# A few questions about the covariant derivative

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• etotheipi
So I believe it's a directional derivative in this sense, but not necessarily in the sense of multivariable calculus.Someone else please correct me if I'm wrong!In summary, the conversation discusses the concept of covariant derivatives on a manifold. The covariant derivative is defined as a (1,1) tensor field that takes a vector field to another vector field and is denoted as ##\nabla V = \left(\frac{\partial V^{\mu}}{\partial x^\rho} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} \right) e_{\mu} \otimes e^{\rho} \equiv {(\nabla V)^{\mu}}_{\
etotheipi
Hey everyone, I was trying to learn in an unrigorous way a bit about making derivatives in the general manifold, but I'm getting confused by a few things. Take a vector field ##V \in \mathfrak{X}(M): M \rightarrow TM##, then in some arbitrary basis ##\{ e_{\mu} \}## of ##\mathfrak{X}(M)## we have$$\frac{\partial V}{\partial x^\rho} = \frac{\partial}{\partial x^\rho} \left( V^{\mu} e_{\mu} \right) = \frac{\partial V^{\mu}}{\partial x^\rho} e_{\mu} + V^{\mu} \Gamma^{\sigma}_{\rho \mu} e_{\sigma} = \frac{\partial V^{\mu}}{\partial x^\rho} e_{\mu} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} e_{\mu} = \left(\frac{\partial V^{\mu}}{\partial x^\rho} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} \right) e_{\mu}$$because of definition of the symbols ##\partial_a e_b = \Gamma^{c}_{ab} e_c##. Then we define a "covariant derivative" as a (1,1) tensor field$$\nabla V = \left(\frac{\partial V^{\mu}}{\partial x^\rho} + V^{\sigma} \Gamma^{\mu}_{\rho \sigma} \right) e_{\mu} \otimes e^{\rho} \equiv {(\nabla V)^{\mu}}_{\rho} e_{\mu} \otimes e^{\rho}$$This tensor can naturally be seen as a map from the tangent space to itself, e.g. if we act the tensor upon an arbitrary basis vector ##e_{\nu}##$$\nabla V(\, \cdot\,, e_{\nu}) = \left(\frac{\partial V^{\mu}}{\partial x^\nu} + V^{\sigma} \Gamma^{\mu}_{\nu \sigma} \right) e_{\mu} = \frac{\partial V}{\partial x^{\nu}}$$My first question is that sometimes I have seen the notation ##\nabla_{\nu} V## used, but never seen it defined. Is this defined to be ##\nabla_{\nu} V \equiv \nabla_{e_{\nu}} V \equiv \partial_{\nu} V##, as above, or something else?

Secondly, say we have some other vector field ##U \in \mathfrak{X}(M): M \rightarrow TM##, then how do we define ##\nabla_{U} V##, which is a function ##\mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow \mathfrak{X}(M)##? Maybe something like$$\nabla(U,V) = \nabla(U^{\mu} e_{\mu}, V)\equiv \nabla_{U} V = \nabla_{U^{\mu} e_{\mu}} V = U^{\mu} \nabla_{e_{\mu}}V = U^{\mu} \partial_{\mu} V$$and I was wondering if this is correct? Also, if ##f## is some function and ##W## a vector field, then why does ##\nabla_{fW} = f \nabla_W##?

Thanks!

Since you said "unrigorous"...
$$\nabla_\rho V^\mu ~:=~ \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma}$$ $$U^\rho \nabla_\rho V^\mu ~=~ U^\rho \left( \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} \right)$$

etotheipi
Right cool thanks, that looks like what I have. Now I feel stupid for writing out so much, oh well

There's one other aspect I wanted to clarify... do we define the covariant derivative of the components of an arbitrary tensor ##T## with respect to some basis to be$$\nabla_V T^{a_1, \dots, a_m}_{b_1,\dots,b_n} = (\nabla_{V} T)^{a_1, \dots, a_m}_{b_1,\dots,b_n}$$where ##\nabla_V = V^{\mu} \nabla_{e_{\mu}}##, as usual? Or, is the notation on the LHS just shorthand for that on the RHS?

The LHS does seem a bit ambiguous, given that a tensor component is a function on a manifold - whose covariant derivative would ordinarily just be a partial derivative.

Last edited by a moderator:
etotheipi said:
do we define the covariant derivative of the components of an arbitrary tensor ##T## with respect to some basis

No, we don't. There is no such thing as "the covariant derivative of the components". The covariant derivative is an operator that acts on tensors. It doesn't act on components.

Also, the covariant derivative is a different operator from the absolute derivative along a curve, which is what you seem to be talking about.

The covariant derivative operator, ##\nabla##, takes an ##(m, n)## tensor (i.e., a tensor with ##m## upper and ##n## lower indexes) to an ##(m, n + 1)## tensor. That's why, in component notation, it is often written as ##\nabla_a##, but it would be more correct to write, not ##\nabla_a T_{bc}## but ##(\nabla T)_{abc}##.

The absolute derivative operator along a curve, ##\nabla_V##, tells how a tensor changes along a curve whose tangent vector is ##V##. If all you know is the values of the tensor on the curve, it makes no sense to write ##\nabla_V## as ##V^a \nabla_a##; to write it in the latter form, you need to know the components of a vector field ##V## in some neighborhood of the curve, not just the tangent vector of the curve itself.

This Insights article might be helpful:

https://www.physicsforums.com/insights/precession-in-special-and-general-relativity/

vanhees71 and etotheipi
Thanks, that clarifies it! In that case the ##\nabla_V T^{a_1, \dots, a_m}_{b_1,\dots,b_n}## is just shorthand.

And yeah, I should have said that ##T## is a tensor field, otherwise it certainly doesn't make sense

vanhees71
strangerep said:
Since you said "unrigorous"...
$$\nabla_\rho V^\mu ~:=~ \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma}$$ $$U^\rho \nabla_\rho V^\mu ~=~ U^\rho \left( \frac{\partial V^\mu}{\partial x^\rho} ~+~ V^\sigma \Gamma^\mu_{~\rho \sigma} \right)$$
Is that second equation a directional derivative?

dsaun777 said:
Is that second equation a directional derivative?

I believe it's the ##\mu## component of the vector field ##\nabla_{U} V##, i.e. the ##\mu## component of the derivative of ##V## along the vector ##U##.

Also, if ##U = \dot{\gamma}## is the tangent vector to a curve ##\gamma##, then it's the ##\mu## component of the derivative of the vector field ##V## along the curve.

## 1. What is the definition of a covariant derivative?

The covariant derivative is a mathematical operation that measures the rate of change of a vector field along a given direction. It takes into account the curvature of the underlying space, making it a more general and accurate measure of change compared to the ordinary derivative.

## 2. How is the covariant derivative different from the ordinary derivative?

The covariant derivative takes into account the curvature of the underlying space, while the ordinary derivative does not. This means that the covariant derivative is a more accurate measure of change for vector fields on curved surfaces, while the ordinary derivative is more suitable for flat surfaces.

## 3. What is the role of the Christoffel symbols in the covariant derivative?

The Christoffel symbols are used in the covariant derivative to account for the curvature of the underlying space. They represent the connection between the tangent spaces at different points on a curved surface, and are essential for calculating the covariant derivative of a vector field.

## 4. How is the covariant derivative used in physics?

The covariant derivative is used in physics to describe the behavior of vector fields in curved space-time. It is an important tool in the field of general relativity, where it is used to define the equations of motion for particles and to describe the curvature of space-time caused by massive objects.

## 5. Are there any real-world applications of the covariant derivative?

Yes, the covariant derivative has many real-world applications, particularly in fields such as physics, engineering, and computer graphics. It is used in various applications, such as calculating the trajectory of a satellite in orbit, simulating fluid flow, and creating realistic 3D animations of objects moving in curved space.

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