I am trying to find geodesics for the FRW metric, $$ d\tau^2 = dt^2 - a(t)^2 \left(d\mathbf{x}^2 + K \frac{(\mathbf{x}\cdot d\mathbf{x})^2}{1-K\mathbf{x}^2} \right), $$ where $\mathbf{x}$ is 3-dimensional and $K=0$, $+1$, or $-1$.

Geodesic equation

Using the Christoffel symbols From Weinberg's Cosmology (Eqs. 1.1.17 - 20) in the geodesic equation I get:

\begin{align} 0 &= \frac{d^2 t}{d\lambda^2} + a\dot{a} \left[ \left( \frac{d\mathbf{x}}{d\lambda} \right)^2 +\frac{K(\mathbf{x}\cdot \frac{d\mathbf{x}}{d\lambda})^2}{1-K \mathbf{x}^2}\right], &\text{($t$ equation)}\\ 0 &= \frac{d^2\mathbf{x}}{d\lambda^2} + 2 \frac{\dot{a}}{a}\frac{dt}{d\lambda}\frac{d\mathbf{x}}{d\lambda} + \left[ \left(\frac{d\mathbf{x}}{d\lambda}\right)^2 + \frac{K(\mathbf{x} \cdot \frac{d\mathbf{x}}{d\lambda})^2}{1-K\mathbf{x}^2} \right]K\mathbf{x}, &\text{($\mathbf{x}$ equation)} \end{align} where $\lambda$ is the affine parameter, and $\dot{a}=da/dt$.

Variational principle

It should also be possible to get the geodesics by finding the paths that extremize the proper time $d\tau$, i.e. using the Euler-Lagrange equations with a Lagrangian equal to the square root of the $d\tau^2$ I wrote above: $$ L = \frac{d\tau}{dp}= \sqrt{ t'^2 - a(t)^2 \left(\mathbf{x}'^2 + K \frac{(\mathbf{x}\cdot \mathbf{x}')^2}{1-K\mathbf{x}^2} \right) }, $$ where a prime is the derivative with respect to the variable $p$ that parameterizes the path.

When I try this $L$ in the E-L equation for $t$ I get the same equation as above. However, when I try the E-L equation for $\mathbf{x}$ my result does not agree with the geodesic equation.

I find $$ \frac{\partial L}{\partial \mathbf{x}} = -\frac{1}{L} \frac{a^2 K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), $$ and

$$ \frac{\partial L}{\partial \mathbf{x}'} = -\frac{a^2}{L} \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right). $$

I write the E-L equation $$\frac{d}{dp}\frac{\partial L}{\partial \mathbf{x}'}=\frac{\partial L}{\partial \mathbf{x}},$$ and then multiply both sides by $dp/d\tau$ to replace $p$ with $\tau$ everywhere and get rid of the $L$'s in the denominators (using the fact that $1/L=dp/d\tau$ and changing the meaning of the primes to mean derivatives with respect to proper time $\tau$).

I get

$$ \frac{d}{d\tau} \left[ a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right) \right] = \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} a^2 \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right). $$

I cannot rearrange this into the formula from the geodesic equation and suspect that the two sets of equations are not equivalent. I've gone through both methods a couple of times but haven't spotted any errors.

Can anyone tell me where the inconsistency (if there actually is one) is coming from?

[Interestingly, the E-L equation can be integrated once with an integrating factor of $\sqrt{1-K\mathbf{x}^2}$, whereas I don't see how to do so with the geodesic equation (not that I am very good at solving differential equations).]

  • 1
    $\begingroup$ I think you are also missing some steps. For example, if $dt/dp$ is not zero, that should also mean that $d(a)/dp$ is not zero since $a(t)$ is a function of $t$. Also, note that these two "methods" must agree because they are actually one method: Euler-Lagrange equation for trajectories are geodesics so they must satisfy geodesic equation, i.e. Euler-Lagrange for this Lagrangian is precisely the geodesic equation. $\endgroup$ Apr 25, 2019 at 6:09
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Apr 25, 2019 at 6:33
  • $\begingroup$ You've definitely gone wrong somewhere if the two aren't coming out as equivalent. Like Everiana said, the geodesic equation results from applying the Euler-Lagrange equation; in the case of timelike worldlines it should extremeise proper time. I am just wondering what is your motivation for defining L with respect to p in the way you have? If it were me I would explicitly parameterise the worldline by propertime so that L = 1. This way you know that extremising L will extremeise proper time. $\endgroup$
    – Ollie113
    Apr 25, 2019 at 9:34
  • $\begingroup$ For clarification, what is the metric signature that you are using? $\endgroup$
    – Ollie113
    Apr 25, 2019 at 9:35
  • $\begingroup$ @Ollie113 I start with $p$ instead of $\tau$ as independent variable in the "action" integral because I'm not comfortable with the integral having fixed values of $\tau$ at the end points. With fixed $\tau$ end points, it doesn't seem like doing the variation will extremize proper time. For me it's clearer to start with an arbitrary parameterization for doing the variation and then switching to using proper time. For the metric signature Weinberg uses $(-,+,+,+)$, but you don't need one for the E-L equations ($d\tau$ is a proper time which is positive for moving objects). $\endgroup$
    – Alex
    Apr 25, 2019 at 11:39

2 Answers 2


I don't think that your geodetic equations are correct?

This is your metric:

$$G= \left[ \begin {array}{cccc} 1&0&0&0\\ 0&- \left( a \left( t \right) \right) ^{2} \left( 1+{\frac {K{x}^{2}}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }} \right) &-{\frac { \left( a \left( t \right) \right) ^{2}Kyx}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }}&-{\frac { \left( a \left( t \right) \right) ^{2}Kzx}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }}\\ 0&-{ \frac { \left( a \left( t \right) \right) ^{2}Kyx}{1-K \left( {x}^{2} +{y}^{2}+{z}^{2} \right) }}&- \left( a \left( t \right) \right) ^{2} \left( 1+{\frac {K{y}^{2}}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }} \right) &-{\frac { \left( a \left( t \right) \right) ^{2} Kzy}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }} \\ 0&-{\frac { \left( a \left( t \right) \right) ^{ 2}Kzx}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }}&-{\frac { \left( a \left( t \right) \right) ^{2}Kzy}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2 } \right) }}&- \left( a \left( t \right) \right) ^{2} \left( 1+{ \frac {K{z}^{2}}{1-K \left( {x}^{2}+{y}^{2}+{z}^{2} \right) }} \right) \end {array} \right] $$

and my program calculate this geodetic

$$ {\frac {d^{2}}{d{\lambda}^{2}}}t \left( \lambda \right) +{\frac {a \left( t \right) \left( -1+K{y}^{2}+K{z}^{2} \right) \left( {\frac {d}{dt}}a \left( t \right) \right) \left( {\frac {d}{d\lambda}}x \left( \lambda \right) \right) ^{2}}{-1+K{x}^{2}+K{y}^{2}+K{z}^{2}}} -2\,{\frac {a \left( t \right) Kyx \left( {\frac {d}{dt}}a \left( t \right) \right) \left( {\frac {d}{d\lambda}}x \left( \lambda \right) \right) {\frac {d}{d\lambda}}y \left( \lambda \right) }{-1+K {x}^{2}+K{y}^{2}+K{z}^{2}}}+{\frac {a \left( t \right) \left( -1+K{x} ^{2}+K{z}^{2} \right) \left( {\frac {d}{dt}}a \left( t \right) \right) \left( {\frac {d}{d\lambda}}y \left( \lambda \right) \right) ^{2}}{-1+K{x}^{2}+K{y}^{2}+K{z}^{2}}}+{\frac {a \left( t \right) \left( -1+K{x}^{2}+K{y}^{2} \right) \left( {\frac {d}{dt}}a \left( t \right) \right) \left( {\frac {d}{d\lambda}}z \left( \lambda \right) \right) ^{2}}{-1+K{x}^{2}+K{y}^{2}+K{z}^{2}}}=0$$

and so on.

2) If you want to calculate the geodetic with E.L. methode you can also use this lagrangian


  • $\begingroup$ I think you're missing minus signs on the last two terms on the diagonal of the metric ($g_{yy}$ and $g_{zz}$). Also the $a(t)$'s should be squared. $\endgroup$
    – Alex
    Apr 25, 2019 at 17:34
  • $\begingroup$ yes i correct it, but is your geodetic for t is o.k ? $\endgroup$
    – Eli
    Apr 25, 2019 at 18:59
  • $\begingroup$ Yes, both EL and geodesic eqn seem to give the same $t$ equation. But the $\mathbf{x}$ equation looks different between the two. Does your program do the $x$ geodesic equations too? $\endgroup$
    – Alex
    Apr 25, 2019 at 19:02
  • $\begingroup$ I still think that you "t equation" not correct? I compare it with my computer calculate equation $\endgroup$
    – Eli
    Apr 25, 2019 at 19:23
  • $\begingroup$ I think you may still have an error in there. The metric is totally symmetric with respect to x, y, and z but your equation doesn't look very symmetric -- e.g. the second line has $Kyx$ but there's no terms with $Kyz$ or $Kzx$ like I would expect. $\endgroup$
    – Alex
    Apr 25, 2019 at 19:28

I think the equations may be consistent after all. First a solution to the EL equation for $\mathbf{x}$ also satisfies the geodesic equation:

Starting with the EL equation I have above: $$ \frac{d}{d\tau} \left[ a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right) \right] = \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} a^2 \left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right), $$

define $\mathbf{f}$ as $$ \mathbf{f} \equiv a^2\left(\mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}}{1-K\mathbf{x}^2}\right),$$

so the EL equation is

$$ \frac{d\mathbf{f}}{d\tau} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} \mathbf{f}=\mathbf{0}. $$

Note that \begin{align} \mathbf{f} \cdot \mathbf{x} &= a^2\left(\mathbf{x} \cdot \mathbf{x}' + \frac{K(\mathbf{x} \cdot \mathbf{x}')\mathbf{x}^2}{1-K\mathbf{x}^2}\right) \\ &= a^2 (\mathbf{x} \cdot \mathbf{x}') \left(1 + \frac{K\mathbf{x}^2}{1-K\mathbf{x}^2}\right) \\ &= \frac{a^2 (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2}, \end{align} and $$ \mathbf{f} \cdot \mathbf{x}' = a^2\left(\mathbf{x}'^2 + \frac{K(\mathbf{x} \cdot \mathbf{x}')^2}{1-K\mathbf{x}^2}\right) \equiv Q, $$ ($Q$ appears in the geodesic equation for $\mathbf{x}$).

Next dot the EL equation with $\mathbf{x}$: \begin{align} 0 &= \frac{d\mathbf{f}}{d\tau} \cdot \mathbf{x} - \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2} \mathbf{f}\cdot \mathbf{x} \\ &=\frac{d}{d\tau}\left(\mathbf{f} \cdot \mathbf{x}\right) - \mathbf{f} \cdot \mathbf{x}' - a^2 \frac{K (\mathbf{x} \cdot \mathbf{x}')^2}{(1-K\mathbf{x}^2)^2}, \end{align} so $$ \frac{d}{d\tau}\left(\mathbf{f} \cdot \mathbf{x}\right) = Q +a^2 \frac{K (\mathbf{x} \cdot \mathbf{x}')^2}{(1-K\mathbf{x}^2)^2}. $$

Now go back to the original EL equation (first equation) and apply the $d/d\tau$ inside the brackets:

$$ \text{EL LHS} = \frac{d}{d\tau}\left(a^2 \mathbf{x}'\right) + \left[\frac{d}{d\tau}\left( a^2 \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2}\right)\right]\mathbf{x} + a^2 \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2}\mathbf{x}'. $$ The last term above cancels with the first term on the right hand side of the EL equation.

Moving everything that's left to one side you get

\begin{align} \mathbf{0} &= \frac{d}{d\tau}\left(a^2 \mathbf{x}'\right) + \left[\frac{d}{d\tau}\left( a^2 \frac{K (\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2}\right)\right]\mathbf{x} - \frac{a^2 K^2(\mathbf{x} \cdot \mathbf{x}')^2\mathbf{x}}{(1-K\mathbf{x}^2)^2} \\ &= \frac{d}{d\tau}\left(a^2 \mathbf{x}'\right) + \left[\frac{d}{d\tau}\left( a^2 \frac{(\mathbf{x} \cdot \mathbf{x}')}{1-K\mathbf{x}^2}\right) - \frac{a^2 K(\mathbf{x} \cdot \mathbf{x}')^2}{(1-K\mathbf{x}^2)^2}\right]K\mathbf{x}\\ &= \frac{d}{d\tau}\left(a^2 \mathbf{x}'\right) + \left[\frac{d}{d\tau}\left( \mathbf{f} \cdot \mathbf{x}\right) - \frac{a^2 K(\mathbf{x} \cdot \mathbf{x}')^2}{(1-K\mathbf{x}^2)^2}\right]K\mathbf{x} \\ &= \frac{d}{d\tau}\left(a^2 \mathbf{x}'\right) + Q K\mathbf{x}, \end{align}

which, after dividing both sides by $a^2$, is exactly the geodesic equation from my original question.

If you want to start with a solution to the geodesic equation and show it satisfies the EL equation you can almost reverse the steps. The only new thing you need to show is the reverse of the very last step, that the geodesic equation implies

$$ Q = \frac{d}{d\tau}\left(\mathbf{f} \cdot\mathbf{x}\right) - \frac{a^2 K(\mathbf{x} \cdot \mathbf{x}')^2}{(1-K\mathbf{x}^2)^2}. $$ You start by dotting the geodesic equation with $\mathbf{x}$, and then start rearranging (using the definitions of $\mathbf{f}$ and $Q$ at some point).


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