0
$\begingroup$

I'm reading a quantum mechanics book, and it has the following equation:

$$ \Delta x \approx \frac{\lambda}{\sin\alpha} \sim \frac{h}{mc\sin\alpha} $$

What is the difference between $\approx$ and $\sim$?

$\endgroup$
  • $\begingroup$ What's the context? It looks like they're using de Broglie's relation $\lambda=h/p$ and then taking $p=mc$ (which could make sense under the right conditions). That kind of detail is important when interpreting notation in a physics text v. a math text. $\endgroup$ – Semiclassical Aug 29 '14 at 0:17
  • $\begingroup$ OK. After looking at it for awhile I know what they are doing... $\Delta x \approx \frac{\lambda}{\sin \alpha}$ is referring to the positional uncertainty of a particle measured by a microscope with angular aperture $2\alpha$. However, due to the Crompton effect, the wavelength will increase proportional to $\frac{h}{mc}$ when a photon hits a particle. $\endgroup$ – daviewales Aug 29 '14 at 2:57
5
$\begingroup$

$\approx$ is used as the mathematical "approximately" symbol $-$ this means $ \Delta x $ has approximately the same value as $ \frac {\lambda}{\sin \alpha} $. However, $\sim$ is used a proportionality symbol, so there is some factor tacked onto $\frac{h}{mc\sin \alpha}$, which may be $1000$ or $2$ or something else.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.