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If one does long calculations in natural units how does one find the right expression in let's say SI units in the end?

I know that natural units make the calculations easier and also help to show physical equations in a clearer way, i.e. it's nice to think of the four-momentum squared as $$p^\mu p_\mu = E^2-p^2= m^2$$ So one can think of the invariant length of a four-momentum as the mass squared. Or it also shows the equality of mass very clearly.

But I have done long calculations in natural units and now I need to calculate final results with numbers. And I've got to say I hate numbers ;). And also all the basic textbooks always work in natural units so it's hard to guess what's the right expression.

For concrete examples... Let's say the Higgs coupling is $m_i/v$ in natural units. Or the propagator for a spin-0 particle is $\frac{1}{p^2-m^2}$ So what are the correct expressions here in SI units. My guess would be that for every mass there should be a $c^2$. But for $\hbar$ it gets more difficult. I.e the total decay rate is related to the mean lifetime of a particle by $\Gamma=\frac{\hbar}{\tau}$. So here one really needs to know the dimensions of the other two quantities to guess the $\hbar$.

Are there any practical tips in day to day life of a particle physicist to guess the correct constants?

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    $\begingroup$ In my opinion, the best way to think about this is to introduce the proper variables. If $x$ is length, then define a new quantity $y \equiv x / c$ which is a time, not a length. Do the calculation in terms of $t$ and $y$, and when you're done, replace $y$ with $x/c$. This is a sane approach because: 1) there's no hand-waving nonsense about "setting constants to 1" (which is an utterly ridiculous statement in my opinion), 2) no need to follow any vague recipes like "put constants in to make the units work out", and 3) no horribly misleading need to even think about "units". $\endgroup$ – DanielSank Apr 22 at 3:48
  • $\begingroup$ @DanielSank That seems more like an answer than like a comment. $\endgroup$ – rob Apr 22 at 14:28
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One interpretation of setting $c=1$ is that we are measuring distances and times in "the same units." For instance, one way to set $c=1$ is to measure durations in years and distances in light-years. To undo such a choice, you multiply by an appropriate power of $c$.

The effect of $\hbar = c = 1$ is to use "the same units" for length, time, mass, and energy. It's pretty common (example, example) to hear people talk about "the dimension" or "the mass dimension" of an expression, which is basically how many factors of energy there are in its $\hbar=c=1$ units. For example, in order for the action $S=\int \mathrm d^4x\,\mathcal L$ to be dimensionless, the Lagrangian density $\mathcal L$ much have dimension four.

I find myself most frequently wanting to compare energies and lengths, for which it's useful to have $\hbar c = 0.197\rm\,GeV\,fm \approx \frac 15\,GeV\,fm$ in the back of my head. For example, I think a lot about the effective range parameter $r_0 = \hbar c / mc^2$ of a Yukawa potential for a force whose mediating particle has mass $m$. This parameter has dimension $-1$ and is the same as the Compton wavelength of the massive particle. There are fewer characters if I type $r_0 = \hbar / mc$, which is algebraically the same; however the cognitive load for me is lighter if I see $mc^2$ and think "energy" and then see $\hbar c$ and think "energy converts to length." The system is not overdetermined, so there is only one nontrivially correct way to convert using powers of $\hbar$ and $c$.

Your propagator $\frac1{p^2-m^2}$ has dimension $-2$, and your decay width $\Gamma = 1/\tau$ has dimension 1. A cross-section (dimensionally an area) must have mass dimension $-2$. You get the idea.

For coupling constants, it's nice if they are dimensionless, but that's not always how it works out. An example.


You ask in a comment about the branching ratio

$$\frac{G_F^2m^9\tau}{m^4\hbar} $$

which should be dimensionless, like any branching ratio. I would process this by turning every sub-expression into energies and lengths:

$$ \left(\frac{G_F}{(\hbar c)^3}\right)^2 \frac{(mc^2)^9}{(mc^2)^5} \frac{c\tau}{\hbar c} $$

The Fermi constant is usually tabulated with the factors of $\hbar c$ which put it into energy units, where it has dimension $-2$. For most particles the rest energy $mc^2$ is easier to locate than the mass $m$ in SI units. And I find it less confusing to identify ${c\tau}/{\hbar c}$ as an expression with energy dimension $-1$ than I do the equivalent expression $\tau/\hbar$ --- perhaps because $\hbar c$ has a reasonable value in energy-length units.

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  • $\begingroup$ The thing with the mass dimension appeals to me. This way can probably get a good intuition for most things. Thanks a lot for that very ood answer. $\endgroup$ – higgshunter Apr 20 at 21:00
  • $\begingroup$ Now I'm confused again, I have branching ratio that is ... $\frac{G_F^2m^9\tau}{m^4\hbar} = \frac{G_F^2m^5 \tau}{\hbar} = \frac{[eV/c^2]^5 [s]}{[eV]^5 [s]}= \frac{1}{c^{10}}$ so since the branching ratio has no dimension I need to multiply this by $c^{10}$? $G_F$ is the Fermi constant and $\tau$ the lifetime btw. But that would give me a result that is by far to big. SO what am I missing here? $\endgroup$ – higgshunter Apr 21 at 17:09
  • $\begingroup$ @higgshunter Edited. $\endgroup$ – rob Apr 21 at 22:06
  • $\begingroup$ I do not agree with this answer. Physics equations are not written in any particular system of units. For example, the equation $F = ma$ is true in a platonic sense that the constraint represented by the equation is true in no matter how we choose to quantify the three physical quantities therein. "Setting $c$ to 1" is simply nonsense and what you're really doing is redefining the meaning of the symbol $x$ to be "distance divided by the speed of light", which has dimensions of time and makes no reference to units whatsoever. $\endgroup$ – DanielSank Apr 22 at 3:51
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    $\begingroup$ @DanielSank Actually I mostly agree with you. However I have to cope with colleagues who process the world differently than I do. I've gotten reasonably good at it, and this answer describes how. Think of relativists who write $E^2 = p^2+ m^2$ as peddlers of a gateway drug. $\endgroup$ – rob Apr 22 at 4:11
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Just put in factors of $c$ and $\hbar$ in such a way that the units make sense. There is always a unique way to do this -- unique in the sense that if you have two options, they're equivalent. For example, the natural-units equation $E=m$ can be made valid in SI as $E=mc^2$ or $E/c^2=m$, but these are equivalent equations.

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  • $\begingroup$ Ok got that, but then again I already need to know where I'm going. What about the Higgs coupling i.e. It's $m/v$ but I don't know what it should be like... $\endgroup$ – higgshunter Apr 20 at 19:45
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    $\begingroup$ @higgshunter: The process doesn't generally work for expressions, only for equations. There is in general no notion of converting an expression from natural units to SI. However, for some expressions you can infer something about what has to be done, since quantities with different units can't be added or subtracted. For example, the quantity $E-m$ is certainly not possible in SI units. It could be converted to $E-mc^2$ or $E/c^2-m$. $\endgroup$ – Ben Crowell Apr 20 at 20:06
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There is a very simple and systematic, if not perhaps the least time-consuming and work-consuming, method to accomplish this.

When you change from one system of units to another, you are, in effect, multiplying all the converted quantities by proportionality constants. Thus, the mathematical answer is very simple: just first affix to every figure a proportionality constant. The value of this constant is the value, in the new unit system, of the size of the units used in the existing unit system. In the case you are mentioning, the "new" system is SI units, the "old" one is Planck units.

For example, consider Newton's law of gravitation in Planck units: it looks like

$$F_G = \frac{m_1 m_2}{r^2}$$

First, you need to identify which quantities have units, i.e. which are dimensional. This is pretty obvious: every variable here is such. Hence you will introduce one proportionality constant for each, representing the new units to measure that quantity in:

$$(k_{F_G} F_G) = \frac{(k_{m_1} m_1) (k_{m_2} m_2)}{(k_r r)^2}$$

Note that we included constants for each term individually - this is because you can, if you like, measure the two masses in different units (which might be useful, e.g. if you're talking a planet and artificial satellite you may measure the planet's mass in something like yottagrams or Earth masses and the satellite in kilograms or megagrams ("tonnes")). Now you just substitute. If we're going to SI base units, then you have the following: assign

$$k_{m_1} := k_{m_2} := m_\mathrm{P} = \sqrt{\frac{\hbar c}{G}}$$ $$k_r := l_\mathrm{P} = \sqrt{\frac{\hbar G}{c^3}}$$ $$k_{F_G} := F_\mathrm{P} = \frac{c^4}{G}$$

where the last one is Planck force. Substituting all these in gives

$$\frac{c^4}{G} F_G = \frac{\left(\sqrt{\frac{\hbar c}{G}} m_1\right) \left(\sqrt{\frac{\hbar c}{G}} m_2\right)}{\left(\sqrt{\frac{\hbar G}{c^3}} r\right)^2}$$

from which we can clear the various roots to get

$$\frac{c^4}{G} F_G = \frac{\left(\frac{\hbar c}{G}\right) m_1 m_2}{\left(\frac{\hbar G}{c^3}\right) r^2}$$

and now we pull that constanty mess out

$$\frac{c^4}{G} F_G = \left(\frac{\frac{\hbar c}{G}}{\frac{\hbar G}{c^3}}\right) \frac{m_1 m_2}{r^2}$$

take the one from the left

$$F_G = \left[\frac{G}{c^4} \left(\frac{\frac{\hbar c}{G}}{\frac{\hbar G}{c^3}}\right)\right] \frac{m_1 m_2}{r^2}$$

and after a ton of cancellations you are left with

$$F_G = G \frac{m_1 m_2}{r^2}$$

which is the familiar form.

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