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Mathematical proof regarding angle mirroring

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$begingroup$

I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):

Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.

Let me show some examples to make this clearer.

Example 1

$beta = 45^circ$, $theta = 30^circ$

The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.

Example 2

$beta = 50^circ$, $theta = 15^circ$

In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.

I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?

geometry proof-writing polar-coordinates angle

share|cite|improve this question

edited Mar 18 at 15:38

Tendero

asked Mar 18 at 13:58

TenderoTendero

378214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
  $endgroup$
  – dmtri
  Mar 18 at 14:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri Yes, $theta$ must lie between $0$ and $beta$.
  $endgroup$
  – Tendero
  Mar 18 at 14:10
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):

Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.

Let me show some examples to make this clearer.

Example 1

$beta = 45^circ$, $theta = 30^circ$

The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.

Example 2

$beta = 50^circ$, $theta = 15^circ$

In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.

I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?

geometry proof-writing polar-coordinates angle

share|cite|improve this question

edited Mar 18 at 15:38

Tendero

asked Mar 18 at 13:58

TenderoTendero

378214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you also assume that the angle domain of $β$ measure contains the point $P$ ?
  $endgroup$
  – dmtri
  Mar 18 at 14:08
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @dmtri Yes, $theta$ must lie between $0$ and $beta$.
  $endgroup$
  – Tendero
  Mar 18 at 14:10
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to find a proof for a statement that is made in Griffiths' Introduction to Electrodynamics. It can be stated as follows (in my own words):

Let $P$ be a point such that its angle in polar coordinates is given by $theta>0$. Let $beta > theta$ be the angle between any two lines in the 2D plane. Then, if we try to find all the points such that both lines act as mirrors, if $beta$ is an integer divisor of $180$, all of these points will fall outside the region delimited by the angular interval $(0, beta)$, independently of the value of $thetain (0,beta)$.

Let me show some examples to make this clearer.

Example 1

$beta = 45^circ$, $theta = 30^circ$

The red dot represents the original $P$ point. Note that the blue points are placed in a way that make the lines with slope $0$ and $beta$ 'mirrors'. Due to the fact that $beta = frac{180^circ}{4}$, no blue dots appear inside the region delimited by the black lines.

Example 2

$beta = 50^circ$, $theta = 15^circ$

In this case, due to the fact that $frac{180^circ}{beta}$ is not an integer, blue dots appear inside the region delimited by the black lines. This is, however, not general. It is possible to find pairs of values of $theta$ and $beta$ such that no blue dots appear in the region, but it is not likely.

I've tried with some values and the statement made by Griffiths seems to be valid. Is there a way to prove this generically?

geometry proof-writing polar-coordinates angle

share|cite|improve this question

edited Mar 18 at 15:38

Tendero

asked Mar 18 at 13:58

TenderoTendero

378214

$endgroup$

share|cite|improve this question

edited Mar 18 at 15:38

Tendero

asked Mar 18 at 13:58

TenderoTendero

378214

share|cite|improve this question

edited Mar 18 at 15:38

Tendero

asked Mar 18 at 13:58

TenderoTendero

378214

asked Mar 18 at 13:58

TenderoTendero

378214

asked Mar 18 at 13:58

TenderoTendero

378214

1 Answer
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oldest

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2

$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.

share|cite|improve this answer

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
  $endgroup$
  – Tendero
  Mar 18 at 16:13
  
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.

share|cite|improve this answer

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
  $endgroup$
  – Tendero
  Mar 18 at 16:13
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.

share|cite|improve this answer

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your answer. It's short and clear. Is there a way to extend the proof to explain why if $beta neq pi /n$, it is posible to find value/s of $theta$ such that blue dots will appear inside the region of interest (which is not posible if $beta = pi /n$, as you have just shown)?
  $endgroup$
  – Tendero
  Mar 18 at 16:13
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $ell_1$ and $ell_2$ be the two mirror lines, and assume $beta={piover n}$. Denote the reflections in the $ell_i$ by $rho_i$. You then obtain blue points from the starting red point $P$ by applying alternatively $rho_1$ and $rho_2$. Now it is easy to see that $rho_2circrho_1$ is a rotation by the angle $alpha=|2beta|={2piover n}$ in one of the two directions. It follows that $(rho_2circrho_1)^n$ is the identity, hence gives back $P$. In all you obtain the $2n$ points $$(rho_2circrho_1)^k (P)quad(0leq kleq n-1),qquad (rho_2circrho_1)^kcircrho_2 (P)quadquad(0leq kleq n-1) ,$$
one in each sector of central angle $beta={2piover 2n}$.

share|cite|improve this answer

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

$endgroup$

share|cite|improve this answer

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327

answered Mar 18 at 15:54

Christian BlatterChristian Blatter

176k8115327