Suffix Notation Help: Nabla Vector Calculation

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Homework Help Overview

The discussion revolves around demonstrating a vector calculus identity involving the curl of a vector field, specifically in the context of suffix notation and tensor calculus. The vector field in question is defined as \(\vec{p} = \frac{\vec{B} \times \vec{r}}{r^3}\), where \(\vec{B}\) is a constant vector and \(\vec{r}\) represents a position vector in three-dimensional space.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the curl operator and the use of identities involving the Levi-Civita symbol. There are attempts to expand the expression for the curl using various mathematical identities, including the product rule and chain rule. Questions arise regarding the correct application of derivatives and the handling of suffix notation.

Discussion Status

Several participants have provided guidance on the mathematical manipulations required, including the need for proper notation and the application of the chain rule. There is an ongoing exploration of the terms generated during the calculations, with some participants expressing uncertainty about the cancellation of terms and the implications of specific identities.

Contextual Notes

Participants note the complexity of suffix notation and its implications for the calculations, indicating that this is a new area of study for some. There are references to missing information and potential misunderstandings regarding the mathematical identities being used.

Matt atkinson
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Homework Statement


Show that the equation \nabla \times \vec{p} = -\frac{\vec{B}}{r^3} + 3\frac{\vec{B} \bullet \vec{r}}{r^5}\vec{r}
Where ;
\vec{p} = \frac{\vec{B} \times \vec{r}}{r^3}
\vec{r}=(x_1 ,x_2 ,x_3)
and \vec{B} is a constant vector.
and r is the magnitude of \vec{r}

Homework Equations


above

The Attempt at a Solution


\nabla \times \vec{p} = \epsilon_{ijk} \epsilon_{klm} \frac{d}{dx_j} B x_m |r|^{-3}
= \epsilon_{ijk} \epsilon_{klj} B |r|^{-3} - 3 \epsilon_{ijk} \epsilon_{klm} B x_m x_j |r|^{-5}
I've tried expanding and using various identities such as;
\epsilon_{ijk} \epsilon_{klm} = \delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}
If someone could give me a push in the right direction or let me know if i went wrong somewhere (i know i skipped a few steps but it took me 20 mins to right out the latex code for this adk if you don't see what I did).
 
Last edited by a moderator:
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Matt atkinson said:

Homework Statement


Show that the equation \nabla \times \vec{p} = -\frac{\vec{B}}{r^3} + 3\frac{\vec{B} \bullet \vec{r}}{r^5}\vec{r}
Where ;
\vec{p} = \frac{\vec{B} \times \vec{r}}{r^3}
\vec{r}=(x_1 ,x_2 ,x_3)
and \vec{B} is a constant vector.
and r is the magnitude of \vec{r}

Homework Equations


above

The Attempt at a Solution


\nabla \times \vec{p} = \epsilon_{ijk} \epsilon_{klm} \frac{d}{dx_j} B x_m |r|^{-3}

You're missing an l suffix on the B and the derivative is partial:
<br /> (\nabla \times \vec p)_i = \epsilon_{ijk} \epsilon_{klm} B_l \frac{\partial}{\partial x_j} \left(\frac{x_m}{r^3}\right)

You will also need
<br /> \frac{\partial r}{\partial x_j} = \frac{x_j}{r} and <br /> \frac{\partial x_m}{\partial x_j} = \delta_{jm}<br />
 
ah okay, hmm.
so I should use the product rule and get
\epsilon_{ijk} \epsilon_{klm} B_l x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}
I just don't see how to use \frac{\partial r}{\partial x_j} maybe I'm being a little stupid.
 
Last edited by a moderator:
Matt atkinson said:
ah okay, hmm.
so I should use the product rule and get
\epsilon_{ijk} \epsilon_{klm} B_l x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}
You are missing some brackets: you should have
<br /> \epsilon_{ijk} \epsilon_{klm} B_l \left(x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}\right)<br />


I just don't see how to use \frac{\partial r}{\partial x_j} maybe I'm being a little stupid.

<br /> \frac{\partial}{\partial x_j} \left(\frac{1}{r^3}\right) = \frac{d (r^{-3})}{dr} \frac{\partial r}{\partial x_j}<br />
 
Thankyou so much, I've been trying to do this for a while now, can't believe it was chain rule i forgot
 
pasmith said:
You are missing some brackets: you should have
<br /> \epsilon_{ijk} \epsilon_{klm} B_l \left(x_m \frac{\partial r^{-3}}{\partial x_j} + r^{-3} \delta_{jm}\right)<br />

<br /> \frac{\partial}{\partial x_j} \left(\frac{1}{r^3}\right) = \frac{d (r^{-3})}{dr} \frac{\partial r}{\partial x_j}<br />

Okay, I am sorry i got stuck again, so i did the math and substitute the two epsilons for deltas uisng the identities i gave before, but i can't seem to cancel;
\delta_{il}\delta_{jj} B_jr^{-3} -\delta_{ij}\delta_{jl} B_jr^{-3} -3\delta_{il}\delta_{jm}B_j x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_j x_m x_j r^{-5}
to the show that answer I am assuming the first and third term are zero? but I am not sure why, sorry about all the questions this suffix notation is very new to me.
 
Last edited:
Matt atkinson said:
Okay, I am sorry i got stuck again, so i did the math and substituent the two epsilons for deltas uisng the identities i gave before, but i can't seem to cancel;
\delta_{il}\delta_{jj} B_jr^{-3} -\delta_{ij}\delta_{jl} B_jr^{-3} -3\delta_{il}\delta_{jm}B_j x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_j x_m x_j r^{-5}
The suffix on B should be an l; try again with
<br /> \delta_{il}\delta_{jj} B_l r^{-3} -\delta_{ij}\delta_{jl} B_l r^{-3} -3\delta_{il}\delta_{jm}B_l x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_l x_m x_j r^{-5}
 
pasmith said:
The suffix on B should be an l; try again with
<br /> \delta_{il}\delta_{jj} B_l r^{-3} -\delta_{ij}\delta_{jl} B_l r^{-3} -3\delta_{il}\delta_{jm}B_l x_m x_j r^{-5}+3\delta_{im}\delta_{jl}B_l x_m x_j r^{-5}

okay, I did that and got;
3B_i r^{-3} - B_i r^{-3} -3B_i x_j x_j r^{-5}+3B_j x_i x_j r^{-5}
I just don't see how two of the terms cancel, not sure what I am missing
 
Last edited:
Matt atkinson said:
okay, I did that and got;
3B_i r^{-3} - B_i r^{-3} -3B_i x_j x_j r^{-5}+3B_j x_i x_j r^{-5}
I just don't see how two of the terms cancel, not sure what I am missing

x_jx_j = r^2.
 
  • #10
pasmith said:
x_jx_j = r^2.

Wow thanks I feel kinda silly now.
Thanks so much for your help.
 

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