Timeline for What is the additional gravitational term from general relativity given by?

8 events

when toggle format	what		by	license	comment
Jun 7, 2016 at 19:18	vote	accept	user3741635
Jun 7, 2016 at 17:54	comment	added	John Rennie		@user3741635: aha, Carroll derives the expression for a unit mass, i.e. $m=1$, so $m$ does not appear in his equation.
Jun 7, 2016 at 17:54	history	edited	John Rennie	CC BY-SA 3.0	Respond to comment
Jun 7, 2016 at 5:01	comment	added	John Rennie		Hmm, yes, OK. I'll have to see if I can dig out a copy of Carroll's book to see exactly what he has derived. The form of $V_\text{eff}$ I've given is the one I was taught, and the one that matches Wikipedia in the limit of $M\gg m$.
Jun 7, 2016 at 4:59	comment	added	user3741635		Sorry to bother you again John but you said that Carroll gives the potential per unit mass, so to get the potential I need to multiply by $m$ but doing that I get the effective potential as $$ V_{eff}(r) = \frac{1}{2}\epsilon\,m-\frac{GMm}{r} + \frac{m L^2}{2r^2} - \frac{GmML^2}{c^2r^3} \tag{1} $$ If you notice the last two terms are different from the last two term in Eq $(1)$ in you answer. Could you explain this please .
Jun 5, 2016 at 18:56	vote	accept	user3741635
Jun 7, 2016 at 4:45
Jun 5, 2016 at 18:49	history	edited	John Rennie	CC BY-SA 3.0	Tweak
Jun 5, 2016 at 18:39	history	answered	John Rennie	CC BY-SA 3.0