Covariant Derivatives

  • #1
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1

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...
 
Last edited:

Answers and Replies

  • #2
martinbn
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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?
 
  • #3
stevendaryl
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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.
 
  • #4
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1
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.
 
  • #5
stevendaryl
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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).
 
Last edited:
  • #6
stevendaryl
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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##
 
  • #7
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1
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.
 
Last edited:
  • #8
64
1
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.
 
  • #9
stevendaryl
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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.
 
  • #10
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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...
 
  • #11
stevendaryl
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##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.
 
  • #12
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1
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!
 

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