Would a body free falling in a gravitational field which has a gradient large enough that it would affect the free falling body 'feel' the effect of the tidal forces on it. I'm curious because would any ruler falling with the body be stretched the same amount as the free falling body, meaning that in the frame of the free fall, there would be no measurable stretching?

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    $\begingroup$ Is the freely falling body more or less rigid than the ruler? $\endgroup$ – DJohnM Aug 14 '14 at 20:22
  • $\begingroup$ Classically, if your ruler is stretched it's no longer a good measuring device. $\endgroup$ – BMS Aug 14 '14 at 23:51

There is a difference between "feeling the force" and "being stretched".

If you imagine two balls connected by a spring, and falling towards a massive object, then the closer ball will experience a greater force and therefore "accelerate away" from the ball that is further away - the spring between them will stretch, and thus provide a force balance.

A ruler that falls alongside these balls may or may not change shape in the same way - in general though, it will not. Unless the ruler was made of the same materials as the object falling, it's likely that its shape would not change in the same way (especially if the object falling has been made especially flexible as in my example).

Note this is a classical answer - there is no need to invoke special relativity for there to be a differential force on the two balls in my example.


The principle of equivalence only applies to objects that are "sufficently" small over a timeframe that is also "sufficently" small. In order to feel tidal forces, an object has to have finite size, and if the tidal forces are measurably "visible", then the object is not sufficiently small.


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