I Is the Killer Crate Paradox Resolved?

[Sagittarius A-Star]

Edit 2: Conclusion: If I am the criminal mastermind plotting this deed then Alice better listen to Bob.

I thought it would be easier for Alice to throw a sphere (with radius $R$) instead of a crate, that fits through the window, with speed $\frac{c}{\sqrt{2}}$ at 45°. But now I am not sure.

By length contraction it becomes an ellipsoid:
$a=R$
$b=R/\gamma(\frac{c}{\sqrt{2}}) = R \frac{\sqrt{2}}{2}$.

$x(\varphi) = R \cos(\varphi)$
$y(\varphi) = R \frac{\sqrt{2}}{2} \sin(\varphi)$

$d = \sqrt{x(135°)^2+y(135°)^2}=R \frac{\sqrt{3}}{2}$

The result for d has the vertical component of the length contraction of the sphere, which fits to the length contraction factor of the window, moving at $\frac{c}{2}$. But this cannot go through the window, because in the ground frame, the sphere must touch the "South" edge of the window earlier than the "North" edge. Reason: Relativity of simultaneity.

My conclusion would be, that Alice is in this case not allowed to use a 45° angle.

Am I correct?

 

[Dale]

I thought it would be easier for Alice to throw a sphere (with radius R) instead of a crate, that fits through the window, with speed c2 at 45°. But now I am not sure.

Look at it in the train frame. If it can fit in the train frame then it can fit in the ground frame too. Just describe it fully in the train frame and then transform to the ground frame.

 

[Sagittarius A-Star]

Look at it in the train frame. If it can fit in the train frame then it can fit in the ground frame too. Just describe it fully in the train frame and then transform to the ground frame.

In the train frame it must fit through the window. Then there must be an error in my calculation.

 

[Orodruin]

$x(\varphi) = R \cos(\varphi)$
$y(\varphi) = R \frac{\sqrt{2}}{2} \sin(\varphi)$

$d = \sqrt{x(135°)^2+y(135°)^2}=R \frac{\sqrt{3}}{2}$

You have an arithmetic error here somewhere. Considering that $R/\sqrt{2} < R$, it follows that
$$
d = R\sqrt{ \cos^2(\varphi) + \sin^2(\varphi)/2} \leq R \sqrt{\cos^2(\varphi) + \sin^2(\varphi)} = R.
$$
You therefore clearly cannot get a result larger than $R$.

That the ball will fit through the window is clear from the fact that it can be inscribed in the box and we know that the box will go through if oriented correctly.

 

[Sagittarius A-Star]

You therefore clearly cannot get a result larger than $R$.

My $d = R \frac{\sqrt{3}}{2}$ is smaller than $R$. The window is length contracted by the same factor. But there must be an arithmetical error.

 

[Orodruin]

My $d = R \frac{\sqrt{3}}{2}$ is smaller than $R$. The window is length contracted by the same factor. But there must be an arithmetical error.

Yes, that was clearly a brain fart on my part. Ok, new try: The argument 135 degrees does not correspond to the line d that you drew. It is the vertical line in the circle rest frame but is tilted in the ground rest frame. This probably has something to do with the resolution. You need the point on the ellipse with the largest vertical projection.

 

[Orodruin]

The vertical projection is $-\cos(\varphi)/\sqrt 2 + \sin(\varphi)/2$ (I normalized R to one). This has amplitude $\sqrt 3/2$, same as the length contraction of the window. Hence, no problem.

 

[jartsa]

The question is not whether Alice and Bob know about Wigner rotation. The question is if the assassins do. If they do then they can make the plot work by launching the box in the appropriate orientation. If they don’t, they will fail. So whether Alice or Bob will be right will depend on the assassins’ knowledge of relativity.

Oh yes, Alice and Bob were police people, not assassins. The story has too many characters for me to follow.

 

[Sagittarius A-Star]

Yes, that was clearly a brain fart on my part. Ok, new try: The argument 135 degrees does not correspond to the line d that you drew. It is the vertical line in the circle rest frame but is tilted in the ground rest frame. This probably has something to do with the resolution. You need the point on the ellipse with the largest vertical projection.

I think I found my error (after a long time). It has nothing to do with relativistic effects, because I calculated only in the ground frame. I simply wrote wrong formulas for an ellipse.

Contains wrong formulas:

$a=R$
$b=R/\gamma(\frac{c}{\sqrt{2}}) = R \frac{\sqrt{2}}{2}$.

$x(\varphi) = R \cos(\varphi)$
$y(\varphi) = R \frac{\sqrt{2}}{2} \sin(\varphi)$

$d = \sqrt{x(135°)^2+y(135°)^2}=R \frac{\sqrt{3}}{2}$

Correct formulas:

$(x, y) = (a \cos(t), b \sin(t))\ \ \ \ \$ Here $0 \leq t \leq 2\pi$ is a parameter, but not the angle to a point on the elipse.

Polar form relative to center:
$r(\varphi) = \frac{ab}{\sqrt{((b \cos(\varphi))^2 + ((a \sin(\varphi))^2}}$

Source:
https://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center

 

[Orodruin]

I think I found my error (after a long time). It has nothing to do with relativistic effects, because I calculated only in the ground frame. I simply wrote wrong formulas for an ellipse.

Contains wrong formulas:Correct formulas:

$(x, y) = (a \cos(t), b \sin(t))\ \ \ \ \$ Here $0 \leq t \leq 2\pi$ is a parameter, but not the angle to a point on the elipse.

Polar form relative to center:
$r(\varphi) = \frac{ab}{\sqrt{((b \cos(\varphi))^2 + ((a \sin(\varphi))^2}}$

Source:
https://en.wikipedia.org/wiki/Ellipse#Polar_form_relative_to_center

As I said earlier, there is nothing really wrong with the formulas. The wrong assumption is that the angle $\varphi = 135^\circ$ corresponds to the vertical line. This is true only in the rest frame of the ellipse.

 

[Sagittarius A-Star]

As I said earlier, there is nothing really wrong with the formulas. The wrong assumption is that the angle $\varphi = 135^\circ$ corresponds to the vertical line. This is true only in the rest frame of the ellipse.

Yes. For calculating $d$, I should have used either another angle (that of the related point on the circle, before "compressing" the circle to an ellipse) or the other formula for $r(\varphi)$ in posting #59.