Prove the reflection and transmission coefficients always sum to 1

[Pendleton]

Homework Statement

Prove the reflection and transmission coefficients of light crossing a sharp boundary between uniform volumes of different refractive indices sum to one.

Relevant Equations

$$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

Consider polarized light crossing a sharp boundary between two volumes, each of a different but uniform refraction index $n_1$ or $n_2$.

Prove that the sum of the transmission and reflection coefficients of this light $R+T=1$, where

$$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

First, we note that the real amplitudes are

$$E_{0_R} = \left|{1-\beta \over 1+\beta}E_{0_I}\right|, \quad\quad E_{0_T} = {2 \over 1+\beta}E_{0_I}$$

Next, we substitute them into the coefficient formulae

$$R={\left(\left|{1-\beta \over 1+\beta}E_{0_I}\right|\right)^2 \over E_{0_I}^2} = \left({1-\beta \over 1+\beta}\right)^2,
\quad\quad
T= {\epsilon_2 v_2 \over \epsilon_1 v_1}{\left({2 \over 1+\beta}E_{0_I}\right)^2 \over E_{0_I}^2} = {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2$$

Therefore,

$$R+T = \left({1-\beta \over 1+\beta}\right)^2 + {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2 = {(1-\beta)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over (1+\beta)^2} $$

Expanding the binomials,

$${1-2\beta +\beta^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\beta+\beta^2}$$

Substituting for beta,

$${1-2\left({\mu_1n_2\over\mu_2n_1}\right) +\left({\mu_1n_2\over\mu_2n_1}\right)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\left({\mu_1n_2\over\mu_2n_1}\right)+\left({\mu_1n_2\over\mu_2n_1}\right)^2}$$

Multiplying the top and bottom by the squared denominator of beta,

$${(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2+4{\epsilon_2 v_2 \over \epsilon_1 v_1}(\mu_2n_1)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Refactoring and restating the $v$ as $n$,

$${(4{\epsilon_2 n_1 \over \epsilon_1 n_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

By definition,

$$n \equiv \sqrt{\epsilon\mu\over\epsilon_0\mu_0}$$

Therefore,

$$\epsilon = n^2\epsilon_0{\mu_0\over\mu}$$

Therefore,
$$4{\epsilon_2 n_1 \over \epsilon_1 n_2} = 4{n_2^2\epsilon_0{\mu_0\over\mu_2} n_1 \over n_1^2\epsilon_0{\mu_0\over\mu_1} n_2}$$
$$= 4{n_2\mu_1 \over n_1\mu_2}$$

Substituting back,
$${(4{n_2\mu_1 \over n_1\mu_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Multiplying the top and bottom by the denominator of the fraction in the numerator,
$${(4n_2\mu_1+n_1\mu_2)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +n_1\mu_2(\mu_1n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+n_1\mu_2(\mu_1n_2)^2}$$

Putting the Greek letters first,
$${(4\mu_1 n_2+\mu_2 n_1)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +\mu_2 n_1(\mu_1 n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+\mu_2n_1(\mu_1n_2)^2}$$

I am out of ideas.

 

[PeroK]

If you start again and express $R + T$ in terms of $\beta$, the answer should fall out in a couple of steps!

 

[Pendleton]

If you start again and express $R + T$ in terms of $\beta$, the answer should fall out in a couple of steps!

(Record of my thoughts as I did this problem...)

Really? I wish you'd told me more. Ugh, fine, I'll try, but I doubt it'll work...

$$R = \left({1-\beta \over 1+\beta}\right)^2, \quad\quad T = {\epsilon_2n_1 \over \epsilon_1n_2}\left({2 \over 1+\beta}\right)^2$$
$$\therefore R+T= \left({1-\beta \over 1+\beta}\right)^2 + {\epsilon_2n_1 \over \epsilon_1n_2}\left({2 \over 1+\beta}\right)^2$$
$$={(1-\beta)^2+4{\epsilon_2n_1 \over \epsilon_1n_2}\over (1+\beta)^2}$$
$$\epsilon = n^2\epsilon_0{\mu_0\over\mu}$$
$$\therefore {\epsilon_2n_1 \over \epsilon_1n_2}={n_2^2\epsilon_0{\mu_0\over\mu_2}n_1 \over n_1^2\epsilon_0{\mu_0\over\mu_1}n_2}={n_2^2\mu_1n_1 \over n_1^2\mu_2n_2}={n_2\mu_1 \over n_1\mu_2}$$
$$\therefore {(1-\beta)^2+4{\epsilon_2n_1 \over \epsilon_1n_2}\over (1+\beta)^2} = {(1-\beta)^2+4{n_2\mu_1 \over n_1\mu_2}\over (1+\beta)^2}$$
See? Right back where we started. Now just to substitute for beta and end up in a mess...
$$\beta = {\mu_1n_2 \over \mu_2n_1}$$
Wait a second... holy moly...
$$\therefore {(1-\beta)^2+4{n_2\mu_1 \over n_1\mu_2}\over (1+\beta)^2} = {(1-\beta)^2+4\beta \over (1+\beta)^2}$$
Now what? Well, nothing to it but to do it...
$$={1 - 2\beta + \beta^2 + 4\beta \over 1 + 2\beta + \beta^2}$$
Woah. Wait another second.
$$={1 + 2\beta + \beta^2 \over 1 + 2\beta + \beta^2}=1$$
HOLY COW. IT REALLY DID FALL RIGHT OUT.
$$QED, \quad\quad R+T = 1$$

I just knew it had to work out like that. That mess of a fraction had to disappear to make it all cancel out. It just never occurred to me that the definition of beta was the key! Thanks so much, PeroK. I'm really grateful.

Adam

 

[PeroK]

The following comes up a lot in physics (in some shape or form): $$(1 - \beta)^2 + 4\beta = (1 + \beta)^2$$

 

[Pendleton]

The following comes up a lot in physics (in some shape or form): $$(1 - \beta)^2 + 4\beta = (1 + \beta)^2$$

Woaaaahhh... yeah, I remember that from my other classes! It never occurred to me how common that property is, but now that you point it out, being able to transform the operator inside a squared binomial is downright amazing. So many problems become easy if you can do that. Thanks so much again!

 

[robphy]

The following comes up a lot in physics (in some shape or form): $$(1 - \beta)^2 + 4\beta = (1 + \beta)^2$$

https://en.wikipedia.org/wiki/Polarization_identity

 

[Charles Link]

This one comes up a lot in Optics, and it is usually done without magnetic materials.
In that case, the intensity, in simplified units, $I=n E^2$.
It is normally done in the general case, for arbitrary $n_1$,
using Fresnel coefficients (for the ratios of the electric field amplitudes) $\rho=(n_1-n_2)/(n_1+n_2)$ and $\tau=2n_1/(n_1+n_2)$.
The energy reflection coefficient $R=\rho^2$, and the energy transmission coefficient $T= (n_2/n_1) \tau^2$.
The algebraic "trick" mentioned by @PeroK will lead to the final result that $R+T=1$.

 

[guv]

I don’t recall doing this in the past, my first reaction reading the title is isn’t this what must follow from conservation of energy.

Homework Statement:: Prove the reflection and transmission coefficients of light crossing a sharp boundary between uniform volumes of different refractive indices sum to one.
Relevant Equations:: $$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

Consider polarized light crossing a sharp boundary between two volumes, each of a different but uniform refraction index $n_1$ or $n_2$.

Prove that the sum of the transmission and reflection coefficients of this light $R+T=1$, where

$$R \equiv {I_R \over I_I} = \left( {E_{0_R} \over E_{0_I}} \right)^2, \quad\quad T \equiv {I_T \over I_I} = {\epsilon_2 v_2 \over \epsilon_1 v_1} \left( {E_{0_T} \over E_{0_I}} \right)^2, \quad\quad \tilde E_{0_R} = {1-\beta \over 1+\beta}E_{0_I}, \quad\quad \tilde E_{0_T} = {2 \over 1+\beta}E_{0_I}, \quad\quad \beta = {\mu_1n_2\over\mu_2n_1}$$

First, we note that the real amplitudes are

$$E_{0_R} = \left|{1-\beta \over 1+\beta}E_{0_I}\right|, \quad\quad E_{0_T} = {2 \over 1+\beta}E_{0_I}$$

Next, we substitute them into the coefficient formulae

$$R={\left(\left|{1-\beta \over 1+\beta}E_{0_I}\right|\right)^2 \over E_{0_I}^2} = \left({1-\beta \over 1+\beta}\right)^2,
\quad\quad
T= {\epsilon_2 v_2 \over \epsilon_1 v_1}{\left({2 \over 1+\beta}E_{0_I}\right)^2 \over E_{0_I}^2} = {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2$$

Therefore,

$$R+T = \left({1-\beta \over 1+\beta}\right)^2 + {\epsilon_2 v_2 \over \epsilon_1 v_1}\left(2 \over 1+\beta\right)^2 = {(1-\beta)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over (1+\beta)^2} $$

Expanding the binomials,

$${1-2\beta +\beta^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\beta+\beta^2}$$

Substituting for beta,

$${1-2\left({\mu_1n_2\over\mu_2n_1}\right) +\left({\mu_1n_2\over\mu_2n_1}\right)^2+4 {\epsilon_2 v_2 \over \epsilon_1 v_1} \over 1+2\left({\mu_1n_2\over\mu_2n_1}\right)+\left({\mu_1n_2\over\mu_2n_1}\right)^2}$$

Multiplying the top and bottom by the squared denominator of beta,

$${(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2+4{\epsilon_2 v_2 \over \epsilon_1 v_1}(\mu_2n_1)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Refactoring and restating the $v$ as $n$,

$${(4{\epsilon_2 n_1 \over \epsilon_1 n_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

By definition,

$$n \equiv \sqrt{\epsilon\mu\over\epsilon_0\mu_0}$$

Therefore,

$$\epsilon = n^2\epsilon_0{\mu_0\over\mu}$$

Therefore,
$$4{\epsilon_2 n_1 \over \epsilon_1 n_2} = 4{n_2^2\epsilon_0{\mu_0\over\mu_2} n_1 \over n_1^2\epsilon_0{\mu_0\over\mu_1} n_2}$$
$$= 4{n_2\mu_1 \over n_1\mu_2}$$

Substituting back,
$${(4{n_2\mu_1 \over n_1\mu_2}+1)(\mu_2n_1)^2-2\mu_1n_2\mu_2n_1 +(\mu_1n_2)^2 \over (\mu_2n_1)^2+2\mu_1n_2\mu_2n_1+(\mu_1n_2)^2}$$

Multiplying the top and bottom by the denominator of the fraction in the numerator,
$${(4n_2\mu_1+n_1\mu_2)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +n_1\mu_2(\mu_1n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+n_1\mu_2(\mu_1n_2)^2}$$

Putting the Greek letters first,
$${(4\mu_1 n_2+\mu_2 n_1)(\mu_2n_1)^3-2\mu_1n_2(\mu_2n_1)^2 +\mu_2 n_1(\mu_1 n_2)^2 \over (\mu_2n_1)^3+2\mu_1n_2(\mu_2n_1)^2+\mu_2n_1(\mu_1n_2)^2}$$

I am out of ideas.

 

[robphy]

Possibly useful:

Suppose
R = \left({1-\beta \over 1+\beta}\right)^2, \qquad\qquad T = \beta\left({2 \over 1+\beta}\right)^2 .
By the above discussion (from the simplified polarization identity, written as (1+\beta)^2 -(1-\beta)^2 \equiv 4\beta ), we have R+T=1.

Now, if \beta=\exp(2x),
then
R = (-\tanh x)^2=\tanh^2 x , \qquad\qquad T = \left( \frac{1}{\cosh x} \right)^2 = \mbox{sech}^2 x
and, if \beta=\exp(-2x) (that is, \frac{1}{\exp(2x)} ), you get...
So, some hyperbolic trigonometry lurks in these expressions,
suggesting that hyperbolic trig identities may help guide algebraic manipulations.

(Similarly, hyperbolic trigonometry lurks in the equations of special relativity
and may also help guide algebraic manipulations:
in special relativity notation, with x as the rapidity, this is
\left( \frac{v}{c}\right )^2 + \left( \frac{1}{\gamma}\right)^2 \equiv 1,
where \gamma =\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}.
By the way, in this relativistic context \exp(2x)
is like the square of the Doppler factor. )

 

[neilparker62]

The following comes up a lot in physics (in some shape or form): $$(1 - \beta)^2 + 4\beta = (1 + \beta)^2$$

Or $$ (\alpha-\beta)^2=(\alpha+\beta)^2 - 4\alpha\beta.$$You can even solve quadratics this way. There are standard expressions for sum of roots (-b/a) and product of roots (c/a). The above algebra let's you get the difference of roots and then the rest is easy.

 

[neilparker62]

Possibly useful:

...

Now, if \beta=\exp(2x),
then
R = (-\tanh x)^2=\tanh^2 x , \qquad\qquad T = \left( \frac{1}{\cosh x} \right)^2 = \mbox{sech}^2 x
and, if \beta=\exp(-2x) (that is, \frac{1}{\exp(2x)} ), you get...

Or if you are a good old Euclidean/Pythagorean 'flat earther' like me , you could let: $$\beta=\tan^2\left(\frac{x}{2}\right).$$ Then $$\frac{1-\beta}{1+\beta}=\cos(x)$$ and $$\frac{2 \sqrt{\beta}}{1+\beta}=\sin(x)$$