David Carroll said:
Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?
[Edit: On further consideration, the contents of this post are wrong (more than just the 42 vs 84 corrections; rather the whole thing is wrong). Sorry for the confusion.]
Allow me to replace the scenario with a new one that hopefully demonstrates the important concepts, without the need to bring in the flying spaghetti monster and its delivery person.
Consider a simple, non-rotating Earth. Drill a hole through the Earth from one side to the other. (If you wish, you can also remove the atmosphere from the Earth so it doesn't leak into the hole. If not, just neglect air resistance).
Now drop a clock down the hole. Also keep another clock (initially synchronized) on the surface of the Earth near one of the ends of the hole. After around
42 minutes [Edit: should be 84 minutes, not 42], the clock that you dropped down will come back up the same hole (the clock will oscillate back and forth through the Earth, very similar to a mass on a spring, with a period of roughly
42 minutes [Edit: should be 84 minutes, not 42]). At that moment, you can easily compare the two clocks.
Which clock has ticked slower?
Before you answer, here are a couple of things to note:
- At the moment you dropped the clock, both clocks were synchronized, and existed in the same place (well close enough) at the same time and had zero relative velocity.
- When the clock comes back up through the hole (after about
42 [Edit: should be 84, not 42] minutes or so), once again both clocks exist in the same place and time and have zero relative velocity; although now they are no longer synchronized. (The clocks are easy to compare though.)
The important point with the above notes is that both clocks share a single point in spacetime at the beginning, and they also share another point in spacetime at the end.
So which path through spacetime extremalizes proper time, and which does not?
Be careful here. A little knowledge about relativity can actually lead you to the wrong answer. Many would follow the following logic and reach the wrong conclusion:
- Well, the clock falling through the Earth is moving back and forth, and thus, like the standard twin paradox, it would tick slower, right?
- Also, the falling clock goes deeper into Earth's gravity well, so that's another reason why the falling clock should tick slower, due to gravitational time dilation, right?
With that, someone might conclude that the clock falling through Earth would tick slower, for multiple reasons. But that someone would be wrong.
The correct answer is that the falling clock will, overall, tick
faster! The clock sitting there at rest by the end of the hole will experience
more time dilation than the falling clock.
There is a little truth to the logic about the gravity-well deepness part, but the comparison to the standard twin paradox is very wrong here.
The falling clock is in
freefall, meaning it is tracing a path through spacetime on a geodesic. And that causes its proper time to be extremal. And it will always tick faster than any other clock that went though spacetime taking a different path. (Recall that both clocks share the same beginning and ending points in spacetime.)
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