tag:blogger.com,1999:blog-4005750306801645234.post8550603445546811945..comments2024-02-02T20:17:24.676+01:00Comments on Spacetime and Geometry: Exercise 5.5 Observer and beacon outside black holeUnknownnoreply@blogger.comBlogger4125tag:blogger.com,1999:blog-4005750306801645234.post-42425112078493105442023-11-06T13:44:42.494+01:002023-11-06T13:44:42.494+01:00Oops! Yes. I corrected it, thanks Lorenzo.
Sorr...Oops! Yes. I corrected it, thanks Lorenzo. <br /><br />Sorry for the delay. Sunday is my day off.Georgehttps://www.blogger.com/profile/04824865122846470839noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-65646852029669370692023-11-04T14:24:22.025+01:002023-11-04T14:24:22.025+01:00Hi, would you kindly review formula (44) of the at...Hi, would you kindly review formula (44) of the attachment? It seems a factor 2 is missing. Thanks! LorenzoAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-22133099601885706142023-02-17T15:22:20.902+01:002023-02-17T15:22:20.902+01:00Two hashes in a row to delimit inline math!
Now in...Two hashes in a row to delimit inline math!<br />Now in a locally inertial coordinate system, where the metric is Minkowski, this quantity is nothing other than ##1/\sqrt{1-\vec{v}^2}##. (This is actually special relativity stuff.) With ##r=r_{obs} = 2GM##, we have ##v^\mu v'_\mu = \infty## and therefore ##\vec{v} = 1##, where ##\vec{v}## is the 3-velocity of the beacon observed by the observer who suddenly decided to fall freely as the beacon passes by (and making its own frame a locally inertial coordinate system).Georgehttps://www.blogger.com/profile/04824865122846470839noreply@blogger.comtag:blogger.com,1999:blog-4005750306801645234.post-31827783237775568192023-02-17T02:29:01.955+01:002023-02-17T02:29:01.955+01:00Here's my solution to part (b). From part (a),...Here's my solution to part (b). From part (a), the time component of the 4-velocity of the beacon is $$v^0 = \left[\frac{1}{1-2GM/r}\frac{r(r_*-2GM)}{ r_*(r-2GM)} \right]^{1/2}.$$ The observer's 4-velocity is $$v'_\mu = ((1-2GM/r_{obs})^{1/2},0,0,0).$$ So we have $$v^\mu v'_\mu = \left[\frac{1-2GM/r_{obs}}{1-2GM/r}\frac{r(r_*-2GM)}{ r_*(r-2GM)} \right]^{1/2}.$$ Now in a locally inertial coordinate system, where the metric is Minkowski, this quantity is nothing other than $1/\sqrt{1-\vec{v}^2}$. (This is actually special relativity stuff.) With $r=r_{obs} = 2GM$, we have $v^\mu v'_\mu = \infty$ and therefore $\vec{v} = 1$, where $\vec{v}$ is the 3-velocity of the beacon observed by the observer who suddenly decided to fall freely as the beacon passes by (and making its own frame a locally inertial coordinate system).Petra Axolotlhttps://www.blogger.com/profile/06597951512037995047noreply@blogger.com