Gravitational DE(?) from Schwartzschild spacetime

  • #1
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Main Question or Discussion Point

Imagining that an object spining around a spherical mass M has angular momentum that has z-component(θ=0) only, then

$$g_{μν}\frac{dx^μ}{dτ}\frac{dx^ν}{dτ}=(1-\frac{r_s}{r})c^2(\frac{dt}{dτ})^2-\frac{1}{1-\frac{r_s}{r}}(\frac{dr}{dτ})^2-r^2(\frac{dθ}{dτ})^2-r^2\sin^2θ(\frac{dφ}{dτ})^2=c^2$$

becomes

$$(1-\frac{r_s}{r})c^2(\frac{dt}{dτ})^2-\frac{1}{1-\frac{r_s}{r}}(\frac{dr}{dτ})^2-r^2(\frac{dφ}{dτ})^2=c^2.$$
(because the object's path follows ##θ=\frac{π}{2}## and ##\frac{dθ}{dτ}=0##.)


According to my professor, by calculus of variation,

$$δ\int ds=δ∫\sqrt{g_{μν}\frac{dx^μ}{dτ}\frac{dx^ν}{dτ}}=0.$$

and if we define ##g_{μν}\frac{dx^μ}{dτ}\frac{dx^ν}{dτ}=c^2≡2T##,

$$δs=δ\int{\sqrt{2T}dτ}=\int{\frac{δT}{\sqrt{2T}}dτ}=\frac{1}{c^2}\int{δT}dτ=0.$$

Therefore, by Euler-Lagrange Equation ##\frac{d}{dτ}(\frac{∂T}{∂\dot{x_μ}})-\frac{∂T}{∂x^μ}=0##, with ##μ=0## we get
$$(1-\frac{r_S}{r})\frac{dt}{dτ}=const.$$


So, here goes my question finally.

By multiplying ##mc## to ##\frac{dt}{dτ}##,
$$mc\frac{dt}{dτ}=m\frac{d(ct)}{dτ}=m\frac{dx^0}{dτ}=mU^0=P^0=\frac{E}{c}$$(##U^0## represents four-velocity and ##P^0## four-momentum)

so if we compare the first and the last term, ##\frac{dt}{dτ}=\frac{E}{mc^2}## is the result that I think is correct. But my textbook tells ##(1-\frac{r_S}{r})\frac{dt}{dτ}=\frac{E}{mc^2}##.

If I use ##\frac{dt}{dτ}=\frac{E}{mc^2}##, I get differential equation below:
$$(\frac{dr}{dτ})^2+\frac{J^2}{m^2r^2}(1-\frac{r_S}{r})-\frac{2GM}{r}=c^2[(1-\frac{r_S}{r})^2(\frac{E}{mc^2})^2-1],$$

but using ##(1-\frac{r_S}{r})\frac{dt}{dτ}=\frac{E}{mc^2}##, I get
$$(\frac{dr}{dτ})^2+\frac{J^2}{m^2r^2}(1-\frac{r_S}{r})-\frac{2GM}{r}=c^2[(\frac{E}{mc^2})^2-1].$$

Could you tell me why ##\frac{dt}{dτ}=\frac{E}{mc^2}## is not correct? Thanks!
 

Answers and Replies

  • #2
PeterDonis
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Therefore, by Euler-Lagrange Equation ##\frac{d}{dτ}(\frac{∂T}{∂\dot{x_μ}})-\frac{∂T}{∂x^μ}=0##, with μ=0 we get

$$
(1-\frac{r_S}{r})\frac{dt}{dτ}=const.
$$
Yes. This constant is usually called "energy at infinity" (more precisely it would be "energy per unit mass at infinity"). There is also another constant of the motion, since the metric is also independent of ##\varphi##.

if we compare the first and the last term, ##\frac{dt}{dτ}=\frac{E}{mc^2}## is the result that I think is correct. But my textbook tells## (1-\frac{r_S}{r})\frac{dt}{dτ}=\frac{E}{mc^2}##.
That's because your textbook is using ##E## to mean "energy at infinity" (and then dividing by the rest mass to get "energy per unit mass at infinity", as I said above). But you are using ##E## to mean "energy as measured by an observer momentarily at rest with respect to the object". Those are two different things, and only the first one (the one your textbook is using ##E## to mean) is a constant of the motion.
 
  • #3
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Oh I totally got it. Thanks a lot!!
 

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