The most general metric for a static and spherically symmetric metric is given by:

$$ ds^2 = e^{2\gamma(u)}dt^2 - e^{2\alpha(u)}du^2 - e^{2\beta(u)}d\Omega^2 $$

I have the freedom of choosing Schwarzschild coordinates, in which:

$$ e^{2\beta(u)} = u^2, $$

and now $u$ becomes the radius of the sphere.

However I have the freedom of choosing other coordinates systems, i.e. gauge freedom, like the harmonic gauge $\alpha(u) = 2\beta(u) + \gamma(u)$ or the quasi-global gauge given by $\alpha(u)=-\gamma(u)$ (those two are the ones I use the most).

My questions are:

  1. How do I prove that those gauges may be used without any loss of generality? (As example, how do I show if $\alpha=-5\gamma + \beta^2$ is a gauge or not?)

  2. I see this as a coordinate freedom, but in my research it's mostly known as "gauge freedom", any differences or just nomenclature?

  • $\begingroup$ Interessting question: I am not sure about 1) but I think in this case all those remaining coordinate changes should be related to the fact that the system is invariant under transformations $u\rightarrow f(u)$. Not sure on 2) either but I think in GR "gauge freedom" and "free choice of coordinates" is the same. But since the coordinates are so important in GR, since they are directly related to curvature, one refers to a coordinate change as gauge ("you gauge the curvature"?). I might be wrong on that too and for sure there is a more formal explanation. $\endgroup$
    – N0va
    Sep 15, 2016 at 21:53
  • $\begingroup$ Related: physics.stackexchange.com/q/21705/2451 $\endgroup$
    – Qmechanic
    Sep 16, 2016 at 6:47
  • $\begingroup$ Qmechanic, I can see that what's happening in this derivation, however I want the other way around, how do I prove that a coordinate condition may be used thoroughly without any loss of generality? Also, in this link you posted, you are already assuming Schwarzschild coordinate system a priori... $\endgroup$ Sep 16, 2016 at 12:43
  • $\begingroup$ What do you mean by "non-standard"? When $e^{2\beta}=u^2$ $u$ is the radial coordinate with $u=0$ the center of the 2-sphere... $\endgroup$ Sep 16, 2016 at 16:34

2 Answers 2


To start with, given the metric ansatz $$ ds^2 = e^{2\gamma(u)}dt^2 - e^{2\alpha(u)}du^2 - e^{2\beta(u)}d\Omega^2,\tag{1} $$ one can apply the Cartan-Karlhede algorithm to check local equivalence with the Schwarzhschild solution (this is best carried out with a computer implementation). Instead of this general solution for the local equivalence of some metric ansatz and a given metric, one can in this case also investigate the demands for (1) to be a vacuum solution, since this will make it the Schwarzschild solution by virtue of being static and spherically symmetric (Birkhoff's theorem); the latter approach should of course also be used if one is primarily interested in the requirements for an ansatz to give a solution rather than some specific soluion.

  1. To prove that $f(\alpha,\beta,\gamma) = 0$ is a "gauge" you can plug it into your ansatz and take the approach outlined above. For example for the ansatz (1) any "gauge" that does not violate the equations below is allowed.

  2. I am new to gauge theory, but I believe the correct use of gauge transformations in general relativity would entail $SO_0(1,3)$-transformations of vectors in a rigid frame. Thus I believe you are correct in that this is a matter of coordinate freedom rather than gauge freedom.

Here I approach the metric ansatz (1) as an example: to that end I will be working in a local Lorentz (orthonormal) frame: \begin{align}\begin{split} \omega^0 &= e^{\gamma(u)}dt, \\ \omega^1 &= e^{\alpha(u)}du, \\ \omega^2 &= e^{\beta(u)}d\vartheta, \\ \omega^3 &= e^{\beta(u)}\sin(\vartheta) d\varphi. \end{split}\tag{2} \end{align} We then work with the first Cartan equation $$ d\omega^i = \gamma^i{}_{jk}\omega^j \wedge \omega^k,\tag{3} $$ where $\gamma^i{}_{jk}$ are the Ricci rotation coefficients (components of the connection forms). Applying (3) to (2) we find \begin{align} \gamma^0{}_{jk}\omega^j\wedge\omega^k &= \gamma'(u)e^{\gamma(u)}du \wedge dt \\ &= \gamma'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^0, \\ \gamma^1{}_{jk}\omega^j\wedge\omega^k &= 0, \\ \gamma^2{}_{jk}\omega^j\wedge\omega^k &= \beta'(u)e^{\beta(u)}du\wedge d\vartheta \\ &= \beta'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^2, \\ \gamma^3{}_{jk}\omega^j\wedge\omega^k &= \beta'(u)e^{\beta(u)}\sin(\vartheta)du\wedge d\varphi + e^{\beta(u)}\cos(\vartheta)d\vartheta\wedge d\varphi \\ &= \beta'(u)e^{-\alpha(u)}\omega^1 \wedge \omega^3 + \cot(\vartheta)e^{-\beta(u)}\omega^2 \wedge \omega^3, \end{align} and by inspection we find \begin{align} \begin{split} \gamma^0{}_{10} &= \gamma'(u)e^{-\alpha(u)}, \\ \gamma^2{}_{12} &= \beta'(u)e^{-\alpha(u)}, \\ \gamma^3{}_{13} &= \beta'(u)e^{-\alpha(u)}, \\ \gamma^3{}_{23} &= \cot(\vartheta)e^{-\beta(u)}, \end{split}\tag{4}\end{align} to be the only non-zero coefficients (up to symmetries: $\gamma_{ijk} = \gamma_{[ij]k}$). Next the second Cartan equation \begin{align} \frac{1}{2}R^i{}_{jk\ell}\omega^k \wedge \omega^\ell &= d\gamma^i{}_j + \gamma^i{}_k \wedge \gamma^k{}_j,\tag{5} \\ &= \left(-\gamma^i{}_{j[k|\ell]} + \gamma^i{}_{jm}\gamma^m{}_{[k\ell]} - \gamma_{mj[k}\gamma^{im}{}_{\ell]}\right)\omega^k \wedge \omega^\ell \end{align} gives, using (4): \begin{align}\begin{split} R^0{}_{101} &= \left(\alpha'(u)\gamma'(u) - \gamma''(u) - \gamma'(u)^2\right)e^{-2\alpha(u)}, \\ R^0{}_{202} &= -\beta'(u)\gamma'(u)e^{-2\alpha(u)}, \\ R^0{}_{303} &= -\beta'(u)\gamma'(u)e^{-2\alpha(u)}, \\ R^1{}_{212} &= \left(\alpha'(u)\beta'(u) - \beta''(u) - \beta'(u)^2\right)e^{-2\alpha(u)}, \\ R^1{}_{313} &= \left(\alpha'(u)\beta'(u) - \beta''(u) - \beta'(u)^2\right)e^{-2\alpha(u)}, \\ R^2{}_{323} &= e^{-2\beta(u)} - \beta'(u)^2e^{-2\alpha(u)}, \end{split}\tag{6}\end{align} as the only non-zero Riemann components, again up to symmetries ($R_{ijk\ell} = R_{([ij][k\ell])}$), or in terms of the Ricci components (suppressing function arguments): \begin{align}\begin{split} R_{00} &= \left(2\beta'\gamma' - \alpha'\gamma' + \gamma'' + \gamma'^2 \right)e^{-2\alpha}, \\ R_{11} &= \left(\alpha'\gamma' + 2\alpha'\beta' - \gamma'' - 2\beta'' - \gamma'^2 - 2\beta'^2 \right)e^{-2\alpha}, \\ R_{22} &= \left(\alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 \right)e^{-2\alpha} + e^{-2\beta}, \\ R_{33} &= \left(\alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 \right)e^{-2\alpha} + e^{-2\beta}, \\ \end{split}\tag{7}\end{align} The Ricci tensor must vanish in a vacuum solution (assuming no cosmological constant for simplicity), so (7) gives us a system of three differential equations: \begin{align} 2\beta'\gamma' - \alpha'\gamma' + \gamma'^2 + \gamma'' &= 0, \tag{A}\\ \alpha'\gamma' + 2\alpha'\beta' - \gamma'' - 2\beta'' - \gamma'^2 - 2\beta'^2 &= 0, \tag{B}\\ \alpha'\beta' - \beta' \gamma' - \beta'' - 2\beta'^2 &= -e^{2(\alpha - \beta)}\tag{C}. \end{align} Notice that equations (A), (B), and (C) can be simplified:

  1. If $\gamma' = 0$ we end up with a single equation: $$ e^{\alpha} = \beta'e^\beta, $$ under which all Riemann components (6) vanish. So this corresponds to flat (Minkowski) space.

  2. Otherwise, by solving for $\gamma''$ in (A) and (B) we find $-\beta'\gamma' = \alpha'\beta' - \beta'' - \beta'^2$, allowing us to write \begin{align} \frac{\beta''}{\beta'} &= \alpha' - \beta' + \gamma', \tag{A*}\\ \frac{\gamma''}{\gamma'} &= \alpha' - 2\beta' - \gamma', \tag{B*}\\ 2\beta'\gamma' + \beta'^2 &= e^{2(\alpha - \beta)},\tag{C*} \end{align} because $\beta' \neq 0$ by necessity, and we can integrate (A*) and (B*) to find \begin{align}\begin{split} \beta' &= C_1e^{\alpha - \beta + \gamma}, \\ \gamma' &= C_2e^{\alpha - 2\beta - \gamma}, \end{split}\tag{D}\end{align} under which (C*) becomes $$ C_1e^{-\alpha + \beta + \gamma}\left(2C_2e^{\alpha - 2\beta - \gamma} + C_1e^{\alpha - \beta + \gamma}\right) = 1, $$ or $$ e^{2\gamma} = C_1^{-2}\left(1 - 2C_1C_2e^{-\beta}\right)\tag{E1}. $$ Note that (D) has already given us: $$ e^{2\alpha} = C_1^{-2}\beta'^2e^{2(\beta - \gamma)}\tag{E2}. $$

Equations (E1) and (E2) allows us to write (1) as $$ ds^2 = \left(1 - 2Me^{-\beta(u)}\right)dt^2 - e^{2\beta(u)}\left(\frac{\beta'(u)^2}{1 - 2Me^{-\beta(u)}}du^2 + d\Omega^2\right),\tag{1.1} $$ both by rescaling coordinates by constant factors if necessary. Notice that we have performed no (non-trivial) coordinate change. Thus (1.1) is just (1) rewritten to conform with the EFEs, excluding flat space, with an aptly chosen name on the integration constant.

We observe that imposing $\alpha(u) = -\gamma(u)$ is equivalent to restricting to $\beta' = e^{-\beta}$ or equivalently $\beta = \log(u + C)$ or $e^{2\beta} = (Cu)^{2}$ (new $C$), by (E2), which is just the Schwarzschild coordinates for (1.1). Similarly, from (B*) we can see that imposing $\alpha(u) = 2\beta(u) + \gamma(u)$ is equivalent to imposing $\gamma = Au + B$, which allows us to rewite (1.1) as (barring any calculation errors) $$ ds^2 = e^{2Au}dt^2 - \frac{1}{\left(1 - C^2e^{2Au}\right)^{2}} \left(\frac{e^{2Au}}{\left(1 - C^2e^{2Au}\right)^2}du^2 + d\Omega^2\right). $$ For example $\alpha(u) = -\gamma(u) + \beta(u)^2$ can similarly be plugged into e.g. (E2) to yield $$ \int C_1^2 du = \int e^{2\beta - \beta^2} d\beta . $$

As another example the condition $\beta = 2\gamma$ can be seen to demand a constant $\beta$, from (E1) and so cannot be used.

To be more general, any condition on the form $\alpha = f(\beta,\gamma)$ gives a first order ODE for $\beta$, by (E1) and (E2). By the standard existence theorem such a condition is consistent with the EFEs. On the other hand a condition on the form $\beta = f(\gamma)$ gives an equation for $\beta$, and so is not consistent with the EFEs. The consistency of imposing more than one condition can also be checked with the above equations. For example $\alpha = \beta = \gamma$ can immediately be seen to not be consistent with the EFEs, by (D).

  • $\begingroup$ I understood every step you did Erik, I got to (1.1) with no problems at all, however I still cannot see when a set coordination condition may be used or not. In (1.1) you have re-written the whole metric depending only on $\beta(u)$, so I have no restriction with any choice of $\gamma(u)$ and $\alpha(u)$? $\endgroup$ Sep 18, 2016 at 15:11
  • $\begingroup$ @Edison Perhaps it was only confusing to add equation (1.1); the idea is to use the alphabetically tagged equations to check for the consistency of any coordinate condition. Primarily one would use (E1) and either (E2) or (D). I added a paragraph where I explained this a bit more explicitly. I reverted the changes to the Ricci tensor components since they were incorrect in the local Lorentz frame. They were correct in the coordinate frame, but given that they were derived in the Lorentz frame I think it's only confusing to revert to a coordinate frame, especially without stating so explicitly $\endgroup$ Sep 19, 2016 at 1:11

What Erik has said is correct, essentially you write down an ODE (for your case) but PDE in general and see if there are solutions for which your 'gauge' choice holds. This is exactly a coordinate-freedom, perhaps this is where the confusion for question 2 comes in.

In GR there is a second kind of gauge freedom known as tetrad invariance, which is invariance under reference frame transformations. Essentailly you want the metric tensor to be invariant under a tetrad transformation. For example, in the Newman-Penrose formalism we write the metric in terms of the null tetrad $(\boldsymbol{l},\boldsymbol{n},\boldsymbol{m},\boldsymbol{\bar{m}})$ \begin{equation} g_{ab}=-l_a n_b - n_a l_b +m_a \bar{m}_b +\bar{m}_a m_b, \end{equation} which we want to be invariant under a transformation of the basis (e.g. $m \rightarrow e^{i \vartheta} m$ and $\bar{m} \rightarrow e^{-i \vartheta} \bar{m}$ for some angle $\vartheta$). Note the absence of coordinates. The term `gauge invariant' is usually used to blanket both the coordinate and tetrad invariance together. A fantastic article which covers this in detail is Stewart and Walker (1974) (unfortunately pay-walled), where they discuss the importance of the distinction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.