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Haorong Wu

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- TL;DR Summary
- How to understand the circular orbits in Schwarzschild metric for photons?

Hello, there. I am learning the chapter, The Schwarzschild Solution, in Spacetime and geometry by Caroll. I could not grasp the idea of circular orbits.

It starts from the equations for ##r##, $$\frac 1 2 (\frac {dr}{d \lambda})^2 +V(r) =\mathcal E$$ where $$V(r)=\frac {L^2}{2r^2}-\frac {GML^2}{r^3}$$ and $$ \mathcal E =\frac 1 2 E^2.$$ ##L## is the angular momentum and ##E## is the conserved energy.

By ##dV/dr=0##, it can be solved to obtain the circular orbits ##r_c=3GM##.

All above are from the textbook. Now I am trying to solve the variables for the circular orbits for photons.

First, in the circular orbits, ##dr/d\lambda## should be zero, so ##V(r)=\mathcal E##. After substituting ##r_c## into ##r##, I have $$\frac {L^2}{27 G^2 M^2}=E^2.$$ Next, for photons, ##L=r_c \cdot p## and ##E=pc##. But now, the equation becomes $$1=\frac {1}{3c^2}.$$ I am not sure what goes wrong.

My guess is that ##E## is not the energy for photons. I mean not the energy just related to the frequency of the photons. From its definition, $$E=(1-\frac {2GM}{r})\frac {dt}{d\lambda},$$ maybe ##\frac {dt}{d\lambda}## is the part from the frequency, while ##-\frac {2GM}{r}\frac {dt}{d\lambda}## is the gravitational potential energy. Should I subtract the potential energy part from the frequency part?

Thanks!

It starts from the equations for ##r##, $$\frac 1 2 (\frac {dr}{d \lambda})^2 +V(r) =\mathcal E$$ where $$V(r)=\frac {L^2}{2r^2}-\frac {GML^2}{r^3}$$ and $$ \mathcal E =\frac 1 2 E^2.$$ ##L## is the angular momentum and ##E## is the conserved energy.

By ##dV/dr=0##, it can be solved to obtain the circular orbits ##r_c=3GM##.

All above are from the textbook. Now I am trying to solve the variables for the circular orbits for photons.

First, in the circular orbits, ##dr/d\lambda## should be zero, so ##V(r)=\mathcal E##. After substituting ##r_c## into ##r##, I have $$\frac {L^2}{27 G^2 M^2}=E^2.$$ Next, for photons, ##L=r_c \cdot p## and ##E=pc##. But now, the equation becomes $$1=\frac {1}{3c^2}.$$ I am not sure what goes wrong.

My guess is that ##E## is not the energy for photons. I mean not the energy just related to the frequency of the photons. From its definition, $$E=(1-\frac {2GM}{r})\frac {dt}{d\lambda},$$ maybe ##\frac {dt}{d\lambda}## is the part from the frequency, while ##-\frac {2GM}{r}\frac {dt}{d\lambda}## is the gravitational potential energy. Should I subtract the potential energy part from the frequency part?

Thanks!