Let $\gamma(s)$ be the worldline of a photon. In coordinates $x$ we have $x(\gamma(s))=(x^0(s),x^1(s),x^2(s),x^3(s))$

The tangent vector $V$ of the photon in this coordinates is given by $V=\dot{x}^a \frac{\partial}{\partial x^a}$. Let $g$ be the metric in this space , since $V$ is null we have that $$g(V,V)=\dot{x}^a\dot{x}^b g\left(\frac{\partial}{\partial x^a},\frac{\partial}{\partial x^b}\right)=\dot{x}^a\dot{x}^b g_{ab}=0 \tag 1$$ Multiplying $(1)$ by $ds$ and solve in relation to $dx^0$ we obtain two solution \begin{array}{l} d x^{0(1)}=\frac{1}{g_{00}}\left(-g_{0 \alpha} d x^{\alpha}-\sqrt{\left.\left(g_{0 \alpha} g_{0 \beta}-g_{\alpha \beta} g_{00}\right) d x^{\alpha} d x^{\beta}\right)}\right. \\ d x^{0(2)}=\frac{1}{g_{00}}\left(-g_{0 \alpha} d x^{\alpha}+\sqrt{\left(g_{0 \alpha} g_{0, \beta}-g_{\alpha \beta} g_{00}\right) d x^{\alpha} d x^{\beta}}\right) \tag 2 \end{array}

Now as is showed in the figure Bob sends light towards Alice and is immediately back reflected back to Bob.If $x_e^0$ is the time coordinates of emission and $x_r^0$ is the time coordinates of reception of the photon by Bob why we have $x_e^0=x^0+ d x^{0(1)}$ and $x_r^0=x^0+ d x^{0(2)}$?

I am extracting this from this Wikipedia page Synchronous frame

Let $G$ be a lie group and $G_i$ its generators. Suppose for a set of fields $\chi_{\alpha}(x)$ we have $$ \left[G_{i}, \chi_{\alpha}(x)\right]=-\left(g_{i}\right)_{\alpha \beta} \chi_{\beta}(x) $$

If the Lagragian $\mathcal{L}$ is invariant under the $G$ then we have the conserved charges $$ G_{i} \equiv Q_{i}=\int d^{3} x J_{i}^{o}=\int d^{3} x\left[\frac{\partial \mathcal{L}}{\partial \partial_{o} \chi_{\alpha}} \frac{1}{i}\left(g_{i}\right)_{\alpha \beta} \chi_{\beta}\right] $$ Suppose that $$ \left\langle 0\left|\left[G_{i}, \chi_{\alpha}(x)\right]\right| 0\right\rangle=-\left(g_{i}\right)_{\alpha \beta}\left\langle 0\left|\chi_{\beta}(x)\right| 0\right\rangle \neq0. $$ then we have $$ \begin{aligned} \left(g_{i}\right)_{\alpha \beta}\left\langle 0\left|\chi_{\beta}(x)\right| 0\right\rangle=& \sum_{n} \int d^{3} y\left\{\left\langle 0\left|e^{-i P y} J_{i}^{o}(0) e^{i P y}\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle\right.\\ &\left.-\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|e^{-i P y} J_{i}^{o}(0) e^{i P y}\right| 0\right\rangle\right\} \\ =& \sum_{n} \int d^{3} y e^{i P_{n} y}\left\langle 0\left|J_{i}^{o}(0)\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle \\ &-\sum_{n} \int d^{3} y e^{-i P_{n} y}\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|J_{i}^{o}(0)\right| 0\right\rangle \\ =& \sum_{n}(2 \pi)^{3} \delta^{3}\left(\vec{p}_{n}\right)\left\{e^{-i P_{n}^{o} y^{o}}\left\langle 0\left|J_{i}^{o}(0)\right| n\right\rangle\left\langle n\left|\chi_{\alpha}(x)\right| 0\right\rangle\right.\\ &\left.-e^{+i P_{n}^{o} y^{o}}\left\langle 0\left|\chi_{\alpha}(x)\right| n\right\rangle\left\langle n\left|J_{i}^{o}(0)\right| 0\right\rangle\right\} \end{aligned} $$ By assumption this expression does not vanish and, furthermore, since the LHS is independent of $y^{o}$ it must also be independent of $y^{o} .$ Clearly this can only happen if in the theory there exist some massless one-particle states $|n\rangle$ and only these states contribute to the sum.

This is the proof that I found of Goldstone theorem .

But the proof only shows that particles are massless . How do we show that these particles are scalars and how the fields $\phi_i$ associated with this $$|p,i\rangle$$ particles transform under the group $G$ ?