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Index Notation - Prove the following

  • Thread starter squire636
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  • #1
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Homework Statement



http://imgur.com/gTapO


Homework Equations




The Attempt at a Solution



The first one is easy, just use the fact that δi = δ/δxi and it reduces to the sum from with i=1,2,3 of δxi/δxi = 1 + 1 + 1

I tried to do a similar thing with the second one, also using the fact that r = |x| = √(x dot x), but for some reason I can't make any progress. Since r is a scalar, can't I pull it out of the original dot product and have the following:

1/r * (∇ dot u)

Well, obviously I can't because that would give me 3/r rather than 2/r.


I've made no progress on the third one so far, so any advice is appreciated.

Thanks!
 

Answers and Replies

  • #2
gabbagabbahey
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The first one is easy, just use the fact that δi = δ/δxi and it reduces to the sum from with i=1,2,3 of δxi/δxi = 1 + 1 + 1
You need to be careful with your notation here. Do you mean you used [itex]\partial_i x_j = \delta_{ij}[/itex]? If so, then yes that is the correct approach.

I tried to do a similar thing with the second one, also using the fact that r = |x| = √(x dot x), but for some reason I can't make any progress. Since r is a scalar, can't I pull it out of the original dot product and have the following:

1/r * (∇ dot u)

Well, obviously I can't because that would give me 3/r rather than 2/r.
Just because [itex]r[/itex] is a scalar doesn't mean [itex]\partial_i \frac{\mathbf{f}(x,y,z)}{r}= \frac{1}{r} \partial_i \mathbf{f}(x,y,z)[/itex]. For that to be true, [itex]r[/itex] would need to be constant with respect to the variable of differentiation.

Instead, use the fact that [itex]r=\sqrt{x^2+y^2+z^2}=\sqrt{x_jx_j}[/itex]. Then just apply the product rule.
 
  • #3
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I'm not sure I follow. Do you mean to do something like the following:

∇ . n = ∇ . x/r
∇ . n = (1/r) * dx/dx + x * d(1/r)/dx

I've been sloppy with the index notation (honestly because I don't know what to do in this situation), but it seems to me that we have already proven that the first term will be 3/r, and then we are adding more to this, so unless the second term is negative, we have a problem.
 
  • #4
gabbagabbahey
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I'm not sure I follow. Do you mean to do something like the following:

∇ . n = ∇ . x/r
∇ . n = (1/r) * dx/dx + x * d(1/r)/dx

I've been sloppy with the index notation (honestly because I don't know what to do in this situation), but it seems to me that we have already proven that the first term will be 3/r, and then we are adding more to this, so unless the second term is negative, we have a problem.
You need to be more careful with your notation.

Start by writing the divergence of an arbitrary vector [itex]\mathbf{v}=v_1\mathbf{i} + v_2\mathbf{j} + v_3\mathbf{k}[/itex] in index notation. What vector are you trying to take the divergence of in this case? What is [itex]v_i[/itex]?
 
  • #5
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The divergence of v is δivi.

In my situation, I'm trying to take the divergence of vector:
n = x/r = x/√(xjxj)

My "vi" in this case would be xi/√(xjxj)

So (∇ . n)i = δixi/√(xjxj)

Is this correct? What next? Thanks so much for your help, by the way.
 
  • #6
gabbagabbahey
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The divergence of v is δivi.
It is better to use ∂ or [itex]\partial[/itex] to represent a partial derivative, to avoid confusion with the kronecker delta [itex]\delta_{ij}[/itex].

In my situation, I'm trying to take the divergence of vector:
n = x/r = x/√(xjxj)

My "vi" in this case would be xi/√(xjxj)

So (∇ . n)i = δixi/√(xjxj)

Is this correct? What next? Thanks so much for your help, by the way.
This is correct all the way up until the last line. There is an implied summation over [itex]i[/itex] (and another over [itex]j[/itex]) in the term [itex]\frac{\partial_i x_i}{\sqrt{x_jx_j}}[/itex] (In the Einstein summation convention, any index which occurs exactly twice in a term is summed over), so it does not represent the [itex]i[/itex]th component of the divergence (divergence is a scalar, so it doesn't even have vector components!), but rather it represents the total divergence. You should also have brackets to make it clear that the derivative needs to be applied to the entire term ( [itex]v_i=\frac{x_i}{\sqrt{x_jx_j}}[/itex] ), not just to [itex]x_i[/itex]:

[tex]\nabla \cdot \mathbf{n} = \partial_i \left( \frac{x_i}{ \sqrt{x_j x_j} }\right)[/tex]

From here, use the product rule and chain rule to calculate the derivative.
 
  • #7
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I must really be missing something. From there, using the product rule to take the derivative, we would get:

∇.n = (1/√(xjxj))*∂ixi + xi*∂i*(1/√(xjxj))

Since we've already shown that ∂ixi = 3, the first term becomes simply:
3/√(xjxj)

Using the chain rule, the second term would become:
xi * (1/2) * (xjxj)-3/2

This doesn't seem right...
 
  • #8
gabbagabbahey
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Using the chain rule, the second term would become:
xi * (1/2) * (xjxj)-3/2
No, using the chain rule, the second term becomes [itex]-\frac{1}{2}x_i(x_jx_j)^{-3/2}\partial_i(x_kx_k)[/itex] (the reason for the appearance of the [itex]k[/itex] index is so that the repeated (dummy) index [itex]j[/itex] does not appear more than twice, to be consistent with the Einstein summation convention). Apply the product rule again to find [itex]\partial_i(x_kx_k)[/itex].
 
Last edited:
  • #9
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Okay I follow you there, and I think I see what is going to happen, but I'm not sure if my math is working correctly.

The product rule for ∂i(xkxk) = xk * 2∂i(xk)

This makes the second term become:
-xi(xjxj)-3/2xki(xk)

If we can somehow combine the xi and xk term, we should be able to multiply to reduce the exponent to (xjxj)-1/2, which would be great. The only problem is the ∂i(xk). If this is equal to 1, then we're all good, but it seems to me that it should be zero.
 
  • #10
gabbagabbahey
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The only problem is the ∂i(xk). If this is equal to 1, then we're all good, but it seems to me that it should be zero.
When in doubt, write out all the possible terms explicitly. Start with [itex]i=1[/itex] and look at when [itex]k=1[/itex] , [itex]k=2[/itex] and [itex]k=3[/itex] Are all the terms zero? Then look at [itex]i=2[/itex] for [itex]k=1[/itex] , [itex]k=2[/itex] and [itex]k=3[/itex] and so on.
 
  • #11
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Looks to me like we can just replace it with the kronecker delta δik, since it is 0 when the indices are different, and 1 when the indices are the same. This also takes care of combining the xi and xk terms like I wanted to.

Thanks so much for your help, this finally makes some sense. Index notation is completely new to me, and I haven't taken calculus in 3.5 years, so I'm a bit rusty with that as well.

I still need to do the third problem...but I don't really think I have much of a chance of getting it.
 
  • #12
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In case you're still around and feel like helping more (it'll be greatly appreciated), I can use the commutative property of the dot product to write the third one as the following:

(∂juj) xk/√(xixi)

How do I take the derivative of u in this case?
 
  • #13
gabbagabbahey
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In case you're still around and feel like helping more (it'll be greatly appreciated), I can use the commutative property of the dot product to write the third one as the following:

(∂juj) xk/√(xixi)

How do I take the derivative of u in this case?
The dot product may commute, but only with vectors (and vector fields). When one of the terms is a differential operator, there is no rule that says it commutes.

Start with the definition of [itex]\mathbf{u} \cdot \mathbf{\nabla}[/itex] in terms of Cartesian components...
 

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