# physics?

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 January 7, 2001, 16:24 physics? #1 Girish Bhandari Guest   Posts: n/a Sponsored Links Hello everyone, If vectors(4x1) are in terms of Qr and Qi, where r is real and i is imaginary, functions of x and y, then they can also be represented in terms of Qr=qcos(theta) and Qi=qsin(theta), where 'q' is the magnitude and 'theta' is the phase.The magnitude 'q' and the phase 'theta' are also functions of x and y. Now,If I take the derivatives of Qr and Qi with x and y, will those derivatives be the same as 'qcos(theta)' and 'qsin(theta)' w.r.t. x and y? Keep in mind that the magnitude and the phase is also function of x and y. One has to expand 'qcos(theta)' or 'qsin(theta)' by chain rule w.r.t. x and y. Would these derivatives be able to represents the same physics? -Girish
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 January 8, 2001, 20:19 Re: physics? #2 John C. Chien Guest   Posts: n/a (1). Basically, you have three sets of coordinate systems. (2). So, a point in the (Qr,Qi) space can also be represented by the (q,r) coordinates (in this case a cylindrical coordinates), and also by the (x,y) coordinates.(which is still unknown in terms of the functional relationship) (3). So, Qr=qcos(theta)=q(x,y)*cos(theta(x,y)). And dQr/dx=d(q(x,y)*cos(theta(x,y)))/dx. (where d/dx: partial differential) Since q(x,y) function and theta(x,y) function are not explicitly specified, even if you expand the expression, d(q(x,y))/dx and d(theta(x,y))/dx are still unknown functions. It does not involve the physics, it is just a problem in calculus.

 January 9, 2001, 19:57 Re: physics? #3 Girish Bhandari Guest   Posts: n/a You didnt get it John. I am not talking about any third coordinate system. My question was, d(Qr)/dx=(?)=d(qmag)/dx*cos(theta)-qmag*sin(theta)*d(theta)/dx. And what do you mean by unknown functions? What I am saying is that if you have Qr and Qi everywhere in domain, you can certainly get the qmag and theta out of them. Now, whether the derivative shown above represents the same nature is the question. I know its a calculus problem but what do you have in CFD other than simple(?) calculus? Anyway thanks. ---GB

 January 9, 2001, 21:21 Re: physics? #4 John C. Chien Guest   Posts: n/a (1). d(Qr)/dx = d(qmag*cos(theta))/dx, because Qr=qmag*cos(theta). But the result is unknown,because qmag and theta are unknow functions of x and y. (2). If qmag=qmag(x,y) and theta=theta(x,y),then d(qmag*cos(theta))/dx=d(qmag(x,y)*cos(theta(x,y)))/dx. (3). This can be expanded as d(qmag(x,y))/dx*cos(theta(x,y))-qmag(x,y)*sin(theta(x,y))*d(theta(x,y))/dx. Up to this point, everything is calculus. And d(qmag(x,y))/dx is unknown, because qmag(x,y) is unknown function of (x,y). The same is true for the theta(x,y) function.

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