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I have been studying GR for sometime and doing exercises from Schutz and I have a question about differentiating along a geodesic. Here is what I know. The equation of geodesic in terms of four momentum is given as,$$p^\alpha p^{\beta}_{;\alpha}$$. Now if I want to differentiate a scalar along the geodesic I figured I have to do this, $$\frac{d\phi}{d\tau}$$ Here, $\tau$ is the proper time which is the parameter along the curve. The change of the scalar $\phi$ along the curve is equal to, $$\frac{d\phi}{d\tau}=\phi,_{\beta}U^\beta$$ Here, $U^\beta$ is the four velocity of the curve. Writing this in covariant derivative form I believe it should just be,$$\frac{d\phi}{d\tau}=\phi_{;\beta}U^\beta$$ So if a scalar (like a dot product between vectors) is constant along the geodesic then I believe it means that,$$\frac{d\phi}{d\tau}=0.$$ Is this correct?

In the question I am trying to solve, the condition is that $p^\alpha\epsilon_\alpha=\text{constant}$ along the geodesic. I am trying to write the condition of what this means to proceed with further calculation.

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So I ended up trying to solve the problem, assuming what I stated above is correct. I will state the problem here for reference, Show that if a vector field $\epsilon^\alpha$ satisfies Killing’s equation then $p^\alpha\epsilon_\alpha$ is constant along the geodesic. So I just took covariant derivative of $\phi$ as, $$\frac{d\phi}{d\tau}=U^\beta\phi_{;\beta}$$ and then set it to zero. Here I defined $\phi = p^\alpha \epsilon_\alpha$. Then basically expanded the covariant derivative of $\epsilon$ and got to the point (using the equation of geodesic $U^\alpha U^\beta_{;\alpha}=0$) where you get $U^\alpha U^\beta \epsilon_{\alpha,\beta}$. Since $\epsilon^\alpha$ is known to be Killing vector field then it satisfies the Killing equation $\epsilon_{\alpha,\beta}=-\epsilon_{\beta,\alpha}$ which basically makes it anti-symmetric hence the sum $U^\alpha U^\beta \epsilon_{\alpha,\beta}$ is zero which proves that, $$\frac{d(p^\alpha\epsilon_\alpha)}{d\tau}=U^\beta(p^\alpha\epsilon_\alpha)_{;\beta}=0$$.

Hope this made sense. I also checked solution manual (which I only check once I am out of all options to figure it out myself) this is how it is done as well.

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