delta dirac propieties

[alejandrito20]

hello

 

in need to find T in the follow equation:

[math]\delta(\phi-\pi)+k=T\delta(\phi-\pi)[/math]

 

where [math]\phi[/math] is a angular coordinate between([math]-\pi,\pi[/math])

 

¿is correct do:

[math]\int^{\pi-\epsilon}_{-\pi+\epsilon}\delta(\phi-\pi)d\phi+\int^{\pi-\epsilon}_{-\pi+\epsilon} k=T\int^{\pi-\epsilon}_{-\pi+\epsilon}\delta(\phi-\pi)[/math]???????????

 

 

¿ [math]\delta(\phi-\pi)[/math] with [math]\phi=-\pi+\epsilon[/math] is infinity??????

[Bob_for_short]

The original equation makes sense only if delta is a constant which is not the case.

 

For any other function you have a contradiction: (T-1)delta = k. A constant cannot be variable so the original equation has the only solution: T=1 if k = 0.

[alejandrito20]

The original equation makes sense only if delta is a constant which is not the case.

 

For any other function you have a contradiction: (T-1)delta = k. A constant cannot be variable so the original equation has the only solution: T=1 if k = 0.

 

but in the delta dirac potential:

 

[math]\frac{-h^2}{2m}f''+\delta(x)f=Ef[/math] (eq1)

 

with [math]f=Ae^{ikx}[/math]

 

then

[math]-\frac{k^2h^2}{2m}+E=\delta(x)[/math]

 

but [math]\delta(x)[/math] is infinity in zero. This problem would not have sence too.

 

If [math]f=Ae^{ik|x|}[/math]

 

then eq 1 is

[math]-\frac{k^2h^2}{m}\delta(x)+\delta(x)=E[/math]

this problem would have sense only if E= 0, but E IS NOT ZERO.

[Bob_for_short]

When delta-potential is different from zero, it is f'' of f itself that compensate the potential term in the equation. So the plane wave cannot be a wave equation solution everywhere. "Inside" potential barrier the wave function is specific, zero, for example.

[alejandrito20]

When delta-potential is different from zero, it is f'' of f itself that compensate the potential term in the equation. So the plane wave cannot be a wave equation solution everywhere. "Inside" potential barrier the wave function is specific, zero, for example.

 

specifically, my problem is (u,v) component of Einstein equation:

 

[math]kf''(x)+k1=\frac{k2}{f(x)^2}(T\delta(x)+T1\delta(x-\pi)[/math]

with k,k1,k2 constant

with x between ([math]-\pi,\pi[/math])

 

i need to find, T and T1.

 

In (5,5) component of einstein equation (this equation don't have deltas of dirac) the solution is

[math]f(x)=C(|x|+C1)^2[/math]

with C and C1 konstant

[math]f''=2C\frac{d|x|}{dx}+2C1\frac{d^2|x|}{dx^2}[/math]

with [math]\frac{d^2|x|}{dx^2}=2(\delta(x)-\delta(x-\pi)[/math]

Edited February 23, 2010 by alejandrito20

[Amr Morsi]

For the former equation you wrote, integrate both sides and make the boundary conditions -Pi and phi. You will get 1+k*phi=T, i.e. phi=(T-1)/k.

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Sorry for this error. Make the boudary conditions phi and pi+epsilon. And, you will get 1+k*(pi - phi)=T.