Covariant Derivative: Understanding Motivation Quickly

[binbagsss]

As we can not meaningfully compare a vector at 2 points acted upon by this operator , because it does not take into account the change due to the coordinate system constantly changing, I conclude that the elementary differential operator must describe a change with respect to space-time,

How do we know that the elementary differential operator does this, instead of describing a change wrt a coordinate system?

Thanks in advance.

 

[Matterwave]

As we can not meaningfully compare a vector at 2 points acted upon by this operator , because it does not take into account the change due to the coordinate system constantly changing, I conclude that the elementary differential operator must describe a change with respect to space-time,

What do you mean by "elementary differential operator"? I do not think this is standard terminology. Also, how do you draw this conclusion? I don't see how "must describe a change with respect to space-time" follows from "can not compare a vector at 2 points acted upon by this operator". In general, there is no way to compare a vector at 2 different points on a manifold without defining either a connection plus a path, or a congruence.

 

[binbagsss]

oh, sorry,it's the other way around. so the elementary partial derivative gives the change with respect to coordinate system, and the aim of the covariant derivative is to give the change with respect to space-time?

 

[Matterwave]

oh, sorry,it's the other way around. so the elementary partial derivative gives the change with respect to coordinate system, and the aim of the covariant derivative is to give the change with respect to space-time?

This is a very imprecise way of looking at things, but is roughly true. A simple partial derivative with respect to some coordinate system $\partial_\mu$ is not a valid tensor operation, and as such can not act on vector fields or tensor fields. What they can do is act on the components, in that coordinate system, of vector/tensor fields. So $\partial_\mu V^\nu$ gives you the partial derivative, in the $\mu$ direction of the $\nu$ component of the vector $V^a$ in the implicitly defined coordinate system. The $n^2$ quantities $\partial_\mu V^\nu$ do not define the components of a rank 2 tensor in any coordinate system. The co variant derivative is a valid tensor operator, and so acts on the vector itself. The co variant derivative tells you how a vector changes locally, in some particular direction, with respect to the (metric compatible in GR) connection defined on the manifold. In other words, it tells you how a vector changes with respect to locally parallel transported (with parallel transport defined by the connection) vectors along any such direction.