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

  • Thread starter Thread starter Blanchdog
  • Start date Start date
  • Tags Tags
    Fresnel Positive
AI Thread Summary
The discussion revolves around simplifying the expression derived from the Fresnel Equation, specifically focusing on ensuring the result remains invariant under cyclic permutations of indices. An initial simplification attempt was deemed incorrect due to a lack of invariance, prompting a request for assistance in identifying the mistake. After recognizing a typo in the expression, a new form was presented that maintained the necessary invariance. The conversation further explores the need to establish relationships among the terms to demonstrate that certain positive terms exceed negative ones, suggesting that additional information about the indices may be crucial for a complete solution. The dialogue emphasizes the importance of maintaining mathematical properties throughout simplifications.
Blanchdog
Messages
56
Reaction score
22
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?
 
Physics news on Phys.org
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.
 
Last edited:
kuruman said:
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?
 
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.
 
Last edited:
kuruman said:
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.
 
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top