2
$\begingroup$

This paper speculates that the EM drive produces thrust with out of phase photons:

http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807

My question is this, do out of phase photons have more momentum than the same set of photons traveling in phase?

Edit: More specifically, I'm interested in whether all the force of the beam is applied in one direction, rather than half on the receiving and half on the sending.

$\endgroup$
  • $\begingroup$ I think your asking a good question from an interesting angle. I also wonder if there are three categories including out of phase, in phase positively and in phase negatively. The last one is never suggested. I write about this here singleedgecertainty.wordpress.com $\endgroup$ – Bill Alsept Jun 16 '16 at 23:00
  • 2
    $\begingroup$ The "paper" is totally nonsensical to begin with, there is consensus about the source of the effect, at least among sensible physicists and within the range of "shoddy measurements" to "outright fraud". $\endgroup$ – CuriousOne Jun 16 '16 at 23:13
  • 1
    $\begingroup$ @CuriousOne I agree with you that the paper is really strange. I am surprised that it got published in this form. $\endgroup$ – hsinghal Jun 18 '16 at 3:31
  • 1
    $\begingroup$ @hsinghal: AIP is one of those pay to publish junk journals. The "reviewers" are the same people who can't publish anywhere else, so what do you expect? $\endgroup$ – CuriousOne Jun 18 '16 at 3:52
1
$\begingroup$

The momentum of the photon is $p=h\nu/c$, so it only depends on its frequency, not its phase. At constant frequency, all photons will have the same $p$ regardless of phase.

$\endgroup$
  • $\begingroup$ My question is rooted in the fact that with a regular beam of light the momentum is transfered to both ends of the beam. When the phases cancel each other out, the receiving end of the beam doesn't feel force (or does it). So that would leave all the force to one end. Which is why I imagine it having twice the momentum on one side. Of course it would still follow the equation, but practically it would behave different (maybe). $\endgroup$ – brysgo Jun 18 '16 at 2:51
1
$\begingroup$

I am not supporting or seconding any finding in the article. They appear to be just incoherent chatting. However the pressure excerted by a electromagnetic wave is

$P_{rad}=\frac{I} {c}$

where I is the intensity of light and c is light velocity. In case of totally reflecting surfaces this will be doubled. Here you can see that the pressure is not of one photon but of whole em wave and it is independent of frequency of photon. It is assumed that the em wave have very large number of photons.

Now if you think of splitting this beam in a Michelson interferometer and overlap it again precisely you can introduce phase difference in the beam. If path difference between two arms is very small, you will see bright and dark fringes, this comes from interference and energy is merely shifted from one point to other point. At the dark portion 'photons' will interfere destructively and in bright parts they will interfere constructively. The momentum is governed by the spatial intensity distribution. Destructive interference means low momentum transfer.

$\endgroup$
  • 1
    $\begingroup$ The momentum of a standing wave is zero. It does cause an internal pressure in the interferometer, though. That's no different from the pressure inside the tank of a rocket, which, as we all know, can be converted to propulsive force, if we open the tank... just like the light in an interferometer can be converted to a much smaller propulsive force... if we open the interferometer. $\endgroup$ – CuriousOne Jun 18 '16 at 3:56
  • $\begingroup$ Radiation pressure is used to accelerate protons please see this article. Modulation in laser intensity profile at focus also change the acceleration properties. The interference pattern is not formed by standing wave. Two co propagating beams overlap on the screen to form the interference pattern. $\endgroup$ – hsinghal Jun 18 '16 at 4:06
  • $\begingroup$ I thought you were talking about interference in an interferometer? That's a closed system. The fringes in a Michelson are basically just interference under small angles, there is nothing special about that light, that's true. In that case we are looking at a prettier open light tank rocket. :-) $\endgroup$ – CuriousOne Jun 18 '16 at 4:12
1
$\begingroup$

They are trying to explain the resonant cavity thrusters , controversial propelling engines.

It seems to me that the paper you quote confuses photons with light waves, attributing a real space wave nature to the photon. From their fig1:

Our reasoning is that when light waves combined with opposite phases, the photons do not vanish for nothing but continue propagating and carrying momentum.

The light wave emerges from a confluence of photons, elementary particles described by quantum electrodynamics. Individually they have energy equal to h*nu and momentum h/lamda . That is all, there are no real phases measurable for an individual photon.

Momentum is a vector quantity and in a classical particle is uniquely defined in direction too. In a quantum mechanical particle where a wavefunction in space gives the probability density for the location of the "particle" the uncertainty in the location spatially will give an uncertainty to the individual photon direction. True the photon does not "vanish", it changes direction and manifests from the dark fringes to the bright fringes during interference.

This blog entry by Lubos helped me understand how the classical beam emerges from the quantum mechanical individual photons. Hand waving: the classical beam emerges from the addition in space of the individual photon complex wave functions , which carry the electromagnetic potential A information in their complex manifestation. As photon-photon interactions are much repressed the potentials in their complex form add spatially and all the various classical interference patterns appear. It is how the quantum mechanical position uncertainty is quantified mathematically, for the given frequency, which is not random but controlled by each photon's wave function; this generates the interference patterns , the addition of probability density amplitudes have the sinusoidal dependence, not the photon in space.

As others have observed in standing waves, by construction, the photon momenta add up to zero beam momentum, as the classical beam emerges from photons propagating backwards and forwards . Any opening will destroy the balance and momentum will be carried off by the photons in the beam.

Their basic premise seems to be out dated and wrong, italics mine:

Field theory, when manipulating photons as virtual particles, has been subject to reservations, because its follows from perturbation theory.19,20 Instead, the old atomistic tenet makes sense to us by regarding the quanta of light as indivisible and indestructible basic building blocks of nature.

I am not commenting on whether the particular controversial drive works or not, but on the explanation offered by the authors assuming it does. From the wiki article there seems to be doubt on the experimental side too.

$\endgroup$
0
$\begingroup$

I am not completely sure, but I would say, it depends what is the exact phase structure of the entire beam. If I was trying to approach the problem, I would use wave formalism and analyze the phase profile vs time. Good question.

$\endgroup$

protected by Qmechanic Jun 18 '16 at 14:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.