# The time component of the geodesic equation for Newtonian gravity

I am working on a simple and popular GR textbook exercise. In Dodelson's Modern Cosmology (p. 54), it is stated thus:

The metric for a particle traveling in the presence of a gravitational field is $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$ where $$h_{00} = -2\phi$$ where $$\phi$$ is the Newtonian gravitational potential; $$h_{i0}=0$$; and $$h_{ij} = -2\phi\delta_{ij}$$. Show that the time component of the geodesic equation implies that energy $$p^0+m\phi$$ is conserved.

I have determined that the nonzero Christoffel symbols are $$\Gamma^0_{00}=\frac{\partial\phi}{\partial t},\quad\Gamma^i_{00}=c^2\frac{\partial\phi}{\partial x^i},\quad\Gamma^0_{ii}=\frac{-1}{c^2}\frac{\partial\phi}{\partial t},$$ by assuming that $$\phi \ll 1$$. From there I use the time component of the geodesic equation $$\frac{d^2t}{d\tau^2} = \Gamma^0_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau},$$ which simplifies considerably when written in terms of $$\gamma=(1-v^2/c^2)^{-1/2}$$: $$\gamma\frac{d\gamma}{dt} = -\frac{\partial\phi}{\partial t} - 2\gamma^2\left(\frac{\partial\phi}{\partial x}\frac{dx}{dt}+\frac{\partial\phi}{\partial y}\frac{dy}{dt} + \frac{\partial\phi}{\partial z}\frac{dz}{dt}\right).$$

According to the problem statement, however, I should be ending up with something like $$\frac{\partial}{\partial t}\left(p^0 +m\phi\right)=0$$ (in natural units with $$c=1$$). Does my result somehow reduce to this one? Thus far, I can't even see how the term $$m\phi$$ would arise.

• No $\dfrac {d\phi }{dt}$ just static field – Eli Jan 2 '19 at 21:30
• @Eli I'm not so sure. Earlier in the question, the author affirms that $\Gamma^0_{00}=\partial\phi/\partial t$ without mentioning that it vanishes. – Doubt Jan 3 '19 at 0:38

## 2 Answers

Starting from the geodesic equation

$$\frac{d^2x^0}{d\tau^2} = \Gamma^0_{\mu \nu} \frac{dx^\mu}{d\tau} \frac{dx^\nu}{d\tau} = \Gamma^0_{0 0} \frac{dx^0}{d\tau} \frac{dx^0}{d\tau} \tag{1}$$

Where the last equality follows from the fact that we are in the low speed limit that you can formulate as Dodelson does in this way: $$p^0 \gg p^i$$ or equivalently in term of the four-velocity $$u^0 \gg u^i$$

Since we are ignoring the space components of the four-velocity we have, from the normalization of the four-velocity

$$-1=u_\mu u^\mu = u^0 u^0 = \frac{dx^0}{d\tau} \frac{dx^0}{d\tau} \tag{2}$$

Using $$\Gamma^0_{0 0}= \frac{\partial{\phi}}{\partial{\tau}}$$, inserting $$(2)$$ in $$(1)$$ and multiplying it for $$m$$ we have

$$m \frac{d^2x^0}{d\tau^2} = \frac{d p^0}{d \tau} = - m \frac{\partial{ \phi}}{\partial{ \tau}} \tag{3}$$

From which we obtain

$$\frac{\partial}{\partial \tau} ( p^0 + m\phi) = 0 \tag{4}$$

And therefore the term between parenthesis in $$(4)$$ is conserved:

$$p^0 + m\phi = constant \tag{5}$$

If you're looking for conserved quantities, don't start with the geodesic equation. Write down the arc length as:

$$I = \int d\tau\left({\dot x}^{a}{\dot x}^{b}g_{ab}\right)$$, and take set the variation with respect to each of the $$x^{a}$$ equal to zero. This will give you the geodesic equations if you expand everything out, but it will give you conserved quantities of geodesic motion for free if you choose the right coordinates.