kronecker delta in stress tensor

kronecker delta in stress tensor

Postby fisher garry » Sat Dec 28, 2019 6:13 pm

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This is taken from a physics book but since it is mathematical I thought I could ask here. In the text they use kronecker delta notation. However I am not sure about the notation.In (8.18) what does the subscript j mean in [tex](\textbf{a}\cdot \overleftrightarrow{\textbf{T}})_j=\displaystyle\sum_{i=x,y,z}a_i T_{ij}[/tex]

Could someone perhaps write out that first equation in (8.18) as well?

And in

[tex](\nabla\cdot \overleftrightarrow{\textbf{T}})_j=\epsilon_0[(\nabla\cdot\textbf{E})E_j+(\textbf{E}\cdot\nabla)E_j -\frac{1}{2}\nabla _j E^2+\frac{1}{\mu_0}[(\nabla\cdot\textbf{B})B_j+(\textbf{B}\cdot\nabla)B_j -\frac{1}{2}\nabla _j B^2][/tex]

Can someone derive this by uing (8.18). I am a bit lost
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Re: kronecker delta in stress tensor

Postby HallsofIvy » Mon Dec 30, 2019 9:42 am

"T" (with the double ended arrow over it) is a tensor with both i and j components. "a" is a vector with i components. Taking the "dot" product of "a" with the i components of "T" (the sum is over "i") leaves a vector with j components. That is what the subscript j indicates.

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Re: kronecker delta in stress tensor

Postby Guest » Mon Dec 30, 2019 12:10 pm

HallsofIvy wrote:"T" (with the double ended arrow over it) is a tensor with both i and j components. "a" is a vector with i components. Taking the "dot" product of "a" with the i components of "T" (the sum is over "i") leaves a vector with j components. That is what the subscript j indicates.


Is this correct?

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Re: kronecker delta in stress tensor

Postby fisher garry » Mon Dec 30, 2019 2:12 pm

I did not know I could answer as guest so I could not update my previous answer as guest. Is this correct?

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Re: kronecker delta in stress tensor

Postby HallsofIvy » Tue Dec 31, 2019 11:21 pm

Yes, except that you are missing the "vector" notation. You should have either
[tex]\left[a_xT_{x1}+ a_yT_{y1}+ a_zT_{z1}, a_xT_{x2}+ a_yT_{y2}+ a_zT_{z2}, a_xT_{x3}+ a_yT_{y3}+ a_zT_{z3}\right][/tex]
or
[tex]\begin{bmatrix}a_xT_{x1}+ a_yT_{y1}+ a_zT_{z1} \\ a_xT_{x2}+ a_yT_{y2}+ a_zT_{z2} \\ a_xT_{x3}+ a_yT_{y3}+ a_zT_{z3}\end{bmatrix}[/tex]

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Re: kronecker delta in stress tensor

Postby fisher garry » Thu Jan 02, 2020 3:41 pm

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I the derivation they start with force per unit volume. After that they say that you kind find components of it by taking derivative of the stress tensor matrix. IT has units F/area which I could understand since taking its position derivative adds one length so you would unitvise go from F/area to F/Volume. But what I dont understand is the physical significance. If I write the derivative as a fraction:

[tex]\frac{F_2/A-F_1/A}{dx}=\frac{F_2-F_1}{dV}[/tex]

But what if
[tex]F_2=F_1[/tex]

then the fraction is 0 but at the same time there is a force in the volume? So the force density should not be 0?
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