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How do I get from $$ \frac {dt} {T} = \frac {dx} {gt} \sqrt { \frac {g} {2h} } $$ to $$ \frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx $$ where $x(t) = \frac {1} {2} gt^2 $ and $ T = \sqrt { \frac {2h} {g}}$.

I'm currently struggling with Griffiths' Introduction to Quantum Mechanics. This is from a worked example on p.11-12 of the 2nd edition.

StubbornAtom
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Philip Witt
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  • this notation is hardly close to a formal mathematical notation. I think it will be better to migrate this question to the stackexchange physics portal, as they will understand the kind of notation used there. By example, rigorously speaking, the notation $\frac{dt}T$ seems some kind of differential form – Masacroso Jan 02 '21 at 00:17
  • $\frac {1} {2 \sqrt {hx}} $ is a probability density derived from $ x(t) = \frac {1} {2} gt^2$. – Philip Witt Jan 02 '21 at 00:22

2 Answers2

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It’s just algebra. I figured out the steps to get the first expression for $dt\over T$ from the second, but you can reverse the steps. Using what you are given for $x(t)$ and $T$, and assuming $t\ge0$,

$$\frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx = \frac {1} {{2 \sqrt {h\left({1\over2}gt^2\right)}} } dx = \frac {1} {t\sqrt {2hg} } dx=\frac {1} {gt\sqrt {2h\over g} } dx=\frac {\sqrt {g\over 2h}dx} {gt }.$$

Steve Kass
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We have $$ x=\frac{1}{2}gt^2 \quad\Leftrightarrow\quad gt^2=2x $$ so \begin{align*} \frac{dt}{T} & =\frac{dx}{gt}\sqrt{\frac{g}{2h}} =\sqrt{\frac{g}{g^2t^2\cdot2h}}\,dx =\sqrt{\frac{1}{gt^2\cdot2h}}\,dx \\& =\sqrt{\frac{1}{2x\cdot2h}}\,dx =\sqrt{\frac{1}{4hx}}\,dx =\frac{1}{\sqrt{4hx}}\,dx =\frac{1}{2\sqrt{hx}}\,dx. \end{align*}

mf67
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