# Hafele and Keating Experiment

1. Feb 6, 2004

### Sammywu

While researching how SR effect could be applied to GPS clock after GR effect, I encountered the experiment made on 1971.

In the section "Kinematic Time Shift Calculation", I saw an totally no sense algorithm, violating simple math.

First,
Ts = T0 * (1+R^2w^2/(2*c^2)), where Ts is the Earth surface time, R is the Earth's radius and w is the Earth's angular velocity of Earth's rotation.

Second,
Ta = T0 * (1+(R^2*w^2+v^2)/(2*c^2)), where Ta is the airplane's time and v is airplane's velocity.

Then,
Ta-Ts = T0 *( 2Rwv+V^2)/(2c^2).

Up to here , it's totally reasonable.

Next thing, it brought in is completely math. incorrect.

Ta-Ts = - Ts * (2Rwv+v^2)/2c^2.

Note the minus sign coming from nowhere. If Ta is larger than Ts from the first two equations, Ta-Ts will never be negative.

If this math. can live in the Physics, there is definitely something wrong.

2. Jan 15, 2007

### mma

I think you mean Kinematic Time Shift Calculation

Yes, this derivation is loose a litle bit.
In the first equation, i.e. in $$T = \frac{T_0}{\sqrt{1 - v^2/c^2}}$$
$$T_0$$ is the time elapsed between two events occuring in the centre of the Earth, while $$T$$ stands for the time elapsed between this two events in an inertial reference frame moving with velocity v with respect the centre of the Earth.

Really, here is nothing to do with this. We need just the opposite. If the proper time elapsed on the airplane is T (this isn't an inertial frame!), then during this, in the centre of the Earth elapses
$$T_0 = \frac{T}{\sqrt{1 - v^2/c^2}}$$
time (the centre of the Earth is regarded now to move inertially)

So, our second equation reads correctly
$$T \approx T_0 \left[ 1 - \frac{v^2}{c^2}\right]$$.

And so on. The sign was missed at the beginning.

3. Jan 15, 2007

### Chris Hillman

4. Jan 16, 2007

Fine! Thx!