
New paper claims that the EM Drive doesn't defy Newton's third law - vajrabum
http://www.sciencealert.com/new-paper-claims-that-the-em-drive-doesn-t-defy-newton-s-3rd-law-after-all
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amluto
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/6/6/10.106...](http://scitation.aip.org/content/aip/journal/adva/6/6/10.1063/1.4953807)]

~~~
Filligree
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
>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
> 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.

Not quite. If both waves are plane waves, the interference could be
destructive everywhere or, depending on the setup, destructive everywhere
except between your sources. If you calculate the time-averaged power needed
to source the waves, it'll sum to zero. And there are no photons in the far
field.

~~~
dogma1138
Yes quite, plane waves physically don't exist in reality they are a
mathematical idealization.

Again you can't make an "anti laser".

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whatshisface
> _" 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
Exactly. I look forward to the improved EMDrive, if they can build it.

~~~
dogma1138
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?".

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tgb
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.

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beevai142
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.

~~~
tgb
Thanks for confirming what I suspected.

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Aelinsaar
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
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.

~~~
gus_massa
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.

