I am trying to understand coordinate transformations in general relativity. This might be very fundamental, but I had a hard time finding this on google.

In special relativity we have the condition

$$\Lambda^T \eta \Lambda = \eta.$$

Does that mean that in general relativity, we get

$$ \Lambda^T g \Lambda = g,$$

so that $ds^2$ stays invariant? So in that case, since $g$ may depend on time and position, the transformations $\Lambda$ are not linear anymore which gives rise to why the second term in the Christoffel symbol transformation is non-zero:

$$ \Gamma ^{\alpha'}_{\; \mu' \nu'} = \Lambda^{\alpha'}_{\;\alpha}\Lambda^{\mu}_{\;\mu'}\Lambda^{\nu}_{\;\nu'} \Gamma ^{\alpha}_{\; \mu \nu} + \frac{\partial^2 x^\beta}{\partial x^{\mu'}\partial x^{\nu'}}\frac{\partial x^{\alpha'}}{\partial x^{\beta}}$$.

So in special relativity, the second term would be zero since the transformation $\Lambda$ is linear. But in GR the group algebra of the transformations may depend on time and position, so we get a non-zero second term. Would that be correct to say?


2 Answers 2


Nice one! Let's start with your first question:

Does that mean that in general relativity, we get $$\Lambda^T g \Lambda$$ so that $ds^2$ stays invariant?

It depends on what you mean by "invariance". I don't know how familiar you are with differential geometry, so I'll try to keep it less technical. As you may already know, in general relativity spacetime is described by a smooth manifold $M$, i.e. a space that "locally looks like" $\mathbb{R}^n$ and has some nice properties allowing us to differentiate and stuff. Take for example the surface of a sphere. You can take $M$ to be the set of positions in spacetime. However, in opposite to $\mathbb{R}$, $M$ is not a vector space. Instead, at each point $p$ of $M$ we can naturally define a vector space $T_pM$ called tangent space at $p$. It can be imagined as a tangent surface attached at each point whose elements are the directional vectors. e.g. velocity of a partical. Now, the metric tensor can be defined as a map $g_p:T_pM\times T_pM\rightarrow\mathbb{R}$ at each point $p\in M$ that behaves smooth under variation of $p$. Also, it should satisfy all the stuff like symmetry and non-degeneracy.

To get to your question: Note that we didn't use any coordinates so far. We see that the metric tensor is invariant under coordinate transformations in the sense that the object $g$ remains the same, independent of the coordinates. However, this is not true for the coordinate representation $g_{\mu\nu}$ of $g$ in a certain basis. But this is something we already have to deal with when using polar coordinates in $\mathbb{R}^2$: $$g_{ij}(r,\phi)=\text{diag}(1, r^2)\neq \text{diag}(1,1).$$

Getting to the next question:

So in that case, since $g$ may depend on time and position, the transformations $\Lambda$ are not linear anymore which gives rise to why the second term in the Christoffel symbol transformation is non-zero.

Indeed, $g$ depends on time and position as it is a tensor field on $M$. Hwoever, this has no influence on the linearity of the transformation. Let's continue our beginning discussion. When doing physics in $\mathbb{R}^n$ we usually dont't really make a difference between positional vectors and directional (tangent) vectors. However, in curved spacetime position cannot be treaten as vector. Again, this is also the case when working in $\mathbb{R}^2$ with polar coordinates where coordinate tuples $(r,\phi)$ connot just be added. As a consequence, we also have to differ between the transformation of spacetime points and tangent vectors, although they are closely connected. A coordinate chart $\psi: U\rightarrow \mathbb{R}^n$, with $U\subseteq M$, naturally induces a basis in the tangent vector spaces, often denoted by the partial derivatives $e_\mu^\psi\equiv\partial_\mu^\psi$ along the coordinate lines. A coordinate transformation between coordinate charts then induces a linear transformation of the basis vectors given by $$e_\mu^\xi|_p = \left(\frac{\partial\xi_\mu}{\partial\psi_\nu}\right)|_p \cdot e_\nu^\psi|_p$$ with $p$ as index because the basis clearly depends on the spacetime point the of the tangent space of consideration. Thus, an arbitrary coordinate transformation leads to a linear transformation in the tangent space basis and consequently of the coordinates of tangent vectors. However, the Christoffel symbols $\Gamma^\lambda_{\mu\nu}$ do not transform linearely as a tensor, because they ain't discribing one. That's a problem of doing general relativity entirely in coordinates with Einstein sum convention. The Christoffel symbols are coefficients of an affine connection instead of a tensor, i.e. an object discribing how "neighboring" tangent spaces are connected by introducing parallel transport between the tangent spaces. By the way, curvature then is a meassure of how much this transport depends on the path you choose to connect two spacetime points. Anyway, the transformation rule of the Christoffel Symbols is just the transformation rule of the coordinate representation of such a connection. As you have already noted the transformation is linear if the second derivatives vanish, but that barley happens when dealing with curved spaces. Sidefact: When workin with $\mathbb{R}^2$ in polar coordinates the vector derivative also gets a Christoffel symbol part due to the spacial dependence of the basis vectors: $$\partial_\phi V = \left(\partial_\phi V^i\right)\cdot e_i + V^i \cdot \partial_\phi e_i. $$ As a rule that works in most cases, Christoffel symbols are non-vanishing in $\mathbb{R}^n$ if the coordinates are non-linear.

Hope this answers your questions!



The distinction between special relativity and general relativity is that special relativity requires spacetime to be flat. You can do SR in any coordinates you like, and there is no requirement that the coordinates be such that the metric has the form $\eta$. The reason the interval $ds^2$ is invariant is not that we pick special coordinates or pick a metric that has a special form. The reason the interval is invariant, in both SR and GR, is that it's a scalar product of tensors, $g_{ij}dx^idx^j$.

Does that mean that in general relativity, we get $ \Lambda^T g \Lambda = g,$ so that $ds^2$ stays invariant?

No, for the reasons given above. And note that there is no such thing as a Lorentz transformation $\Lambda$ in general relativity.

Re the Christoffel symbols, there is no requirement that coordinate transformations be linear in SR, and you can have Christoffel symbols that don't vanish. For example, you can do SR in spherical coordinates. The restriction is that we have to have zero curvature, which depends on the derivatives of the Christoffel symbols.


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.