Question
A good approximation to the metric outside the surface of the Earth is provided by$$
{ds}^2=-\left(1+2\Phi\right){dt}^2+\left(1-2\Phi\right){dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)
$$where$$
\Phi=-\frac{GM}{r}
$$may be thought of as the familiar Newtonian gravitational potential. Here ## G## is Newton's constant and ## M## is the mass of the Earth. For this problem ## \Phi## may be assumed to be small.
a) Imagine a clock on the surface of the Earth at a distance ##R_1## from the Earth's center, and another clock on a tall building at distance ##R_2## from the Earth's center. Calculate the time elapsed on each clock as a function of the coordinate time ## t##. Which clock moves ticks faster?
b) Solve for a geodesic corresponding to a circular orbit around the equator of the Earth (##\theta=\pi/2## ). What is ## d\phi/dt##?
c) How much proper time elapses while a satellite at radius ##R_1## (skimming along the surface of the Earth, neglecting air resistance) completes one orbit? You can work to first order in ## \Phi## if you like. Plug in the actual numbers for the radius of the Earth and so on (don't forget to restore the speed of light) to get an answer in seconds. How does this number compare to the proper time elapsed on the clock stationary on the surface?
Answers
Apparently the metric given is the Newtonian metric. I learnt how to reinstate the speed of light ## c## and so do real calculations. I was brutally reminded that ##{ds}^2=-{d\tau}^2## where ## \tau## is the proper time and that the proper time is what appears on ideal clocks. I also found the Schwarzschild radius - the radius of the event horizon of a black hole with a given mass. Some time ago I had found two sample solutions which had been left lying around on the internet by careless professors. They have now gone but luckily I had downloaded them before. I found that they were both littered with errors! The worst was probably UCSB's who popped an extra question and got totally the wrong answer for gravitational time dilation on a GPS satellite. Lucky they weren't involved in the construction! Finally I had a look to see if I could do anything with elliptical orbits but then decided I had spent enough time on this exercise.
Full answer and bonus questions at
Ex 3.06 Metric outside Earth.pdf (16 pages including links to other answers)
For (a), there shouldn't be a minus inside the square root in (3). So proper time is real. (It would be crazy for us to measure our own everyday time using imaginary numbers, wouldn't it?) For time to remain time (since time is the dimension that has negative signature), we must have GM/r < 1. Also, the author clearly stated that GM/r should be assumed to be small. So your discussion about the scenario in which 2GM > R1, R2 should be better left out.
ReplyDeleteNow some comments on (b). You were very methodical. It is good for beginners. But as you make more progress, you will one day realize that that laziness is a virtue. (BTW, the UCSB solution is also not lazy enough.)
ReplyDeleteHere is a fast way to get the correct solution. From a conservation of energy point of view, you can immediately realize that both dt/dτ and dφ/dτ must be a constant. (The object should have constant speed.) To make your life easier, you can simply set λ=t because dt/dτ is a constant.
Now to keep something in orbit, the tricky part is really that r must be a constant. So the only interesting geodesic equation is the one for r. The first term of (28) must vanish. So we have a simple equation Γ^r_{tt} (dt/dt)^2 + Γ^r_{φφ} (dφ/dt)^2 = 0. (The other terms vanish.) But dt/dt = 1. So we have dφ/dt = ±√ (-Γ^r_{tt} / Γ^r_{φφ}) and we are done. (There is probably a sign error in one of your Γ's.)
And with the hashhash or dollardollar delimiters:
DeleteSo we have a simple equation $$Γ^r_{tt} (dt/dt)^2 + Γ^r_{φφ} (dφ/dt)^2 = 0$$. (The other terms vanish.) But ##dt/dt = 1##. So we have $$dφ/dt = ±√ (-Γ^r_{tt} / Γ^r_{φφ})$$ and we are done. (There is probably a sign error in one of your ##Γ##'s.)
Thank's Petra!
Hi, I think you make a mistake when you calculate the proper time of the clock that is stationary on the Earth in part c) and in part d) ( the extra UCSB question). Actually in part c) I don't see where you calculate the proper time elapsed by a clock that is stationary on the Earth? You see to assume that time is t? To get that proper time you need the relationship between the proper time and the coordinate time that you found in part b). So you need a factor of sqrt{1- 2GM/R_1}.
ReplyDeleteIn part d) you write a factor sqrt{1- 3GM/R_1} in equation (77) but it should be sqrt{1- 2GM/R_1} for the same reason. That's why you should replace the 3 with a 2!
Also the "Is there a significant effect from special relativity?" part seems like you have not understood that GR extends SR you don't need to incorporate effects from SR into GR. In GR there you can think that there are two effects one comes from the fact that the clocks are in a gravitational field and the other effect coming from the motion relative to the gravitational field. The sqrt{1- 3GM/R_2} in (76) takes both of these into account which you can see clearly in equation (72) . But since the clock on the Earth is stationary there is no effect from it moving so there should only be the factor sqrt{1- 2GM/R_1}. You see that from equation (72) since in the case of the stationary clock you have to put d phi/dt =0.