To take the newton limit of the geodesic equation:

$$\frac{\mathrm{d}^2x^\mu}{\mathrm{d}\tau^2}+\Gamma^\mu_{\nu\rho}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\rho}{\mathrm{d}\tau}=0 $$

We assume obeject is moving slowly so :

$$\frac{\mathrm{d}x^i}{\mathrm{d}\tau}\ll \frac{\mathrm{d}x^0}{\mathrm{d}\tau} \tag{1} \text{ }i = 1,2,3$$

Therefore, we only take the first component of the second term in the geodesic equation.

My question is, does equation (1) necessarily mean that $\Gamma^\mu_{00}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}$ is the dominant term of all possible combinations? Because in the affine connection:

\begin{equation} \Gamma^\mu_{\nu\rho} = \frac{\partial{x^{\mu}}}{\partial{\xi^{\alpha}}} \frac{\partial^2{\xi^\alpha}}{\partial{x^\rho}\partial{x^\nu}} \end{equation}

the c terms in $x^0$ which is considered to dominate will be cancelled in the dominator of the affine connection. What makes us still take $\Gamma^\mu_{00}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}$ to be the dominate term?


You cannot expect so many $\Gamma^\mu_{\nu\rho}\frac{dx^\nu}{d\tau}\frac{dx^\rho}{d\tau}$ terms to become neglectable small in the non-relativistic limit. Because you have curvilinear coordinates, the situation is not so simple.

As a simple example consider a homogeneous gravity field (with strength $g$ in $z$-direction) in cylindrical coordinates $(r,\phi,z)$. From Newtonian/Lagrangian mechanics you have the non-relativistic equations of motion: $$\begin{align} \frac{d^2 r}{dt^2}-r\left(\frac{d\phi}{dt}\right)^2&=0 \\ \frac{d^2\phi}{dt^2}+\frac{2}{r}\frac{dr}{dt}\frac{d\phi}{dt}&=0 \\ \frac{d^2 z}{dt^2}+g&=0 \end{align}$$

From these equations you see that, for the non-relativistic limit of the geodesic equations (in coordinates $t,r,\phi,z$), you would expect dominant $\Gamma^\mu_{\nu\rho}\frac{dx^\nu}{d\tau}\frac{dx^\rho}{d\tau}$ terms from the following elements: $$\begin{align} &\Gamma^r_{\phi\phi} &=-r \\ &\Gamma^\phi_{r\phi}=\Gamma^\phi_{\phi r}&=\frac{1}{r} \\ &\Gamma^z_{tt}&=g \end{align}$$ You see $\Gamma^z_{tt}\frac{dt}{d\tau}\frac{dt}{d\tau}$ is not the only dominant term. The other terms arising from the curvilinear space coordinates are dominant as well. These other terms would only vanish if you we would have started with cartesian space coordinates $(x,y,z)$ instead of $(r,\phi,z)$.

  • $\begingroup$ Sorry, I chose the wrong word 'eliminate'. Take the example of A1B1+A2B2+A3B3 = c, if we know A1B1>>A2B2,A3B3 we can take the limit A1B1 = C. My question is, in the recovery of the newtonian potential shown in my post, we only know that B1>>B2,B3, how can we say that A1B1 is the dominant term? $\endgroup$
    – Winniebear
    Feb 2 '20 at 23:14
  • $\begingroup$ I think the answer is still valid. The $\Gamma^\mu_{00}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}\frac{\mathrm{d}x^0}{\mathrm{d}\tau}$ term is not the only dominant term. $\endgroup$ Feb 2 '20 at 23:20
  • $\begingroup$ If it is not the only dominant term, why in the derivation we only keep this term to find the Newtonian limit? $\endgroup$
    – Winniebear
    Feb 2 '20 at 23:45
  • 1
    $\begingroup$ May be the authors of this derivation (silently) assumed the space-like coordinates to be cartesian. $\endgroup$ Feb 3 '20 at 0:04

I think you are getting confused by the difference between the coordinate variables and the coordinates of a particular trajectory through spacetime.

It's helpful to rename your variables. Say you have a spacetime that you are describing with four coordinates called $x^0, x^1, \dots$. Then say you have a particular trajectory through that spacetime. The trajectory is a function that maps the object's proper time to points in spacetime, so there are four functions that describe the spacetime position as a function of proper time. We'll call them $y^0(\tau), y^1(\tau), \dots$ and so on. That is, at proper time $\tau$ the object has $x^0$ coordinate equal to $y^0(\tau)$, etc.

The important thing is that the functions that tell you the values of the coordinates of the object are different than the coordinate themselves.

In this notation the geodesic equation is $$\frac{d^2 y^\mu}{d\tau^2}= -\Gamma^\mu_{\rho\sigma} \frac{dy^\rho}{d\tau}\frac{dy^\sigma}{d\tau}$$ $$ \sim - g^{\mu \alpha}\frac{\partial g_{\alpha \rho}}{\partial x^\sigma} \frac{dy^\rho}{d\tau}\frac{dy^\sigma}{d\tau} $$

where it is key that the $\partial x$ in $\partial g^{\alpha\rho}/\partial x^\sigma$ is entirely different than the $dy^\sigma$. They are unrelated. One tells you how much spacetime is curved at a point, the other tells you how fast a particular object is moving past that point. Changing one has no bearing on the other. I can consider a whole continuum of trajectories where $dy^0/d\tau$ is as large as I want and it will not change the fact that the gradient of the metric at the point is $1/r$ or whatever.

And that's what's happening. We are comparing what happens to different trajectories at the same point. We keep the metric and Christhoffel symbols fixed and then study what happens to trajectories where $dy^0/d\tau$ becomes large compared to the other components of the trajectory.


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