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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!

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

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!