On 04/01/2010 02:06 PM, OrionWorks - Steven V Johnson wrote: >>From Mr. Lawrence > > ... > >> For example, if we dig a spherical chamber in the center >> of a planet, there will be *no* gravitational "field" >> within that chamber caused by the mass of the planet. >> However, the gravitational potential is lower in that >> chamber than it is on the surface, and clocks in the >> chamber will run SLOWER than clocks on the surface. > > ...SLOWER than clocks on the surface ??? > > You sure about that???
Yes, positive. > > What have precise atomic clocks revealed when positioned within the > deepest mine shafts of our own planet. Do they run faster or slower > than atomic clocks positioned at sea-level, where the effects of > gravity should in theory be greatest. Misconception -- gravitational time dilation depends on the potential, not on the field strength. (Believe it; it's true; I'll show why at the end of this note.) So, the clocks in mine shafts should run slower than clocks on mountains. I don't know if that's been demonstrated; the effect is pretty small. What *has* been demonstrated is that the "corrections" which GR said would be needed for the GPS were applied and are correct. The GPS satellites are going kind of fast, which implies an SR correction, but they're also far, far above the Earth, which is where the GR correction comes in. > > Can someone refresh my memory about the precise time measurements > conducted with atomic clocks positioned at different elevations on the > surface of Earth. Weren't the clocks positioned at the highest > elevations (mountains), where the effects of gravity were slightly > less, experiencing the passage of time more quickly that their > siblings positioned closer to sea-level, where gravity is slightly > stronger? ...or have I got it switched around. You've got that right but for the wrong reasons. Again, it's *potential*, not *strength*, which matters. Here's why, explained in a gedanken experiment: Einstein stands on a stepladder, holding a rock of mass M. His assistant sits at the foot of the ladder. Einstein drops the rock. The rock accelerates downward, accumulating kinetic energy. His assistant catches it, turning the kinetic energy into heat, and warming the rock. It now masses M + epsilon. The assistant, who is an apprentice sorcerer, turns the rock into electrical energy and uses it to power a laser. He shines the laser beam up at Einstein. The beam contains P photons, of frequency V, with total energy M+epsilon. Einstein, the master sorcerer, *catches* the beam in the palm of his hand (just like Darth Vader) and turns it back into a rock. But beware: If COE is to be observed, the rock Einstein creates must mass just M .... *not* M+epsilon! The beam Einstein catches must also contain P photons, of course -- the same number must arrive at the top of the ladder as departed at the bottom. However, total energy received by Einstein is smaller than that emitted by the assistant's laser. Therefore, the frequency must be smaller (since photon energy is proportional to frequency). So, frequency is V-delta. Note that WAVE CRESTS in a laser beam are physical entities, and we can count them. Einstein must receive the same number of WAVE CRESTS as were emitted. This can only work out if the arrival time of the beam is "stretched out", with fewer crests per second arriving at Einstein's hand than were emitted by the assistant's laser. And that implies the assistant's clock is running slower than Einstein's. ***** And this exact experiment may be performed using any two points, as long as an object can "fall" from one of the points to the other. For the spherical chamber in the planet, for example, we dig a very skinny shaft down from the surface to the chamber, and drop the rock down the shaft.
