Prove the following identity [Einstein notation]

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Homework Help Overview

The discussion revolves around proving an identity involving vector calculus and Einstein notation, specifically the expression for the gradient of the dot product of two vectors, \(\vec{A}\) and \(\vec{B}\). Participants are exploring the mathematical reasoning behind the identity and the implications of using Einstein notation in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts to prove the identity, including decomposing the right-hand side and expressing the left-hand side using partial derivatives. There is a focus on the potential efficiency of using Einstein notation to simplify the process.

Discussion Status

Some participants have provided insights into the challenges of using Einstein notation, noting that it may not necessarily aid in deriving the identity. There are ongoing questions about the correct application of indices and the interpretation of terms in the expression.

Contextual Notes

Concerns are raised about the use of the same index multiple times in expressions, which is noted as problematic. Additionally, there is a caution against misinterpreting the notation when functions of coordinates are involved.

Mulz
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Homework Statement


[/B]
Prove the following identity:

\vec{\nabla}(\vec{A} \cdot \vec{B}) = (\vec{A} \cdot \vec{\nabla})\vec{B} + (\vec{B} \cdot \vec{\nabla})\vec{A} + \vec{A} \times (\vec{\nabla} \times \vec{B}) + \vec{B} \times (\vec{\nabla} \times \vec{A})

Homework Equations



Kronecker's delta, levi-civita tensor

The Attempt at a Solution


[/B]
My solution consisted of simply solving the RHS by decomposition. I wrote \vec{A} = A_{1}A_{\hat{x}} + A_{2}A_{\hat{y}} + A_{3}A_{\hat{z}} likewise with B. I manually solved for each term in RHS and I did get the correct result which was LHS. My problem is that there is probably a more efficient way of doing this, perhaps by incorporating Einstein notation and solve LHS immediately without having to look at RHS.

How should I do this?

I tried \vec{\nabla}(\vec{A} \cdot \vec{B}) = \partial_{x_{i}}A_{i}B_{i} = \frac{\partial A_{i}}{\partial_{x_{i}}}B_{i} + A_{i} \frac{\partial B_{i}}{\partial_{x_{i}}} but then I got stuck. How to continiue? I'm not sure how all those \times appear using einstein notation.
 
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Mulz said:

Homework Statement


[/B]
Prove the following identity:

\vec{\nabla}(\vec{A} \cdot \vec{B}) = (\vec{A} \cdot \vec{\nabla})\vec{B} + (\vec{B} \cdot \vec{\nabla})\vec{A} + \vec{A} \times (\vec{\nabla} \times \vec{B}) + \vec{B} \times (\vec{\nabla} \times \vec{A})

Homework Equations



Kronecker's delta, levi-civita tensor

The Attempt at a Solution


[/B]
My solution consisted of simply solving the RHS by decomposition. I wrote \vec{A} = A_{1}A_{\hat{x}} + A_{2}A_{\hat{y}} + A_{3}A_{\hat{z}} likewise with B. I manually solved for each term in RHS and I did get the correct result which was LHS. My problem is that there is probably a more efficient way of doing this, perhaps by incorporating Einstein notation and solve LHS immediately without having to look at RHS.

How should I do this?

I tried \vec{\nabla}(\vec{A} \cdot \vec{B}) = \partial_{x_{i}}A_{i}B_{i} = \frac{\partial A_{i}}{\partial_{x_{i}}}B_{i} + A_{i} \frac{\partial B_{i}}{\partial_{x_{i}}} but then I got stuck. How to continiue? I'm not sure how all those \times appear using einstein notation.
You should not have the same index (here the "i") used three times in an expression, this is a big no-no. Also notice that your left side is a vector an your right side is a scalar, so that cannot be right. What you need is

$$ \hat{e}_j ~\partial_j (A_i B_i ) $$
where by ##\partial_j## I mean
$$ \partial_j \equiv \frac{\partial}{\partial x_j} $$.
 
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Einstein notation is just a shorthand -- doesn't help you derive something. Might even confuse some folks ... :rolleyes:

This any use ? (where you may need the liberty to read ##(a\cdot\nabla)## as ##(\nabla\cdot a)## by virtue of ##a\cdot b = b \cdot a ## ... )

Must admit I got stuck with factors of 2, though :wideeyed:
 
BvU said:
Einstein notation is just a shorthand -- doesn't help you derive something. Might even confuse some folks ... :rolleyes:

This any use ? (where you may need the liberty to read ##(a\cdot\nabla)## as ##(\nabla\cdot a)## by virtue of ##a\cdot b = b \cdot a ## ... )

Must admit I got stuck with factors of 2, though :wideeyed:
Just to not confuse the OP, we cannot write ##(a\cdot\nabla)## as ##(\nabla\cdot a)## when ##a## is a function of the coordinates.
 
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