How Do You Convert Complex Tensor Notation to Vector Notation?

lostidentity
Messages
16
Reaction score
0
Hi,

I have the following term in tensor notation

\frac{\partial{c}}{\partial{x_i}}\frac{\partial{u_i}}{\partial{x_j}}\frac{\partial{c}}{\partial{x_j}}

I'm not sure how to write this in vector notation.

Would it be?

\nabla{c}\cdot\nabla\boldsymbol{u}\cdot{c}

The problem I have is \nabla\boldsymbol{u} is a tensor, whereas \nabla{c} is a vector. Not sure what type of multiplication it would be between a vector and a tensor. Surely not a simple dot product?

Thanks.
 
Mathematics news on Phys.org
This doesn't look like a proper tensor expression to me. If you have a repeated index, it should be repeated on "top" and on "bottom", whereas you have it repeated on the bottom both times here.
 
Hi,

Sorry I don't think I defined the problem correctly. c is a scalar field and \vec{u} is a vector field.

I checked with a paper and it seems that what I've got for the vector notation is correct. However, I'm having difficulty with the following term

\frac{\partial^2{c}}{\partial{x_k}\partial{x_i}}\frac{\partial^2{c}}{\partial{x_k}\partial{x_i}} -----(2)

At first site I thought this could be the product of two Laplacian of a scalar field, however then I found that the correct form for the Laplacian in index notation is

\frac{\partial^2{c}}{\partial{x_i}\partial{x_i}}

So how would I write the above term (2) in vector notation?

Thanks.
 
Could \frac{\partial^2{c}}{\partial{x_k}\partial{x_i}} be a second order tensor?

Since \frac{\partial{c}}{\partial{x_i}} is the gradient of c (i.e. a vector), therefore \frac{\partial}{\partial{x_k}}\left(\frac{\partial{c}}{\partial{x_i}}\right) would be the gradient of a vector field, i.e. a second order tensor?

If I were to write this is in tensor notation would it be

\nabla(\nabla{c})?

Thanks.
 
lostidentity said:
\frac{\partial{c}}{\partial{x_i}}\frac{\partial{u_i}}{\partial{x_j}}\frac{\partial{c}}{\partial{x_j}}

I think nicksauce is right - I don't know how to read this to make it come out as a tensor equation. If we're summing over repeated indices, then some of the indices are in the wrong position. If there's no summation, but it's meant for a particular i j, it's not a tensor equation. Perhaps you could give us the whole problem or tell us the summation convention you're using.

In rectangular components, \nabla is \partial /\partial x_1 + \partial /\partial x_2 + \partial /\partial x_3

but the equations you're writing it looks like you're focussing on one component only - unless there's some kind of implicit summation over indices you've got in mind. Perhaps for now write out the summation explicitly so we know what you've got in mind.
 
Thanks for the reply.

Actually I'm working with a transport equation for a scalar variable N that has the following form (I've ignored a number of terms and constant coefficients as I don't think they are relevant).

{u_j}\frac{\partial{N}}{\partial{x_j}} = \frac{\partial^2{c}}{\partial{x_k}\partial{x_i}}\frac{\partial^2{c}}{\partial{x_k}\partial{x_i}} - \frac{\partial{c}}{\partial{x_i}}\frac{\partial{u_ i}}{\partial{x_j}}\frac{\partial{c}}{\partial{x_j} }

So perhaps the summation is done for the j-th component since that's what's on the LHS?

I was wondering if I could write this whole equation in tensor notation rather than in the index notation as above. So far what I've got is something like\vec{u}\cdot\nabla{N} =[ \nabla(\nabla{c})]^2 - \nabla{c}\cdot\nabla\vec{u}\cdot\nabla{c}
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top