Special Relativity and atomic clocks

[keeperofthekeys]

The question is have is as follows:

In 1971 four portable atomic clocks were flown around the world in jet aircraft, two east bound and two westbound, to test the times dilation predictions of relativity. a) If the westbound plane flew at an average speed of 1500 km/h relative to the surface, how long would it have to fly for the clock on board to lose 1s relative to the reference clock on the ground? b) In the actual experiment the plane circumflew Earth once and the observed discrepancy of the clocks was 273ns. What was the plane's average speed?

For the first part, I convereted 1500km/h to 416.6m/s, then put it terms of c, or 1.389x10^-6c. I then took the standard equation t'/(sqrt(1-(1.389x10^-6c)^2)) = t.
I then made t-t'=1, solved for t'=-1+t and put that into the equation.
Solving, I got 1x10^12 s, or 31,688 years. Is this a reasonable answer?

I think my method would be wrong, because I followed similar steps to get b. After removing the t' prime from the equation as I did in part a, I set t = 40075160m/v, 40075160m being the circumference of the earth. My equation was as follows:
(1/(sqrt(1-(v^2/c^2)))*(-273x10^-9s + 40075160/v) = 40075160/v

I solved for v, but got 1226.2 m/s, which is a speed I don't believe we've even held for a sustained flight now, let alone in 1971. Where am I going wrong?

 

[Andrew Mason]

The question is have is as follows:

In 1971 four portable atomic clocks were flown around the world in jet aircraft, two east bound and two westbound, to test the times dilation predictions of relativity. a) If the westbound plane flew at an average speed of 1500 km/h relative to the surface, how long would it have to fly for the clock on board to lose 1s relative to the reference clock on the ground? b) In the actual experiment the plane circumflew Earth once and the observed discrepancy of the clocks was 273ns. What was the plane's average speed?

For the first part, I convereted 1500km/h to 416.6m/s, then put it terms of c, or 1.389x10^-6c. I then took the standard equation t'/(sqrt(1-(1.389x10^-6c)^2)) = t.
I then made t-t'=1, solved for t'=-1+t and put that into the equation.
Solving, I got 1x10^12 s, or 31,688 years. Is this a reasonable answer?

I think my method would be wrong, because I followed similar steps to get b. After removing the t' prime from the equation as I did in part a, I set t = 40075160m/v, 40075160m being the circumference of the earth. My equation was as follows:
(1/(sqrt(1-(v^2/c^2)))*(-273x10^-9s + 40075160/v) = 40075160/v

I solved for v, but got 1226.2 m/s, which is a speed I don't believe we've even held for a sustained flight now, let alone in 1971. Where am I going wrong?

Just a suggestion that would make it easier for you and for others: you should show the equations in algebraic form and then plug in numbers.

I think you are supposed to ignore the effects of gravity on time and treat it as the airplane moving in uniform motion relative to the ground clock. The problem is that gravity has a real effect, so the numbers based on SR only will not correspond to reality.

As you realize, you have to use the Lorentz transformation for time between two inertial frames. That transformation is:

$$t' = t_o\gamma$$ where

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{1}{\sqrt{1-\beta^2}}$$

For small velocities you can use the binomial expansion and ignore higher order terms so it works out to:

$$t' = t_0(1 + \frac{v^2}{2c^2})$$

$$t' - t_0 = t_0\frac{v^2}{2c^2}$$


Now the complicated part here is to take into account the fact that the surface of the Earth is moving. What about 'it is all relative' you say? Well, it is all relative to another inertial frame but neither the plane nor the surface are really inertial. So think of everything happening in the rest frame of the centre of the earth. You can see that both the Earth surface and the airplane are continually moving with respect to that frame (which is almost inertial). So work out the time dilations for the plane and the Earth surface relative to that frame. You will see why it is significant when you do the math.

The starting point on the surface is moving from east to west at a speed of 40,000,000 m/24 hr = 463 m/sec or about 1667 km/hr.

If the plane is moving from east to west at 1500 km/hr relative to this point, it is moving at 3167 km/hr (v_p) relative to the inertial rest frame. So its time dilation relative to the rest frame is:

$$t_p - t_0 = t_0\frac{v_p^2}{2c^2}$$

The time dilation of the starting point on the Earth is:

$$t_e - t_0 = t_0\frac{v_e^2}{2c^2}$$ where v_e = 1667 km/hr

The difference is:

$$t_p - t_e = t_0(\frac{v_p^2}{2c^2} - \frac{v_e^2}{2c^2})= \frac{t_0}{2c^2}(v_p^2 - v_e^2)$$

Since v/c is very small, t_e is approximately equal to t₀ so:

$$t_p - t_e \approx t_e(\frac{v_p^2}{2c^2} - \frac{v_e^2}{2c^2})= t_e(\frac{v_p^2 - v_e^2}{2c^2})$$


See if you can work that out. As I say, you should get a speed that is higher than actual because part of that observed time dilation was due to general relativity effects.

AM

 

[lubuntu]

Sorry to revive a very old thread but I'm working on the same problem. I get the same answers as the OP and am convinced I am right. I'm not sure why you'd have to account for the motion of the Earth about it is axis because we are comparing the clocks on the planes to the clock on the Earth surface and considering the Earth's surface as the inertial reference frame not comparing it some clock that claims it is a rest compared to the Earth's surface.