Any notation for componentbycomponent vector multiplication?by Curl Tags: componentbycomponent, multiplication, notation, vector 

#1
Apr2311, 06:09 PM

P: 751

I have a scalar function and a vector function and I need to make a scalar function as so:
k=[kx ky kz] T=T(x,y,z) Function I want: div(k ? gradT) where "?" would be some operator that multiplies each component of k and gradT to make the vector [kx ∂T/dx , ky ∂T/∂y , kz ∂T/∂z] that way I can apply the divergence operator and get: ∂/∂x (kx ∂T/∂x) + ∂/∂y (ky ∂T/∂y) + ∂/∂z (kz ∂T/∂z) So is there some way to express this using elementary notation? 



#2
Apr2311, 10:09 PM

P: 3

Certeinly there is a notation of that operator, in general if you want to have a operator of that kind you have to put:
[tex]k\nabla\cdot(A)[/tex] Where A is the vector field that you want to operate. Greetings 



#4
Apr2311, 10:51 PM

P: 284

Any notation for componentbycomponent vector multiplication?
I believe it will be
[tex] \vec \nabla \cdot \left( ( \vec \nabla T)^T \cdot I_3 \vec k \right)^T [/tex] I think this works 



#5
Apr2311, 10:53 PM

P: 284

Oh yes... the superscript T means transpose. Treat grad T as a 3 X 1 matrix, and k also as a 3 X 1 matrix. I_{3} is just the identity matrix for R^{3}.




#6
Apr2311, 11:18 PM

P: 284

Actually...
[tex] \left( ( \mathbf{ \nabla } T)^T \mathbf{ I_3 } \textbf{ k } \right) \mathbf{ \nabla } [/tex] This may be a better way to put it. If we interpret nabla as a 3 X 1 matrix and k is also a 3 X 1 matrix 



#7
Apr2411, 12:34 AM

P: 751

I can't see what you are trying to do. The Identity matrix does nothing, and multiplying a 1x3 matrix by a 1x3 is not defined.




#8
Apr2411, 09:23 AM

P: 284





#9
Apr2411, 09:25 AM

P: 284





#10
Apr2411, 09:59 AM

P: 284

If you can find an operation that will transform k into a diagonal matrix, then the above function should work if you replace I3 k with that operation.




#11
Apr2411, 11:37 AM

P: 1,400

You could express it with the dyadic product as
[tex]\nabla \cdot \text{diag}(\textbf{k} \otimes \nabla T) = \nabla \cdot \text{diag}([k] [\nabla T]^T),[/tex] taking "diag" to mean "form a vector whose components are the diagonal entries of this matrix". Here [k] is a 3x1 matrix (a column vector), and the transpose of [del T] a 1x3 matrix (row vector), so that their product is a 3x3 matrix. Or simply use the summation sign: [tex]\sum_{i=1}^{n} \partial_i (k_i \partial_i T) = \sum_{i=1}^{n} \frac{\partial }{\partial x_i}\left ( k_i \frac{\partial T}{\partial x_i} \right ).[/tex] 



#12
Apr2411, 01:04 PM

P: 284





#13
Apr2511, 06:46 AM

P: 1,400

I don't know. I'm learning this stuff from books and the internet, and have never formally taken a course in linear algebra, so I don't know where and when such things would normally be taught. There's a section on dyadics in Snider & Davis: Vector Analysis, 6th edition. Gilbert Strang demonstrates the technique of making an n x n matrix out of vectors by multiplying a column vector by its transpose on the right in one of his MIT linear algebra lectures, which are online here
http://ocw.mit.edu/courses/mathemati...ideolectures/ Looking at the titles, it's probably Lecture 16: Projection matrices and least squares, or perhaps Lecture 15: Projections onto subspaces. Another place I've seen something similar is in presentations of the relativistic velocityaddition formula, and that did involve the identity matrix. I thought there was an example on the Wikipedia page: http://en.wikipedia.org/wiki/Velocity_addition But it looks like they've replaced it now; or maybe it was a different page where I saw it. I can post details if you're interested. The projection matrix idea works like this. Suppose we want to project a vector x onto a vectot a, then we can this as a matrix equation: [tex]\frac{a^Tx}{a^T a} \; a = \frac{aa^T}{a^T a} \; x = Px.[/tex] (If a is unit length, we don't need to worry about the denominator.) 


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