Using the Divergence Theorem to Solve Vector Calculus Problems

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The discussion revolves around applying the Divergence Theorem to tensor calculus, specifically the expression v ⊗ n. The original poster struggles to derive useful results from their application of the theorem, noting that their calculations do not lead to the expected gradient of v. A participant clarifies that the divergence of the tensor can be expressed as n · (∇v) + n (∇·v), emphasizing that it results in a vector, not a scalar. The conversation highlights the importance of correctly handling terms in tensor calculus to achieve valid conclusions. Ultimately, the thread underscores the complexities involved in applying the Divergence Theorem to tensor fields.
SP90
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Homework Statement



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Homework Equations



So I have that v \otimes n = \left( \begin{array}{ccc}<br /> v_{1}n_{1} &amp; v_{1}n_{2} &amp; v_{1}n_{3} \\<br /> v_{2}n_{1} &amp; v_{2}n_{2} &amp; v_{2}n_{3} \\<br /> v_{3}n_{1} &amp; v_{3}n_{2} &amp; v_{3}n_{3} \end{array} \right) <br />

The Attempt at a Solution



I've tried applying the Divergence theorem for Tensors:
<br /> \int_{\partial B} ( v \otimes n )n dA = \int_{B} \nabla \cdot ( v \otimes n ) dV<br />

But that doesn't lead anywhere particularly useful. I thought it might be worth noting that \nabla \cdot ( v \otimes n ) = \frac{dv_{1}}{dx_{1}}n_{1}+\frac{dv_{2}}{dx_{2}}n_{2} + \frac{dv_{3}}{dx_{3}}n_{3} but I can't seem to get anywhere near \nabla v

And this problem isn't homework, it's just an optional exercise, but it's frustrated me for a while and I figured I should get some pointers.
 

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SP90 said:
I thought it might be worth noting that \nabla \cdot ( v \otimes n ) = \frac{dv_{1}}{dx_{1}}n_{1}+\frac{dv_{2}}{dx_{2}}n_{2} + \frac{dv_{3}}{dx_{3}}n_{3} but I can't seem to get anywhere near \nabla v

No it's not! I think you will find (assuming n is constant) that \nabla \cdot ( v \otimes n ) = n \cdot (\nabla v) + n (\nabla \cdot v)

Especially note that it is a vector and not a scalar
 
Isn't \nabla \cdot v \otimes n = (n_{1}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{2}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{3}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}))

Which is n \cdot \nabla v?

This would make sense since it gives that

\int_{\partial B} ( v \otimes n )n dA = \int_{B} (\nabla v) ndV

And then since n is just so constant vector, the result follows.

Is that right? Or am I missing something?
 
SP90 said:
Isn't \nabla \cdot v \otimes n = (n_{1}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{2}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{3}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}))

Which is n \cdot \nabla v?

This would make sense since it gives that

\int_{\partial B} ( v \otimes n )n dA = \int_{B} (\nabla v) ndV

And then since n is just so constant vector, the result follows.

Is that right? Or am I missing something?

Of course not; you can't just get rid of the terms that give you trouble... This is what you need to show, but the formula I gave above is still correct.
 
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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