# Faster than light in Quantum mechanics?

1. Oct 20, 2011

### stgdf01

In <Engineering Electromagnetics> written by W.H.Hayt and J.A.Buck( 6th edition,McGraw-Hill,p372), the phase velocity vp of electromagnetic waves in copper at 60Hz (commercial electric power) is 3.2m/s. Substituting the value into vpv = c2 of de Broglie theory, v=108c should be much faster than light speed c in vacuum?

2. Oct 21, 2011

### PhilDSP

Hi stgdf01 and welcome to the forum,

The analysis you posted sounds incorrect. Phase velocity (of the EM wave) must always be greater or equal to c. The analysis seems to have been done with the assumption that the AC voltage wave represents the actual movement of energy. It doesn't. The energy travels far quicker than the swinging of the voltage. But maybe the person doing the analysis meant the velocity of the conduction current? If so, then group velocity vg rather than phase velocity is the correct term to use.

vp vg= c2

Last edited: Oct 21, 2011
3. Oct 21, 2011

### stgdf01

The study to vp in conductors is based on first principle(Maxwell's equations) and has nothing to do with parameters in circuit theory such as voltage,inductance and capacitance. For example, J.D.Jackson's <Classical Electrodynamics> and D.J.Griffiths's <Introduction to Electrodynamics>.

The group velocity vg=dw/dk of low-frequency electromagnetic field in conductors is twice as much as vp=w/k (P.Lorrain, D.R.Corson, F.Lorrain, <Electromagnetic Fields and Waves>, 3rd edition). For commercial power in copper, vp=3.2m/s and vg=6.4m/s. In this case, neither vp nor vg can play the role of c2/vp in de Broglie's theory. Actually, it is v=p/m (m=E/c2) in point mechanis.

4. Oct 25, 2011

### PhilDSP

It sounds likely that the analysis involved integration over macroscopic areas of space and large increments of time. In that case the average variation of the EM values would have washed out the high frequency effects. That's what I was trying to express in the first response. The analysis apparently ignores quantum effects.

5. Oct 25, 2011

### Cthugha

This result of de Broglie theory is usually explicitly derived for matter waves. You will of course get strange results when you try to use it without proper modification for particles without rest mass and em waves.

6. Oct 26, 2011

### navilon

neutrinos as being said faster than light,,have very small mass and could be compound of gravity also accelerating speed such as dark energy particles..therefore interacting with light much like gravity does..as a mediator..a boson wave..already there..QED

7. Oct 27, 2011

### stgdf01

E=hw and p=hk is of course tenable to photons otherwise the Compton effect, photoelectric effect, blackbody radiation and others cannot be interpreted. In history, these two relations are concluded from experiments to massless photons and then de Broglie extended to massive matter waves(electrons,say).

8. Oct 27, 2011

### Cthugha

Sure, but that does not have anything to do with $$v_p v=c^2.$$ This result is derived explicitly in the nonrelativistivc limit for matter waves and requires w and p to depend on mass. Some good textbooks on this topic state this explicitly. Even the really basic about about basic knowledge for engineers by Hennecke mentions this.

Last edited: Oct 27, 2011
9. Oct 27, 2011

### PhilDSP

Actually the history behind that for waves in general extends much further back in time:

http://en.wikipedia.org/wiki/Group_velocity
http://en.wikipedia.org/wiki/John_William_Strutt,_3rd_Baron_Rayleigh

10. Oct 27, 2011

### stgdf01

But in optics and electromagnetism, $$v_p v=c^2.$$ is still valid to a field(photons) in vacuum. On the other hand, the photon can be described in relativistic mechanics as a special case of V=c and $$m_0=0$$

11. Oct 27, 2011

### stgdf01

I regards this as a counter-example of $$v_p v_g=c^2$$,although $$v_p v=c^2$$ is universal.That is to say, the group velocity vg is not always equal to V. It can be applied to explain why those experiments of superluminal group velocities vg>c are not really faster than light v>c.

12. Oct 28, 2011

### Cthugha

Yes, in vacuum. You were interested in light in a material, no?

You can treat a photon as having $$m_0=0$$ if you include the full relativistic mass $$E=\sqrt{(m_0 c^2)^2 +(p c)^2}$$ in the derivation.

However, you get the $$v$$ in $$v_p v=c^2$$ in common de Broglie theory by using $$p=m v$$. For photons you do not have this explicit dependence and just use $$p=\frac{h}{\lambda}$$, so you only arrive at the trivial $$v_p =\frac{c}{n}.$$

13. Oct 28, 2011

### PhilDSP

If we know or can determine the dispersion equation (which doesn't include a mass term) then we can find the phase velocity algebraically and the group velocity by taking the derivative of the dispersion equation (or its roots) with respect to wave number.

$$v_p = \frac {\omega}{k} \ \ \ \ \ \ v_g = \frac {\partial \omega}{\partial k}$$

For example: the FT of $\ \ \ (\nabla^2 - \frac {1}{c^2} \frac {\partial^2}{\partial t^2}) \phi = 0 \ \ \$ is $\ \ \ (-k^2 + \frac {\omega^2}{c^2}) \phi = 0 \ \ \$ so that $\ \ \ D(\omega, k) = -k^2 + \frac {\omega^2}{c^2}$

then $\ \ \ \ \omega = ±ck \ \ \ \$ giving $\ \ \ \ v_p = ±c \ \ \ \$ and $\ \ \ \ v_g = ±c \ \ \ \$

Last edited: Oct 28, 2011