Relativistic de Broglie Wavelength

[metrictensor]

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

 

[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.

 

[jtbell]

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!

 

[metrictensor]

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.

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.

 

[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.

 

[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.

 

[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.

 

[jtbell]

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

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

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 $v_{particle}$.

 

[metrictensor]

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 $v_{particle}$.

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.

 

[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

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

 

[mingshey]

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 $v_{particle}$.

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?

 

[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.

 

[Berislav]

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.

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

 

[Hans de Vries]

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?

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

 

[KFC]

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!

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?

 

[malawi_glenn]

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?

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

 

[Helios]

Of interest might be a post on the relativistic index of refraction

https://www.physicsforums.com/showthread.php?t=192736

my question is if

v$$_{phase}$$ = c/n

holds true?