Relativity drive

CraigD · 10-02-2008

Quote:

Originally Posted by wade_b

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 $M$ and a left and a right wall separated by distance $D$ .
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 $p$ .
Per conservation of momentum, the box has a momentum of $–p$ , and therefore, a velocity $\frac{–p}{M}$ .
The photon is absorbed by the right wall after time $\frac{D}{c}$ .
The box has traveled $\frac{–pD}{cM}$ in that time.
Its momentum is again 0.

Therefore, it has shifted its position $\frac{–pD}{cM}$ , 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, $pc$ . When the right wall absorbed it, its mass increased by the same amount.
$pc = E = mc^2$ , so $m_{\gamma}= \frac{pc}{c^2} = \frac{p}{c}$ .

Center of mass equations are similar to torque equations.
Assuming the left and right walls were initially of equal mass, the initial center is $\frac{D}2$ .
The new center of mass $d_1$ is given by the equation
$\left(\frac{D}2-\frac{p}{c}\right) \left(D-d_1 \right) = \left(\frac{D}2-\frac{p}{c}\right)d_1$
which gives
$d_1 = \left( \frac{M}2 -\frac{p}{c}\right)\frac{D}{M} = \frac{D}2 -\frac{pD}{cM}$

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

wade_b · 10-02-2008

Quote:

Originally Posted by CraigD

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

The paradox:

The box has mass $M$ and a left and a right wall separated by distance $D$ .
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 $p$ .
Per conservation of momentum, the box has a momentum of $–p$ , and therefore, a velocity $\frac{–p}{M}$ .
The photon is absorbed by the right wall after time $\frac{D}{c}$ .
The box has traveled $\frac{–pD}{cM}$ in that time.
Its momentum is again 0.

Therefore, it has shifted its position $\frac{–pD}{cM}$ , 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, $pc$ . When the right wall absorbed it, its mass increased by the same amount.
$pc = E = mc^2$ , so $m_{\gamma}= \frac{pc}{c^2} = \frac{p}{c}$ .

Center of mass equations are similar to torque equations.
Assuming the left and right walls were initially of equal mass, the initial center is $\frac{D}2$ .
The new center of mass $d_1$ is given by the equation
$\left(\frac{D}2-\frac{p}{c}\right) \left(D-d_1 \right) = \left(\frac{D}2-\frac{p}{c}\right)d_1$
which gives
$d_1 = \left( \frac{M}2 -\frac{p}{c}\right)\frac{D}{M} = \frac{D}2 -\frac{pD}{cM}$

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?

CraigD · 10-03-2008

Quote:

Originally Posted by wade_b

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.

Karnuvap · 10-05-2008

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 · 10-15-2008

Quote:

Originally Posted by Karnuvap

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.

Quote:

Originally Posted by Karnuvap

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 · 10-15-2008

Quote:

Originally Posted by Qfwfq

: 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 · 10-15-2008

Quote:

Originally Posted by KickAssClown

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 · 10-16-2008

Quote:

Originally Posted by Karnuvap

...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.

KickAssClown · 12-02-2008

Chinese are building it, apparently.

I figured I would update on this topic.

Relativity drive

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