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Conserving algebraic quantities

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1

$begingroup$

I have the relations

$$r_0^pm = frac12left (frachat rho_0zeta_1^1/2hat rho_1zeta_0^1/2 pm fraczeta_0^1/2zeta_1^1/2 right )$$

$$r_1^pm = frac12left (frachat rho_1zeta_2^1/2hat rho_2zeta_1^1/2 pm fraczeta_1^1/2zeta_2^1/2right )$$

$$T_0,1 = frac1r_0,1^+$$
$$R_0,1= -fracr_0,1^-r_0,1^+$$

and I'm trying to show that

$$R_0,1^2+T_0,1^2 = 1.$$

if $hat rho_0 = hat rho_2$ and $zeta_0=zeta_2.$

So far I've failed to show this. When I compute the above, I get

$$R_0^2+T_0^2 = frachat rho_1^2 zeta_0^2 + 4hat rho_1^2 zeta_0 zeta_1 - 2 hat rho_1 hat rho_0 zeta_0 zeta_1 + hat rho_0^2 zeta_1^2(hat rho_1 zeta_0 + hat rho_0 zeta_1)^2$$

If I could somehow find a way to factor this to achieve $1$, it would be great. Could anyone cast an eye over this to see if I'm not spotting the obvious?

Thanks in advance.

algebra-precalculus

share|cite|improve this question

asked yesterday

rodger_kicksrodger_kicks

8110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $r_0,1^+$ and $r_0,1^-$?
  $endgroup$
  – someguy
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They are linking terms for waves moving through a medium
  $endgroup$
  – rodger_kicks
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm asking how they relate to the first two equations.
  $endgroup$
  – someguy
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand your question. $T_0,1$ and $R_0,1$ depend on the $r$ expressions.
  $endgroup$
  – rodger_kicks
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what might be different notations, e.g. $r_0^+$, $r_1^+$ and $r_0,1^+$. Is the latter the product of the first two?
  $endgroup$
  – someguy
  yesterday
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

I have the relations

$$r_0^pm = frac12left (frachat rho_0zeta_1^1/2hat rho_1zeta_0^1/2 pm fraczeta_0^1/2zeta_1^1/2 right )$$

$$r_1^pm = frac12left (frachat rho_1zeta_2^1/2hat rho_2zeta_1^1/2 pm fraczeta_1^1/2zeta_2^1/2right )$$

$$T_0,1 = frac1r_0,1^+$$
$$R_0,1= -fracr_0,1^-r_0,1^+$$

and I'm trying to show that

$$R_0,1^2+T_0,1^2 = 1.$$

if $hat rho_0 = hat rho_2$ and $zeta_0=zeta_2.$

So far I've failed to show this. When I compute the above, I get

$$R_0^2+T_0^2 = frachat rho_1^2 zeta_0^2 + 4hat rho_1^2 zeta_0 zeta_1 - 2 hat rho_1 hat rho_0 zeta_0 zeta_1 + hat rho_0^2 zeta_1^2(hat rho_1 zeta_0 + hat rho_0 zeta_1)^2$$

If I could somehow find a way to factor this to achieve $1$, it would be great. Could anyone cast an eye over this to see if I'm not spotting the obvious?

Thanks in advance.

algebra-precalculus

share|cite|improve this question

asked yesterday

rodger_kicksrodger_kicks

8110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $r_0,1^+$ and $r_0,1^-$?
  $endgroup$
  – someguy
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  They are linking terms for waves moving through a medium
  $endgroup$
  – rodger_kicks
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm asking how they relate to the first two equations.
  $endgroup$
  – someguy
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't understand your question. $T_0,1$ and $R_0,1$ depend on the $r$ expressions.
  $endgroup$
  – rodger_kicks
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what might be different notations, e.g. $r_0^+$, $r_1^+$ and $r_0,1^+$. Is the latter the product of the first two?
  $endgroup$
  – someguy
  yesterday
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I have the relations

$$r_0^pm = frac12left (frachat rho_0zeta_1^1/2hat rho_1zeta_0^1/2 pm fraczeta_0^1/2zeta_1^1/2 right )$$

$$r_1^pm = frac12left (frachat rho_1zeta_2^1/2hat rho_2zeta_1^1/2 pm fraczeta_1^1/2zeta_2^1/2right )$$

$$T_0,1 = frac1r_0,1^+$$
$$R_0,1= -fracr_0,1^-r_0,1^+$$

and I'm trying to show that

$$R_0,1^2+T_0,1^2 = 1.$$

if $hat rho_0 = hat rho_2$ and $zeta_0=zeta_2.$

So far I've failed to show this. When I compute the above, I get

$$R_0^2+T_0^2 = frachat rho_1^2 zeta_0^2 + 4hat rho_1^2 zeta_0 zeta_1 - 2 hat rho_1 hat rho_0 zeta_0 zeta_1 + hat rho_0^2 zeta_1^2(hat rho_1 zeta_0 + hat rho_0 zeta_1)^2$$

If I could somehow find a way to factor this to achieve $1$, it would be great. Could anyone cast an eye over this to see if I'm not spotting the obvious?

Thanks in advance.

algebra-precalculus

share|cite|improve this question

asked yesterday

rodger_kicksrodger_kicks

8110

$endgroup$

share|cite|improve this question

asked yesterday

rodger_kicksrodger_kicks

8110

share|cite|improve this question

asked yesterday

rodger_kicksrodger_kicks

8110

asked yesterday

rodger_kicksrodger_kicks

8110

asked yesterday

rodger_kicksrodger_kicks

8110

1 Answer
1

active

oldest

votes

0

$begingroup$

For $iin0,1$ you want to show that $R_i^2+T_i^2=1$, or equivalently
$$(r_i^+)^2-(r_i^-)^2=1.$$
For the sake of legibility let $a_k:=hatrho_k$ and $b_k:=zeta_k^1/2$. Then for $i=0$ we have
$$r_0^pm=frac12left(fraca_0b_1a_1b_0pm fracb_0b_1right)
=fraca_0b_1^2pm a_1b_0^22a_1b_0b_1,$$
and so the standard identity $(x+y)^2-(x-y)^2=4xy$ shows that
begineqnarray*
(r_0^+)^2-(r_0^-)^2
&=&frac(a_0b_1^2+a_1b_0^2)^24a_1^2b_0^2b_1^2
-frac(a_0b_1^2-a_1b_0^2)^24a_1^2b_0^2b_1^2\
&=&4fraca_0a_1b_0^2b_1^24a_1^2b_0^2b_1^2=fraca_0a_1,
endeqnarray*
and so the identity holds if and only if $a_0=a_1$. Because $a_2=a_0$ and $b_2=b_0$, the exact same proof works for $i=1$ by changing the indices; change all the $0$'s to $1$'s and all the $1$'s to $0$'s.

share|cite|improve this answer

edited 15 hours ago

answered yesterday

ServaesServaes

27.5k34098

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, I think you have a made a mistake, on the second terms in the last line. For this to hold we would need $4a_0a_1b_0^2 b_1^2$ on the numerator? So this only holds if $a_0=a_1$
  $endgroup$
  – rodger_kicks
  16 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was thinking of some way to achieve the relation purely in terms of $a_0,2,b_0,2$ so that I can then apply $a_2=a_0, b_0=b_2.$ The problem arises with the $a_1,b_1$ terms...
  $endgroup$
  – rodger_kicks
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rodger_kicks You are absolutely right, I have adjusted my answer accordingly. I don't understand what you mean by your second comment.
  $endgroup$
  – Servaes
  15 hours ago
  
  

  
  
  
  

add a comment | 

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0

$begingroup$

For $iin0,1$ you want to show that $R_i^2+T_i^2=1$, or equivalently
$$(r_i^+)^2-(r_i^-)^2=1.$$
For the sake of legibility let $a_k:=hatrho_k$ and $b_k:=zeta_k^1/2$. Then for $i=0$ we have
$$r_0^pm=frac12left(fraca_0b_1a_1b_0pm fracb_0b_1right)
=fraca_0b_1^2pm a_1b_0^22a_1b_0b_1,$$
and so the standard identity $(x+y)^2-(x-y)^2=4xy$ shows that
begineqnarray*
(r_0^+)^2-(r_0^-)^2
&=&frac(a_0b_1^2+a_1b_0^2)^24a_1^2b_0^2b_1^2
-frac(a_0b_1^2-a_1b_0^2)^24a_1^2b_0^2b_1^2\
&=&4fraca_0a_1b_0^2b_1^24a_1^2b_0^2b_1^2=fraca_0a_1,
endeqnarray*
and so the identity holds if and only if $a_0=a_1$. Because $a_2=a_0$ and $b_2=b_0$, the exact same proof works for $i=1$ by changing the indices; change all the $0$'s to $1$'s and all the $1$'s to $0$'s.

share|cite|improve this answer

edited 15 hours ago

answered yesterday

ServaesServaes

27.5k34098

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, I think you have a made a mistake, on the second terms in the last line. For this to hold we would need $4a_0a_1b_0^2 b_1^2$ on the numerator? So this only holds if $a_0=a_1$
  $endgroup$
  – rodger_kicks
  16 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was thinking of some way to achieve the relation purely in terms of $a_0,2,b_0,2$ so that I can then apply $a_2=a_0, b_0=b_2.$ The problem arises with the $a_1,b_1$ terms...
  $endgroup$
  – rodger_kicks
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rodger_kicks You are absolutely right, I have adjusted my answer accordingly. I don't understand what you mean by your second comment.
  $endgroup$
  – Servaes
  15 hours ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

For $iin0,1$ you want to show that $R_i^2+T_i^2=1$, or equivalently
$$(r_i^+)^2-(r_i^-)^2=1.$$
For the sake of legibility let $a_k:=hatrho_k$ and $b_k:=zeta_k^1/2$. Then for $i=0$ we have
$$r_0^pm=frac12left(fraca_0b_1a_1b_0pm fracb_0b_1right)
=fraca_0b_1^2pm a_1b_0^22a_1b_0b_1,$$
and so the standard identity $(x+y)^2-(x-y)^2=4xy$ shows that
begineqnarray*
(r_0^+)^2-(r_0^-)^2
&=&frac(a_0b_1^2+a_1b_0^2)^24a_1^2b_0^2b_1^2
-frac(a_0b_1^2-a_1b_0^2)^24a_1^2b_0^2b_1^2\
&=&4fraca_0a_1b_0^2b_1^24a_1^2b_0^2b_1^2=fraca_0a_1,
endeqnarray*
and so the identity holds if and only if $a_0=a_1$. Because $a_2=a_0$ and $b_2=b_0$, the exact same proof works for $i=1$ by changing the indices; change all the $0$'s to $1$'s and all the $1$'s to $0$'s.

share|cite|improve this answer

edited 15 hours ago

answered yesterday

ServaesServaes

27.5k34098

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, I think you have a made a mistake, on the second terms in the last line. For this to hold we would need $4a_0a_1b_0^2 b_1^2$ on the numerator? So this only holds if $a_0=a_1$
  $endgroup$
  – rodger_kicks
  16 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I was thinking of some way to achieve the relation purely in terms of $a_0,2,b_0,2$ so that I can then apply $a_2=a_0, b_0=b_2.$ The problem arises with the $a_1,b_1$ terms...
  $endgroup$
  – rodger_kicks
  15 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @rodger_kicks You are absolutely right, I have adjusted my answer accordingly. I don't understand what you mean by your second comment.
  $endgroup$
  – Servaes
  15 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

For $iin0,1$ you want to show that $R_i^2+T_i^2=1$, or equivalently
$$(r_i^+)^2-(r_i^-)^2=1.$$
For the sake of legibility let $a_k:=hatrho_k$ and $b_k:=zeta_k^1/2$. Then for $i=0$ we have
$$r_0^pm=frac12left(fraca_0b_1a_1b_0pm fracb_0b_1right)
=fraca_0b_1^2pm a_1b_0^22a_1b_0b_1,$$
and so the standard identity $(x+y)^2-(x-y)^2=4xy$ shows that
begineqnarray*
(r_0^+)^2-(r_0^-)^2
&=&frac(a_0b_1^2+a_1b_0^2)^24a_1^2b_0^2b_1^2
-frac(a_0b_1^2-a_1b_0^2)^24a_1^2b_0^2b_1^2\
&=&4fraca_0a_1b_0^2b_1^24a_1^2b_0^2b_1^2=fraca_0a_1,
endeqnarray*
and so the identity holds if and only if $a_0=a_1$. Because $a_2=a_0$ and $b_2=b_0$, the exact same proof works for $i=1$ by changing the indices; change all the $0$'s to $1$'s and all the $1$'s to $0$'s.

share|cite|improve this answer

edited 15 hours ago

answered yesterday

ServaesServaes

27.5k34098

$endgroup$

share|cite|improve this answer

edited 15 hours ago

answered yesterday

ServaesServaes

27.5k34098

answered yesterday

ServaesServaes

27.5k34098

answered yesterday

ServaesServaes

27.5k34098