I apologize for the subsequent headache. There is a person who claims quite adamantly that Einstein is wrong, using the following reasoning:
A clock on a train slows down the faster the train moves. Mechanical clocks are valid clocks too. The wheels of the train are mechanical clocks. Do they slow down?
The train moves at $0.5c$. The diameter of its wheels is $1$ meter for simplicity. The top of the wheel moves at $0.5c$ relative to the train, the lower point at $-0.5c$. The sides move vertically. From an observer on the ground the lower point of the wheel is stationary, just about to move upwards. The upper part has velocity $$\frac{0.5c + 0.5c}{1+0.25c^2/c^2} = \frac{c}{1.25} = 0.8c.$$ With that said, we can see that the top of the wheel rotates at 80% of the angular velocity required to keep the train moving at $0.5c$ - the wheel is deformed.
What this shows further is that the wheels as a whole don't slow down. They have to keep spinning. So time is not running any slower in this sense. It's only deformed.
Also, this shows that this thing that we call time is not a real thing - because the real thing would be a lower rate of events, as in a lower angular velocity of the wheel. Instead, it's just a mathematical variable we can play around with.
I know he is wrong but I'm not an expert in this and I would enjoy an answer.
Answer
Yes, in the reference frame of the rail rotation of the wheel must slow down, that seems as a paradox. It seems that the faster the train moves, the slower its wheels must rotate.
If the train moves at velocity close to $c$, rotation of the wheels must slow down until almost complete stop. This way the train appears to be “slipping” on the rail in the reference frame of the rail, though it is impossible, since “smooth rotation” is an absolute effect and it cannot depend on chosen reference frame.
Resolution of the paradox is in relativistic kinematics. The rim of the wheel Lorentz – contracts as velocity of the train increases.
The rest length of the rim of the wheel must remain constant. This means that the rim Lorentz contracts, and that the radial extension of the wheel contracts accordingly. The result is that the wheel becomes infinitely small in the limit that the train moves with the velocity of light.
If $v$ is velocity on the rim in the rest frame $K$ of the wheel, we have $\Omega=v/R$, where $R=R_0/\gamma$ is the contracted radius of the rotating wheel, and $R_0$ is their radius when they are at rest. The angular velocity of the rotating wheel is then
$$\Omega = \gamma v /R_0$$
Hence, in this case the angular velocity $\Omega$ must approach an infinitely great value in $K$ when the speed of the train approaches that of light. As observed in the rail frame $K'$, the distance between the marks on the path each time a point on a rim of the wheel leaves it is
$$l'=\gamma 2 \pi R = 2\pi R_0$$
and this distance is independent of the speed of the train, even if the radius of the wheel decreases with increasing velocity, because the distance between the marks depends upon the rest length of the rim of the wheel and not their Lorentz contracted length. Also in this frame the angular velocity of the wheel remains finite even if the wheel has a vanishing radius when the velocity of the train approaches that of light,
$$\Omega'=\gamma^{-1} \Omega = v/R_0$$
and hence $\lim\limits_{v \to c} \Omega' =c/R_0$, which is finite.
As you correctly noted, bottom part of the wheel is stretched and upper part is contracted. Spokes of rolling wheel have unusual shape; they are “inclined” upward. Please look for: VI. The shape of a rolling wheel, fig 6. And fig.7 https://oda.hioa.no/nb/a-relativistic-trolley-paradox.
Also: K. Voyenli “Alternative derivation of the circumference of a relativistic rotating disk” Am. J. Phys. 45, 876-877 (1977)
Also this work, Fig. 8 and Fig. 9 at page 39 https://www.researchgate.net/publication/252135276_Space_Geometry_in_Rotating_Reference_Frames_A_Historical_Appraisal
No comments:
Post a Comment