Why do physics professionals often use various different systems of units instead of SI units. Especially I ask about when constants like $c$ or $\hbar$ are put to 1....what is the advantage of this?
$\begingroup$
$\endgroup$
3
-
2$\begingroup$ Possible duplicate of How does natural unit make sense? $\endgroup$– jacob1729Commented Nov 25, 2019 at 16:16
-
1$\begingroup$ To make life easy such that you don't have carry these constants in calculations and eventually missing them. $\endgroup$– MSTCommented Nov 25, 2019 at 16:22
-
$\begingroup$ Stripping these meaningless unit-conversion constants from physics equations makes the meaning of the equations more obvious. For example, $m^2=E^2-\mathbf{p}^2$ is a statement about mass being invariant because it is the length of the energy-momentum 4-vector in Minkowski spacetime. The $c$’s in $(mc^2)^2=E^2-(\mathbf{p}c)^2$ just muddle that meaning. $\endgroup$– G. SmithCommented Nov 25, 2019 at 16:51
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Everything!!! Natural units just make your life simpler, is like removing every unnecessary conversion factors which need to be added, helps by making the equation look smaller, and can easily give back the final equation just by dimensional analysis.
Okay you might be wondering what dimensional analysis has got to do with natural units. Well let's see an example. The Lorentz transformation factor is given as $$\gamma =\frac{1}{\sqrt{1-v^2}}$$ Now obviously, this equation needs to be converted back to some units that people are familiar with to make sense. So you know that the speed of light in natural units is 1, so you can multiply it anywhere, with any exponent. So if you analyse the square root, it must be unitless, but 1-v² is definitely not unitless soyou make the v² as v²/c², and behold, you get the Lorentz transformation factor as
$$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$
-
$\begingroup$ Can you expand on dimensional analysis? $\endgroup$– user248141Commented Nov 25, 2019 at 16:57