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August 23, 2007, 00:01 
A few papers from momentum_waves, diaw, desA

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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... 

August 23, 2007, 05:57 
Re: A few papers from momentum_waves, diaw, desA

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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. 

August 24, 2007, 09:02 
Re: A few papers from momentum_waves, diaw, desA

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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 predotting 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... 

August 27, 2007, 14:19 
Re: A few papers from momentum_waves, diaw, desA

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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  1dimension in space. This equation has as its mechanism, in many cases, a diffusiontype process  but, it essentially describes the motion of u in a wave motion. Actually, this can be seen in simple convectiondiffusion 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 mindbending. 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 wavenature of the momentum equations. Any equation with a d()/dt component will transport in some wavelike manner. I terms this wavetype as a viscous wave (oozytype flow pattern). I'll add in a note to the paper, to include the 'reaction diffusion equation' reference. mw... 

August 28, 2007, 10:51 
Update 3 : A few papers from momentum_waves

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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... 

August 28, 2007, 12:54 
Re: Update 3 : A few papers from momentum_waves

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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.


August 29, 2007, 01:56 
Re: Update 3 : A few papers from momentum_waves

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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 nonionproduct 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 vectordyad 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... 

August 29, 2007, 04:33 
Re: Update 3 : Errata + additional massaging

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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 ndimensional 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 threedimensional 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}.  

August 29, 2007, 06:45 
Re: Update 3 : A few papers from momentum_waves

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"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 lefthandside is a column vector while the righthandside 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. 

August 29, 2007, 07:09 
Re: Update 3 : A few papers from momentum_waves

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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 lefthandside is a column vector while the righthandside 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 vectortensor work. How would one then write nabla2_v in a pure vectortensor 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 contravectors / covectors & 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 layup, 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... 

August 29, 2007, 07:42 
Re: Update 3 : A few papers from momentum_waves

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"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 antisymmetric 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! 

August 29, 2007, 09:17 
Re: Update 3 : A few papers from momentum_waves

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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. Oddlyenough 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 NS, 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 laydown. 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> 

August 29, 2007, 13:28 
Re: Update 3 : A few papers from momentum_waves

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"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"? 

August 29, 2007, 13:44 
Re: Update 3 : A few papers from momentum_waves

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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... 

August 29, 2007, 16:26 
Update 4 : Settled...

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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 NS, 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 crosscheck my work.  If this is settled, then I'm still on track in my endgoal, 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... 

August 31, 2007, 04:06 
Re: An old question revisited...

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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 lowspeed, 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 rewording: Can incompressible fluid flow within confined domains express velocity jumps, or vortex sheets? mw... 

August 31, 2007, 05:26 
Re: An old question revisited...

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"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). 

September 17, 2007, 09:18 
Re: Update 4 : Settled...

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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.


September 27, 2007, 09:27 
Re: Update 4 : Settled...

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Many thanks, Rami.
mw... 

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