TheEtherWind said:
I don't know exactly what I'm looking for in this question so I'll ask it in a vague way. What is the connection between a particle's proper acceleration and the christoffel symbol of the second kind (single contravariant and double covariant) ? Is this correct?
\frac{∂^2x^\alpha}{∂\tau^2}=-\Gamma^\alpha{}{}_\beta\gamma\frac{∂x^\beta}{∂\tau}\frac{∂x^\gamma}{∂\tau}
What are the physical meanings behind these quantities?
The way I prefer to think about the covariant derivative, and the geodesic equation, and proper acceleration and all that is to start with VECTOR equations (or more generally, tensor equations) and then look at specific components, rather than starting off with components.
So if you have a particle that is at a postion P(s) as a function of a path parameter s (proper time, typically), then there is a corresponding velocity 4-vector U(s) defined by
U = \dfrac{d}{ds} P
Then there is a corresponding acceleration 4-vector A(s) defined by:
A = \dfrac{d}{ds} U
It's completely straight-forward. The complexity comes in when you start looking at the components of the acceleration 4-vector. To do this, we write:
U = \sum U^{\alpha} e_{\alpha}<br />
= U^{\alpha} e_{\alpha} (sum is implicit).
where e_{\alpha} is a set of basis vectors, and the U^{\alpha} are the components of U in that basis. Then we use the product rule to compute \dfrac{d}{ds}U:
\dfrac{d}{ds}U = (\dfrac{d}{ds}U^{\alpha}) e_\alpha + U^{\alpha} (\dfrac{d}{ds}e_{\alpha})
Now we apply the chain rule: \dfrac{d}{ds} f(g(s)) = \dfrac{df}{dg} \dfrac{dg}{ds}
\dfrac{d}{ds}e_{\alpha} = \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} \dfrac{d x^{\beta}}{ds} = \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} U^{\beta}
So what is the meaning of the expression \dfrac{\partial e_{\alpha}}{\partial x^{\beta}}? This expression shows how the basis vector e_{\alpha} changes as one moves in the direction of increasing x^{\beta}. For example, think about polar coordinates for the 2D plane. The two basis vectors are e_r which points in the direction of increasing radial distance r and e_\theta, which points in the direction of increasing angle \theta. These two vectors are
not constant; when you are to the left of the origin (\theta = \pi), e_r points to the left, and when you are to the right of the origin (\theta = 0), e_r points to the right. So changing \theta changes the basis vector e_{r} (and e_\theta as well).
So in general, \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} will be non-zero. Since for a specific choice of coordinates, and for specific \alpha and \beta, \dfrac{\partial e_{\alpha}}{\partial x^{\beta}} is the derivative of a vector e_\alpha with respect to a scalar, x^{/beta}, the result will be another vector.
Since we can write any vector as a linear combination of basis vectors, we can write:
\dfrac{\partial e_{\alpha}}{\partial x^{\beta}} = \Gamma^{\lambda}_{\alpha \beta} e_{\lambda}
So the coefficients \Gamma^{\lambda}_{\alpha \beta} are just the components of the vector \dfrac{\partial e_{\alpha}}{\partial x^{\beta}}. They just describe how the basis vectors e_\alpha vary from place to place.
Putting it all together, we have:
\dfrac{d}{ds} U = (\dfrac{d}{ds}U^{\alpha}) e_{\alpha} + (\Gamma^{\lambda}_{\alpha \beta} \ e_{\lambda} \ U^{\beta}) U^{\alpha}
The first term on the right side of the = takes into account how U^\alpha varies as s increases, and the second term takes into account how e_{\alpha} varies. Since \lambda and \alpha are dummy indices, we can switch them without changing anything:
(\Gamma^{\lambda}_{\alpha \beta} \ e_{\lambda} \ U^{\beta}) U^{\alpha}= (\Gamma^{\alpha}_{\lambda \beta} \ e_{\alpha} \ U^{\beta}) U^{\lambda}. So we can factor out e_\alpha and write:
\dfrac{d}{ds} U = (\dfrac{d}{ds}U^{\alpha} + \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}) e_{\alpha}
So \dfrac{d}{ds} U is a 4-vector A with components
A^{\alpha} = \dfrac{d}{ds}U^{\alpha} + \Gamma^{\alpha}_{\lambda\beta} \ U^{\beta} U^{\lambda}
The notation
A^{\alpha} = \dfrac{D U^{\alpha}}{ds} is sometimes used, but it should be remembered that this is an operation on the entire 4-vector U, not just on the component U^{\alpha}. A more accurate notation would be
A^{\alpha} = (\dfrac{d U}{ds})^\alpha. It is the \alpha component of \dfrac{d U}{ds}, not any kind of derivative of U^{\alpha}.