# Covariant Derivatives

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## Main Question or Discussion Point

I've just learned about the covariant derivatives ($\nabla_i$ and $\delta/\delta t$) and I have a doubt.
We should be able to say that $$J^i = \frac{\delta A^i}{\delta t} = \frac{\delta^2 V^i}{\delta^2 t} = \frac{\delta^3 Z^i}{\delta^3 t}$$ where $J$ is the jolt. This should mean that $$\frac{\delta Z^i}{\delta t} = \frac{d Z^i}{dt}$$ but I can't prove it. This should be equivalent to proving that $$Z^j \Gamma_{jk}^i V^k = 0$$ Here's my heroic attempt: \begin{align*} Z^j \Gamma_{jk}^i V^k &= Z^j V^k \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\ &= Z^j \frac{d Z^k}{dt} \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\ &= Z^j \frac{d \pmb{Z}_j}{dt}\cdot \pmb{Z}^i \\ \end{align*} Now what? I also tried using the product rule but it doesn't seem to improve the situation...

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martinbn
This should mean that

$\frac{\delta Z^i}{\delta t} = \frac{d Z^i}{dt}$

but I can't prove it.
Can you clarify your notations, and say why you think this should be true?

stevendaryl
Staff Emeritus
I've just learned about the covariant derivatives ($\nabla_i$ and $\delta/\delta t$) and I have a doubt.
We should be able to say that $$J^i = \frac{\delta A^i}{\delta t} = \frac{\delta^2 V^i}{\delta^2 t} = \frac{\delta^3 Z^i}{\delta^3 t}$$ where $J$ is the jolt. This should mean that $$\frac{\delta Z^i}{\delta t} = \frac{d Z^i}{dt}$$ but I can't prove it. This should be equivalent to proving that $$Z^j \Gamma_{jk}^i V^k = 0$$ Here's my heroic attempt: \begin{align*} Z^j \Gamma_{jk}^i V^k &= Z^j V^k \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\ &= Z^j \frac{d Z^k}{dt} \frac{\partial \pmb{Z}_j}{\partial Z^k}\cdot \pmb{Z}^i \\ &= Z^j \frac{d \pmb{Z}_j}{dt}\cdot \pmb{Z}^i \\ \end{align*} Now what? I also tried using the product rule but it doesn't seem to improve the situation...
Even though it is written that way, a position $Z^k$ is not a vector. It's just an indexed collection of coordinate values. So there is no distinction between the covariant derivative of a position and the ordinary derivative of a position.

Even though it is written that way, a position $Z^k$ is not a vector. It's just an indexed collection of coordinate values. So there is no distinction between the covariant derivative of a position and the ordinary derivative of a position.
That's what I'm trying to prove! I'm watching the videos by Pavel Grinfeld and he defines $\delta/\delta t$ as a way to rewrite the derivative of the velocity vector in a cleaner way. (One can then prove that this definition makes sense (e.g. the Leibniz rule applies)).
When he introduced the covariant derivative $\nabla_i$ he defined it on the contravariant components of a vector but then applied it to the vectors of the covariant basis itself, so I surmised that his definitions are "algorithmic" i.e. the same "rule" can be applied to different objects (and everything magically works out!)
Therefore, I thought that his writing $J^i = \delta^3 Z^i / (\delta t^3)$ implied that the same expression worked out even when applied to $Z^i$ and I was expecting to recover the equivalence with the ordinary derivative.

stevendaryl
Staff Emeritus
Well, proving that $\Gamma^i_{jk} Z^j \frac{dZ^k}{dt} = 0$ is not the way to prove that $Z^j$ is not a vector. The expression $\Gamma^i_{jk} Z^j \frac{dZ^k}{dt}$ only makes sense if $Z^j$ IS a vector (and also it only makes sense if $\frac{dZ^k}{dt}$ is a vector). But you don't need to prove that $Z^j$ is or is not a vector. What you need is to know that $\frac{dZ^j}{dt}$ is a vector. That's the velocity vector (or the components of the velocity vector).

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stevendaryl
Staff Emeritus
When he introduced the covariant derivative $\nabla_i$ he defined it on the contravariant components of a vector but then applied it to the vectors of the covariant basis itself, so I surmised that his definitions are "algorithmic" i.e. the same "rule" can be applied to different objects (and everything magically works out!)
Even though people write $\nabla_j V^k$, and it looks like it is acting on the components of vector $V$, that's not what is happening. Covariant derivatives act on vectors and return vectors. So strictly speaking, it should be written this way: $(\nabla_j V)^k$. Its meaning is "Component $k$ of the covariant derivative of $V$", not "The covariant derivative of component $k$ of $V$".

A basis vector is a vector, so you can take the covariant derivative of it. The connection coefficient $\Gamma^i_{jk}$ is defined in terms of the basis vectors:

$\Gamma^i_{jk} = (\nabla_j e_k)^i$

Even though people write $\nabla_j V^k$, and it looks like it is acting on the components of vector $V$, that's not what is happening. Covariant derivatives act on vectors and return vectors. So strictly speaking, it should be written this way: $(\nabla_j V)^k$. Its meaning is "Component $k$ of the covariant derivative of $V$", not "The covariant derivative of component $k$ of $V$".

A basis vector is a vector, so you can take the covariant derivative of it. The connection coefficient $\Gamma^i_{jk}$ is defined in terms of the basis vectors:

$\Gamma^i_{jk} = (\nabla_j e_k)^i$
That makes more sense. Unfortunately, Grinfeld explicitly defined it on the components and that confused me a little. But what about general tensors?
His full definition works for any tensor $T^{ijk\cdots}_{rst\cdots}$. Basically, there's the ordinary derivative and then an additional $+\Gamma$ term for every upper index and a $-\Gamma$ term for every lower index.

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Well, proving that $\Gamma^i_{jk} Z^j \frac{dZ^k}{dt} = 0$ is not the way to prove that $Z^j$ is not a vector. The expression $\Gamma^i_{jk} Z^j \frac{dZ^k}{dt}$ only makes sense if $Z^j$ IS a vector (and also it only makes sense if $\frac{dZ^k}{dt}$ is a vector). But you don't need to prove that $Z^j$ is or is not a vector. What you need is to know that $\frac{dZ^j}{dt}$ is a vector. That's the velocity vector (or the components of the velocity vector).
I'm not trying to prove that $Z^j$ is not a vector. I already know that. I was basically trying to prove that $\delta \pmb{R} / \delta t = d\pmb{R} / dt$ by using the formula for the covariant derivative. What are the components of the position $\pmb{R}$ vector, by the way? $\pmb{R}$ seems somewhat special because the covariant basis "originates" from it: $$\pmb{Z}_i = \frac{\partial \pmb{R}}{\partial Z^k}$$ In conclusion, should the definition of the covariant derivative be expanded to handle $R$ as well by basically saying that it's equivalent to the ordinary derivative? I thought I could prove that by using the general formula given to me.

stevendaryl
Staff Emeritus
I'm not trying to prove that $Z^j$ is not a vector. I already know that. I was basically trying to prove that $\delta \pmb{R} / \delta t = d\pmb{R} / dt$ by using the formula for the covariant derivative.
I know. I'm saying that that doesn't make any sense. $R$ is not a vector, so you can't take the covariant derivative of it.

I know. I'm saying that that doesn't make any sense. $R$ is not a vector, so you can't take the covariant derivative of it.
$R$ is the position vector, so it's a vector by definition. To introduce $R$, Grinfeld chose an arbitrary origin. The problem is that it can't be expressed through a covariant basis for obvious reasons so the covariant derivative is not needed and doesn't make sense. I thought that it was still applicable and the gamma term became 0.

Oh I see what you mean. I remember that in differential geometry the vectors are the elements of tangent space and R is merely a point on the manifold. Grinfeld uses a different approach...

stevendaryl
Staff Emeritus
$R$ is the position vector, so it's a vector by definition.
No, it's not a vector. In a flat geometry, it can be associated with a vector, but position is not a vector. The time derivative of the position is a vector, though.

If you have a vector $V$, then you can relate its components in one coordinate system, $V^j$ to its coordinates in another coordinate system, $V^a$ as follows:

$V^j = \frac{\partial x^j}{\partial x^a} V^a$

But let's see if $x^j$ itself satisfies this rule:

$x^j = \frac{\partial x^j}{\partial x^a} x^a$

That's only true if the relationship between $x^j$ and $x^a$ is linear. It's not true if you're transforming between (say) Cartesian coordinates and polar coordinates.

No, it's not a vector. In a flat geometry, it can be associated with a vector, but position is not a vector. The time derivative of the position is a vector, though.

If you have a vector $V$, then you can relate its components in one coordinate system, $V^j$ to its coordinates in another coordinate system, $V^a$ as follows:

$V^j = \frac{\partial x^j}{\partial x^a} V^a$

But let's see if $x^j$ itself satisfies this rule:

$x^j = \frac{\partial x^j}{\partial x^a} x^a$

That's only true if the relationship between $x^j$ and $x^a$ is linear. It's not true if you're transforming between (say) Cartesian coordinates and polar coordinates.
OK, I think I understand now. The components $V^j$ and $V^a$ are related by the Jacobian matrix only because they're themselves derivatives and we can apply the chain rule. Thank you!