A few papers from momentum_waves, diaw, desA
 

Hi everyone,

As promised quite some time ago, I've now published a few of my research papers & idea flow, available at the following link:

<http://adthermtech.com/wordpress3/?page_id=5>

Please consider the documents on 'momentum waves' to be works in progress. Of course, there will always be arguments for & against any concept, but this the the joy of research & debate. I trust that the documents will continue to fuel debate. Please feel free to leave your comments on the site.

I would like to thank the many participants who have engaged in open debate over the past few years. Tom, you have been a wonderful sounding board.

I plan to publish a number of papers in a number of more meaningful Journals & probably ariv.org. I've got mountains of numeric studies I need to get into print. (If anyone knows any decent promoters on the arxiv forum, I'd be grateful).

Most of all - have fun... :)

mw...

Re: A few papers from momentum_waves, diaw, desA
 

I haven't read the papers but I had a quick glance (I haven't much time due to work at the moment).

But one thing that jumped out at me was the equation

grad.(grad u) = grad^2 u

which is wrong! It should be grad.( (grad u)^T ) = grad^2 u. For an incompressible fluid this is important because grad.(grad u) = 0.

This is easy to verify since

grad.(grad u)_j = (u_i,j),i = (u_i,i),j = 0

but

grad.( (grad u)^T )_j = (u_j,i),i = (grad^2 u_j).

The other point is that the "first order in time second order in space" equation is not a wave equation it's a reaction diffusion equation.

Re: A few papers from momentum_waves, diaw, desA
 

Hi Tom,

I'm out of town at the moment & will comment upon my return.

It's not grad of grad(), but grad dot (nabla()). These are different. Think of it this way, as first developing the nabla v tensor then pre-dotting it with nabla, instead of first developing nabla dot nabla operator, then operating on an article.

In tensor form, the argument holds.

I'll comment in more detail later.

mw...

Re: A few papers from momentum_waves, diaw, desA
 

Tom wrote:

The other point is that the "first order in time second order in space" equation is not a wave equation it's a reaction diffusion equation.

momentum_waves replies:

I'll simply refer here to "Traveling Wave Solutions of Parabolic Systems", Volpert A.I., Volpert V.A, Volpert A.V., Vol 140, AMS, 1994.

They review wave solutions of parabolic systems of equations, of the form:

du/dt = A.d2u/dx2 + F(u) (all partials)

This reduces to the form I mentioned in the paper. Volpert terms this a stationary, planar wave - 1-dimension in space. This equation has as its mechanism, in many cases, a diffusion-type process - but, it essentially describes the motion of u in a wave motion.

Actually, this can be seen in simple convection-diffusion processes eg. temperature form of energy equation, if the convection velocity is small & the thermal diffusivity is large. The conveyance of T, via diffusion, is an obvious one. Conveyance of u (if velocity), via diffusion, takes some mind-bending.

Basically, it depends on one's frame of reference. The form in the paper was specifically selected to progressively build up a notion of a larger wave-nature of the momentum equations. Any equation with a d()/dt component will transport in some wavelike manner. I terms this wave-type as a viscous wave (oozy-type flow pattern).

I'll add in a note to the paper, to include the 'reaction diffusion equation' reference.

mw...

Update 3 : A few papers from momentum_waves
 

Tom wrote:

But one thing that jumped out at me was the equation

grad.(grad u) = grad^2 u

which is wrong! It should be grad.( (grad u)^T ) = grad^2 u. For an incompressible fluid this is important because grad.(grad u) = 0.

This is easy to verify since

grad.(grad u)_j = (u_i,j),i = (u_i,i),j = 0

but

grad.( (grad u)^T )_j = (u_j,i),i = (grad^2 u_j).

mw replies:

Tom, I'm trying to understand the terminology you've used. I'm wondering if the tensor notation I've used, has perhaps clouded things somewhat.

In the paper, is written, by convention:

nabla dot (nabla_v) !=! nabla dot (nabla nonion product v) = nabla2_v

Nonion product !=! tensor outer product (refer p10 'Continuum Mechanics for Engineers', Mase & Mase, 1999) - written as nabla_v by convention.

Is this what you are referring to as:

grad.( (grad u)^T ) = grad^2 u

If so, then we are notations apart. If not, then I'll continue writing up the proofs of my statement. These have been developed from known relationships, as proof - & are performed at least 3 different ways.

mw...

Re: Update 3 : A few papers from momentum_waves
 

Simply put grad(u) is the dyadic product of the two vectors grad and u. If you now take the divergence of this you contract along the first index (and hence obtain 0 by continuity as in case 1). If you transpose the dyad before applying the div operator you obtain the Laplacian as in my second case! You should always remember that the viscous part of the stress tensor is not grad(u) but grad(u) + grad(u)^T; i.e. it's symmetric.

Re: Update 3 : A few papers from momentum_waves
 

Tom wrote:

Simply put grad(u) is the dyadic product of the two vectors grad and u. If you now take the divergence of this you contract along the first index (and hence obtain 0 by continuity as in case 1). If you transpose the dyad before applying the div operator you obtain the Laplacian as in my second case! You should always remember that the viscous part of the stress tensor is not grad(u) but grad(u) + grad(u)^T; i.e. it's symmetric.

mw replies:

Lets step back a little & take the matrix representation of the tensor out of the picture - for a moment.

Expanding the expression I have written provides the same final result in tensor form - which is, after all, a string of terms, which we later elect to collect into vector format, or split along component lines as desired. I have concerns that the matrix interpretation may be introducing issues here, since along the way, the unit vectors & groups of unit vectors are discarded using this procedure.

I repeat that, in the pure tensor form, expansions of:

nabla dot (nabla nonion-product v) = nabla2_v

give the same end result, after expansion.

nabla2_v = v_i,jj*e'i ; i,j=1,2,3

Think of the relation for vector-dyad products:

u.vw = (u.v)w = u.(vw) = u_j*v_j*w_i*e'i [1]

then substitute:

u => nabla = d,i*e'i

v => nabla = d,j*e'j

The dot of e'i*e'j = kronecker delta, converts v,j -> v,i

This is a bit difficult to show up in this text medium, but I'll try:

Direct substitution in [1]: [b]u.vw = u_j*v_j*w_i*e'i = (d,j)*(d,j)*(w_i)*e'i = w_i,jj*e'i

Direct development: [b] N.Nw = (d,i*e'i)*(d,j*e'j)*(w_k*e'k) = (d,i*e'i*e'j*d,j)*(w_k*e'k) = (d,i*d,i)*(w_k*e'k) = w_k,ii*e'k = w_k,jj*e'k = w_i,jj*e'i

(Whew!!!)

You can expand the other parts of the expression seperately & check the final results - in tensor form. After this is established, we can debate the matrix/vector representation of the tensor expression as a separate matter. It is vital to retain all the e'i terms & groups to make proper sense of tensor expansions - shortcuts, matrix rules & so forth can cloud the issue a lot.

If there should be a difference, it should stem from expression [1].

mw...

Re: Update 3 : Errata + additional massaging
 

Knew I'd miss a dot somewhere:

Direct development: N.Nw = (d,i*e'i)o(d,j*e'j)*(w_k*e'k) = (d,i* e'i o e'j *d,j)*(w_k*e'k) = (d,i*d,i)*(w_k*e'k) = w_k,ii*e'k = w_k,jj*e'k = w_i,jj*e'i

It is obvious that, in this manner, the following expansions can also be performed:

(N.N)w

N.(Nw)

Expansion 1 leads to the direct form of the Laplacian, the 2nd form leads to 'div (nabla_w)' in old parliance, or 'nabla dot (Nabla_w). Here Nabla_w is a dyadic with all the useful properties of a tensor.

Two ways to skin the same cat, but each offering different insights into the terms involved.

--------------------

<http://en.wikipedia.org/wiki/Laplace_operator> is a useful link also referring to:

The Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence of the gradient:

\Delta = \nabla^2 = \nabla \cdot \nabla.

Equivalently, the Laplacian is the sum of all the unmixed second partial derivatives:

\Delta = \sum_{i=1}^n \frac {\partial^2}{\partial x^2_i}.

Here, it is understood that the xi are Cartesian coordinates on the space; the equation takes a different form in spherical coordinates and cylindrical coordinates, as shown below.

In the three-dimensional space the Laplacian is commonly written as

\Delta = \frac{\partial^2} {\partial x^2} + \frac{\partial^2} {\partial y^2} + \frac{\partial^2} {\partial z^2}.

-------------

Re: Update 3 : A few papers from momentum_waves
 

"After this is established, we can debate the matrix/vector representation of the tensor expression as a separate matter."

The problem is that it is not a separate matter (a vector is a tensor of rank 1).

(u.v)w /= u.(v\ocross w) (\ocross is the dyadic product symbol and /= stands for not equal to). The left-hand-side is a column vector while the right-hand-side is a row vector. The components may be equal but that is not equivalent to the "entities" being equal. This is crucial when it comes to manipulating the equations in vector form (and hence the need for a transpose).

What you need to do is reconcile my two (correct) formulae with what you're trying to do. You also need to be more careful about your brackets in the above so that terms are grouped correctly and there is no potential for misuse of operations. As an example if you had the triple vector product a.bxc you should write a.(bxc) since (a.b)xc does not make sense.

Re: Update 3 : A few papers from momentum_waves
 

Thanks Tom, for your input. I'll ask a few questions here:

Tom wrote:

(u.v)w /= u.(v\ocross w) (\ocross is the dyadic product symbol and /= stands for not equal to). The left-hand-side is a column vector while the right-hand-side is a row vector. The components may be equal but that is not equivalent to the "entities" being equal. This is crucial when it comes to manipulating the equations in vector form (and hence the need for a transpose).

mw writes:

A point could be to understand where u,v,w are declared as column vectors. What if, as in many texts on dynamics, they are taken as row vectors? Per definition, Nabla is a row vector (or no?).

Under this interpretation, (v\ocross w) would still be defined, as would u.(v\ocross w) as a row vector (or no?).

Why I'm asking this is because, this seems to be a rather loose area in vector-tensor work. How would one then write nabla2_v in a pure vector-tensor form <u>without the use of transposes (matrix cludges)</u> so that the final result reflects the elegance of tensor theory - only?

In the expansions I've shown, no assumption has been made of the lay of the vectors - as I understand.

I would like to have an elegant interpretation which does not rely on contra-vectors / co-vectors & all that, as I am led to understand that for cartesian coordinates, both are equivalent. When we work on numerical solutions, we twist these things any which way to provide the known equation lay-up, but tensor expressions should be able, in cartesian coords at least, to reflect a consistent vector definition (no matrices, or transposes).

------------------

What you need to do is reconcile my two (correct) formulae with what you're trying to do. You also need to be more careful about your brackets in the above so that terms are grouped correctly and there is no potential for misuse of operations. As an example if you had the triple vector product a.bxc you should write a.(bxc) since (a.b)xc does not make sense.

Thanks tom, I'll continue on the road. :)

mw...

Re: Update 3 : A few papers from momentum_waves
 

"A point could be to understand where u,v,w are declared as column vectors. What if, as in many texts on dynamics, they are taken as row vectors? Per definition, Nabla is a row vector (or no?)."

The transpose has to be there; i.e. consider the dot product a.b. In column form this is (a^T)b while in row form it is a(b^T). An important aspect of tensors is the requirement that we can apply the integral theorems to them (in particular Guass' divergence theorem). If we denote the stress tensor by S then the surface integral of this contribution in the clasical derivation of the continuum equations is

surface_integral (S^T)n ds = volume_integral div(S) dv (*)

where n is the unit outward normal to the surface. For a symmetric tensor S^T = S and so there is no need for the transpose (for a fluid S = -pI + nu.D where D is the symmetric part of the velocity gradient - the anti-symmetric part is the spin tensor is not invariant under reflections and so should not contribute to the stress).

For nonsymmetric tensors the transpose is crucial in (*)!

Have you looked at Chadwick's book on continuum mechanics? It covers most of what you need to know about tensors.

"tensor expressions should be able, in cartesian coords at least, to reflect a consistent vector definition (no matrices, or transposes)."

But a Tensor is just a special form of a matrix!

Re: Update 3 : A few papers from momentum_waves
 

Hi Tom,

Many thanks for your pointers.

I'm pretty comfortable with the reasoning behind the symmetry of the stress tensor, & lack of symmetry of the nabla_v & velocity gradient tensors.

Oddly-enough Chadwick's book is in the air as I write this, as is a legitimate copy of Aris's work - I'd been searching through the Dover series for useful references - this is a goldmine series. If you have any other useful references around Tensor theory & practical expansions, & so forth, I'd be most grateful.

I find that many of the continuum mechanics books are too light in this area & tend to jump into the specifics of continuum mechanics a little too soon.

I have in my mind's eye a solid link between bifurcation theory & tensor structures - specifically in terms of the eigenvalues. For the Euler eqns this seems reasonably straightforward, but for the N-S, I obviously need to clean up a little further. I'll make the amendments once my head is a little more clear on the exact lay-down.

Thanks again for your input. In the meantime though, don't let the nabla2_v debate distract from the rest of the discussion in the papers I made available - it was a tiny, tiny point in the overall thought process. I'd love to hear your views on the rest. Actually, practically, this only represents a small portion of my research - a lot more useful stuff is to follow eg. Cox, Hui & so forth.

mw...

<http://adthermtech.com/wordpress3/?page_id=5>

Re: Update 3 : A few papers from momentum_waves
 

"If you have any other useful references around Tensor theory & practical expansions, & so forth, I'd be most grateful."

The only book I ever really use for vectors & tensors is, apart from Chadwick,

"vector analysis" by M.R. Speigel (Schaum's series).

Apart from Aris the only other books I have on this subject you'd probably find too difficult (i.e. tensors on manifolds, various works on differential geometry, and the absolute differential calculus).

I'm not sure what you mean by "practical expansions, and so forth"?

Re: Update 3 : A few papers from momentum_waves
 

Tom wrote:

I'm not sure what you mean by "practical expansions, and so forth"?

mw replies:

Here, I was thinking of a text which was loaded heavily towards development of various tensor expressions - from fundamental principles. Much like say Mase's book, but with many more examples.

Thanks so much for your input.

mw...

Update 4 : Settled...
 

Hi Tom,

Ok. I worked through your logic & settled the matter in my mind. I see where I erred.

So to confirm, basically, in my paper, I need to revise the nabla2_v term to read:

grad.( (grad u)^T ) = grad^2 u.

nabla2_v = nabla dot ((nabla_v)^T)

I concur with your development & implication of the v_i,i=0 (continuity, isochoric flow)

Basically, I had not checked thoroughly directly at v.T & v.(T)^T level & had missed the transpose. In Mase, in that section for N-S, they work only in indicial notation & I had no reference back to symbolic notation, for a check. The folks at home base also missed that one despite numerous requests to cross-check my work.

--------------

If this is settled, then I'm still on track in my end-goal, since I work with the principle value of the tensor & as such (nabla_v)* = ((nabla_v)^T)*

Hopefully I did not miss a trick on that one. :)

Thanks for your patience.

mw...

Re: An old question re-visited...
 

mw wrote in another thread - hitherto still unanswered:

Jump conditions - incompressible fluid flow Posted By: momentum_waves <Send Email> Date: Tue, 14 Aug 2007, 9:31 p.m.

Many thanks to Tom for the reference to "High Speed Flow", Chapman C.J. This excellent text covers the mass conservation, momentum conservation & energy conservation equations in depth - together with the relevant jump conditions. His review of characteristic surfaces is also excellent.

I have been working through Chapman's work & have been converting the results of the jump conditions for low-speed, incompressible flows. This has been rather interesting & has begun to present some solid ground for the wave phenomena (periodic solutions) I have been researching.

The question I have at the moment regards references & papers that apply the jump conditions & characteristic surfaces, for incompressible flows within confined domains eg. ducts, pipes etc. If anyone can point me towards additional reference information, I would be extremely grateful.

At this point, the information trail seems to be becoming increasingly rarefied. Many thanks.

mw...

--------------

Chadwick also refers to jump conditions & the options of shock jumps, or vortex sheet jumps. Given this information, I have begun to understand the observed 'critical lines' in some of the numeric solutions presented in the papers listed, as a type of vortex jump, or even a 'reflective jump' where the jump occurs in only one direction, with the other unaffected.

Does anyone have additional references to similar concepts in the literature?

Perhaps it could be a good time to rephrase an ancient question I brought up a few years ago - with slight re-wording:

Can incompressible fluid flow within confined domains express velocity jumps, or vortex sheets?

mw...

Re: An old question re-visited...
 

"Can incompressible fluid flow within confined domains express velocity jumps, or vortex sheets?"

Depends on the initial condition. If you have a continuous twice differentiable initial condition I would say no - it's believed that the viscous equations cannot spontaneously form a discontinuity (but there is no rigourous proof of this as yet in 3D).

However if you insert a discontinuity into the initial state in an inviscid fluid it will persist (i.e. a vortex sheet is a material surface). In a viscous fluid the vortex sheet will diffuse out which will smooth out the discontinuity (the analytical solution for the diffusion of a vortex line shows this).

A simple example is the shock in Burgers equation - the invisid result requires jump conditions to define the discontinuity but the viscous solution does not (because the jump is smoothed out by the diffusion operator).

Re: Update 4 : Settled...
 

Sorry for interfering your dialog, but I think you may find treatment of tensors in many solid mechanics text, mainly in shell theory. The one I used (in my solid analysis days) was the first few chapters of F.I. Niordson, Introduction to Shell Theory. Definitely there are many more.

Re: Update 4 : Settled...
 

Many thanks, Rami.

mw... :)