I Covariant Derivatives

64
1
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:

martinbn

Science Advisor
1,565
408
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
Science Advisor
Insights Author
8,400
2,573
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.
 
64
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.
 

stevendaryl

Staff Emeritus
Science Advisor
Insights Author
8,400
2,573
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:

stevendaryl

Staff Emeritus
Science Advisor
Insights Author
8,400
2,573
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##
 
64
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:
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.
 

stevendaryl

Staff Emeritus
Science Advisor
Insights Author
8,400
2,573
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.
 
64
1
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
Science Advisor
Insights Author
8,400
2,573
##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.
 
64
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!
 

Related Threads for: Covariant Derivatives

Replies
3
Views
12K
Replies
5
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
8
Views
2K
Replies
9
Views
7K
Replies
37
Views
2K
Replies
0
Views
2K
Replies
7
Views
831
Top