Vector calculus identity using index and comma notation

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The discussion focuses on demonstrating the identity div(curl(v)) = 0 using index and comma notation. Participants explore the substitution of the curl expression into the divergence formula, leading to the expression involving the Levi-Civita tensor. The challenge lies in interpreting this complex expression without expanding it into individual terms, with emphasis on the anti-symmetric properties of the tensor and the symmetric nature of mixed partial derivatives. There is a consensus that mastering index-comma notation requires practice and familiarity with tensor analysis, and participants express a desire for more tutorials and examples to aid understanding. The conversation highlights the confusion surrounding terminology and notation in the field, particularly regarding the tensor product.
hotvette
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Homework Statement


Use index and comma notation to show:
\begin{equation*}\text{div }(\text{curl } \underline{\bf{v}}) = 0\end{equation*}

Homework Equations


\begin{align*}
& \text{(1) div } \underline{\bf{v}} = v_{i,i} \\
& \text{(2) curl } \underline{\bf{v}} = \epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k
\end{align*}

The Attempt at a Solution


Substituting (2) into (1) I get:
\begin{equation*}
\text{(3) div }(\text{curl } \underline{\bf{v}}) = (\epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k)_{p,p}
\end{equation*}
Don't know what to do next. I know the right hand side of (3) is correct because if I write out the individual terms the result is zero. But, we're not suppose to write out the terms. So, how does one tell that the right hand side of (3) is zero? Is there some special trick to interpret the expression?

P.S. I also have to show that \text{curl(grad }\phi)=\underline{\bf{0}} but the question is really the same. How to interpret a (complex) index-comma expression without writing out the terms?
 
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I think what's bogging you down is awful notation - I don't know why people use the comma notation but I find it much harder to spot and do index gymnastics with it as compared to more clear-cut differential operators.

hotvette said:
Substituting (2) into (1) I get:
\begin{equation*}
\text{(3) div }(\text{curl } \underline{\bf{v}}) = (\epsilon_{ijk} v_{j,i} \underline{\bf{e}}_k)_{p,p}
\end{equation*}
This can be simplified to ##(\epsilon_{ijp} v_{j,i} )_{,p}##, because the p-th component of the the curl is ##(\epsilon_{ijp} v_{j,i} )##. The next step then is to use the anti-symmetric property of the Levi-Civita tensor together with the symmetric nature of mixed partial derivatives to do some index relabelling to show that the term must be zero.
 
Thanks, I can sort of see it. I guess it takes a lot of practice to begin to see how these things work! Sure wish there were some tutorials in internet land illustrating how to simplify index-comma expressions. Seems to me we need to see lots of examples in order to get the hang of things.
 
Yes, the notation can be very powerful - but it takes time to figure out the tricks involved in performing the index gymnastics. Perhaps you could try picking up a book on tensor analysis (preferably geared towards physicists) or GR? Usually the first chapter or two will have plenty of examples and exercises on these.
 
This subject is extremely confusing, and the terminology doesn't help. I think whoever coined \underline{\bf{e}}_i \otimes \underline{\bf{e}}_j as "tensor product" purposely wanted to confuse beginners of the subject. For days I thought "tensor product" was some mysterious operation on two vectors but could not find a definition of what the operation does. Even my professor couldn't easily explain what it meant (to someone unfamiliar with the subject). I slowly came to realize (still not sure) that it is simply a way to represent a collection of basis vectors, thus \underline{\bf{e}}_i \otimes \underline{\bf{e}}_j is a thing rather than an operation on two vectors.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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