I have a cumbersome problem with Vector calculus

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Homework Help Overview

The discussion revolves around a problem in vector calculus, specifically related to the application of the BAC-CAB rule and the use of the nabla operator in physics. Participants express varying levels of familiarity with the concepts and techniques involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the BAC-CAB rule but struggles with understanding the underlying concepts of vector calculus. Some participants suggest using index notation and the Levi-Civita symbol as alternatives. Others question the necessity of complex calculations and advocate for a more straightforward approach.

Discussion Status

Participants are exploring different methods to tackle the problem, with some suggesting that using index notation may simplify the calculations. There is a mix of opinions on the best approach, and while some guidance has been offered, no consensus has been reached on a single method.

Contextual Notes

There is a mention of the time constraints faced by physics majors in studying mathematics, which may influence their understanding of vector calculus. Additionally, the discussion includes references to specific mathematical symbols and conventions that may not be universally familiar to all participants.

fdbjruitoirew
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I am unfamiliar with Vector calculus, a tool for learning Physics
I select a homework I did not solve yet, then hope a help from you guys, in attachment pdf file

My attempt: I tried to use BAC-CAB rule, but the key hardness of mine is I still do not know the concepts clearly (as you know a physics-majored student could not have a lot of time to study Math)

Thank you in advance
 

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I always found the calculus using the nabla symbol a bit cumbersome and "unsafe". For such calculations I prefer the Ricci-index formalism. For the curl of a vector field you write in components
(\vec{\nabla} \times \vec{V})_j=\epsilon_{jkl} \partial_k V_l,
where \epsilon_{ijk} is the fully antisymmetric 3rd-rank tensor with \epsilon_{123}=1, also known as the Levi-Civita symbol.

In the index calculus the bac-cab rule is reflected in the following identity for the Levi-Civita symbol,
\epsilon_{ijk} \epsilon_{ilm}=\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl},
where we used the Einstein summation convention, i.e., one always has to sum over repeated indices. Further
\delta_{ij}=\begin{cases}<br /> 1 &amp; \text{if} \quad i=j,\\0&amp; \text{if} \quad i \neq j.<br /> \end{cases}<br />

I don't know, what's to "calculate" much with your first expression, but in index calculus it's simply
[(\vec{a} \cdot \vec{\nabla}) \vec{b}]_j=a_i \partial_i b_j.
Here, \partial_i=\frac{\partial}{\partial x_i}.

The second term on your problem list is
[(\vec{a} \times \vec{\nabla}) \times \vec{b}]_j=\epsilon_{klm} a_l \partial_m \epsilon_{jkn} b_n=-\epsilon_{klm} \epsilon_{kjn} a_l \partial_m b_n = -(\delta_{lj} \delta_{mn} - \delta_{ln} \delta_{mj}) a_l \partial_m b_n.
Now you only have to evaluate this a bit further and translate back into the nabla-operator notation.
 
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Thanks
I still prefer a direct method than using quite complicated calculation, then find out a meaning behind an expression. Maybe later I would do it smoothly but now I am just newbie with Vector calculus.
 
fdbjruitoirew said:
I am unfamiliar with Vector calculus, a tool for learning Physics
I select a homework I did not solve yet, then hope a help from you guys, in attachment pdf file

My attempt: I tried to use BAC-CAB rule, but the key hardness of mine is I still do not know the concepts clearly (as you know a physics-majored student could not have a lot of time to study Math)
What do you mean a physics major doesn't have a lot of time to study math? Mathematical techniques are a big part of what you're supposed to be learning!

vanhees71's suggestion is actually the most straightforward and least complicated way to do those calculations. It's definitely worth spending a little effort to learn how to use index notation and the Levi-Civita symbol.

Nevertheless, you can definitely do the problem by writing it all out, component by component. Show us what you've done. Don't just describe what you did in general terms. That's pretty useless. Show us your actual work so we can see where you're getting stuck.
 
vanhees71 said:
I always found the calculus using the nabla symbol a bit cumbersome and "unsafe". For such calculations I prefer the Ricci-index formalism. For the curl of a vector field you write in components
(\vec{\nabla} \times \vec{V})_j=\epsilon_{jkl} \partial_k V_l,
where \epsilon_{ijk} is the fully antisymmetric 3rd-rank tensor with \epsilon_{123}=1, also known as the Levi-Civita symbol.

In the index calculus the bac-cab rule is reflected in the following identity for the Levi-Civita symbol,
\epsilon_{ijk} \epsilon_{ilm}=\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl},
where we used the Einstein summation convention, i.e., one always has to sum over repeated indices. Further
\delta_{ij}=\begin{cases}<br /> 1 &amp; \text{if} \quad i=j,\\0&amp; \text{if} \quad i \neq j.<br /> \end{cases}<br />

I don't know, what's to "calculate" much with your first expression, but in index calculus it's simply
[(\vec{a} \cdot \vec{\nabla}) \vec{b}]_j=a_i \partial_i b_j.
Here, \partial_i=\frac{\partial}{\partial x_i}.

The second term on your problem list is
[(\vec{a} \times \vec{\nabla}) \times \vec{b}]_j=\epsilon_{klm} a_l \partial_m \epsilon_{jkn} b_n=-\epsilon_{klm} \epsilon_{kjn} a_l \partial_m b_n = -(\delta_{lj} \delta_{mn} - \delta_{ln} \delta_{mj}) a_l \partial_m b_n.
Now you only have to evaluate this a bit further and translate back into the nabla-operator notation.

Hi. I'll be curious. I've started a course on continuum mechanics, and we are using this notation for tensor calculus. Can you give a demonstration for these formulas you've posted? or tell me where to find'em?

Thanks.
 
The formula is pretty easy to justify. Take the first term in the sum, i.e.,
\epsilon_{1jk} \epsilon_{1lm}.
Obviously this can only be different from 0 if j,k \in \{2,3\} and at the same time l,m \in \{2,3 \}. Thus you either have
j=l \quad \text{and} \quad k=m
or
j=m \quad \text{and} \quad k=l.
In the first case the two epsilon symbols are both +1 or both -1, and their product thus always 1. This gives you
\delta_{jl} \delta_{km}
In the other case you get
-\delta_{jm} \delta_{kl}.
This same argument works of course for the other two values of the summation index i=2 and i=3, and this proves the formulat.
 
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