alejandrito20 Posted February 22, 2010 Share Posted February 22, 2010 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?????? Link to comment Share on other sites More sharing options...
Bob_for_short Posted February 22, 2010 Share Posted February 22, 2010 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. Link to comment Share on other sites More sharing options...
alejandrito20 Posted February 22, 2010 Author Share Posted February 22, 2010 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. Link to comment Share on other sites More sharing options...
Bob_for_short Posted February 23, 2010 Share Posted February 23, 2010 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. Link to comment Share on other sites More sharing options...
alejandrito20 Posted February 23, 2010 Author Share Posted February 23, 2010 (edited) 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 Link to comment Share on other sites More sharing options...
Amr Morsi Posted March 25, 2010 Share Posted March 25, 2010 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. Merged post follows: Consecutive posts mergedSorry for this error. Make the boudary conditions phi and pi+epsilon. And, you will get 1+k*(pi - phi)=T. Link to comment Share on other sites More sharing options...
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