how to calculate covariant derivative of vector field

[logistic_guy]

here is the question

Given two vector fields $\displaystyle V$ and $\displaystyle W$ such that $\displaystyle V = x^2yU_1 + e^xU_2 + z^3U_3$ and $\displaystyle W = x^3U_1 + y\sin(z)U_2 - \cos(xz^2\log[xz])U_3$. Calculate the covariant derivative of $\displaystyle W$ with respect to $\displaystyle V$.

i know how to calculate to some point. the covariant derivative $\displaystyle \nabla_VW = \sum V[w_i]U_i$

$\displaystyle \sum_{i=1}^{3} V[w_i]U_i = V[w_1]U_1 + V[w_2]U_2 + V[w_3]U_3 = V[x^3]U_1 + V[y\sin(z)]U_2 + V[\cos(xz^2\log[xz])]U_3$

here my problem. how to calculate $\displaystyle V[\cos(xz^2\log[xz])]$

 

[mario99]

here my problem. how to calculate $\displaystyle V[\cos(xz^2\log[xz])]$

I am not familiar with the word covariant in this context, but I am guessing that you have to take the derivative with respect to $\displaystyle z$ since this expression is placed in the third term. Do you know how to use the chain and product rules?

If you write the definition of the covariant derivative in this context, we might provide a better hint.

 

[blamocur]

If you write the definition of the covariant derivative in this context, we might provide a better hint.

My familiarity with the topic is very sketchy and superficial, and the statement of the problem in the op is rather terse and incomplete, so here are my assumptions both about the problem and the definition of the covariant derivative in the euclidean space:

$U_i$ is an orthonormal basis in $\mathbb R^3$, and if $W = \sum_i w_i U_i$ and $V = \sum_i v_i U_i$ then the definition of the covariant derivative is:

$$\nabla_V W = \sum_{ij} v_j \frac{\partial w_i}{\partial x_j} U_i = \sum_i\left( v_1 \frac{\partial w_i}{\partial x} + v_2 \frac{\partial w_i}{\partial y} + v_3 \frac{\partial w_i}{\partial z} \right) U_i$$
I'll wait for the confirmation from the OP before going any further.

 

[logistic_guy]

I am not familiar with the word covariant in this context, but I am guessing that you have to take the derivative with respect to $\displaystyle z$ since this expression is placed in the third term. Do you know how to use the chain and product rules?

If you write the definition of the covariant derivative in this context, we might provide a better hint.

thank mario99

My familiarity with the topic is very sketchy and superficial, and the statement of the problem in the op is rather terse and incomplete, so here are my assumptions both about the problem and the definition of the covariant derivative in the euclidean space:

$U_i$ is an orthonormal basis in $\mathbb R^3$, and if $W = \sum_i w_i U_i$ and $V = \sum_i v_i U_i$ then the definition of the covariant derivative is:

$$\nabla_V W = \sum_{ij} v_j \frac{\partial w_i}{\partial x_j} U_i = \sum_i\left( v_1 \frac{\partial w_i}{\partial x} + v_2 \frac{\partial w_i}{\partial y} + v_3 \frac{\partial w_i}{\partial z} \right) U_i$$
I'll wait for the confirmation from the OP before going any further.

thank blamocur. there lots of definitions i can't handle them. i'll confirm this with my teacher. probably it won't reply

hold on blamocur, i'll get back to you

 

[blamocur]

thank blamocur. there lots of definitions i can't handle them.

I don't really understand the definition you posted ($\nabla_V W=\sum V[w_i]U_i$). Particularly, what does $V[w_i]$ stand for?

 

[logistic_guy]

i'm back

I don't really understand the definition you posted ($\nabla_V W=\sum V[w_i]U_i$). Particularly, what does $V[w_i]$ stand for?

i think this $\displaystyle \nabla_V W$ and this $\displaystyle V[w_i]$ both derivatives. the result of the first is another vector field while the result of the second is a coordinate function

My familiarity with the topic is very sketchy and superficial, and the statement of the problem in the op is rather terse and incomplete, so here are my assumptions both about the problem and the definition of the covariant derivative in the euclidean space:

$U_i$ is an orthonormal basis in $\mathbb R^3$, and if $W = \sum_i w_i U_i$ and $V = \sum_i v_i U_i$ then the definition of the covariant derivative is:

$$\nabla_V W = \sum_{ij} v_j \frac{\partial w_i}{\partial x_j} U_i = \sum_i\left( v_1 \frac{\partial w_i}{\partial x} + v_2 \frac{\partial w_i}{\partial y} + v_3 \frac{\partial w_i}{\partial z} \right) U_i$$
I'll wait for the confirmation from the OP before going any further.

i show your sum to my teacher it say alter the sum to $\displaystyle \nabla_V W = \sum_{i} v_i \frac{\partial w_i}{\partial x_i} U_i$

it say i can also do this $\displaystyle V[w_i] = \sum v_i U_i[w_i] = \sum v_i \frac{\partial w_i}{\partial x_i}$

I am not familiar with the word covariant in this context, but I am guessing that you have to take the derivative with respect to $\displaystyle z$ since this expression is placed in the third term. Do you know how to use the chain and product rules?

If you write the definition of the covariant derivative in this context, we might provide a better hint.

it also advice me to read about the product rule and the chain rule to solve the problem, so i think mario99 is correct

one thing i don't understand why do i need the product rule and the chain rule? is it wrong to say the derivative of $\displaystyle \cos$ is $\displaystyle -\sin$?

 

[blamocur]

i show your sum to my teacher it say alter the sum to...

Consider the simplest case: $V = U_1$ is a constant field (i.e. $v_1 = 1$, $v_2=v_3=0$), and $W = x^2 U_2$ (i.e., $w_1 = w_3 = 0$ and $w_2 = x^2$). For these two vector fields we would have $\nabla_V W = 0$ if we use your teacher's definition, but to me it seems counterintuitive. On the other hand:

My familiarity with the topic is very sketchy and superficial

 

[logistic_guy]

Consider the simplest case: $V = U_1$ is a constant field (i.e. $v_1 = 1$, $v_2=v_3=0$), and $W = x^2 U_2$ (i.e., $w_1 = w_3 = 0$ and $w_2 = x^2$). For these two vector fields we would have $\nabla_V W = 0$ if we use your teacher's definition, but to me it seems counterintuitive. On the other hand:

i don't see any problem if we follow the definition and get zero. i think it is so normal to get zero for your example

 

[blamocur]

i don't see any problem if we follow the definition and get zero. i think it is so normal to get zero for your example

Just not the answer I'd expect, but, to repeat, I don't claim to really understand the subject. Still, would be interesting to hear if your teacher agrees.

 

[mario99]

one thing i don't understand why do i need the product rule and the chain rule? is it wrong to say the derivative of $\displaystyle \cos$ is $\displaystyle -\sin$?

No, it is not wrong. But the derivative will also depend on the parameters of cosine. Let us first understand when to apply the product rule, then we will go to the chain rule. Consider this example, $xe^x$ and we want to find $\displaystyle \frac{d[ \ xe^x \ ]}{dx}$.

Now imagine we have two objects:

$\text{object}_1 = x$
$\text{object}_2 = e^x$

Whenever you have two objects that share the same variable and are multiplied by each other, you apply the product rule.

A second scenario to complete the picture:

$\displaystyle \frac{d[ \ xe^x\ln x \ ]}{dx}$

$\text{object}_1 = x$
$\text{object}_2 = e^x$
$\text{object}_3 = \ln x$

The product rule is designed for two objects but now we have three objects. So what to do?

The idea is to combine two objects together:

$\text{object}_1 = xe^x$
$\text{object}_2 = \ln x$

You apply the product rule for the two objects above, and when you finish, you apply the product rule again for the two objects below:

$\text{object}_1 = x$
$\text{object}_2 = e^x$

Now, you have the idea of when to differentiate a function by the product rule even if this function contains $1000$ objects (kidding).

Try to solve this:

$\displaystyle \frac{d[ \ xe^x \ ]}{dx} = \ ?$

If you got it correct, we would move to the chain rule.

 

[logistic_guy]

Just not the answer I'd expect, but, to repeat, I don't claim to really understand the subject. Still, would be interesting to hear if your teacher agrees.

let me show you something i'm study. you may be convince

Definition: $\displaystyle \nabla_vW = W(\bold{p} + t\bold{v})'(0)$

$\displaystyle \bold{p}$ point and $\displaystyle \bold{v}$ tangent vector and $\displaystyle 0$ mean $\displaystyle t = 0$

$\displaystyle \nabla_vW$ mean the covariant derivative of the vector field $\displaystyle W$ with respect to the tangent vector $\displaystyle \bold{v}$ at the point $\displaystyle \bold{p}$

Lemma: $\displaystyle \nabla_vW = \sum \bold{v}[w_i]U_i(\bold{p})$

$\displaystyle \bold{v}[w_i]$ mean the derivative of the coordinate functions of the vector field $\displaystyle W$ with respect to the tangent vector $\displaystyle \bold{v}$

$\displaystyle U_i(\bold{p})$ mean natural frame field or unit vectors at the point $\displaystyle \bold{p}$

this lemma can be proof

by looking at all of the above the pointwise principle tell us if the covariant derivative can be apply with respect to a single tangent vector it can also apply with respect to a vector field, so

$\displaystyle \nabla_VW = \sum V[w_i]U_i(\bold{p})$

finally

let $\displaystyle f$ be a real valued function

if $\displaystyle V = \sum v_i U_i$

it is obvious $\displaystyle V[f] = \sum v_i U_i[f] = \sum v_i \frac{\partial f}{\partial x_i}$

if instead of the function $\displaystyle f$ we've vector field coordinate funtions $\displaystyle w_i$ you'll get the original form $\displaystyle V[w_i]$

all remain is to show you $\displaystyle U_i[f] = \frac{\partial f}{\partial x_i}$ which if you get convince it's not necessary

it would be more interesting if you're still not convince

No, it is not wrong. But the derivative will also depend on the parameters of cosine. Let us first understand when to apply the product rule, then we will go to the chain rule. Consider this example, $xe^x$ and we want to find $\displaystyle \frac{d[ \ xe^x \ ]}{dx}$.

Now imagine we have two objects:

$\text{object}_1 = x$
$\text{object}_2 = e^x$

Whenever you have two objects that share the same variable and are multiplied by each other, you apply the product rule.

A second scenario to complete the picture:

$\displaystyle \frac{d[ \ xe^x\ln x \ ]}{dx}$

$\text{object}_1 = x$
$\text{object}_2 = e^x$
$\text{object}_3 = \ln x$

The product rule is designed for two objects but now we have three objects. So what to do?

The idea is to combine two objects together:

$\text{object}_1 = xe^x$
$\text{object}_2 = \ln x$

You apply the product rule for the two objects above, and when you finish, you apply the product rule again for the two objects below:

$\text{object}_1 = x$
$\text{object}_2 = e^x$

Now, you have the idea of when to differentiate a function by the product rule even if this function contains $1000$ objects (kidding).

Try to solve this:

$\displaystyle \frac{d[ \ xe^x \ ]}{dx} = \ ?$

If you got it correct, we would move to the chain rule.

thank mario99

i know the product rule. i asked why do i have to use it?

$\displaystyle xe^x = (1)e^x + xe^x = e^x + xe^x = e^x(1 + x)$

the derivative correct?

 

[blamocur]

it would be more interesting if you're still not convince

No, not yet. I assume that in your definition
$$\nabla_v \mathbf W = W(\mathbf p + t\mathbf v)^\prime(0) = \frac{d}{dt} W(\mathbf p + t\mathbf v)(0)$$If we plugin my example from post #7 we get:
$$\nabla_v W = \frac{d}{dt} W((x_p+t,y_p, z_p))(0) = \frac{d}{dt} (0, (x_p+t)^2, 0)(0) = (0, 2x_p, 0)$$I.e. $\nabla_V W \neq 0$ when $x_p \neq 0$ !

 

[logistic_guy]

No, not yet. I assume that in your definition
$$\nabla_v \mathbf W = W(\mathbf p + t\mathbf v)^\prime(0) = \frac{d}{dt} W(\mathbf p + t\mathbf v)(0)$$If we plugin my example from post #7 we get:
$$\nabla_v W = \frac{d}{dt} W((x_p+t,y_p, z_p))(0) = \frac{d}{dt} (0, (x_p+t)^2, 0)(0) = (0, 2x_p, 0)$$I.e. $\nabla_V W \neq 0$ when $x_p \neq 0$ !

this definition is for vector field $\displaystyle W$ and tangent vector $\displaystyle \bold{v}$, but you apply it for vector field $\displaystyle W$ and vector field $\displaystyle V$. this is why you got wrong result.

 

[blamocur]

this definition is for vector field W\displaystyle WW and tangent vector v\displaystyle \bold{v}v, but you apply it for vector field W\displaystyle WW and vector field V\displaystyle VV. this is why you got wrong result.

Don't see any difference here: $\nabla_V W(p) = \nabla_v W$ where $v = V(p)$.

 

[logistic_guy]

Don't see any difference here: $\nabla_V W(p) = \nabla_v W$ where $v = V(p)$.

there are big difference

a tangent vector $\displaystyle \bold{v_p}$ consists of two points, starting $\displaystyle \bold{p}$ and ending $\displaystyle \bold{p} + \bold{v}$. even the vector part $\displaystyle \bold{v}$ of $\displaystyle \bold{v_p}$ is just a point like $\displaystyle \bold{p}$ but with different letter

if $\displaystyle f$ is a function, the derivative of $\displaystyle f$ with respect to $\displaystyle \bold{v_p} = \bold{v_p}[f] = \sum v_i\frac{\partial f}{\partial x_i}(\bold{p})$

here $\displaystyle v_i$ will always be a constant

while vector field $\displaystyle V$ consists of coordinate functions $\displaystyle v_i$, not necessary constants, attahched with unit vectors $\displaystyle U_i$.

if $\displaystyle f$ is a function, the derivative of $\displaystyle f$ with respect to $\displaystyle V = V[f] = \sum v_iU_i[f] = \sum v_i\frac{\partial f}{\partial x_i}(\bold{p})$

here $\displaystyle v_i$ can be anything

the final result looks similar but the middle step is so crucial and without it you'll never know the difference between them

for the covariant derivatives just change $\displaystyle f$ to the coordinate functions of vector field to match them with the above examples

$\displaystyle \nabla_vW = \sum \bold{v}[w_i]U_i(\bold{p})$

$\displaystyle \nabla_VW = \sum V[w_i]U_i(\bold{p})$

applying the definition when $\displaystyle t = 0$ assuming $\displaystyle \bold{v}$ = $\displaystyle V$ gives wrong result is a proof the two covariant derivatives are different

 

[blamocur]

applying the definition when t=0\displaystyle t = 0t=0 assuming v\displaystyle \bold{v}v = V\displaystyle VV gives wrong result is a proof the two covariant derivatives are different

Don't see a proof here. Which results do you get in my simple example from post #7?

 

[logistic_guy]

Don't see a proof here. Which results do you get in my simple example from post #7?

me and Dr.Peterson discuss some proof and when i give one example Dr.Peterson tell me proof depends on the general case, not one example

if i can proof $\displaystyle \nabla_vW = W(\bold{p} + t\bold{v})'(0) = \sum \bold{v}[w_i]U_i(\bold{p})$

for arbitrary $\displaystyle \bold{p} = (p_1,p_2,p_3)$ and $\displaystyle \bold{v} = (v_1,v_2,v_3)$ on $\displaystyle \bold{R}^3$ we're done!

do you agree with @Dr.Peterson ?

i can proof the above statement in general as true. can you proof your #7 in general as true? if you can't your example #7 is discarded.

also your example have $\displaystyle \bold{p} = (p_1,0,0)$ and $\displaystyle \bold{v} = (v_1,0,0)$ which seams $\displaystyle \bold{p}$ and $\displaystyle \bold{v}$ isn't in $\displaystyle \bold{R}^3$

(at least $\displaystyle \bold{v}$ isn't in $\displaystyle \bold{R}^3$)

 

[mario99]

i know the product rule. i asked why do i have to use it?

Because you have two objects that are multiplied by each other and share the same variable $z$.

the derivative correct?

Yes.

Note: You are discussing high level mathematics which I myself don't understand, with professor blamocur, but in the same time you do not understand why do you have to use the product rule! This looks awkward.

I think that you will not need me to make another story about the chain rule!

 

[blamocur]

me and Dr.Peterson discuss some proof and when i give one example Dr.Peterson tell me proof depends on the general case, not one example

No argument here. But special case can be useful as counter-examples.

if i can proof ∇vW=W(p+tv)′(0)=∑v[wi]Ui(p)\displaystyle \nabla_vW = W(\bold{p} + t\bold{v})'(0) = \sum \bold{v}[w_i]U_i(\bold{p})∇vW=W(p+tv)′(0)=∑v[wi]Ui(p)

for arbitrary p=(p1,p2,p3)\displaystyle \bold{p} = (p_1,p_2,p_3)p=(p1,p2,p3) and v=(v1,v2,v3)\displaystyle \bold{v} = (v_1,v_2,v_3)v=(v1,v2,v3) on R3\displaystyle \bold{R}^3R3 we're done!

I don't see what this would prove. You claim that for $v = V(p)$ derivatives $\nabla_v W (p)$ and $\nabla_V W (p)$ are different, and I don't see why this should be so. For example, if $V=v$ at every point in $\mathbb R^3$ would you still expect different derivatives?

BTW, you use $v[w_i]$ and $V[w_i]$ a lot but never define them explicitly. Would you like posting formulas for both? For example, if $V = \sum_k v_k U_k$ what is the expression for $V[w_i]$ ?

 

[logistic_guy]

Because you have two objects that are multiplied by each other and share the same variable $z$.




Yes.

Note: You are discussing high level mathematics which I myself don't understand, with professor blamocur, but in the same time you do not understand why do you have to use the product rule! This looks awkward.

I think that you will not need me to make another story about the chain rule!

i'm sorry if i can't express my question correctly, but thank for the help you give me

No argument here. But special case can be useful as counter-examples.

I don't see what this would prove. You claim that for $v = V(p)$ derivatives $\nabla_v W (p)$ and $\nabla_V W (p)$ are different, and I don't see why this should be so. For example, if $V=v$ at every point in $\mathbb R^3$ would you still expect different derivatives?

BTW, you use $v[w_i]$ and $V[w_i]$ a lot but never define them explicitly. Would you like posting formulas for both? For example, if $V = \sum_k v_k U_k$ what is the expression for $V[w_i]$ ?

it's a cumulative process

first you start with a differentiable function $\displaystyle f$ on $\displaystyle \bold{R}^3$ and a tangent vector $\displaystyle \bold{v}_p$ to $\displaystyle \bold{R}^3$

Definition: $\displaystyle \bold{v}_p[f] = \frac{d}{dt}(f(\bold{p} + t\bold{v}))|_{t=0}$

then you get lemma which can be proof by the definition

Lemma: $\displaystyle \bold{v}_p[f] = \sum v_i\frac{\partial f}{\partial x_i}(\bold{p})$

then the pointwise principle allow you to extend the derivative of the function to vector field $\displaystyle V[f]$

then an immediate consequence of the lemma above you get $\displaystyle U_i[f] = \frac{\partial f}{\partial x_i}$ which can be proof by looking at the derivative of the function with respect to one unit vector $\displaystyle U_1(\bold{p})[f] = \frac{d}{dt}(f(p_1 + t, p_2,p_3))|_{t=0}$ as $\displaystyle U_1(\bold{p}) = (1,0,0)_p$

the vector field $\displaystyle V$ and the derivative of $\displaystyle f$ with respect to $\displaystyle V$ are defined like this

$\displaystyle V = \sum v_i U_i$ and $\displaystyle V[f] = \sum v_i U_i[f]$ respectively

then we start with covariant derivative with this definition

Definition: $\displaystyle \nabla_vW = W(\bold{p} + t\bold{v})'(0)$

then the process repeats itself again and again

which give new formulas that involve $\displaystyle \bold{v}_p[w_i]$ and $\displaystyle V[w_i]$ instead of $\displaystyle \bold{v}_p[f]$ and $\displaystyle V[f]$. the same concept apply