: : It also depends on whether you are in free fall there because tidal forces are not invariant.

: What does that mean? Does it just mean that as you : approach a gravitating sperical mass that the tidal : forces increase?

It means that the amount of tidal force calculated between the top of your space probe and the bottom of the probe will vary depending on whose coordinate frame is used to calculate it. It is not frame invariant. Sometimes a hybridizing of frames is even used to do such calculations. For instance take the tidal acceleration near the event horizon of a black hole given by Da = 16p3GLMh/C3 at the following web sight

Black hole equations.

This is most easily derived by taking a gradient with respect to "remote frame coordinates" of the proper acceleration. Thus L is a remote frame length. The proper acceleration given by a = -GM/r2. This acceleration is a hybrid of remote frame coordinate r and local free fall time t. a = d2r/dt2. The problem with calling that the tidal force for an object in free fall is not in the hybridizing of the acceleration, but in the common mistake in the use of the proper length of the ship for L. Because of special relativistic length contraction, if you have a 100m proper length for the ship you should use a much smaller number for L depending on initial conditions of speed and location for the fall.

• Re: Tidal forces for those into the math - Zephram_Cochrane/">Zephram Cochrane - October 19, 1999 - 06:31 UTC
• Re: blackhole - Walk_your_own_tracks./">Walk your own tracks. - October 19, 1999 - 06:41 UTC