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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.

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