tag:blogger.com,1999:blog-4005750306801645234.post1080484890756059171..comments2024-09-14T23:45:59.807+02:00Comments on Spacetime and Geometry: Exercise 3.06 Metric outside EarthUnknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4005750306801645234.post-65283900972264584052023-04-09T23:43:20.969+02:002023-04-09T23:43:20.969+02:00Hi, I think you make a mistake when you calculate ...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}. <br /><br />In 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!<br /><br />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. <br /><br />Finnoreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-84137481627731174282023-01-06T13:53:13.707+01:002023-01-06T13:53:13.707+01:00And with the hashhash or dollardollar delimiters: ...And with the hashhash or dollardollar delimiters: <br />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.)<br /><br />Thank's Petra!Georgehttps://www.blogger.com/profile/04824865122846470839noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-39689917962273423732023-01-06T03:55:52.160+01:002023-01-06T03:55:52.160+01:00Now some comments on (b). You were very methodical...Now 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.)<br />Here 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.<br />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.)Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-74604565978200459982023-01-06T03:08:11.739+01:002023-01-06T03:08:11.739+01:00For (a), there shouldn't be a minus inside the...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.Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.com