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Old   April 9, 2007, 12:34
Default Could someone please assist? Plane Waves
  #1
Bob
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I'm currently reading a paper where the author discusses the propagation of errors in terms of plane waves. The paper aims to reduce the reflection of such errors by including appropriate boundary conditions at the domain boundaries.

The equation we're trying to solve is:

2*u_xt = K*u_xx + u_yy (where K is a constant and I have borrowed the notation "_x" to mean differentiation w.r.t x and so on)

The author states that the solution to such an equation is the set of plane waves f( ct - x*w1 - y*w2 ) which satisfies this equation if

c= -( K*w1^2 + w2^2 )/( 2*w1 ) so far fairly easy )

The author then states that this value 'c' is the "local wave speed" in the direction of the unit vector (w1,w2). Could someone please advise me if this is correct? my knowledge of wave mechanics is somewhat rudimentary and I am a tad confused.

My understanding is that the phase speeds are c/w1 & c/w2 in the x and y directions respectively. For the particular case I'm dealing with w1 is far smaller than w2 which means the phase speed should be much larger in the x-direction than in the y-direction. If the authors statement is correct, however, then I'd expect the wave to travel with local speed 'c' in the direction (w1,w2) which would be, more or less, in the y-direction.

I fear I've missed something important. If someone could please point out my error then I'd be very grateful.

Thanks for reading Bob
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Old   April 9, 2007, 20:51
Default Re: Could someone please assist? Plane Waves
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Ananda Himansu
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Nothing inconsistent there. Just a matter of choice of representation and interpretation. Take, for example, the situation you mention, when w1 << w2. A "plane wave" (which is actually a line in the x,y space, though it is a plane in the x,y,t space) has a normal vector (w1,w2). You will notice that if it is moving in the (w1,w2) direction with a speed c, the intersection of the "plane" with the x axis (the y=0 axis) will move much faster than c, and thus the phase speed in the x direction c/w1 will be much larger than c. This is because the value of f (and thus of u) is constant along the entire line, and it appears as if this information is propagating extremely fast in the x direction. In the limiting case of w1 = 0, the information will appear to propagate with infinite speed (instantaneously) in the x direction, but the f = constant line wave will just be moving with speed c in the y direction. Just picture a line almost parallel to the x axis and move it normal to itself with a unit speed, and you will notice that its intersection with the x axis moves at much faster than unit speed. The confusion arises because in this situation you are considering not the motion of a point, but the motion of a line wave (which is a contour of the function f in the x,y plane). The speed of propagation in (x,y) space of a line in the direction normal to itself is easy to understand. You have to ask yourself, what is its speed of propagation in the direction parallel to itself, if points on the line cannot be distinguished from each other (f is constant along the line)?

To understand this in mathematical terms, in the (x,y) plane, draw a line with unit normal (w1,w2) and at normal distance c from the origin (in the positive quadrant). Notice that it intersects the x axis at a distance c/w1 from the origin, and that it intersects the y axis at c/w2 from the origin. From this, and from the fact that the phase speeds are of the form c/w1 and c/w2, you can see that the vector (w1,w2) has the nature of a covariant vector like function gradient rather than the nature of a contravariant vector like the displacement vector of a particle, though of course it is normalized to be of unit length. There is a triangle rule for addition of covariant vectors, but it is a little different from the familiar one for the usual contravariant vectors. The fear of tensor analysis is the beginning of wisdom.
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Old   April 10, 2007, 14:25
Default Re: Could someone please assist? Plane Waves
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Bob
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Thank you Ananda your reply was very helpful.

I do have one small question, however, if you don't mind. I had a look at the situation you described in your second paragraph but found slightly different results. I'm genuinely confused as to what my results mean.

I draw a line (line A) between the points (0,y0) and (x0,0) in the +ve quadrant as you suggest. The line A is represented by the vector (x0,-y0). I then draw a second line 'B' from the origin which strikes the line A at a right angle; the length of the new line is 's'. Line B is represented by the vector (w1,w2) containing the 'w' terms from the plane waves we mentioned earlier. By considering the fact that the vectors are perpendicular and using pythagoras' theorem I found that the intersections of the plane surfaces with the x & y axis are s^2/w1 and s^2/w2 respectively.

I'm a little confused now. I've double checked the algebra and am pretty sure I've not made a mistake ... but it wouldn't be the first time! )

Does this mean, then, that the phase speed of the waves in x&y direction is actually s^2/w1 & s^2/w2 respectively?

Many thanks, Bob
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Old   April 10, 2007, 15:12
Default How to specify the inflow turbulence?
  #4
EJAZ
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Answer: The turbulent kinetic energy can be estimated as follows k=0.5(x.V)^2b, where x is the velocity pulsation amplitude relative to the mean flow velocity V;

The rate of the turbulent energy dissipation is estimated from 0.09 k^2/e = N&mu; , where &mu; is the dynamic molecular viscosity, N = 100 is a parameter. for more details http://www.fv-tech.com/index.php?id=73

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Old   April 10, 2007, 17:54
Default Re: Could someone please assist? Plane Waves
  #5
Ananda Himansu
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Draw a unit vector (w1,w2) through the origin O of the (x,y) plane, with endpoint S. Thus the vector is the line segment OS and the length(OS) = |OS| = 1. Also, the projection of S onto the x-axis is U and onto the Y axis is V, so that |OU|=w1 and |OV|=w2. Consider a line wave P'Q', perpendicular to OS, and passing through the origin O at time t0. One unit of time later (say, 1 second later, if you are in SI units, or one unit of characteristic time later for a nondimensionalized problem), at time t0 + 1, the wave P'Q' has propagated in the direction OS for a distance c, the given wave speed. So draw a line PQ perpendicular to OS, intersecting the x-axis in P (the point (0,x0) in your description) and the y-axis in Q (the point (y0,0) in your description). The perpendicular from O onto PQ intersects PQ in the point R, so that OSR is a straight line, and |OR|=c. Then, the triangles OUS and ORP are geometrically similar, and |OR|/|OP| = |OU|/|OS|, from which |OP| = c/w1. Similarly, |OQ| = c/w2. Note that in unit time the wave has moved along OS by distance c, and its intersection with the x-axis has moved from O to P by a distance c/w1, and its intersection with the y-axis has moved from O to Q by a distance c/w2. Thus, the wave and phase speeds are c, c/w1 and c/w2.

All the confusion arises because here you are representing the motion by an implicit mathematical function f, rather than by the usual parametrized particle velocity dx/dt = u(t) and dy/dt = v(t). Mathematically speaking, the motion of a wavefront during a time interval dt is expressed by the condition f(c(t0+dt)-w1(x0+dx)-w2(y0+dy)) = f(c*t0-w1*x0-w2*y0) for general f. Therefore, in the (x,y,t) space, you require that the vector (dx,dy,dt) be orthogonal to the vector (w1,w2,c), i.e. c*dt - w1*dx - w2*dy = 0. Substitute this condition in the LHS of the equations relating f at t0 and t0+dt, above, and you will see that it becomes an identity. You can see that the vector (dx,dy,dt) is the contravariant sort of vector typified by coordinate displacement vectors, and therefore the other vector (w1,w2,c) must be the covariant sort of vector typified by gradients such as grad(f). So that the "orthogonality" of their product is an invariant. Geometrically, the "vector" (w1,w2,c) is better thought of as a family of hyperplanes, with the spacing between the hyperplanes being represented as the magnitude of the "vector". Note, the statements in this second paragraph are not necessarily accurate. I am thinking them up on the spot based on the tensor geometry that I know from long ago. But they give you the essence of why the "velocities" c, c/w1 and c/w2 do not seem to behave algebraically like ordinary velocity vectors and components.
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Old   April 10, 2007, 18:08
Default Re: Could someone please assist? Plane Waves
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Ananda Himansu
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oops, in the second para, I meant the INVERSE of the spacing of the hyperplanes represents the magnitude of the "vector". The closer together they are spaced, the greater the magnitude.
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Old   April 10, 2007, 19:26
Default Excellent explanation
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Bob
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Thanks very much Ananda, that was a very good explanation and helped me a great deal.

Cheers, Bob
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Old   April 10, 2007, 20:20
Default You are welcome
  #8
Ananda Himansu
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You are welcome, Bob
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Old   April 11, 2007, 01:31
Default Correction Re: Plane Waves
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Ananda Himansu
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As I suspected, the second paragraph of my 3:54 pm post was inaccurate, because I was dreaming it up as I was writing it. More precisely, as the line wavefront P'Q' at time t0 moves in the spatial direction (w1,w2), i.e. OS, it traces out a plane segment P'Q'Q"P" in the spacetime (x,y,t), where PQ is the projection of P"Q" (which is at time t0+dt) onto the (x,y) plane (which is at time t0). This two-dimensional plane is also a (three minus one)-dimensional hyperplane in the three-dimensional spacetime (x,y,t). You can check that the vector (w1,w2,-c) (note the minus sign which I conveniently ignored before!) is orthogonal to this hyperplane. That is why c*dt - w1*dx -w2*dy = 0, given that the displacement (dx,dy,dt) lies in that hyperplane. Thus this condition which yields f at (x0+dx,y0+dy,t0+dt) equals f at (x0,y0,t0) is simply the expression that any vector that lies in the hyperplane is orthogonal to the normal vector to the hyperplane. The representation of a hyperplane by a vector normal to the hyperplane (similar to the way grad(f) is normal to hypersurfaces of constant f) is what gives that vector its covariant rather than contravariant nature. The vector (w1,w2) is the projection of that normal vector onto the (x,y) plane, and retains the covariant nature. I apologize for the partially misleading second paragraph of the earlier post.
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