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What is the physical meaning of the components of the affine connection in General relativity?

I read in a GR book that the connection describes the gravitational field, so what does each of its components represent and what is the difference between each component?

Please note that I am only just beginning to study GR.

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/2447/2451 , physics.stackexchange.com/q/226031/2451 and links therein. $\endgroup$ – Qmechanic Oct 28 '16 at 16:23
  • $\begingroup$ On the surface, they differ in that I ask about each coefficient. But, thanks for the link, I will check it out and if I find an answer there, I will delete this question. $\endgroup$ – TheQuantumMan Oct 28 '16 at 16:30
  • $\begingroup$ I have attempted an answer, but on rereading your question, I think you really want to know the physical meaning of each Christoffel symbol. Let me know if the answer is not what you want and I will delete it, pronto. $\endgroup$ – user108787 Oct 28 '16 at 16:35
  • $\begingroup$ I advise that you think what connection (as an object) means, and not what are the individual components. Components are extremely misleading in GR. And now for the question: connection tells you what does it mean to parallel transport a vector along a curve. In particular it defines what are straight lines and inertial observers. $\endgroup$ – Blazej Oct 28 '16 at 18:39
  • $\begingroup$ @Qmechanic not 10? $\endgroup$ – TheQuantumMan Oct 28 '16 at 21:10
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What is the physical meaning of the (10 independent) components of the affine connection in General relativity? I read in a GR book that the connection describes the gravitational field, so what does each of its components represent

The metric tensor has 10 independent components, as you say, now what happens when we differentiate them? We don't get a tensor, which is we really want, to ensure compatability with GR related equations in every frame of reference.

But the affine connection, added or subtracted depending on context, to the diffentiated metric tensor, produces a combination that is a tensor, so we are back in business as regards putting GR equations on an equal footing in all frames. That's what I would interpret the physical implications as meaning.

And what is the difference between each component?

I don't think that's as important as knowing that, taken together, they help describe space time curvature. Also, literally I can't answer this question in math terms, as it varies in different situations and I don't think that one Christoffel symbol taken on it's own, or compared to other ones, really means very much. My experience on this is limited, so I hope you get a better answer.

There are more aspects to it than this, but that is my basic understanding, which I will update shortly.

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There are lines called geodesics. This lines have the following property:

If you take some vector say $\vec{\xi}$, which is tangent to this line in this point and then transport it along this line, $\vec{\xi}$ stays tangent to this line in every point. Geodesics are like straight lines in flat space. Geodesics are defined by equation: $$\frac{d^2x^{\lambda}}{ds^2} +\Gamma_{\mu\nu}^{\lambda}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds} = 0,$$ where $x^\mu$ are curvilinear coordinates and $s$ is just a parameter along the curve.

Look at this equation. What if I "change" it in some sense? $$\frac{d^2x^{\lambda}}{ds^2} = -\Gamma_{\mu\nu}^{\lambda}\frac{dx^\mu}{ds}\frac{dx^\nu}{ds}$$ It seems like

$$\frac{d^2x}{dt^2} = \frac{F}{m}$$ doesn't it? So in general relativity objects move along geodesics if there are no forces. But what is those geodesics? It depends on gravitational structure in which objects are. So gravity is not an ordinary force in general relativity. It forms geodesics or curvature of space-time continuum. Note the case when it is no gravity so $\Gamma^{\lambda}_{\mu\nu} = 0$ and trajectories of particles are just straight lines: $$\frac{d^2x^{\lambda}}{ds^2} = 0 \Rightarrow x^\mu(s) = A^{\mu}s + B^\mu.$$ where $A^\mu, B^{\mu}$ are just numbers.

So how can you prove that in flat space all $\Gamma$'s disappear. I hope that you will see it many times on your way of studying GR. But I'll give you some formal and probably not the most simple way. NOTE. I assume Einstein's summation agreement everywhere.

There is (just believe me) equation connecting components of the affine connection $\Gamma$ and metric tensor $g_{\mu\nu}$. $$\Gamma_{\mu\nu}^{\alpha} = \frac{g^{\alpha\lambda}}{2}\bigl(\frac{\partial g_{\lambda\mu}}{\partial{x^\nu}}+\frac{\partial g_{\lambda\nu}}{\partial{x^\mu}}+\frac{\partial g_{\mu\nu}}{\partial{x^\lambda}}\bigr),$$ where as usual $x^\mu$ are coordinates and $g_{\mu\nu}$ and $g^{\mu\nu}$ are metric and its inverse respectively. Metric is responsible for scalar product and lengths in our space in the following manner: $$(A, B) = g_{\mu\nu}A^\mu B^\nu$$ and $$ds^2 = g_{\mu\nu}dx^\mu dx^\nu,$$ where $ds^2$ is squared length associated with $dx^\nu$, $dx^\mu$, $A$ and $B$ some vectors.

But in flat space time Pythagorean theorem is working so you can calculate length usually $ds^2 = dx^{\nu}dx^{\nu}$. Therefore \begin{equation} g_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 0\\ 0 & 0 & 1 & ... & 0\\ & & ... & & \\ 0 & 0 & 0 & ... & 1 \\ \end{pmatrix} = \rm{diag(1)} \end{equation} Or at least with constant components. So you can see each $\Gamma$ disappear therefore.

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  • $\begingroup$ How can I prove that the connection coefficients are zero in the case of a flat spacetime? $\endgroup$ – TheQuantumMan Oct 28 '16 at 18:07
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    $\begingroup$ This question requires definition of parallel transport. But let me be more precise: In flat space (affine space) we can introduce curvilinear coordinates in which components of the affine connection will not be zero. BUT. In that space we always can choose another coordinate system in which all $\Gamma$'s disappear in every point. $\endgroup$ – LRDPRDX Oct 28 '16 at 19:18
  • $\begingroup$ Let me edit my previous answer in order to avoid formulas in comments. $\endgroup$ – LRDPRDX Oct 28 '16 at 19:19
  • $\begingroup$ Okay, good answer, but how does it answer what the connection physically represents? $\endgroup$ – TheQuantumMan Oct 28 '16 at 21:10
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There's a different tack than the ones mentioned so far. In introductory physics we treat space itself as a vector space, and talk about adding displacements, subtracting positions, etc. One of the fundamental properties we tell students about vectors is that you can pick them up, move them, and plop them down anywhere and they're still the same vector. There are a whole bunch of assumptions that go into doing that. Without going into too much detail, we can generalize that process by saying that picking up and moving a vector, $\mathbf{v}'$ from position $\mathbf{r}'$ to $\mathbf{r}$ along the path $\gamma$ then we can say that the result of that process is: $$\mathbf{v} = U_\gamma(\mathbf{r},\mathbf{r}') \mathbf{v}',$$ where $U_\gamma(\mathbf{r},\mathbf{r}')$ is a linear operator (matrix) valued function that depends, in general, on both the starting point, ending point, and shape of the path. A property we may want to enforce is for this process to preserve the angles between vectors and the lengths of vectors. When that is the case, $U_\gamma(\mathbf{r},\mathbf{r}')$ will be an element of one of the classical Lie groups, usually $\operatorname{SU}(N)$ or $\operatorname{SO}(N)$, depending on whether $\mathbf{v}$ is in an $N$ dimensional complex vector space, or real valued one, respectively. Note that this treats every point in spacetime has having its own vector space attached to it, known as a tangent space. When $\mathbf{r}$ and $\mathbf{r}'$ are separated by an infinitesimal $\mathbf{r} - \mathbf{r}' = \operatorname{d}\mathbf{r}$ then, if we require $U$ to be continuous, we can write: $$\mathbf{v} = \left(I - \sum_{i} \operatorname{d}\mathbf{r}_i A_i(\mathbf{r})\right)\mathbf{v}',$$ where $I$ is the identity operator, and $A_i(\mathbf{r})$ are anti-Hermitian operator valued functions of position. In general relativity, then $[A_i]^j_{\hphantom{j}k} = \Gamma^j_{ik}$, so the meaning of an individual component of $\Gamma$ is complicated to tease out, $\Gamma^\mu_{\lambda \nu}(\mathbf{r})$ is the matrix that relates the representation of a vector in a particular basis at $\mathbf{r}$ to a representation of that same vector at $[\mathbf{r} \pm \operatorname{d}\mathbf{r}]_\lambda$ (sign decided by convention) in another basis that is infinitesimally close to the basis at $\mathbf{r}$, and the components of the matrix are labeled by $\mu$ and $\nu$.

For those interested in technical details, this is a comparison of gravity to a non-abelian gauge theory. Thus the Christoffel symbols play the role most closely analogous to the vector potential of the gauge field, with all of the gauge freedom ambiguity that implies. The main differences between gravity and the Yang-Mills theories is that with gravity the tangent vector space is the same one as the infinitesimal translation vectors, and in gravity the metric, the relationship between the vector space and the covector space, is allowed to vary, too. These complications alter some of the details of what's going on, with things like the requirement that the Christoffel symbols be metric compatible, but not the essence.

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  • $\begingroup$ So, to relate it to gravity, it gives the acceleration vector that one observer "sees" with the acceleration than another observer "sees" provided that they represent the same acceleration vector with two coordinates(one for each) that are nearly identical(they only have infinitesimal differences)? $\endgroup$ – TheQuantumMan Oct 29 '16 at 0:32

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