4 ms·

New paper claims that the EM Drive doesn't defy Newton's third law

30 points by vajrabum 10y ago

whatshisface 10y ago
>"But if scientists can verify that these paired photons really are being pushed out the back, sh/t's going to get real for EM drives, because it'll help engineers design better cavities and produce even more thrust." That sounds to me like an easier way to test the theory than building an interferometer sensitive enough to detect gravity from photons (as the article suggests). Just build a cavity according to models based on the theory and see if it works better!
- ChuckMcM 10y ago
  Exactly. I look forward to the improved EMDrive, if they can build it.
  - dogma1138 10y ago
    If that's the case why not just use a flashlight? The thrust that is given out by photons is over 1 order of magnitude less than what we are getting from the EM drive. Unless these are some kind of special photons which would mean much bigger news to the world of physics than a slightly more efficient photon rocket. I would love to be able to dismiss this paper but the University of Helsinki is a fairly reputable university, and this paper just makes me want to scratch my head because the physics is more or less on the level i can comprehend but I've read it twice now and I still keep thinking "Whaaaa?".
jddjdbdbd3 10y ago
So its the same as sticking any old light source on the back of space ship. May as well use a light sail.
amluto 10y ago
This paper is highly, highly questionable. First, if the exhaust is photons, you have a photon rocket. These are easy to build: any light source will do. In special relativity, a photon, like any other massless particle, has momentum p = E/c, or roughly p = E * 0.003 N/MW. The article claims the EM drive gets 0.4 Newton/kW, so the numbers are an order of magnitude off. Second, the paper says "also when photons attain thermodynamic balance with matter in the cavity, their number is changing, but the photons themselves do not vanish for nothing at absorption to the walls and they do not emerge from nothing at emission from atoms constituting the walls." Photons are bosons, and bosons do get freely created and destroyed. Where do you think the photons in the light from a flashlight (incandescent or otherwise) comes from? Third, if you don't like perturbation theory and you don't like thermodynamics, you can calculate the force carried by an electromagnetic field directly. Zero field gives zero force. So, sure, if you're willing to ignore a lot of very well-tested physics, you're welcome to believe this paper. [the actual paper is at http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807 http://scitation.aip.org/content/aip/journal/adva/6/6/10.106...]
- Filligree 10y ago
  I would add that photons, like all other particles, don't have identity. They're 'ripples' in the electromagnetic field; if you cancel them out, then you don't get a ghost photon. You just get no photon.
  - dogma1138 10y ago
    >cancel them out Do you mean destructive interference? because the photons don't go anywhere, they don't disappear, the energy of the wave function and hence the photons remains the same. Other than a photon being absorbed there is no way to "cancel out" a photon, you also don't really have a way to cancel out light with destructive interference because some where else there will be a constructive interference, if we could just "cancel out" light we would have cloaking devices by now ;)
    - amluto 10y ago
Aelinsaar 10y ago
I can't read this in any way that doesn't just mean, "Destructive interference". Given that... NO EXHAUST, no reaction mass, no momentum exchange.
- T-A 10y ago
  I've only scanned the paper quickly (and it does look a bit unusual for a physics paper) but I think there is an obvious answer to that: it's not perfect destructive interference. You can imagine two photons with opposite polarization starting out together and propagating in almost exactly the same direction. As they go along their merry way, they diverge. Eventually, far away, you have two well-separated, ordinary photons. But close to the source, you have near-perfect cancellation, and so negligible interaction with the cavity walls, as the authors would have it.
  - amluto 10y ago
    Sorry for the accidental down vote. Button too small for finger :(
  - Aelinsaar 10y ago
    ...Then they should be detectable, and this thing should be producing light as exhaust, which it isn't. The only way this works is if the cancellation is total, which again...
  - gus_massa 10y ago
    If this were the explanation, they should get a thrust to powered ratio that is lower than the theoretical maximum with the currently accepted physics laws. The theoretical maximum is 1/c = 0.0033 mN/KW, but all the "interesting" experiments get more, for example 0.4 N/KW = 400 mN/KW.
tgb 10y ago
So this paper's foundation seems to be the idea that 2 photons can propagate in the same direction with the same wavelength and phase exactly 180 degrees off. As such, the net effect on the EM field would be 0, so they wouldn't interact with anything through the EM force and could pass through the walls of the chamber, escaping the device and acting as propellant. Importantly, the photon pair, they say, still caries energy even though it doesn't interact with anything. My question for anyone with a background in this is, is this bunk? It doesn't square with my intuition. Suppose we have two plane waves in the EM field propagating like these photons, exactly out of phase, how would this be different from no plane waves in a completely empty space? Where's the energy in this photon pair scenario? If it doesn't interact with anything, in what sense could it possibly have energy? Moreover, what would the production of a pair look like? They're claim of it having energy is important only in so far as it is a release of momentum for the drive, so the formation is the key part. Suppose there's just the first photon - is emitting the second of the pair equivalent to just absorbing the first one? In which case, their claims that photons don't just "vanish for nothing" aren't valid. Edit: Oh and that stuff about Gauss Bonnet ("Geometry at any point along an action can be recapped by signed curvature.") - I do research in related areas of math and could recite Gauss-Bonnet without reference, but I have no idea what they're trying to say. Gauss-Bonnet is a statement relating the global topology of a surface to its curvature, what surface would they even be considering here and how would its topology matter? Honest question: is this representative of how physics papers are written in terms of style? Sentences like "It is no new idea that inertial effects are immediate, because the vacuum resists changes by embracing everything" make me wonder how physicists communicate.
- beevai142 10y ago
  No, it is not typical physics paper (physicist here). Little is clearly defined. E.g. phase shift in what --- if the meaning is the same as in standard EM/QED, what is it the authors exactly propose (probably they don't work in this framework if they reject photon number change), and can the replacement they propose predict something correctly? Overall, the paper reads like a word salad, as the explanations are all over the place. At least the second author of the paper is known also for other bunk, so grains of salt recommended.

30 points by vajrabum 10y ago

>"But if scientists can verify that these paired photons really are being pushed out the back, sh/t's going to get real for EM drives, because it'll help engineers design better cavities and produce even more thrust." That sounds to me like an easier way to test the theory than building an interferometer sensitive enough to detect gravity from photons (as the article suggests). Just build a cavity according to models based on the theory and see if it works better!

ChuckMcM 10y ago
Exactly. I look forward to the improved EMDrive, if they can build it.
- dogma1138 10y ago
  If that's the case why not just use a flashlight? The thrust that is given out by photons is over 1 order of magnitude less than what we are getting from the EM drive. Unless these are some kind of special photons which would mean much bigger news to the world of physics than a slightly more efficient photon rocket. I would love to be able to dismiss this paper but the University of Helsinki is a fairly reputable university, and this paper just makes me want to scratch my head because the physics is more or less on the level i can comprehend but I've read it twice now and I still keep thinking "Whaaaa?".

jddjdbdbd3 10y ago

So its the same as sticking any old light source on the back of space ship. May as well use a light sail.

amluto 10y ago

This paper is highly, highly questionable. First, if the exhaust is photons, you have a photon rocket. These are easy to build: any light source will do. In special relativity, a photon, like any other massless particle, has momentum p = E/c, or roughly p = E * 0.003 N/MW. The article claims the EM drive gets 0.4 Newton/kW, so the numbers are an order of magnitude off. Second, the paper says "also when photons attain thermodynamic balance with matter in the cavity, their number is changing, but the photons themselves do not vanish for nothing at absorption to the walls and they do not emerge from nothing at emission from atoms constituting the walls." Photons are bosons, and bosons do get freely created and destroyed. Where do you think the photons in the light from a flashlight (incandescent or otherwise) comes from? Third, if you don't like perturbation theory and you don't like thermodynamics, you can calculate the force carried by an electromagnetic field directly. Zero field gives zero force. So, sure, if you're willing to ignore a lot of very well-tested physics, you're welcome to believe this paper. [the actual paper is at http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807 http://scitation.aip.org/content/aip/journal/adva/6/6/10.106...]

Filligree 10y ago
I would add that photons, like all other particles, don't have identity. They're 'ripples' in the electromagnetic field; if you cancel them out, then you don't get a ghost photon. You just get no photon.
- dogma1138 10y ago
  >cancel them out Do you mean destructive interference? because the photons don't go anywhere, they don't disappear, the energy of the wave function and hence the photons remains the same. Other than a photon being absorbed there is no way to "cancel out" a photon, you also don't really have a way to cancel out light with destructive interference because some where else there will be a constructive interference, if we could just "cancel out" light we would have cloaking devices by now ;)
  - amluto 10y ago

Aelinsaar 10y ago

I can't read this in any way that doesn't just mean, "Destructive interference". Given that... NO EXHAUST, no reaction mass, no momentum exchange.

T-A 10y ago
I've only scanned the paper quickly (and it does look a bit unusual for a physics paper) but I think there is an obvious answer to that: it's not perfect destructive interference. You can imagine two photons with opposite polarization starting out together and propagating in almost exactly the same direction. As they go along their merry way, they diverge. Eventually, far away, you have two well-separated, ordinary photons. But close to the source, you have near-perfect cancellation, and so negligible interaction with the cavity walls, as the authors would have it.
- amluto 10y ago
  Sorry for the accidental down vote. Button too small for finger :(
- Aelinsaar 10y ago
  ...Then they should be detectable, and this thing should be producing light as exhaust, which it isn't. The only way this works is if the cancellation is total, which again...
- gus_massa 10y ago
  If this were the explanation, they should get a thrust to powered ratio that is lower than the theoretical maximum with the currently accepted physics laws. The theoretical maximum is 1/c = 0.0033 mN/KW, but all the "interesting" experiments get more, for example 0.4 N/KW = 400 mN/KW.

tgb 10y ago

So this paper's foundation seems to be the idea that 2 photons can propagate in the same direction with the same wavelength and phase exactly 180 degrees off. As such, the net effect on the EM field would be 0, so they wouldn't interact with anything through the EM force and could pass through the walls of the chamber, escaping the device and acting as propellant. Importantly, the photon pair, they say, still caries energy even though it doesn't interact with anything. My question for anyone with a background in this is, is this bunk? It doesn't square with my intuition. Suppose we have two plane waves in the EM field propagating like these photons, exactly out of phase, how would this be different from no plane waves in a completely empty space? Where's the energy in this photon pair scenario? If it doesn't interact with anything, in what sense could it possibly have energy? Moreover, what would the production of a pair look like? They're claim of it having energy is important only in so far as it is a release of momentum for the drive, so the formation is the key part. Suppose there's just the first photon - is emitting the second of the pair equivalent to just absorbing the first one? In which case, their claims that photons don't just "vanish for nothing" aren't valid. Edit: Oh and that stuff about Gauss Bonnet ("Geometry at any point along an action can be recapped by signed curvature.") - I do research in related areas of math and could recite Gauss-Bonnet without reference, but I have no idea what they're trying to say. Gauss-Bonnet is a statement relating the global topology of a surface to its curvature, what surface would they even be considering here and how would its topology matter? Honest question: is this representative of how physics papers are written in terms of style? Sentences like "It is no new idea that inertial effects are immediate, because the vacuum resists changes by embracing everything" make me wonder how physicists communicate.

beevai142 10y ago
No, it is not typical physics paper (physicist here). Little is clearly defined. E.g. phase shift in what --- if the meaning is the same as in standard EM/QED, what is it the authors exactly propose (probably they don't work in this framework if they reject photon number change), and can the replacement they propose predict something correctly? Overall, the paper reads like a word salad, as the explanations are all over the place. At least the second author of the paper is known also for other bunk, so grains of salt recommended.