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So I have been thinking about dimensional analysis and I have been thinking about quantities with components that have negative and positive exponents in the same expression.

Two examples:

seconds/second, T T-1, also known as time drift. It's the dimension of the leap day, the inaccuracy of clocks (atomic or otherwise), among other things.

meters/meter3, L L-3, also known as fuel efficiency, or how far you can go per volume of fuel.

Now I have some questions. The laws of algebra would say that it is legal to reduce T T-1 into a "dimensionless" (we'll get to why it's in quotes soon) quantity. Which makes sense to me, that change in time over time would not have a dimension, per se.

So that would also mean that L L-3 would reduce to L-2, otherwise known as inverse area or 1/meter2. That is interesting to me. I'm not quite sure how to visualize that, or even if there is a physical representation. But Wolfram Alpha says it's true. So how would I visualize that and what is it's physical representation of fuel efficiency being inverse area? My guesses are probably nowhere near the mark, so I'll refrain.

Also, are there quantities that are not just "dimensionless", but precisely of dimension zero, other than the trivial ones like pi and phi? Since I cannot say that time drift has the zeroth dimension, simply that the two parts are still there, but in that expression they in a sense "cover over" each other. Meaning they cancel each other out, for the sake of making the paper equations simpler, but are still part of the representation in explicit form.

Aistis Raulinaitis
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  • Formally, $L/L^{-3} = L^{-2}$. If you want to keep a physically sensible interpretation of the unit, better stay with $L/L^{-3}$. I'm not sure why you think every unit combination needs to have a visualization or physical interpretation. – ACuriousMind Mar 24 '15 at 01:02
  • @ACuriousMind Right, I thought as much, thanks. I guess what I was trying to think about is physical sensibility. I'm not saying that every unit combination necessarily needs either. Great, so, formally it is so. Let's ignore physical sensibility for a second and maybe see if $$L^{-2}$$ is still useful in some way. – Aistis Raulinaitis Mar 24 '15 at 01:10
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    XKCD on fuel economy : https://what-if.xkcd.com/11/ – zeldredge Mar 24 '15 at 01:10

2 Answers2

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Dimensional analysis is ultimately just a scaling argument in disguise. You can write down all the equations of physics in a dimensionless form (by using natural units where hbar = c = G = 1), never introduce any dimensional quantities and still reproduce all the results that are conventionally obtained by dimensional analysis. What you do then is rescale certain variable in the equations to study the limiting behavior of the theory when some quantities become infinitely large or small. The standard case of the classical limit is automatically implemented by conventional dimensional analysis.

But you are free to study any particular scaling limit you desire. E.g. if two different lengths $L_1$ and $L_2$ appear in a formula but $L_1$ will normally be much smaller than $L_2$, then it may be useful to study the scaling limit where $L_2$ becomes infinitely larger than $L_1$. In that case $L_1$ and $L_2$ become de-facto dimensionally incompatible, just like the fact that in conventional units time and distances have incompatible dimensions.

JBR
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Count Iblis
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    Interesting! Absolutely you are right, you can switch out the unit with natural units and there is an isomorphism between the system of natural units and the system with different units. I think your most interesting point was your point about the two lengths becoming different dimensions at a limit. Is there any literature that rigourizes that notion? – Aistis Raulinaitis Mar 24 '15 at 02:47
  • About the two lenghts, you can look at boundary layer theory. Here you have second order (partial) differential equations where the second derivatives are multiplied by a small constant (e.g. proportional to viscocity, or heat conduction coeffient). For small values of the constant, you get a boundary layer, outside this boundary layer you can set the constant equal to zero with negligible error, but then the differential equation is of first order and with that differential equation you cannot impose the boundary conditions. – Count Iblis Mar 24 '15 at 03:06
  • The equations valid inside the boundary layer are obtained by rescaling the length variable, and then you can consider the ideal case where the constant tends to zero and the rescaling is infinite. Then in that scaling limit, you have some physical variable (e.g. temperature) as a function of position in the bulk and also in the boundary layer. The variable you use to denote the position in the boundary layer is then incompatible with the one you use for the bulk in the scaling limit. – Count Iblis Mar 24 '15 at 03:11
  • In the literature you'll rarely find authors making comments about dimensions, the issue is typically considered trivial, some authors will religiously stick to some standard convention. John Cardy has written in one of his books about the application of the renormalization group to percolation theory that domensional analysis is just a trivial case of the renormalization group, which is pretty much the point I made here: you have some physical theory and you want to study how it behaves under some scaling of some parameters. In statistical mechanics and QFT, the rescaling can be non-trivial. – Count Iblis Mar 24 '15 at 03:17
  • Non-trivial because if you rescale a field in a path integral with some cut-off that regularizes the path integral then changing the length scale will also change the cut-off , so you then also need to compute the path integral and integrate out the fluctuations of the field that are smaller than the new cut-off length. This can lead to physical varableles to pick up a "non-classical" exponent under scaling. – Count Iblis Mar 24 '15 at 03:23
  • Might I suggest you incorporate these comments into the answer? – Emilio Pisanty Apr 06 '15 at 21:34
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I will attempt a different perspective as compared to Count Ilbis'.

From the point of view of dimensional analysis, any quantity would either have dimensions, e.g. your

fuel efficiency ... meters/meter3, $L L^{-3}$, ... or how far you can go per volume of fuel.

or would be dimensionless, e.g. your

time drift ... seconds/second, $T T^{-1}$. It's the dimension of the leap day, the inaccuracy of clocks

Why these dimensions matter is easy to see. Suppose I ask the question:

1) What happens to your first quantity if the world decides to introduce a new unit 'peter' to replace the standard 'meter', with the two being related as 1 peter = $0.5937$ meter?

Well there is a disproportionate change in the numerator and the denominator. Suppose your fuel efficiency read $X \ {\rm meters \ per\ }{\rm meter}^3$. With this change of units, the numerator would change by a factor of $1/0.5937$, while the denominator would change by the cube of this factor. So, in the peter system of units, the same fuel efficiency would read $(0.5937)^2X$ units.

That's the fate of dimensionful quantities under unit change - numerical magnitudes change. The fact that a quantity is dimensionful represents that such a change would occur. Now, the point is - whether or not it is sensible to measure fuel efficiency in units of inverse area, the net exponent tells you the scaling factor under a change of units. You can deduce that the numbers would change by the conversion factor squared, just by observing that the dimensions are $L^{-2}$. (This is the point Count Ilbis is trying to make with his answer.)

My next question is:

2) What happens to your second quantity if the world decides to introduce a new unit 'zecond' to replace the standard 'second', with the two being related as 1 zecond = $0.5937$ second?

Nothing happens. Your "time drift", or "leap day" will be a number, but it will be dimensionless number and won't scale. The argument is - both numerator and denominator are dimensionful, but the conversion factors cancel out because both are modified equally, and the net quantity is invariant.

There are various examples of such dimensionless numbers, or Dimensionless Physical Constants, with the fine structure constant of atomic physics being the most famous example. The number is about $1/137$, obviously irrespective of which units you use. because these numbers don't change dimensions, the explanation of why these numbers take the values they take is an important problem. An explanation of why the fundamental coupling constants for the four types of interactions in nature, follow a hierarchy of differing strengths, highest for strong and weakest for gravity, is an open, unsolved problem in Physics. The reason for these importance is - they can't be deduced from theory, they only have to be measured, and because these numbers are so important in dictating the Physics of nature, it becomes crucial to know why they take only these values and not anything else.

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