8
$\begingroup$

Theorists frequently use convenient units like $\hbar=1$ or $m=2$ or whatever is useful to simplify the notation in the problem. And after all the calculations are done the units are recovered based on what the unit of the answer needs to be.

I can definitely see why those units are convenient, but I really don't feel comfortable with the recovering step. So far I have only seen one example and it is not quite enough. If somebody could provide some examples of how to recover the units that would be great. As for as I understand that should only need to involve the initial assumptions and the final answer.

Or if somebody knows a good explanatory text that would also be very much appreciated.

Are there any caveats/limitations while using convenient units?

EDIT:

To further clarify why I am confused about this entire procedure:

  • Let's assume that I have a problem involving a trap length $L$ and a wave length $\lambda_0$. For convenience I set $L=1$ and $\lambda_0=1$. My final answer needs to be in dimensions of meters. How do I know whether my final answer is supposed to be multiplied by $L$ or divided by $\lambda_0$?
  • Lets say I need my final answer in units of angular momentum times capacitance per volume (just as purely hypothetical example). And I started of by setting constants like $\epsilon_0=1$,$L=1$(some length scale),$p0=1$(some momemtum scale). This problem is easy enough that I can still figure it out. But what if I had to deal with constants like the bohr magneton or the conductance quanta. It could become very hard to figure out how I need to combine certain constants to produce the right units. Especially once the number of constants increases. Is there some kind of procedure one can follow that will always spit out the right combination of constants?
$\endgroup$
11
$\begingroup$

This is a good question - as in the example with $L,\lambda$ you provide, not every rescaling and not every set of constants is valid.

The recipe for the set of good natural units is the following: take all the units that appear in your theory and create a space with one dimension for every one of them. Say we have a theory with time, length and energy - then we have a three-dimensional space. Then you can classify a quantity of units $[ \rm length^2 \cdot time]$ as a vector in this space $(2,1,0)$, you take the power as the length of the vector in the respective direction. Now consider this statement: every set of constants defining a set of dimensionless units must form a basis of this space. In this case we would probably pick $[c]=[\rm m\cdot s^{-1}] \to (1,-1,0)$, $[\hbar]=[\rm J\cdot s] \to (0,1,1)$ and a specific unit of time, it could be even a second $[1 \rm s]=[ \rm s] \to (1,0,0)$.

Why linearly independent? Because then when we get a dimensionless result which should be say $\rm[m]\to (0,1,0)$, there is a unique way to compose this vector out of the basis - this is a very basic property of a basis. But instead of adding and subtracting the vectors we multiply and divide. E.g. to get $(0,1,0)$ we have to subtract $(1,-1,0)$ and add $(1,0,0)$, so your result would be multiplied by $1 \,{\rm s}/c$.


Maybe this is too abstract, but consider this - you certainly know that two constants of dimensions length are not linearly independent in this sense. So you cannot use them. Etc. You just have to be careful not to introduce a degeneracy of this type but otherwise you are fine. This is because surprisingly enough, the relevant physical constants do not seem to have this kind of degeneracy! This is an interesting fact - it seems that there is only one fundamental layer of interaction, there seem to be no real fundamental "second scales" to physics.

$\endgroup$
  • $\begingroup$ What to do for cases with e.g. two or more timescales? What if I have to deal with formulas for e.g. magnetic resonance, where there is $T_2 >> T_1$, and both show up in the final result, but there is not much common ground between the physics that causes $T_1$ vs. that of $T_2$? $\endgroup$ – CuriousOne Sep 12 '14 at 22:59
  • 2
    $\begingroup$ @CuriousOne you can express your results in terms of the ratio $T_2/T_1$ in such cases. But this ratio is not something you can set to 1 the way you would set a unit or constant to 1. (Which is clear if you keep in mind that "setting a constant to 1" really means choosing a system of units such that the given constant has a numerical value of 1 and then leaving the units implicit.) $\endgroup$ – David Z Sep 12 '14 at 23:12
  • $\begingroup$ @DavidZ: You are right... that was a poorly chosen example. Maybe I am more thinking along the lines or an example where a constant is fundamental vs. the result of an internal dynamic. I'll try to sort this out for myself before embarking on another poor example. $\endgroup$ – CuriousOne Sep 13 '14 at 0:06
  • $\begingroup$ awesome, answered my question to the point, thanks a lot! $\endgroup$ – ftiaronsem Sep 15 '14 at 17:30
-2
$\begingroup$

"Noah!", said the Lord, "Build me an ark 300 cubits long (137.16 m, 450 ft), 50 cubits wide (22.86 m, 75 ft), and 30 cubits high (13.716 m, 45 ft). How many cubiccubits of volume does your ark measure? How many squarecubits of timber does your ark need on the outside? Do your math wisely before you order wood from the local lumbar yard and make sure you do not overpay, the owner is a goniff and will steal your shekel by charging you for feet!"

I suppose this is along the line of the question, right? The same ratio between feet and cubits will appear three times, once to the power of one, two and three. So how do we make sure, that we don't lose track of that? By keeping the unit length in the equation. 300 cubits x 50 cubits x 30 cubits is 450,000 cubitcubits. The answer 450,000 is not correct. Similarly, if we calculate with $\bar{h}$ and c, even if $\bar{h}=c=1$, the answer should be given as $1\bar{h}c$ rather than 1! In this case we can insert whatever other unit conversions we need by using the values of $\bar{h}$ and c into the answer and arrive at the correct numerical value in those new units.

$\endgroup$

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.