[CraigD]

wade_b said:

What to make of Einstein's box? ...

Here’s a summary of the “Einstein’s box paradox”, and its resolution.

The paradox:

• The box has mass and a left and a right wall separated by distance .
  
• Start with the box stationary, that is, its momentum is 0
  
• The “gun” on the left wall of the box emits a photon of momentum .
  
• Per conservation of momentum, the box has a momentum of , and therefore, a velocity .
  
• The photon is absorbed by the right wall after time .
  
• The box has traveled in that time.
  
• Its momentum is again 0.

Therefore, it has shifted its position , without having been subject to an external force, violating the 3rd law of motion.

The resolution of the paradox is that, when the left wall emitted the photon, its mass decreased by the mass equivalent of the energy of the photon, . When the right wall absorbed it, its mass increased by the same amount.
, so .

Center of mass equations are similar to torque equations.
Assuming the left and right walls were initially of equal mass, the initial center is .
The new center of mass is given by the equation

which gives


So the change in center of mass is equal to the shift in position, resolving the paradox.

[wade_b]

CraigD said:

Here’s a summary of the “Einstein’s box paradox”, and its resolution.

The paradox:

• The box has mass and a left and a right wall separated by distance .
  
• Start with the box stationary, that is, its momentum is 0
  
• The “gun” on the left wall of the box emits a photon of momentum .
  
• Per conservation of momentum, the box has a momentum of , and therefore, a velocity .
  
• The photon is absorbed by the right wall after time .
  
• The box has traveled in that time.
  
• Its momentum is again 0.

Therefore, it has shifted its position , without having been subject to an external force, violating the 3rd law of motion.

The resolution of the paradox is that, when the left wall emitted the photon, its mass decreased by the mass equivalent of the energy of the photon, . When the right wall absorbed it, its mass increased by the same amount.
, so .

Center of mass equations are similar to torque equations.
Assuming the left and right walls were initially of equal mass, the initial center is .
The new center of mass is given by the equation

which gives


So the change in center of mass is equal to the shift in position, resolving the paradox.

Which sounds like it COULD apply to the emdrive? :shrug:

[CraigD]

wade_b said:

Which sounds like it [Einstein’s box] COULD apply to the emdrive?

No, not according to a usual resolution such as the one in post #31

The Einstein’s box thought experiment (this particular one – there are actually several that commonly go by the name) – which can, with very precise equipment and instrumentation, by physically performed – is exactly analogous to the simpler example of a box inside a freely moving box, which I’ve sketched in the image attached below. It moves the outer box a fixed distance, then the system is “used up”, and before it can be used again, must either be reset, which returns it to exactly its starting position, or mass removed and added to it from outside, which defeats the premise of it being a closed system.

What makes Einstein’s box a trickier riddle is that it’s less intuitive that shining emitting light from one object onto another which absorbs the light (a more difficult feat than it sounds at first hearing, as most objects ultimately emit about the same amount of light they absorb) actually transfers mass between them, than it is that moving a small massive box from one end to another of a large hollow box transfers mass within it.

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[Karnuvap]

At first I thought that this was just a spaceship with an open ended microwave cavity on the back thus microwaves out the back = thrust towards the front and thus acceleration (however slight) that would eventually build-up to pretty decent speeds.
But, on closer inspection, the cavity is sealed at the back too and it relies on the fallacious assumption that because the wall at the 'back' has a different surface area then the wall near the 'front' there would be more 'pressure' on the back than at the front. OK that is slightly an oversimplification of what they are trying to achieve and the reality of their theory is slightly more subtle.

The real secret to where they are bamboozling us lies in the expression "Group Velocity". This is not a real velocity that should be used in the subsequent equations. It is a mathematical construct to help solve the equations and not anything real. To give you a clue as to why group velocity isn't anything like real velocity - it is perfectly possible to have a group velocity faster than that of the light photons or waves that make it up (which, obviously, travel at the speed of light themselves).

So, unless something is emitted out of the back of your spacecraft it cannot accelerate.

Of course I might be wrong and it could be that the microwave photons are selectively destructively interfered at one end of the chamber (and, presumably) they constructively interfere at the other end. The destructively destroyed photons re-appear at infinity - or so my physics lecturer told me when I asked him where they would go if I shined two identical frequency lasers at a single point on a screen and then adjusted one of them so that it was precisely one half wavelength further away from the screen than the other. Where would the light go in this perfect destructive interference set-up? - they disappear and re-appear at infinity was his reply. I'd love to know what the real answer was.

[Qfwfq]

Karnuvap said:

The real secret to where they are bamboozling us lies in the expression "Group Velocity". This is not a real velocity that should be used in the subsequent equations. It is a mathematical construct to help solve the equations and not anything real. To give you a clue as to why group velocity isn't anything like real velocity - it is perfectly possible to have a group velocity faster than that of the light photons or waves that make it up (which, obviously, travel at the speed of light themselves).

First, you're exchanging group and phase velocity, the former is that of the wave "packets" (and hence of the particles) and, second, it will be less than c; in a dispersive medium, due to the separation of the velocities, it will be even a bit less than c divided by the refractive index.

Of course, I'm not saying that the contraption could work. It's every bit as pointless as hoping to propel a spaceship with a jar of the same shape filled with hot gas.

Karnuvap said:

I'd love to know what the real answer was.

The real answer is that you could never get the two beams to completely have exactly destructive interference from thence onward along the direction of propagation without getting any reflection. I'd say that physics lecturer hadn't thought it all the way through.

[Karnuvap]

Qfwfq said:

: I'd say that physics lecturer hadn't thought it all the way through.

To give him his due he did mutter something about having to turn the thing on such that the beam's fourrier series wouldn't be pure and that this would result in there not being a perfect match for destruction so I think I may have besmirched him unduly.

I was only asking this because I was thinking about how it might be possible to make a real-life light sabre. My laser would be shon into a crystal and six or eight beams would emerge from the facets at slightly converging angles (such that they crossed about 1metre away). The clever bit was that half the crystal facets would be slightly thicker than the others such that the path length for the beams emerging from these ones would be a half wave longer than that of the beams that emerged from the other ones at the point where they crossed. Thus destructively interfering at precisely one metre away from the handle of my laser sword. Thus my sword would be 1 metre long and no more. << This doesn't work either>> Thanks for your comments though.

The Vap

[Pyrotex]

KickAssClown said:

Could someone explain in plain english how this thing works? I have a ok idea, but I want to hear some other angles on it...

The engine apparantly works because the round side at one end of the truncated cone is larger than the round side at the other end. Therefore the radiation pressure on it will be larger.

The author totally ignores the "other" side: the conical cylinder that connects the top and bottom ends. If you calculate just that portion of the conical cylinder that is in the same plane as the two ends, I'm sure you will find that the area on top is equal to the area on bottom.

Perhaps he should call it, the "Null Drive" -- because it has no net force at all.

[Qfwfq]

Karnuvap said:

...he did mutter something about having to turn the thing on such that the beam's fourrier series wouldn't be pure...

actually that wouldn't be the problem in steady state. Even considering the perfect monochromaticity approximation (which is fine for the brunt of interferometric design, between a few cycles after reaching steady state and a few cycles before losing it) it still isn't possible without some weird non-linear propagation. As you're talking about photons in air and no material object at the sabre's tip, it's just nononono.

[IDMclean]

Chinese are building it, apparently.

I figured I would update on this topic.

[DustinTheWind]

I didn't really buy the group wave velocity idea either. I thought about it some and this is what I think if any ideas how it might work. The tube works best off reflectivity. The Tube is shaped with a larger disk at one end and a smaller disk at the other with angled walls. The photons bouncing off the top plate impart all their momentum at once and in a shorter amount of time. It takes longer for the photons to impart their momentum through the side walls. There for you get a force in one direction for a short amount of time before it is canceled out. Now imagine this on a much larger scale and increase the scale by now allowing the reflections to continue more so as the Q of the cavity increases due to superconductivity. So basically the thrust may be due to a time delay in canceling out the momentum on the back reflection. Of course this is just an idea I had after having asked God how this thing might work.

I believe that group velocity can not transfer momentum as stated in the paper and this also was what bothered me.

[Pyrotex]

You guys are making this so complicated.
The presumed net force created my the Microwave Thruster is proportional to:
Area(of big circle) - Area(of small circle).

The MT assumes that the microwaves do not apply any force at all to the conical sides of the chamber.
And this is patently false.

Here is how to calculate the forces. However the microwaves are piped into the chamber, we assume that after some small time, t, the microwaves are in some sort of equilibrium in the chamber.

Now we add up all the upward vertical components of the forces on the chamber walls.

Then we will add up all the downward vertical components of the forces.

As drawn, we can assume the microwaves are piped in from anywhere. (The answer will be unchanged) So, for symmetry, we assume they are piped in from the center of the chamber.

Waves going toward the large circular side hit it, and have a vertical component that depends upon:
energy flux of emitted waves;
1 over-distance-squared from the center of the chamber;
cosine of the angle between wave and flat upper circular side.

If you integrate the tiny elements of force on every small area over the entire upper circular side, and over all angles between wave and flat upper circular side, You will get a total force that is proportional to:
The energy flux of emitted waves;
the cross-sectional area of the chamber as seen from above. (a large circle)

Waves going toward the lower circular side AND the conical sides, hit them, and have a vertical component that depends upon:
energy flux of emitted waves;
1 over-distance-squared from the center of the chamber;
cosine of the angle between wave and flat lower circular side & between wave and conical sides;
cosine of the angle between conical sides and flat lower circular side.

If you integrate the tiny elements of force on every small area over the entire lower circular side AND the conical sides, and over all angles between wave and flat lower circular side, You will get a total force that is proportional to:
The energy flux of emitted waves;
the cross-sectional area of the chamber as seen from below. (a large circle)

Net force = force up minus force down.

BUT, the chamber has the SAME CROSS-SECTIONAL AREA seen from above AND below!!!!

The lower circular side may be smaller than the upper circle, but it's irrelevant. What is relevant is the total cross-sectional area of the entire chamber as seen from top and bottom -- and they are identical.

The force up is exactly equal to force down. changing the shape of the chamber CANNOT violate this simple fact (unless you put holes in the chamber)

Once you "GET IT" that the partial force upward or downward (actually, in ANY arbitrary pair of directions) is determined by the total cross-sectional area, and not the area of any select piece of the chamber, then it becomes dirt-obvious that no net force can arise.

[DustinTheWind]

Pyrotex said:

You guys are making this so complicated.
The presumed net force created my the Microwave Thruster is proportional to:
Area(of big circle) - Area(of small circle).

The MT assumes that the microwaves do not apply any force at all to the conical sides of the chamber.
And this is patently false.

Here is how to calculate the forces. However the microwaves are piped into the chamber, we assume that after some small time, t, the microwaves are in some sort of equilibrium in the chamber.

Now we add up all the upward vertical components of the forces on the chamber walls.

Then we will add up all the downward vertical components of the forces.

As drawn, we can assume the microwaves are piped in from anywhere. (The answer will be unchanged) So, for symmetry, we assume they are piped in from the center of the chamber.

Waves going toward the large circular side hit it, and have a vertical component that depends upon:
energy flux of emitted waves;
1 over-distance-squared from the center of the chamber;
cosine of the angle between wave and flat upper circular side.

If you integrate the tiny elements of force on every small area over the entire upper circular side, and over all angles between wave and flat upper circular side, You will get a total force that is proportional to:
The energy flux of emitted waves;
the cross-sectional area of the chamber as seen from above. (a large circle)

Waves going toward the lower circular side AND the conical sides, hit them, and have a vertical component that depends upon:
energy flux of emitted waves;
1 over-distance-squared from the center of the chamber;
cosine of the angle between wave and flat lower circular side & between wave and conical sides;
cosine of the angle between conical sides and flat lower circular side.

If you integrate the tiny elements of force on every small area over the entire lower circular side AND the conical sides, and over all angles between wave and flat lower circular side, You will get a total force that is proportional to:
The energy flux of emitted waves;
the cross-sectional area of the chamber as seen from below. (a large circle)

Net force = force up minus force down.

BUT, the chamber has the SAME CROSS-SECTIONAL AREA seen from above AND below!!!!

The lower circular side may be smaller than the upper circle, but it's irrelevant. What is relevant is the total cross-sectional area of the entire chamber as seen from top and bottom -- and they are identical.

The force up is exactly equal to force down. changing the shape of the chamber CANNOT violate this simple fact (unless you put holes in the chamber)

Once you "GET IT" that the partial force upward or downward (actually, in ANY arbitrary pair of directions) is determined by the total cross-sectional area, and not the area of any select piece of the chamber, then it becomes dirt-obvious that no net force can arise.

What I just pointed out in the post just before you which you might have missed is that there may be a time delay in the time it takes to cancel out that momentum. It takes 2 or more reflections to cancel out the momentum on the back reflection than it does for the top plate (1 reflection for top plate). There for you might get a force for a short duration towards the large plate before the force cancels out. Now imagine if you increase the amount of reflections before the photons are converted to heat. This is what the author is doing by using superconductivity to increase the Q or reflectivity of the metal. Not only that it would also be proportional to the amount of photons bouncing around in the sealed chamber.

The only problem with my hypothesis is that I don't know that this explains what is going on in the wave nature of light but from a particle perspective this seems like it might be a plausible explanation of what is going on.

So yes if you don't consider time it would seem the forces would cancel out. However if you consider the passage of time then there is a chance this might work. If so the force might be at least 1/2 that of the force of light reflecting. Reflecting light provides 2 times its force since the particle is being caught and thrown back. But if you increase the % of photons in that state and increase the reflections you would take that force generated by the power output times the number of reflections. ... I think.

[Qfwfq]

DustinTheWind said:

So yes if you don't consider time it would seem the forces would cancel out. However if you consider the passage of time then there is a chance this might work.

uh, brownian motion?

[DustinTheWind]

I don't believe it would be Brownian motion.

[Pyrotex]

DustinTheWind said:

.... However if you consider the passage of time then there is a chance this might work. If so the force might be at least 1/2 that of the force of light reflecting. Reflecting light provides 2 times its force since the particle is being caught and thrown back. But if you increase the % of photons in that state and increase the reflections you would take that force generated by the power output times the number of reflections. ... I think.

Nope. Still won't work.

Consider the Relativity Drive as a closed system. Say, it's built inside a spherical hull of your spaceship. The hull has no holes. Now turn your Drive on. Microwaves are generated and get reflected and absorbed and slapped around and bounced and jiggled and any thing else you care to do to them, inside the Drive chamber which can have any geometry that you can think of.

Now, let's go back outside the ship. Before Drive on, the ship is in free fall in deep space. No accelleration in any direction. Now turn Drive on. Nothing leaves the ship. Nothing gets through the hull. Therefore the total momentum of all radiation in the ship, stays in the ship. For every dF (tiny force) in one direction there is a -dF. Every photon is generated inside the Drive and is absorbed in the Drive. Net total momentum is zero.

The ship experiences no accelleration.

If it DID, say it experienced an accelleration up to a velocity V(d) in the d direction, and the ship has mass M(s) then it gains momentum equal to
V(d) * M(s) in the d direction.

Therefore, something ELSE, a particle let's say with mass M(p) must have been emitted in the -d direction so that
V(d) * M(s) + V(-d) * M(p) = 0

This is conservation of momentum, which is a Law in most civilized universes.
At least, those universes worth talking about.

But since nothing gets out of the Relativity Drive or the ship's hull, then there IS NO "particle".
Which is to say, M(p) = 0 and therefore the resulting velocity of the ship, V(d) must be zero.