The Christoffel symbols at the origin -- Why zero?

[haushofer]

Just another example: within the Newtonian limit of General Relativity, one has gauge-fixed the coordinate system in such a way to obtain Newton's second law with the gravitational force in the following form:

<br /> \ddot{x}^i + \Gamma^i_{00} = 0 \ \ \ \ \ (1)<br />

With the group of Galilei transformations, this equation transforms as a tensor (or: vector equation). Physically, this means no Galilei transformation will introduce a fictitious force. But if you e.g. apply a time-dependent rotation,

<br /> x^{&#039;i} = R^i{}_j (t)x^j \ \ \ with \ \ \ R^i{}_j (t)\ \ \in SO(3) \ \ \forall t<br />

then eqn.(1) does not transform as a tensor anymore, but inhomogenously. The inhomogeneous terms are the fictitious forces. I don't see the problem to speak about eqn.(1) this way, and why it would be so bad to call eqn.(1) a tensor equation under the group of Galilei transformations.

 

[PeterDonis]

Poincaré transformations connect different inertial coordinate systems

Yes. But in a curved spacetime, these inertial coordinate systems are on the tangent space at a particular point, not on the spacetime itself. See post #30. So "translations" in the usual sense of that term make no sense. A "translation" in the context of this thread would be, as I said in post #30, a transformation between Riemann normal charts on a local patch of the spacetime, which is not a Poincare transformation.

 

[PeterDonis]

once you've chosen your spatial coordinates (Cartesian, spherical,...), you can calculate your christoffel symbol. If you then apply a Poincaré transformation (Lorentz transformation, rotation or spacetime translation), this symbol you've calculated transforms tensorially (because the inhomogeneous term in the transformation drops out).

Is this true for translations in spherical coordinates? I don't think such a transformation is linear.

 

[haushofer]

Yes. But in a curved spacetime, these inertial coordinate systems are on the tangent space at a particular point, not on the spacetime itself. See post #30. So "translations" in the usual sense of that term make no sense. A "translation" in the context of this thread would be, as I said in post #30, a transformation between Riemann normal charts on a local patch of the spacetime, which is not a Poincare transformation.

I agree.

 

[Dale]

So once you've chosen your spatial coordinates (Cartesian, spherical,...), you can calculate your christoffel symbol. If you then apply a Poincaré transformation (Lorentz transformation, rotation or spacetime translation), this symbol you've calculated transforms tensorially (because the inhomogeneous term in the transformation drops out).

Hmm, is that true? I have never tried that, and I am not even sure exactly how to do it, but I am rather skeptical about it.

 

[renormalize]

Hmm, is that true? I have never tried that, and I am not even sure exactly how to do it, but I am rather skeptical about it.

It follows directly from the coordinate transformation law for the connection (see my post #22 above). But it's really just a formal result because, other than in rectangular coordinates, a such a transform doesn't correspond to any useful Poincaré transformation. In Schwarzschild coordinates $(t,r,\theta,\phi)$ a linear transform allows us, for example, to rotate through an angle $\alpha$ in the $r-\theta$ plane to get the new coordinates $(t^{^{\prime}},r^{\prime},\theta^{\prime},\varphi^{\prime})=(t,r\cos\alpha-\theta\sin\alpha,r\sin\alpha+\theta\cos\alpha,\varphi)$. And the connection will dutifully transform like a tensor into the primed coordinates. But why and to what end would we do this?

 

[PeterDonis]

It follows directly from the coordinate transformation law for the connection (see my post #22 above).

If the transformation is linear, yes. But, as I noted in post #33, I don't think translations in spherical coordinates (i.e., transforming from one spherical coordinate chart to another with a different origin) are linear. And translations are the type of transformation under discussion in this thread.

 

[renormalize]

I don't think translations in spherical coordinates (i.e., transforming from one spherical coordinate chart to another with a different origin) are linear.

Yes, that's exactly right and its why I consider the linear transforms in my post #22 to be formal. Your use of "translate" in spherical coordinates has the operational meaning: convert to rectangular coordinates, translate the origin, then convert back to a spherical system centered on the new origin. The result is a physically sensible but complicated non-linear transform of the original spherical coordinates. In contrast, "translate" in the context of post #22 simply means adding constants to the set of general coordinates, e.g., $(r,\theta,\varphi)\rightarrow(r+r_{0},\theta+\theta_{0},\varphi+\varphi_{0})$. The connection will then transform as a tensor (i.e., be unchanged in this case) under this linear translation, an observation that is formally true but seemingly rather useless. To me, any non-trivial utilization of the fact that the connection is a tensor under linear transforms is likely restricted to rectangular coordinate systems in curved spacetime.

 

[PeterDonis]

any non-trivial utilization of the fact that the connection is a tensor under linear transforms is likely restricted to rectangular coordinate systems in curved spacetime

No, to rectangular coordinate systems in flat spacetime. In a curved spacetime, even "rectangular" coordinate systems don't transform linearly into each other.

 

[renormalize]

No, to rectangular coordinate systems in flat spacetime. In a curved spacetime, even "rectangular" coordinate systems don't transform linearly into each other.

PeterDonis can you help me understand this statement?

Are you asserting here that on a curved manifold the linear transformation $x^{\prime\sigma}(x)=M{}^{\sigma}{}_{\tau}\:x^{\tau}+\xi^{\sigma}$, with $(M,\xi)$ constant, is mathematically inconsistent, i.e., it is a disallowed general coordinate transform?

Or is it your claim that, while a linear coordinate transform can indeed be performed on a curved manifold, it is, in your judgement, non-useful or unphysical or otherwise ill-advised to do so?

Can you please clarify?

 

[PeterDonis]

is it your claim that, while a linear coordinate transform can indeed be performed on a curved manifold, it is, in your judgement, non-useful or unphysical or otherwise ill-advised to do so?

This, yes. It's similar to the point you made in an earlier post, that while such transformations can be made mathematically, they aren't really useful for anything. The usual justification for using rectangular coordinates is that they make the metric look very simple; in flat spacetime that is of course the case. But in curved spacetime it rarely is. (The only case I can think of offhand in which it is is FRW spacetime.)

 

[Cepatiti]

I'm not sure what you mean exactly by "geometrical object".

Just to briefly come back to the notion of "geometrical object": I don't think this notion is rigorously defined in physics, that's why I put it between quotation marks. However, I believe it is attached to the notion of "physical" degrees of freedom, i.e. degrees of freedom that cannot be annihilated by a coordinate transformation.

For example, if you Taylor-expand the metric around a point in spacetime, then a coordinate transformation provides you with enough degrees of freedom to annihilate the first derivative (and the Christoffel symbols) but not enough to annihilate the second and higher orders. It means that the physical information is not contained in the first derivatives (or the Christoffel symbols) but in the second order, related to the curvature of spacetime, and higher orders. Thus if you place yourself in a frame of free-fall, it looks like a inertial frame of reference up to first order-effects. But physical effects due to the second an higher orders will let you know that you are not in a genuine inertial frame.

Of course, if you restrict the coordinate transformations to a subgroup, at some point the subgroup will not have enough degrees of freedom to annihilate the first derivatives or Christoffel symbols anymore, and the Christoffel symbols may also transform like a tensor under this subgroup. I can understand that you want to use the shortcut "it is a tensor under" instead of "it transforms like a tensor under", but personally, I would not state it that way, because (it's weird! and) you lose the information that it is not a physical quantity in the sense of GR. But truly, it is just a matter of definition/convention/habit.

 

[haushofer]

Hmm, is that true? I have never tried that, and I am not even sure exactly how to do it, but I am rather skeptical about it.

See post #22 of Renormalize.

 

[renormalize]

This, yes. It's similar to the point you made in an earlier post, that while such transformations can be made mathematically, they aren't really useful for anything.

OK, thanks for your clarification.

 

[vanhees71]

Once more the entire discussion becomes much clearer when saying that we discuss the transformation of tensor components, not tensors, because tensors are by definition invariant under the transformations of coordinates under discussion.

Then you have to distinguish, in which framework you are working. Here one has to distinguish whether one discusses general diffeomorphism invariance on differentiable manifolds and here for the special case of a (pseudo-)Riemannian manifold with the implied torsion free connection, which is uniquely determined by the fundamental form (aka the (pseuo-) metric) of this manifold or under certain subgroups of transformation laws.

In GR the general diffeomorphism invariance is a gauge invariance (to be distinguished clearly from a physical symmetry as any gauge invariance). It makes the Poincare symmetry of Minkowski space local in a special way, choosing a pseudo-Riemannian manifold for its realization (more generally when taking into account spin one has to extend this to a Einstein-Cartan manifold with both a pseudo-metric and torsion). There is of course no meaning of Poincare or Lorentz symmetry on the global manifold.