Problem in understanding composition of covariant derivatives

[Mike_bb]

Hello, PF!

When we use covariant derivative ($\nabla_a$ or $\nabla_b$) we have that $\nabla_a$ or $\nabla_b$ shows how vector field changes in $a$ or $b$ directions. It's clear to me.

Next, I think so: if we need to find how vector changes in $a$ and $b$ directions consequently we should use sum of changes of vector along $a$ and $b$ directions.

But I confuse because we use composition of covariant derivatives ($\nabla_a\nabla_b$) instead.

$\nabla_a$ shows rate of change of a vector field. But if we use composition $\nabla_b \nabla_a$ we find rate of rate of change of vector field in direction $b$. It confused me.

Could anyone explain why do we use composition of covariant derivatives?

Thanks.

 

[jedishrfu]

The del A gives you the rate of change in the A direction. When you apply del B to that you get the rate of change of del A in the B direction.

If A and B were perpendicular then you’d get a zero rate of change. If A and B are at 45 degrees then you’d get the component of del A in the B direction.

 

[Mike_bb]

If A and B were perpendicular then you’d get a zero rate of change.

Why?

 

[Matterwave]

If A and B were perpendicular then you’d get a zero rate of change.

This is not true in general, if it were so you'd have $\partial_x \partial_y f =0$ generically and it's obviously not generically true for all $f$.

 

[Matterwave]

I can post a more complete answer later today if nobody else has done so already by then but it'll be a while.

 

[Mike_bb]

I can post a more complete answer later today if nobody else has done so already by then but it'll be a while.

That would be great.

 

[wrobel]

if we need to find how vector changes in a and b directions

what does 'vector changes in a and b directions' mean?

 

[Matterwave]

When we use covariant derivative ($\nabla_a$ or $\nabla_b$) we have that $\nabla_a$ or $\nabla_b$ shows how vector field changes in $a$ or $b$ directions. It's clear to me.

Could you explain your notation? Are you using abstract index notation? That is the notation I know that looks like what you used.

In that notation, as the name suggests, the indices $a,b,...$ are abstract in the sense that they just tell you what type of object you're dealing with and should not be read as actually indexing a particular component.

In this notation, a directional derivative is written as a contraction with a vector such as: $\xi^a \nabla_a$. This object should be read as "the covariant derivative in the direction of $\xi^a$".

Next, I think so: if we need to find how vector changes in $a$ and $b$ directions consequently we should use sum of changes of vector along $a$ and $b$ directions.

As mentioned above, in the abstract index notation $a,b,...$ are simply abstract indices, they don't specify directions.

Vectors by themselves don't change, vector fields do. The covariant derivative operates on (scalar, vector, tensor) fields.

If you want the directional covariant derivative of a vector field $u^a$ along the direction given by the vector $\xi^a$ you write $\xi^a\nabla_a u^b$.

Now it should be obvious that if all you want is a directional derivative, you don't need two separate $\nabla_a$, you simply use a vector to point you in the right direction. Further, if you want to do "directional derivative in the $\xi^a+\chi^a$ direction" you do indeed just add: $(\xi^a+\chi^a)\nabla_a u^b=\xi^a\nabla_a u^b+\chi^a\nabla_a u^b$

But I confuse because we use composition of covariant derivatives ($\nabla_a\nabla_b$) instead.

$\nabla_a$ shows rate of change of a vector field. But if we use composition $\nabla_b \nabla_a$ we find rate of rate of change of vector field in direction $b$. It confused me.

Could anyone explain why do we use composition of covariant derivatives?

The second (covariant) derivative is used for different things than the first derivative. There's nothing mysterious about just taking another derivative.

Just as in regular calculus, the first derivative gives you a rate of change (e.g. velocity, slope) while the second derivative gives you the rate of change of the rate of change (e.g. acceleration, curvature).

You use the right tool for the task.

 

[Mike_bb]

Matterwave, big thanks!!

I found such interpretation of measuring curvature:

Could you explain why "The difference between these two paths is precisely what the commutator calculates"?
Thx.

 

[Ibix]

I had a suspicion this was about the Riemann tensor.

Start with a vector $V^a$ and parallel transport it along a small vector $\delta x^a$ and it becomes $V'^a=V^a+\delta x^b\nabla_bV^a$. Transport $V'^a$ along a small vector $\delta y^a$ and it becomes $V''^a=V'^a+\delta y^b\nabla_bV'^a$. Expand that out (being careful about dummy indices) and you'll find that $V''^a$ depends on first and second derivatives of $V^a$.

Now repeat the process in the other order (transport along $\delta y^a$ first) and you will get a different final result. Taking the difference between the two results will have the original vector and its first derivatives cancel leaving only the commutator of the two derivative operators.

So you are correct that second derivatives aren't the same as two composited transports of a vector, but that isn't all that's happening here. This is why it's generally useful to provide some context when you're asking a question.

 

[Mike_bb]

Ibix, That's what I want! Thanks!

 

[wrobel]

Start with a vector Va and parallel transport it along a small vector δxa and it becomes V′a=Va+δxb∇bVa. T

Rigorously speaking,

$$V^k(x+\delta x) = V^k(x) + \delta x^s \partial_s V^k(x) + \ldots$$

Result of parallel transporting the vector $V(x)$ to the point $x+\delta x$:
$$V'^k(x+\delta x) = V^k(x) - \Gamma^k_{sj}(x) V^j(x) \delta x^s + \ldots$$

The difference yields the covariant derivative:
$$V^k(x+\delta x) - V'^k(x+\delta x) = \delta x^s \left( \partial_s V^k(x) + \Gamma^k_{sj}(x) V^j(x) \right) $$
$$= \delta x^s \nabla_s V^k(x)+\ldots$$