Gravitational Time Dilation: A Layman's Explanation

[Hihello]

I think I have a layman's understanding of time and length contraction. However, I don't really understand gravitational time delay or distance contraction. If we were to put two clocks at the front and end of a rocket, the one on the top would experience time more quickly than would the one on the bottom. Why?

 

[rootone]

I don't think that's the case unless the ship is in a really deep gravity well, basically a black hole, and the ship is the process of being 'spaghettified' by it.
If the ship is traveling close to light speed relative to a distant observer, THEY would notice time dilation for the ship's clock(s).
A clock aboard the ship would appear to them to run slower than their own clock, (whether it's at the front of the ship or at the back).
The distant observer would witness the same thing if the ship is in a very deep gravity well, (although not yet deep enough in for tidal forces to be ripping it apart, - that's the spaghettification bit).
Assuming the ship is of a normal scale, a person inside it could go to one end of it to read a clock, then go to the other end and read the other clock and taking account of the time it takes to walk, they would not notice any discrepancy.

 

[jbriggs444]

I think I have a layman's understanding of time and length contraction. However, I don't really understand gravitational time delay or distance contraction. If we were to put two clocks at the front and end of a rocket, the one on the top would experience time more quickly than would the one on the bottom. Why?

As the rocket accelerates, the notion of simultaneity (as judged from an instantaneous co-moving inertial frame) is changing. Even if the two clocks start synchronized and follow identical acceleration profiles (as judged from some inertial frame), neither will remain synchronized with the other according to any of the succession of tangent frames in which they are momentarily at rest.

 

[Nugatory]

If we were to put two clocks at the front and end of a rocket, the one on the top would experience time more quickly than would the one on the bottom. Why?

Imagine that the clock at the tail of the rocket is a strobe light, flashing once per second according to someone sitting next to it on the rocket. How much time passes between the arrival of each flash at the nose of the rocket? If the rocket is accelerating, each flash has to travel slightly farther than its predecessor, so there will be more than one second between each arrival.

Once you understand this phenomenon clearly, the equivalence principle will get you to gravitational time dilation.

 

[harrylin]

I think I have a layman's understanding of time and length contraction. However, I don't really understand gravitational time delay or distance contraction. If we were to put two clocks at the front and end of a rocket, the one on the top would experience time more quickly than would the one on the bottom. Why?

It sounds as if you received a confused explanation... I'll try to clarify it.

1. For the effect of an accelerating rocket on clock readings, only special relativity is required if it is far from gravitational influences.
If we neglect length contraction* (and at first, with acceleration from v=0, that is very accurate) then the clocks are going at the same rate, according to measurements with a standard inertial reference system.
Inside the rocket, the signals form the rear clock that reach the front clock will be more and more delayed, and similarly the signals from the front clock that reach the rear clock will be less and less delayed. In other words, the frequency of the rear clock's second indicator that is received at the front clock will be less than 1 Hz.

* If in addition we account for length contraction, you may understand that due too the slightly bigger acceleration, the rear clock is always moving very slightly faster than the front clock, and thus it will even tick very slightly slower than the front clock according to measurements with a standard inertial reference system.

2. According to the equivalence principle, the observations should be the same inside a rocket that is resting on the ground in the Earth's gravitational field (with the same accelerometer readings as in the accelerating rocket).
Simplifying for the case of the Earth in rest, the delay time between the clocks is constant. Nevertheless the same frequency difference will be observed as in an accelerating rocket. Einstein concluded* that in a gravitational field the bottom clock is ticking slower than the top clock.

* See p.197, 198 of the English translation of the 1916 paper, here: http://web.archive.org/web/20060829045130/http://www.Alberteinstein.info/gallery/gtext3.html

 

[stevendaryl]

I wrote a post about this a long time ago, but I don't know what the URL is. There are two different effects at work at the same time in time dilation on board an accelerating rocket:

1. From the point of view of the "launch" frame (the frame where the rocket is originally at rest), the length of the rocket contracts continuously (Lorentz contraction). Because of this length contraction, the rear of the rocket is always moving slightly faster than the front of the rocket. That means that relativistic time dilation affects the rear slightly more than the front. So in the launch frame, the rear clock runs a little slower than the front clock.
2. Because of relativity of simultaneity, even if the front and rear clocks were synchronized in the "launch" frame, they would NOT be synchronized in the instantaneous rest frame of the rocket.

Let's identify three different events:

1. $e_1$ = the event at the rear clock when it shows time $T_{rear}$
2. $e_2$ = the event at the front clock that is simultaneous with $e_1$, according to the "launch" frame. Let $T_{front, 1}$ be the time on the front clock at this event.
   
3. $e_3$ = the event at the front clock that is simultaneous with $e_1$, according to the instantaneous rest frame of the rocket. Let $T_{front, 2}$ be the time on the front clock at this event.

So in the rocket frame, the discrepancy between the front and rear clocks is given by:

$\delta T = \delta T_1 + \delta T_2$

where $\delta T_1 = T_{front, 1} - T_{rear}$ and $\delta T_2 = T_{front,2} - T_{front, 1}$

$delta T_1$ is due to length contraction, while $\delta T_2$ is due to relativity of simultaneity.