- #1
Prologue
- 185
- 1
I am reading both Griffiths and Gasiorowicz and I can't get either of them to tell me why the continuity of the derivative of the natural log of the amplitude
[tex]\frac{d(ln(u(x)))}{dx}=\frac{1}{u(x)}\frac{du(x)}{dx}[/tex]
or put a different way
[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]
at a boundary a implies that
[tex]u(a^{-})=u(a^{+})[/tex]
and
[tex]\frac{du(a^{-})}{dx}=\frac{du(a^{+})}{dx}[/tex]Why is this true?
It looks to me that there still is some ambiguity (as in the top equation is necessary but not sufficient) because
[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]
can be reconfigured to
[tex]\frac{u(a^{+})}{u(a^{-})}=\frac{\frac{du(a^{+})}{dx}}{\frac{du(a^{-})}{dx}}[/tex]
Which just means that the ratios have to be equal. But say for instance that [tex]u(a^{+})[/tex] is twice as big as [tex]u(a^{-})[/tex] so that the ratio is 2. Then all that means is the ratio of the slopes at a has to be 2. So, I don't see where they are limited to be identical.
[tex]\frac{d(ln(u(x)))}{dx}=\frac{1}{u(x)}\frac{du(x)}{dx}[/tex]
or put a different way
[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]
at a boundary a implies that
[tex]u(a^{-})=u(a^{+})[/tex]
and
[tex]\frac{du(a^{-})}{dx}=\frac{du(a^{+})}{dx}[/tex]Why is this true?
It looks to me that there still is some ambiguity (as in the top equation is necessary but not sufficient) because
[tex]\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}[/tex]
can be reconfigured to
[tex]\frac{u(a^{+})}{u(a^{-})}=\frac{\frac{du(a^{+})}{dx}}{\frac{du(a^{-})}{dx}}[/tex]
Which just means that the ratios have to be equal. But say for instance that [tex]u(a^{+})[/tex] is twice as big as [tex]u(a^{-})[/tex] so that the ratio is 2. Then all that means is the ratio of the slopes at a has to be 2. So, I don't see where they are limited to be identical.
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