Covariant Derivative: Understanding Motivation Quickly

• binbagsss
In summary, the elementary differential operator describes a change with respect to space-time, as it cannot meaningfully compare a vector at 2 points acted upon by this operator due to the constantly changing coordinate system. The covariant derivative is used to describe the change with respect to space-time, while the elementary partial derivative gives the change with respect to the coordinate system. However, the partial derivative is not a valid tensor operator and can only act on the components of vector/tensor fields.
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?

binbagsss said:
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.

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?

binbagsss said:
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.

I can assure you that the covariant derivative is a mathematical concept used in differential geometry to describe the change of a vector in a curved space, such as space-time. It takes into account the changing coordinate system and allows us to compare vectors at different points in a meaningful way. This is essential in understanding the dynamics of objects moving in a curved space, such as planets orbiting around a star.

The elementary differential operator, on the other hand, is a mathematical tool used to describe the change of a function with respect to a particular coordinate system. It is not designed to take into account the curvature of space-time and therefore cannot accurately describe the change of a vector in a curved space.

To summarize, the covariant derivative is specifically designed to describe changes in a curved space, while the elementary differential operator is used to describe changes in a particular coordinate system. This fundamental difference is what allows us to use the covariant derivative to understand the dynamics of objects in a curved space, while the elementary differential operator is limited to flat spaces.

What is a covariant derivative?

A covariant derivative is a mathematical tool used in differential geometry to measure how a vector field changes as it moves along a curved manifold. It takes into account the curvature of the manifold and allows us to calculate how the vector field changes in a way that is independent of the coordinate system used.

Why is understanding covariant derivative important?

Covariant derivatives are crucial in many areas of physics, particularly in general relativity and electromagnetism. They allow us to describe and understand the behavior of physical systems on curved spaces, which is essential for understanding the universe on a large scale.

How is the covariant derivative different from the ordinary derivative?

The ordinary derivative measures the rate of change of a function with respect to its independent variables. The covariant derivative, on the other hand, measures the rate of change of a vector field as it moves along a curved manifold. It takes into account the curvature of the manifold, which the ordinary derivative does not.

What is the motivation behind using the covariant derivative?

The motivation behind using the covariant derivative is to have a mathematical tool that is independent of the coordinate system used to describe a manifold. This allows us to make calculations and predictions that are consistent and do not depend on the choice of coordinates.

How can one quickly understand the concept of covariant derivative?

One way to quickly understand the concept of covariant derivative is to think of it as a way to "correct" the ordinary derivative for curved spaces. It takes into account the curvature of the manifold and allows us to calculate the rate of change of a vector field in a way that is consistent and independent of the coordinate system used.

• Special and General Relativity
Replies
7
Views
149
• Special and General Relativity
Replies
5
Views
2K
• Special and General Relativity
Replies
16
Views
2K
• Special and General Relativity
Replies
124
Views
6K
• Special and General Relativity
Replies
7
Views
2K
• Special and General Relativity
Replies
10
Views
2K
• Special and General Relativity
Replies
2
Views
1K
• Special and General Relativity
Replies
36
Views
5K
• Special and General Relativity
Replies
1
Views
1K
• Special and General Relativity
Replies
20
Views
5K