I think the answer is no. My reasoning is the following:

We have the Schwartzchild metric with signature $(+,-,-,-)$.

Then we can use the Lagrange-Euler equations

$$\frac{d}{dt}\frac{\partial L}{\partial\dot{x}^{\lambda}}-\frac{\partial L}{\partial x^{\lambda}} = 0$$

With the corresponding Lagrangian.


Along with the geodesic equation (which is equivalent to the Euler-Lagrange equations)


To obtain the connection coefficients $\Gamma_{\mu\nu}^\lambda$ by simple inspection.

On the other hand, if we have the metric signature $(-,+,+,+)$, the new lagrangian is going to be

$$ L' = -L$$

If we replace in the E-L equations:

$$\frac{d}{dt}\frac{\partial L'}{\partial\dot{x}^{\lambda}}-\frac{\partial L'}{\partial x^{\lambda}} = 0 \Leftrightarrow \frac{d}{dt}\frac{\partial L}{\partial\dot{x}^{\lambda}}-\frac{\partial L}{\partial x^{\lambda}} = 0$$

Given that we can factorize the minus sign, we get the same form.

Therefore, if we compare the resulting equation for $L'$ with the geodesic equation outlined above, we will get that the new connection coefficients ${\Gamma'}_{\mu\nu}^\lambda$ are identical to $\Gamma_{\mu\nu}^\lambda$, even in sign.

Is this correct? I'm asking because I'm getting different signs in a few coefficients, compared to the following blog: https://profoundphysics.com/christoffel-symbols-a-complete-guide-with-examples/

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    $\begingroup$ Q. is ambiguous. Are you asking whether the numerical values of the coefficients change sign, or whether a sign appears in one or more formulae involving the coefficients? If the former then look at the formula relating coefficients directly to metric (for Christoffel connection). $\endgroup$ May 22, 2023 at 17:49
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    $\begingroup$ There's actually a subtlety here: The Christoffel symbols of the first kind ($\Gamma_{\lambda \mu \nu}$) will change sign, since they're proportional to the derivatives of the metric. But Christoffel symbols of the second kind ($\Gamma^{\lambda}\ _{\mu \nu}$) will not change sign, because there's another factor of the inverse metric multiplying the entire expression of the first kind which will cancel any overall metric sign. Please correct me anyone if I'm wrong. $\endgroup$
    – Amit
    May 22, 2023 at 17:49
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    $\begingroup$ There is a related question on physics.stackexchange.com/questions/73534, where I gave an answer explaining, why the sign does not change as you suspected. You can find more about sign conventions on en.wikipedia.org/wiki/Sign_convention and there is also an overview given in "Gravitation" by Misner, Thorne and Wheeler directly on the first pages found here: xdel.ru/downloads/lgbooks/… $\endgroup$ May 23, 2023 at 11:24

1 Answer 1


Might as well write an answer for this, though I wrote the idea in a comment:

The Christoffel Symbols of the first kind are given by:

$$\Gamma_{\rho\mu\nu} = \frac{1}{2} (g_{\rho \mu,\nu}+g_{\rho\nu,\mu}-g_{\mu\nu,\rho})$$

So clearly, if we flip the metric's sign ($g_{\mu\nu} \rightarrow -g_{\mu\nu}$), every $\Gamma_{\rho\mu\nu}$ will also pick up an overall sign.

However, the Christoffel symbols of the second kind, which are also the ones known as the connection coefficients will not change sign because they are defined as: $$\Gamma^{\lambda}{}_{\mu\nu} \equiv g^{\lambda \rho}\Gamma_{\rho\mu\nu} = \frac{1}{2} g^{\lambda \rho}(g_{\rho \mu,\nu}+g_{\rho\nu,\mu}-g_{\mu\nu,\rho})$$

The additional factor of $g^{\lambda\rho}$ will cancel any overall sign, so they are not sensitive to a sign flip of the metric.

  • 1
    $\begingroup$ Thank you so much for expanding your comment into this nice answer. That clarifies it all. $\endgroup$
    – StefanGill
    May 24, 2023 at 6:01

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