energy of photons in FLRW universe

Note that for photons $p_{\mu}p^{\mu} = 0$, unlike particles for which $p_{\mu}p^{\mu} = m^2c^2$

$$p_{\mu}p^{\mu} = 0$$
$$p_{t}p^{t} +p_{r}p^{r} +p_{\theta}p^{\theta} +p_{\phi}p^{\phi} = 0$$

$$g^{tt}{p_{t}}^2 +g_{rr}{p^{r}}^2 +g_{\theta\theta}{p^{\theta}}^2 +g^{\phi\phi}{p_{\phi}}^2= 0$$

we know that $p_t = \frac{E}{c}$, $p^r = \frac{dr}{d\lambda}$, where $\lambda$ is an affine parameter. Also, $p^{\theta} = mc\frac{d\theta}{d\lambda} = 0$. And $p_{\phi} = L$

$$\frac{1}{1-\frac{2GM}{c^2R}}[\frac{E}{c}]^2 -\frac{1}{1-\frac{2GM}{c^2R}}[\frac{dr}{d\lambda}]^2 -\frac{L^2}{r^2} =0$$

calling $\frac{GM}{c^2} = \mu$

$$\frac{1}{1-\frac{2\mu}{R}}[\frac{E}{c}]^2 -\frac{1}{1-\frac{2\mu}{R}}[ \frac{dr}{d\lambda}]^2 -\frac{L^2}{r^2} =0$$

$$[\frac{dr}{d\lambda}]^2 = [\frac{E}{c}]^2 -\frac{L^2}{r^2}(1-\frac{2\mu}{R})$$

$$[\frac{dr}{d\lambda}]^2 +V_{eff} = [\frac{E}{c}]^2$$
$$V_{eff} =\frac{L^2}{r^2}(1-\frac{2\mu}{R})$$

$$[\frac{dr}{d\phi}\frac{d\phi}{d\lambda}]^2 = [\frac{E}{c}]^2 -\frac{L^2}{r^2}(1-\frac{2\mu}{R})$$
$$[\frac{dr}{d\phi}]^2[\frac{L}{r^2}]^2 = [\frac{E}{c}]^2 -\frac{L^2}{r^2}(1-\frac{2\mu}{R})$$

say $u = \frac{1}{r}$ implying that $du = -\frac{dr}{r^2}$
and let’s call $\frac{E}{Lc}= \frac{1}{b}$
$$[\frac{du}{d\phi}]^2L^2 = [\frac{E}{c}]^2 -L^2u^2(1-2\mu u)$$
$$[\frac{du}{d\phi}]^2 = [\frac{1}{b}]^2 -u^2 +2\mu u^3$$

differentiating this equation, we get

$$2 \frac{du}{d\phi}\frac{d^2u}{d\phi^2} = -2u +2*3\mu u^2$$
$$\frac{du}{d\phi}\frac{d^2u}{d\phi^2} +u\frac{du}{d\phi} =3\mu u^2\frac{du}{d\phi}$$
$$\frac{d^2u}{d\phi^2} +u =3\mu u^2$$

we need to solve this perturbatively i.e $u = u_0 +\epsilon u_1$ where $\epsilon = \frac{3\mu}{b}$

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