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Proving Dirac delta identity

0

$begingroup$

I'm having a hard proving an identity which includes the Dirac delta function and the Heaviside function. I want to show that for $minmathbbR,kinmathbbR^4$ and $omega_bf k equiv sqrtbf k^2+m^2$ and $k^2= eta_munuk^mu k^nu$ we have
$$deltaleft(k^2-m^2right)=frac12 omega_mathbfkleft[thetaleft(k_0right) deltaleft(k_0-omega_mathbfkright)+thetaleft(-k_0right) deltaleft(k_0+omega_mathbfkright)right],$$
where $eta_munu$ is the Minkowski metric with signature $(+,-,-,-)$.

---

My attempt here was to use the fact that we have
$$delta(g(x))=sum_i fracdelta(x-x_i),$$
where the $x_i$ are the roots of $g$. By setting $g(k_0)=k^2-m^2= k_0^2-omega_bf k^2$ we find $g'(k_0)=2k_0$ and the roots $pmomega_bf k$. Inserting this into the identity for the delta function leads to
$$delta(k^2-m^2)=frac12(delta(k_0-omega_bf k) + delta(k_0+omega_bf k)),$$
which comes close to what I want to prove but is not quite there yet since the Heaviside function is still missing. The problem is now that I don't see how you can get it in there...

Can somebody maybe help me in that last step?

physics dirac-delta

share|cite|improve this question

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Those Heaviside's functions can be freely added because $omega_bfrm k>0$ and therefore $$theta(k_0)delta(k_0-omega_bfrm k) = theta(omega_bfrm k)delta(k_0-omega_bfrm k) = delta(k_0-omega_bfrm k)$$ $$theta(-k_0)delta(k_0+omega_bfrm k) = theta(omega_bfrm k)delta(k_0+omega_bfrm k) = delta(k_0+omega_bfrm k)$$
  $endgroup$
  – Adam Latosiński
  Mar 27 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "I'm having a hard"... one or time? :D
  $endgroup$
  – amsmath
  Mar 27 at 20:37
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm having a hard proving an identity which includes the Dirac delta function and the Heaviside function. I want to show that for $minmathbbR,kinmathbbR^4$ and $omega_bf k equiv sqrtbf k^2+m^2$ and $k^2= eta_munuk^mu k^nu$ we have
$$deltaleft(k^2-m^2right)=frac12 omega_mathbfkleft[thetaleft(k_0right) deltaleft(k_0-omega_mathbfkright)+thetaleft(-k_0right) deltaleft(k_0+omega_mathbfkright)right],$$
where $eta_munu$ is the Minkowski metric with signature $(+,-,-,-)$.

---

My attempt here was to use the fact that we have
$$delta(g(x))=sum_i fracdelta(x-x_i),$$
where the $x_i$ are the roots of $g$. By setting $g(k_0)=k^2-m^2= k_0^2-omega_bf k^2$ we find $g'(k_0)=2k_0$ and the roots $pmomega_bf k$. Inserting this into the identity for the delta function leads to
$$delta(k^2-m^2)=frac12(delta(k_0-omega_bf k) + delta(k_0+omega_bf k)),$$
which comes close to what I want to prove but is not quite there yet since the Heaviside function is still missing. The problem is now that I don't see how you can get it in there...

Can somebody maybe help me in that last step?

physics dirac-delta

share|cite|improve this question

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Those Heaviside's functions can be freely added because $omega_bfrm k>0$ and therefore $$theta(k_0)delta(k_0-omega_bfrm k) = theta(omega_bfrm k)delta(k_0-omega_bfrm k) = delta(k_0-omega_bfrm k)$$ $$theta(-k_0)delta(k_0+omega_bfrm k) = theta(omega_bfrm k)delta(k_0+omega_bfrm k) = delta(k_0+omega_bfrm k)$$
  $endgroup$
  – Adam Latosiński
  Mar 27 at 20:35
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "I'm having a hard"... one or time? :D
  $endgroup$
  – amsmath
  Mar 27 at 20:37
  

  
  
  
  

add a comment | 

$begingroup$

I'm having a hard proving an identity which includes the Dirac delta function and the Heaviside function. I want to show that for $minmathbbR,kinmathbbR^4$ and $omega_bf k equiv sqrtbf k^2+m^2$ and $k^2= eta_munuk^mu k^nu$ we have
$$deltaleft(k^2-m^2right)=frac12 omega_mathbfkleft[thetaleft(k_0right) deltaleft(k_0-omega_mathbfkright)+thetaleft(-k_0right) deltaleft(k_0+omega_mathbfkright)right],$$
where $eta_munu$ is the Minkowski metric with signature $(+,-,-,-)$.

---

My attempt here was to use the fact that we have
$$delta(g(x))=sum_i fracdelta(x-x_i),$$
where the $x_i$ are the roots of $g$. By setting $g(k_0)=k^2-m^2= k_0^2-omega_bf k^2$ we find $g'(k_0)=2k_0$ and the roots $pmomega_bf k$. Inserting this into the identity for the delta function leads to
$$delta(k^2-m^2)=frac12(delta(k_0-omega_bf k) + delta(k_0+omega_bf k)),$$
which comes close to what I want to prove but is not quite there yet since the Heaviside function is still missing. The problem is now that I don't see how you can get it in there...

Can somebody maybe help me in that last step?

physics dirac-delta

share|cite|improve this question

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

$endgroup$

share|cite|improve this question

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

share|cite|improve this question

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61

asked Mar 27 at 20:26

Marius JaegerMarius Jaeger

61