In the mathematical derivation of equations for physics, and involving wave propagation in particular, the propagation speed at the start of the derivation is often set to one (c = 1).

I am working with long derivations where the resulting final equations assume c = 1 and would like to convert them to have propagation speeds that are represented by a variable, say v, and do this without going through the complete derivation from the start to the finish. It is not at all obvious where the c's would reappear if they were not set to 1.

My question is: Are there any methods or 'tricks' I can use for this?

Maybe dimensional analysis would help. Note that after setting c = 1 the equations have dimensional analysis problems--there is a missing [L]/[T].

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    $\begingroup$ I think you just have to add any number of $c$s, multiplying or dividing, until your quantity has the desired dimensions. $\endgroup$ – MBolin Jul 23 '19 at 20:37
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    $\begingroup$ This is precisely why I don't like the c=1 convention that strikes me as sloppy and promotes sloppy thinking. $\endgroup$ – ohwilleke Jul 23 '19 at 22:12
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    $\begingroup$ @ohwilleke Nondimensionalization may well be the exact opposite; it rides, indeed, flies on precision. Both HEP and GR rely on it. $\endgroup$ – Cosmas Zachos Jul 24 '19 at 13:58
  • $\begingroup$ @ohwilleke: Taking $c\ne1$ just means taking a silly, inconvenient set of units. It would be like using different units for work and heat, which people used to do before they understood thermodynamics. $\endgroup$ – user4552 Jul 24 '19 at 16:17
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    $\begingroup$ @BenCrowell The setting a numerical value for c=1 isn't the thing I take issue with (that no different than expressing it in parsecs per years or miles per second), it is the omission of units from terms in an equation in a manner that isn't visibly noted that I take issue with. One of the most economical way to display the units in a term is to display the physical constant with a letter. Another is to display the units involved in a term adjacent to it to be clear about the units of that term. Either way you are visibly dispaying everything that is there. The letter is IMHO the prettier. $\endgroup$ – ohwilleke Jul 24 '19 at 18:05

Redimensionalization, as you were probably taught when introduced to nondimensionalization. You must know the dimensions of everything involved, however, otherwise you cannot do much.

If your quantity F had dimensions $[M]^\alpha [L]^\beta [T]^\gamma$, but the partially nondimensionalized expression $F'$ is $[M]^\alpha [L]^\delta$, then you can be sure that, uniquely, $$ F=F' /c^\gamma $$ and $\beta=\delta-\gamma$.

For example, the nondimensionalized $$ E=\sqrt{m^2 +p^2} \mapsto E=\sqrt{(mc^2)^2 +(pc)^2} ~, $$ since each of the squared quantities under the square root must be redimensionalized to energy, $[M][L]^2[T]^{-2}$; so you owe $c^2$ to m and c to p.

In high energy physics, masses are measured in nondimensionalized units of energy, GeV, so they redimensionalize to units of $GeV/c^2$.In fact, instead of [L], [M], [T] it pays to switch to [action], [speed], [energy], so you can set both c and ℏ to 1, ie, measure speed and action in these units. Retaining superfluous clutter, instead, would challenge one's sanity.


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