Show that the solutions to the Fresnel Equation are real and positive

[Blanchdog]

Homework Statement

Put Fresnel's Equation in quadratic form. Solutions for $n^2$ are real and positive when $n_x, n_y, n_z$ are real and $B^2 - 4AC \geq 0$. Assume that all n are real and show that $B^2 - 4AC \geq 0$ in the special case of $u_x = u_y = u_z = \frac{1}{\sqrt{3}}$

Relevant Equations

Quadratic form of Fresnel Equation (confirmed correct barring typos): $$A n^4 -B n^2 + C = 0,$$ where $$A = u_x^2 n_x^2 + u_y^2 n_y^2 + u_z^2 n_z^2,$$ $$B = u_x^2 n_x^2 (n_y^2+n_z^2) + u_y^2 n_y^2 (n_x^2 + n_z^2) + u_z^2 n_z^2 (n_x^2 + n_y^2),$$ $$C = n_x^2 n_y^2 n_z^2$$

I got as far as simplifying the expression to $$\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$

But that doesn't seem to be a form that is necessarily positive and satisfies the criteria of the homework statement. Little help with this algebra?

 

[kuruman]

Your simplified expression does not look right. One can tell because $A$, $B$ and $C$ are invariant under a cyclic permutation of indices $x \rightarrow y \rightarrow z \rightarrow x$. Any simplification should also be invariant under such a permutation. Yours is not.

 

[Blanchdog]

Your simplified expression does not look right. One can tell because $A$, $B$ and $C$ are invariant under a cyclic permutation of indices $x \rightarrow y \rightarrow z \rightarrow x$. Any simplification should also be invariant under such a permutation. Yours is not.

I'm not sure what a cyclic permutation of indices is, but I pretty much figured that my simplification had a mistake. Any chance you can help me find it?

 

[kuruman]

Here is what I mean by cyclic permutation
Start with the original version of the expression: $$B.0 = u_x^2 n_x^2 (n_y^2+n_z^2) + u_y^2 n_y^2 (n_x^2 + n_z^2) + u_z^2 n_z^2 (n_x^2 + n_y^2)$$Permute the indices according the scheme $x \rightarrow y \rightarrow z \rightarrow x$ to get the first version:$$B.1 = u_y^2 n_y^2 (n_z^2+n_x^2) + u_z^2 n_z^2 (n_y^2 + n_x^2) + u_x^2 n_x^2 (n_y^2 + n_z^2)$$ Permute the first version to get the second version:$$B.2 = u_z^2 n_z^2 (n_x^2+n_y^2) + u_x^2 n_x^2 (n_z^2 + n_y^2) + u_y^2 n_y^2 (n_z^2 + n_x^2)$$ Permute the second version to get:$$B.3 = u_x^2 n_x^2 (n_y^2+n_z^2) + u_y^2 n_y^2 (n_x^2 + n_z^2) + u_z^2 n_z^2 (n_x^2 + n_y^2)$$Note that (a) the last permutation results in the exact form of original expression, i.e. applying the cyclic permutation three times is the identity operation; (b) because addition and multiplication are commutative, the intermediate versions $B.1$ and $B.2$ are equal to the original expression. If I subtract either one from $B.0$, I get zero.

Your expression is $$E.0=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$ Permute once to get $$E.1=\frac{4}{9}(n_y^4 n_z^4 + n_y^4 n_x^4 + n_z^4 + n_x^4 - n_y^4 n_z^2 n_x^2 - n_y^2 n_z^4 n_x^2 - n_y^2 n_z^2 n_x^4)$$If I subtract the second expression from the first, I get $$E.0-E.1=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4-n_y^4 n_z^4 - n_y^4 n_x^4 - n_x^4)\neq 0$$Now $A$, $B$ and $C$ are invariant (do not change) under a cyclic permutation, therefore $B^2-4AC$ must also be invariant. However, as we have seen, your simplified version of that expression is not. You can perhaps find where you went wrong by verifying that your expressions from one step to the next remain invariant as you simplify.

Also note that $A$ goes as $n^2$ and $C$ goes as $n^6$. Their product goes as $n^8$. Now $B$ goes as $n^4$ which means that $B^2$ goes as $n^8$. Why is this useful? Because if you put together $B^2-4AC$ the sum of the exponents in products of $n_i$ must always be 8. Your simplified expression has mostly eights, but I see two fours. This is a complementary way to pinpointing where you took a wrong turn in your simplification.

Short of actually doing it myself to see what is involved, this is all the help that I can offer at the moment. It looks like a tedious job, but there might a shortcut, I don't know.

 

[Blanchdog]

Here is what I mean by cyclic permutation
Start with the original version of the expression: $$B.0 = u_x^2 n_x^2 (n_y^2+n_z^2) + u_y^2 n_y^2 (n_x^2 + n_z^2) + u_z^2 n_z^2 (n_x^2 + n_y^2)$$Permute the indices according the scheme $x \rightarrow y \rightarrow z \rightarrow x$ to get the first version:$$B.1 = u_y^2 n_y^2 (n_z^2+n_x^2) + u_z^2 n_z^2 (n_y^2 + n_x^2) + u_x^2 n_x^2 (n_y^2 + n_z^2)$$ Permute the first version to get the second version:$$B.2 = u_z^2 n_z^2 (n_x^2+n_y^2) + u_x^2 n_x^2 (n_z^2 + n_y^2) + u_y^2 n_y^2 (n_z^2 + n_x^2)$$ Permute the second version to get:$$B.3 = u_x^2 n_x^2 (n_y^2+n_z^2) + u_y^2 n_y^2 (n_x^2 + n_z^2) + u_z^2 n_z^2 (n_x^2 + n_y^2)$$Note that (a) the last permutation results in the exact form of original expression, i.e. applying the cyclic permutation three times is the identity operation; (b) because addition and multiplication are commutative, the intermediate versions $B.1$ and $B.2$ are equal to the original expression. If I subtract either one from $B.0$, I get zero.

Your expression is $$E.0=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$ Permute once to get $$E.1=\frac{4}{9}(n_y^4 n_z^4 + n_y^4 n_x^4 + n_z^4 + n_x^4 - n_y^4 n_z^2 n_x^2 - n_y^2 n_z^4 n_x^2 - n_y^2 n_z^2 n_x^4)$$If I subtract the second expression from the first, I get $$E.0-E.1=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 + n_z^4-n_y^4 n_z^4 - n_y^4 n_x^4 - n_x^4)\neq 0$$Now $A$, $B$ and $C$ are invariant (do not change) under a cyclic permutation, therefore $B^2-4AC$ must also be invariant. However, as we have seen, your simplified version of that expression is not. You can perhaps find where you went wrong by verifying that your expressions from one step to the next remain invariant as you simplify.

Also note that $A$ goes as $n^2$ and $C$ goes as $n^6$. Their product goes as $n^8$. Now $B$ goes as $n^4$ which means that $B^2$ goes as $n^8$. Why is this useful? Because if you put together $B^2-4AC$ the sum of the exponents in products of $n_i$ must always be 8. Your simplified expression has mostly eights, but I see two fours. This is a complementary way to pinpointing where you took a wrong turn in your simplification.

Short of actually doing it myself to see what is involved, this is all the help that I can offer at the moment. It looks like a tedious job, but there might a shortcut, I don't know.

Ah I see the issue, I have a typo. The expression I simplified it to should be $$B^2-4AC=\frac{4}{9}(n_x^4 n_y^4 + n_x^4 n_z^4 + n_y^4 n_z^4 - n_x^4 n_y^2 n_z^2 - n_x^2 n_y^4 n_z^2 - n_x^2 n_y^2 n_z^4)$$

This expression is invariate under a cyclic permutation, and I can show 2 of the positive terms are greater than 2 of the negative terms, but one of the negative terms seems to be bigger than one of the positive terms.

 

[kuruman]

Which two are greater than which two? Is there additional information about the $n_i$ that you did not mention? Is there a relation that sets the squares of their sums equal to a constant? I have forgotten all that stuff that I saw a very long time ago.

It seems to me that if you can show that one positive term is greater than one negative term, then you can by cyclic permutation that this true for the other two pairs. Unless, of course there is additional information that I don't know.