Prove the following identity [Einstein notation]

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The discussion focuses on proving the identity involving the gradient of the dot product of two vector fields, \(\vec{A}\) and \(\vec{B}\). Participants explore using Einstein notation for a more efficient solution but encounter difficulties, particularly with index notation and vector versus scalar discrepancies. One contributor emphasizes the importance of avoiding repeated indices in expressions and clarifies that the left side represents a vector while the right side is a scalar. The conversation highlights the potential confusion that can arise from using shorthand notation in tensor calculus. Ultimately, the discussion underscores the need for careful application of notation to avoid errors in derivation.
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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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