Hmm, let me focus on what you said:
Our surface integral I=\oint\vec{p} d\vec{S}, integrating over S, if \vec{p} equals the spherical unit vector and S is the unit sphere, then I=\oint\vec{p} \vec{n}{dS}, (since dS=\vec{n}dS, where \vec{n} is the unit vector normal to the small surface dS but...
Thanks for the fast reply. Ok, let me be more specific, let's say \vec{p} is the unit vector in spherical coordinates of a unit sphere S, so \vec{p}=(sin(phi)cos(theta),sin(theta)sin(phi),cos(phi)). Will the\oint \vec{p}dp, over p, be equal to the surface area of the sphere? Meaning its the same...
Hello,
We know that surface integrals come to the form of a surface integral of a scalar function over a surface and a vector field over a surface. First one is \oint F(x,y,z)d{S} and the second one is \oint \vec{F}(x,y,z)\cdot d\vec{S}=\oint \vec{F}(x,y,z)\cdot \vec{n}dS, where n is the unit...
Double dot is used extensively in continuum mechanics, even in 2014! For example, 4th order tensors represent orientation of rigid fibers in a 3D space, and 2nd order tensor is the velocity gradient of a flow field.
So should I move my question to the linear algebra section?
My basic question...
What exactly haven't you seen been used before? The double dot product of a tensor of n=4 with one of n=2? You mean you have only seen it being used for tensors of equal order?
The properties I am referring to, is actually expanding the double dot product to two single dot products!
Hello,
Been a long time lurker, but first time poster. I hope I can be very thorough and descriptive. So, I have been battling with a double inner product of a 2nd order tensor with a 4th order one. So my question is:
How do we expand (using tensor properties) a double dot product of the...