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I'm watching this introductory lecture on general relativity by Leonard Susskind (see 31:50 onwards for the relevant part) [Video title: General Relativity Lecture 1, channel: Stanford].

To give context, till that point he'd explained that a uniform gravitational field can be simulated or is equivalent to going to another coordinate frame that's accelerating (constant accelaration in the opposite direction) irrespective of the position.

He then moves on to the case of a gravitational field pointing radially inward towards the sun:

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There's a 2000 ft man who's falling legs first towards the sun. He'll feel a stretch since the force on legs is greater than that on his head. The claim is that no coordinate transformation can simulate this effect.

The way I'm trying to understand it is: if that person were in an elevator in vacuum, there'd be no kind of motion that the elevator can follow such that he'd feel a similar stretch across his body. My question is: why are we sure that there isn't some sort of super-weird motion that the elevator can follow, such that he feels the same stretching effect?

As an example, let's say the person was in the middle of a giant, hollowed-out cylinder (oriented along the cylinder axis). Then he'd feel a radially outward pulling force on his body. If we try to simulate this with the person in the vacuum elevator, no matter what translations we do (zero/non-zero, constant/non-constant accelaration), we can't replicate the radially outward pulling. But now I come to know of this very new kind of motion called rotation, and realize that if the elevator is rotated about its vertical axis, the person inside will feel the same as when he was in the cylinder.

So to reiterate (and I'm certain that I'm missing something) why do we say so surely that the radially inward example (where the person gets stretched) can't be replicated by a vacuum elevator?

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    $\begingroup$ When we say, "elevator" we are talking about a platform (a floor) that pushes upward against the soles of your feet, and walls and a ceiling whose only purpose is to block your view so that you cannot tell whether the upward force is due to the elevator resting on the surface of a planet, or uniformly accelerating through empty space. The only place where an elevator touches you is at the soles of your feet. In order to "stretch" you, something would have to pull your feet one way while pulling your head the opposite way. Whatever that thing is, it doesn't sound much like an elevator to me. $\endgroup$ – Solomon Slow Dec 9 '19 at 23:29
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Warning: this is a mathematical answer.

We know that it's not possible to make tidal forces disappear because mathematically they are governed by the Riemann curvature tensor, which is (in a way) independent of the coordinates.

In more detail: say we have an observer with four-velocity $U^\mu$. If two points (say, the observer's head and feet) are separated by a vector $\xi^\mu$, they will feel a tidal acceleration given by

$$\ddot{\xi}^\mu = R^\mu{}_{\nu\rho\sigma} U^\nu U^\rho \xi^\sigma.$$

Essentially, this says that the tidal acceleration (or relative acceleration) is a linear function of the observer's four-velocity (which isn't really important here, we can treat it as fixed) and the separation. The linear function connecting them is $R$, the Riemann curvature tensor. And the thing about tensors is that if they are zero in one coordinate system, they are zero in all of them. So a non-zero $R$ can't be an artifact of the coordinates; you can't make it go away with a transformation.

In contrast, the acceleration (just acceleration, not relative) of something moving in free fall is given by

$$\dot{U}^\mu = - \Gamma^\mu{}_{\nu\rho} U^\nu U^\rho.$$

And the $\Gamma^\mu{}_{\nu\rho}$, which are known as the Christoffel symbols, do not form a tensor, even though the expressions look similar. So it's possible to make the gravitational acceleration (which is proportional to $\Gamma$) equal to zero by choosing the coordinate system appropriately.

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  • $\begingroup$ Thanks a lot! Admittedly I don't understand the math yet and I'm guessing it's not possible to intuitively justify why it's impossible to make tidal forces disappear. I'll definitely revisit the answer when I've gone through some tensor calculus. $\endgroup$ – Shirish Kulhari Dec 9 '19 at 18:48
  • $\begingroup$ @ShirishKulhari I might try to add a more physical explanation later, no promises though. $\endgroup$ – Javier Dec 9 '19 at 18:51

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