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B About interference, momentum and thrust...

  1. Nov 18, 2016 #1
    [Mentors' note - this post has been edited to remove some discussion of the EM-drive. See https://www.physicsforums.com/threads/nasas-em-drive.884753/ for our current policy with regard to em-drive discussions]

    Hello.
    Imagine a closed box with reflecting walls. Inside you have two antennas emiting electromagnetic waves of certain wavelength. Both antennas are separated 1/4 of the wavelength (or some N wavelengths + 1/4) and one antenna is 90º phase-shifted (1/4 wavelength delayed) with respect to the other.

    I attach a picture of the superposition we will get inside the box. We will get one side of destructive interference, one region between the antennas where a standing wave develops, and another side of constructive interference. You can see that the wall on the left is on the destructive interference region, so it almost doesn't receive any photons, but the wall on the right will receive a wave created by constructive interference:

    example.PNG

    I also attach a top-view animation of the box and the different regions and waves that will develop. You can see there'll be more photons impacting the right wall:

    wave.gif

    So my question is: if we have more photons impacting one wall than the other, why shouldn't the box start moving to that direction?
     
    Last edited by a moderator: Nov 18, 2016
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  3. Nov 18, 2016 #2

    andrewkirk

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    The coloured diagram, which is of a wave in a 1D medium (eg a wave in a stretched wire) does not take account of the reduction in amplitude in proportion to the inverse square of the distance, that occurs with EM waves. Because it assumes no reduction, one of the antennas is always an exact offset and the other is an exact duplication of amplitude. That is incorrect.

    I suspect that when the 3D nature of the waves is taken into account the difference in photon flux between two opposite walls will disappear.
     
  4. Nov 18, 2016 #3

    Dale

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    If you have asymmetric external radiation (either emitted or absorbed) then you will have a net force.
     
    Last edited: Nov 18, 2016
  5. Nov 18, 2016 #4

    Nugatory

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    You've left out the reaction forces on the antennas as they emit radiation.
     
  6. Nov 18, 2016 #5
    Having nothing to add I still can't help jumping in and seconding what Dale said. I can believe that the photons are pushing harder on the right wall of the box. However, I can't believe that without also believing that the antennae are being pushed to the left. You can't have it both ways. If the antennae aren't attached to the box, then this could push the box to the right, but that doesn't violate any conservation laws. If the antennae are inside and attached to the box then there is no net force. If the antennae are inside and NOT attached to the box, then the box will move to the right up until the moment the antennae fly into the left wall!
     
  7. Nov 19, 2016 #6
    Hello.

    The antennas are attached to the box.

    The signals will be not exactly added or cancelled, but the effect will still take place (there will be an imbalance).

    I thought also about possible reaction forces on the antennas, but the antennas are firing equally to all directions, so I think there should be no reaction, but can someone explain how the reaction forces on the antennas would develop if that's the case?
     
  8. Nov 19, 2016 #7

    andrewkirk

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    I suspect that is not the case. What the antennas are sending out symmetrically in all directions is waves, not photons. If it is photons and not waves that transfer momentum, then that symmetric emission of waves need not correspond to a symmetric emission of photons. If more photons are hitting the front of the box than the rear then it seems to make sense that more were emitted in a frontwards than in a rearwards direction, so that the net momentum transfer at the antennas from emission will oppose the net momentum transfer at the walls from impact.

    Of course, this is all just words. To make a concrete claim about impulse one has to write equations, and to contest that claim one has to critique the equations. I haven't seen any equations suggesting the creation of a net nonzero impulse on the box from photons, so at this stage there's no concrete claim to critique.
     
  9. Nov 19, 2016 #8
    Yes, I thought about that difference between waves and photons. But if the antennas are able to know were the photons will be impacting (in order to "counterbalance" the momentum exchanges that will happen at the right wall), then you're saying we could know where a photon will arrive since emitted, and that is something that is not allowed on quantum mechanics (that's where the probability wave comes from).

    Imagine you have a photon source surrounded by a metallic sphere. When a photon leaves the source it's a probability wave until it hits the ball, because by definition you can not know where the photon will hit until it's detected.

    If you could measure the momentum at the source as some kind of force in opposite direction to the direction where the photon will land, then quantum mechanics wouldn't need the probability wave function concept.

    So if there's a reaction on the antennas, then I don't understand why quantum mechanics needs the wave-particle duality to explain photons.
     
  10. Nov 19, 2016 #9
    I made another animation separating the antennas, to show the different regions better.
    You can see that the interference pattern changes a bit, and that the imbalance on both sides tends to dissapear as the distance between the antennas gets bigger (the constructive and destructive interference regions begin to distribute more evenly on the available space).

    So the antennas should be as close as possible (but keeping the wavelength ratios and delays to work properly).

    wave 2.gif
     
  11. Nov 19, 2016 #10

    sophiecentaur

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    The effect of radiation pressure is a very well established (classical) bit of theory and it has been detected with a (proper) radiometer. I don't understand how there seems to be doubt about it applying when a 'certain' arrangement is involved. If momentum is conserved and photons leave in one direction then is there any point in trying to show that there will be no overall force on the 'craft'?
    I would have thought the reverse would be the case. A large aperture would produce a more directional beam, which would waste less energy.
     
  12. Nov 19, 2016 #11
    The point is that I think the radiation pressure on the antennas will manifest equally all around them (they emit waves in all directions, so any possible forces at emission will express as pressure or tension around the whole antenna, so no displacement), yet the patterns the propagating waves will create at each atom position on the walls dictate the probability of energy being transfered to the atoms on that specific positions.

    So each point in space will see two waves propagating, but the energy available for absortion at each point will depend on the pattern both waves create there (and matter presence at that point).

    What I mean is that the electromagnetic waves are created simmetrically around the antennas, yet the probability of energy absortion at the walls is asymmetric.
     
  13. Nov 19, 2016 #12

    Nugatory

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    The reaction force on an antenna or other emitter of electromagnetic radiation depends on the interaction between the antenna and the electromagnetic field around it. You've set things up so that the electromagnetic fields are different on different sides of the two antennas, so the forces on the two sides of the antennas are also different. (One way to see this is to calculate the momentum flux through vertical surfaces on each side of and arbitrarily close to the two antennae).

    I think that you may be misleading yourself by thinking in terms of "Firing equally in all directions". That phrase tempts us to think that the antennas are shooting out photons like little bullets, transferring momentum equally in all directions and making it very hard to see where the net reaction force comes from. However, this image of electromagnetic radiation as photons being fired off in various directions is seriously misleading - photons are not what you're thinking when you hear the term "particle of light" and they don't move through space from source to target like little bullets. A better way of visualizing it is that the photons go where the fluctuations of the electromagnetic field are greatest; thus no photons are emitted in the leftwards direction at all, so there is nothing to balance the reaction force from the photons being emitted to the right.
    (Note: This "better way" is a math-free waving of the hands suitable only for a B level thread. In an I-level thread, we'd be telling you to analyze this setup using classical electrodynamics and Maxwell's equations - no photons required because there are no significant quantum effects involved).
     
  14. Nov 19, 2016 #13

    sophiecentaur

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    If you think of the ship as one large antenna, which only fires on one direction, you have the answer in one.
     
  15. Nov 19, 2016 #14
    I get that in order to object the generation of an asymmetric energy distribution that could provide momentum mainly to one side of the box, we need a balancing force somewhere, and that place seems to be the antennas.

    But if I were talking just about the antenna system alone (with no enclosing box), would you then tell that the antennas would displace to the left just by the effects of the radiation pressure of their own fields?

    If the antennas somehow create a counterbalancing force using just the fields they generate, then we could remove the box altogether and just have the antennas displacing to the left by the fields they create on each other!

    I mean that I don't know which mechanism would make the antennas know that there's a metallic box picking up the energy they emit. The antennas don't care if there's a box around to absorb the energy or not, so they should behave the same way with or without the box around. So if the antennas are somehow counteracting the photon imbalance and momentum transfer present on the right wall, then they would displace to the left when left alone.

    So either way, we end up with a system that is able to displace on its own!
     
  16. Nov 19, 2016 #15

    sophiecentaur

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    It's the whole system that you need to be considering. The 'box' is no different from a parabolic reflector, in principle (part of a directive array) - except in the actual efficiency of the propulsion unit.
    It is not true to say that the antennae would not 'know about' the box. The box can easily have an effect. If there is significant energy reflected by the walls, it will interact with the metal of the antennae and change its input impedance. If the walls are perfect absorbers, no energy would be reflected back to the antenna.
     
  17. Nov 19, 2016 #16
    Yes, exactly, and there is nothing wrong with that. That is just like a rocket. You fire something off to the right and as a reaction get pushed to the left. The center of mass, well ok, make that the net momentum of the sytem doesn't change.
     
  18. Nov 19, 2016 #17
    Yes, after reviewing my own answer I saw that the antenna system cold be working like a normal engine, but with a photon exhaust.
    So in a sense it's like a lantern left in space, but with a complicated and inneficient way of letting the photons scape (or appear) on one direction.

    I thought that reflecting waves with mirrors was fundamentally different to making fields interfere themselves, but it seems the antennas work like mirrors (at least in this case).

    As the antennas seem to be the problem, is there any way we could get the antennas out of the interference area of the fields?
     
  19. Nov 19, 2016 #18

    Nugatory

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    Yes. You've built a (somewhat complicated and hard to analyze) photon rocket: https://en.wikipedia.org/wiki/Photon_rocket
     
  20. Nov 19, 2016 #19

    Nugatory

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    No. That would require an electromagnetic field that doesn't conserve momentum and energy and there is no such thing.
     
  21. Nov 19, 2016 #20

    andrewkirk

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    On what do you base that 'is not allowed' claim? It doesn't sound right to me, although it's not clear to me exactly what it is that you are saying is not allowed. Can you write an equation to make the claim clear and then prove it?

    Apparent retrospectivity (note the emphasis on 'apparent') is easy to achieve in QM - eg delayed choice quantum erasers. In this case the waves generate probability functions that result in certain photon impacts against certain walls. Each photon impact implies a photon emission from one of the antennas in a certain direction. We don't know which antenna but that doesn't matter because, whichever one it was, the momentum transfer to the box, via the antenna and its supporting arm, upon emission, would be the same.

    The waves specify the probability distributions for momentum transfers from impacts. The photon impacts, and their corresponding emissions, are realisations of that distribution.

    I have to reiterate though that without equations this discussion is really just a vague hand-wave.
     
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