# Which clock runs faster? [duplicate]

Can someone help me giving a qualitative answer to this problem in General Relativity:

Imagine you are on earth with two perfectly synchronized clock's. If you hold on in your hand but you throw the other one in the air and catch it after a certain time interval, would they still give the same time?

I think that the one in the air would run faster but I cannot explain why.

Thank you!

## marked as duplicate by AccidentalFourierTransform, Henning Makholm, John Rennie general-relativity StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 22 '17 at 6:43

The clock you throw into the air would run faster.

Doing the calculation rigorously is quite involved, but there is a simple agrument we can use to see why the thrown clock records more time that the stationary clock, and we do this by comparing it to the twin paradox.

Suppose twin $A$ is stationary and twin $B$ is in a rocket. At time zero twin $B$ passes twin $A$ at some velocity $v$, and at that moment they synchronise their clocks. Twin $B$ now starts a constant acceleration towards twin $A$. This constant acceleration gradually slows twin $B$, brings them to a halt then accelerates them back towards twin $A$. When they meet again $A$'s clock shows a time $t_A$ and $B$'s clock shows a time $t_B$. The twin paradox is that $t_B \lt t_A$ i.e. the accelerating twin's clock shows less time.

And this is the key point:

the accelerating twin's clock shows less time than the non-accelerating twin's clock

(If you're interested this type of motion is discussed in What is the proper way to explain the twin paradox?)

In your situation we also have a constant acceleration, the gravitational acceleration $g$, and we have two observers. The key point is that it is the observer who is stationary on Earth who is the accelerating one, therefore it is the observer stationary on Earth who records less time i.e. whose clock runs slower.

Thios may seem odd, because after all it seems like the thrown object that is doing the accelerating. However the acceleration that matters here is the proper acceleration. The proper acceleration of you standing on the Earth's surface is non zero while the proper acceleration of the thrown clock is zero (give or take a bit of air resistance).

The thrown clock is in free fall, which according to GR means that it describes a timelike geodesic through spacetime.

A timelike geodesic is characterized by a variational principle: It is a local maximum of experienced (i.e. proper) time among neighboring timelike curves through spacetime with the same endpoints.

The clock in your hand describes a curve that is close enough to the free-call curve that it must experience less time than the thrown one.

So after you catch the thrown clock, the one that stayed in your hand will show an earlier time, because it didn't experience as much proper time.

• So you mean that "holding the clock" is an additional force on it and that's why it is not in free fall while the other one is in free fall (only gravity is "acceleratong" it)? – APORIL Jan 21 '17 at 13:45
• @Itachi: Yes, correct. – Henning Makholm Jan 21 '17 at 14:28