$

which does turn out to equal $v_{particle}$.

9. Mar 26, 2005

### metrictensor

Very interesting and helpful. I textbooks the velocity in

$$\lambda = \frac{h}{mv}$$

is the particle velocity. There was an exercise that if we define velocity as

$$v=\lambda f$$

and make the substitution into the momentum equation there are some contradictions. According to what you say this is an incorrect step because the two velocities represent different things.

10. Mar 26, 2005

### metrictensor

The problem as I see it with defining the relativistic wave momentum as

$$\gamma mv = \frac{h}{\lambda}$$

is that the right hand side has no term like $$\gamma$$ that enforces the postulates of SR. If we define a "rest wavelength" $$\lambda_0$$ then we could express the momentum as

$$p=\gamma \frac{h}{\lambda_0}$$

11. Jul 7, 2005

### mingshey

Comparing this with the non-relativistic phase velocity:
$$E={p^2\over 2m}$$
$$v_{ph} = {\omega\over k}={E\over p}={p\over 2m} = {1\over2} v_{particle}$$
$$v_{gr} = {d\omega\over dk}={p\over m}=v_{particle}$$

The relativistic phase velocity does not converge to the non-relativistic one even then the velocity is very slow(compared to the spped of light, of course), not quite like other examples showing relativistic mechanics converging to the Newtonian mechanics. This is because in non-relativistic case the rest mass is not considered in the Energy term. So, if you can find a method for measuring the phase velocity of a particle, it would be another criterion for the validity of relativistic theory. Or, is there one?

12. Jul 7, 2005

### dextercioby

You simply can't measure the phase velocity of a particle. IIRC, it's $\frac{c^{2}}{v}$ ,therefore hyperluminal, so it can't be measured.

Daniel.

13. Jul 7, 2005

### Berislav

There's also no self-adjoint operator for phase velocity, if I'm not mistaken, hence no observable.

14. Jul 7, 2005

### Hans de Vries

There are many ways to measure the phase velocity although indirect.

I wrote another section (6) to this document here:

"The Relativistic kinematics of the wave packet"
http://www.chip-architect.com/physics/deBroglie.pdf

which shows how the general rule that the wavefront of matter waves
is always at 90 degrees angles with the physical speed implies a super-
luminal (non-physical) phase velocity of $c^2/v$.

I've included a number of computer simulation images to illustrate things.

Regards, Hans

15. Dec 22, 2008

### KFC

With this result, so if the particle move extremely fast (close to speed of light), what's the result of de Broglie wavelength? zero?

What I confuse is if the momentum in relativistic limit change to $$p = \gamma mv$$, so do we need to apply the length contraction rule to wavelength? Why if not?

Last edited: Dec 22, 2008
16. Dec 23, 2008

### malawi_glenn

With "length contraction rule to wavelength" you mean the relativistic doppler effect right?

17. Dec 23, 2008