4-acceleration definition

1. Jan 20, 2014

TrickyDicky

In the wikipedia page a distinction is made in the definition of the SR 4-acceleration between the case with rectilinear vs. curvilinear coordinates, the latter requiring the use of Christoffel symbols of the coordinates wrt the Minkowski space and therefore an additional term.
Finally it presents the formula of 4-acceleration in the context of GR.
Is it ok to say that there is no formal difference in the formula between the GR 4-acceleration and the SR one in curvilinear coordinates?
Of course we now one refers to curved space and the other to flat space in curved coordinates, so the Christoffel symbols have different meaning in both cases, I'm referring to the formal mathematical operations.

2. Jan 20, 2014

Staff: Mentor

Technically that isn't correct. The formula always "requires the use of Christoffel symbols"; it's just that in rectilinear coordinates in Minkowski spacetime all the Christoffel symbols are zero. So I don't really buy the distinction the Wiki page is making here.

I would say there's no formal difference, period, even in SR in rectilinear coordinates. Formally I can always put in the Christoffel symbols in rectilinear coordinates, since they're zero anyway; and doing that makes things simpler, since the same formula applies in all cases. See above.

I'm not sure I buy this distinction either. You can make the Christoffel symbols vanish at a point in curved spacetime, whereas in flat spacetime you can make them vanish everywhere; but I would say that's because of the curvature, not because of any difference in meaning of the Christoffel symbols.

3. Jan 20, 2014

WannabeNewton

Given a space-time $(M,g_{\mu\nu})$, there necessarily exists a unique torsion-free derivative operator $\nabla_{\mu}$ associated with $g_{\mu\nu}$. If $M$ is Minkowski space-time then regardless of whatever coordinates you represent $g_{\mu\nu}$ in, $\nabla_{\mu}$ is necessarily a flat derivative operator so this is the geometric (invariant) characterization of $\nabla_{\mu}$.

But this doesn't change the function of $\nabla_{\mu}$ when operating on tensor fields irrespective of $M$ being Minkowski space-time (i.e. irrespective of $\nabla_{\mu}$ being flat). If $g_{\mu\nu}$ is represented in an arbitrary coordinate system $\{x^{\mu}\}$ on Minkowski space-time then, for example, $\nabla_{\mu}V^{\nu} = \partial_{\mu}V^{\nu} + \Gamma^{\nu}_{\mu\gamma}V^{\gamma}$ just like in curved space-time. In other words, the Christoffel symbols have the exact same meaning regardless of whether or not we are in Minkowski space-time or curved space-time, as Peter noted. The actual difference is simply that $\nabla_{[\gamma}\nabla_{\mu]}V^{\nu} = 0$ for all vector fields in Minkowski space-time.

4. Jan 20, 2014

George Jones

Staff Emeritus
Adding my 2 cents to what Peter and WannabeNewton wrote.

Let $\left( M . g \right)$ be a spacetime, and let $\gamma$ be the worldline of an observer parametrized by proper time $\tau$. If, along $\gamma$, $u$ is the tangent vector to $\gamma$ (observer's 4-velocity), then the 4-accleration of the observer is given by

$$a = \frac{Du}{D\tau},$$

where $D/D\tau$ is the absolute derivative along the curve (called "covariant derivative along the curve" by Lee in "Riemannian Manifolds") induced by the metric-compatible torsion-free connection.

This is true for all spacetimes, including Minkowski spacetime.

5. Jan 20, 2014

TrickyDicky

Thanks for the answers.
Fully agree. Didn't express myself correctly as I was precisely trying to convey the Christoffel symbols act equally in all cases, just didn't want to be misinterpreted as saying curved and flat is the same thing.