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I am trying to find, analytically, the $\theta,\phi$ that minimize the function U($\theta,\phi$)

Clear[h, d]
m = {Sin[θ] Cos[ϕ], Sin[θ] Sin[ϕ], Cos[θ]};
n = {Sin[θn] Cos[ϕn], Sin[θn] Sin[ϕn],Cos[θn]};
H = h {0, 0, 1};
U = -H.m - d (n.m)^2

where h and d are positive. θn and ϕn are fixed variables.

How can I find the value of $\theta,\phi$ that minimizes U($\theta,\phi$)?

I tried to use

Minimize[{U, h > 0, d > 0}, {θ, ϕ}]

It is not successful, Mathematica just returns the input as the output.

What is the proper way of doing this problem?


Update

I solved it manually. $dU/d\phi=0$ leads to $\phi=\phi_n$ or $\pi+\phi_n/\pi$, and, therefore, $dU/d\theta=0$ gives the equation for $\theta$ as

$\qquad h\sin\theta+d\sin[2(\theta\pm\theta_n)]=0$

Mathmematica is able to give a long analytical solution. I have to use a numerical result for given values of d,h,$\theta_n$, $\phi_n$.

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  • $\begingroup$ Is: θn or: θ*n ? $\endgroup$ Commented Nov 8, 2019 at 20:41
  • $\begingroup$ θn is $\theta_n$, θn and ϕn are fixed variables $\endgroup$
    – p.s
    Commented Nov 8, 2019 at 20:47
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    $\begingroup$ The minimum should be a root of $\nabla U$, and you can see that there is no closed-form solution in general. $\endgroup$
    – anderstood
    Commented Nov 8, 2019 at 22:22
  • $\begingroup$ you mean there is no analytical solution? $\endgroup$
    – p.s
    Commented Nov 8, 2019 at 22:43
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    $\begingroup$ How can you get a numerical result if some parameters are symbolic? $\endgroup$ Commented Nov 9, 2019 at 15:41

1 Answer 1

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Evidently it's having trouble with the transcendental equation. Plug in values and it can do it numerically.

d = .5
h = .5
θn = 45 °
ϕn = 30 °

Minimize[U, {θ, ϕ}]
(*{-0.899519, {θ -> 0.523599, ϕ -> 0.523599}}*)

You need to test your own values, since I don't know what reasonable input value are.

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  • $\begingroup$ Ideally, I hope to get an analytical solution. If there is no analytical one, I will have to use the numerical one. $\endgroup$
    – p.s
    Commented Nov 8, 2019 at 22:42
  • $\begingroup$ I can't see why this should have been down-voted $\endgroup$
    – mikado
    Commented Nov 9, 2019 at 11:50
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    $\begingroup$ If you use exact values for d and h (i.e., 1/2) the minimum is {1/8 (-2 - 3 Sqrt[3]), {\[Theta] -> \[Pi]/6, \[Phi] -> \[Pi]/6}} and {1/8 (-2 - 3 Sqrt[3]), {\[Theta] -> (11 \[Pi])/6, \[Phi] -> (7 \[Pi])/ 6}} $\endgroup$
    – Bob Hanlon
    Commented Nov 9, 2019 at 14:32

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