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Suppose that $f$ is a function from unit $A$ to $B$, then what is the unit of $f'(x)$?. We can do $f'(x)\Delta x$ to get an estimate of $f(x + \Delta x)$. Since the latter has unit $B$, so has the former. $\Delta x$ has unit $A$, so $f'(x)$ has unit $B/A$. So far so good, but what if $A$ is the unit of temperature? Intuitively it does not make sense to give $\Delta x$ the same unit as temperature, because a temperature difference cannot be used interchangeably with an absolute temperature. In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature. Are there extensions to the system of units that let you recognize that as an error just by looking at the units, i.e. a system of units where $\Delta x$ has a different unit than $x$? Similarly, adding two absolute temperatures is invalid. Is there a system of units that lets you only add two relative temperatures to get a new relative temperature or a relative plus an absolute tempurature to get a new absolute temperature?

Now for a more general and more vague question:

I was reading the page on extensions to dimensional analysis on Wikipedia. It describes a system of units that seems to capture the rotational invariance of physical laws. All physical laws should be rotationally invariant, and Siano's system of units makes sure that if a law is not rotationally invariant then it has a unit error (i.e. it does not matter what we define as our zero angle). Is this a correct intuition? Are there similar extensions to the unit system that capture other physical invariances? It seems like the standard units capture scale invariance (i.e. it does not matter what mass we call 1 kg as long as we do it consistently), and the system that I'm after in the first paragraph should capture translation invariance (it does not matter which point we define as zero). Of course there are other symmetries. Is there a system of units that lets you easily check from the units whether a law satisfies these symmetries?

Another limitation of our current unit system is that some mathematical laws are not expressible. For example $e^{ab}=(e^a)^b$ is not expressible if $a$ and $b$ have units. Is there a way to solve this problem?

Qmechanic
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Jules
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1 Answers1

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As you correctly stated at the beginning, the units of $f'(x)$ are easily seen from writing the derivative as $$ f'(x) = \frac{df}{dx} $$ so the units are the same as units of $f/x$. However, your concerns about the temperature can't be justified. The units of absolute temperature and the temperature difference are the same. In particular, the international system of units, SI, uses 1 kelvin for both.

One may also work with temperature scales that are not absolute, e.g. the Celsius degrees. The absolute zero doesn't correspond to 0 °C; in this sense, these scales are "nonlinear". However, a Celsius degree is still a unit of temperature as well as temperature difference. As a unit of temperature difference, 1 °C and 1 K are the very same thing. You can surely never forget or omit units such as degrees (of temperature) from physical quantities, whether they are computed as derivatives or not.

Similarly, adding two absolute temperatures is invalid.

It may be unnatural or useless in most physical situations (see Feynman's "Judging Books By Their Cover") but it is a valid procedure when it comes to the units. After all, absolute temperatures are just energies per degree of freedom so adding absolute temperatures isn't much different from adding energies which is clearly OK.

In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature.

It makes perfect sense. Thermodynamics is full of such expressions. For example consider $f(x)=S(t)$, the entropy as a function of time. Then $S'(t)\cdot T$ is a term that appears in the rate of change of some energy according to the first law of thermodynamics.

Quite generally, it is not sensible to single out temperature in these discussions. The same comments hold for distances, times, or pretty much any other physical quantities. Take time. One may consider the "current year". It's some quantity whose unit is 1 year. (Similarly, the position of something in meters.) And one may consider durations of some events whose units may also be years. It is the same unit. It is obvious that the difference $A-B$ i.e. any difference has the same units as $A$ as well as $B$. In my "current year" analogy, $Y=0$ corresponds to the birth of Jesus Christ, a random moment in the history of the Universe. That's totally analogous to $t=0$ °C, the melting point of ice. But in both cases, the time differences or temperature differences have the same units as the quantities from which the differences were calculated – such as temperature (whether they're absolute or not) or dates.

It wouldn't make any sense to have different units for quantities and their differences because dimensional analysis would cease to hold: one could no longer say, among other things, that the units of $A-B$ are the same as units of $A$ or $B$ separately.

It is very correct that one cannot calculate a sensible value of $\exp(a)$ if $a$ is dimensionful i.e. if it has some nontrivial units. Such an exponential would be adding apples and oranges, literally. Express $\exp(a)$ as the Taylor expansion, $1+a+a^2/2+a^3/6+\dots$. If $a$ fails to be dimensionless, each term has different units so it's not dimensionally correct to add them. For this reason, all exponentials (and, with a somewhat greater tolerance, logarithms) in physics are exponentials of dimensionless quantities (which have no units). The desire to avoid physically (and mathematically) meaningless quantities such as exponentials of dimensionful quantities is one of the very reasons why we use units and dimensional analysis at all. It is not a "problem"; it is a virtue and the very point of these methods.

Luboš Motl
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  • The point of a unit system is that it lets us recognize invalid laws. For example $t + E = 0$ is not a valid law where t is time and E is energy. My point is that there are other laws that are not valid, like $T_1 + T_2 = T_3$ where $T_{1,2,3}$ are absolute temperatures. It may well be that some laws do involve multiplying $S'(t)T$, but that's just because in Kelvin the zero point has special significance. In a similar way $t_1 + t_2 = t_3$ could well be a valid equation where $t_{1,2,3}$ are times relative the the birth of Jesus Christ, if that birth has special significance in that context. – Jules May 25 '12 at 08:24
  • [previous comment continued] I understand that the current system of dimensional analysis is not able to weed out such invalid equations, so my question is whether there is an extension that does capture the fact that laws have to be independent of whatever point we call zero, or whatever angle, and other symmetries. As for your argument that $a$ cannot be dimensional in $exp(a)$, I agree in the current system of units. Perhaps I should explain my motivation better. I would like to design a program that automatically checks the validity of the units in numerical simulations. – Jules May 25 '12 at 08:29
  • [continued further] In a numerical expression it sometimes makes sense to express $e^{ab}$ as $(e^a)^b$ for numerical reasons. So I do not want to disallow such expressions, even though in the conventional unit system that is illegal. Hence my question whether there is a unit system that allows $e^a$ to have a special kind of unit such that $(e^a)^b$ is a valid expression and is unitless again if the unit of $a$ is the inverse of the unit of $b$. – Jules May 25 '12 at 08:32
  • Dear James, units and dimensional analysis may identify some wrong/meaningless equations – those that are dimensionally incorrect – but they cannot identify all errors in our formulae. $E=3mc^2/4$ is dimensionally correct (OK units) but it's still wrong (BTW it is a relationship for the energy and mass of a model of the electron that was believed for a year before special relativity). ... I agree that 0 K has a every special significance but 0 °C still has some significance. It's the melting point of ice. The amount of heat you may get from water before it freezes is $T$ in °C, for example. – Luboš Motl May 25 '12 at 09:02
  • My answer wasn't just about the "current" system of units. It was about any conceivable system of units that still allows one to do dimensional analysis. A system that would assign different units to $A-B$ than the units of $A,B$ separately would be meaningless because it wouldn't allow us to say anything about the units - any dimensional analysis - after the single simplest operation, namely subtraction (or addition, for that matter). And again, there can't exist any system of units that is capable of finding any error in an equation. – Luboš Motl May 25 '12 at 09:04
  • Also, I don't believe that my answer concerning $\exp(ab)$ left any room for any meaningful extra questions. $\exp(ab)$ is always the same thing as $\exp(a)^b$. But if $ab$ is dimensionful, both expressions are meaningless and non-kosher in physics (both because both are still the same thing). If $ab$ is dimensionless but $a$ and $b$ are not, $\exp(ab)$ is well-defined but it's still silly to write it as a $b$-th power of something although it's formally the same thing. Exponents in physics simply shouldn't be dimensionful. – Luboš Motl May 25 '12 at 09:08
  • In this situation, rewriting $\exp(ab)$ as $\exp(a)^b$ would be OK whenever both expressions make sense but $\exp(a)$ itself still doesn't make sense. It's like writing $mc^2$ as $(mc^2+U)-U$ where $U$ is a voltage. Well, $U$ cancels so formally one could say that it is $mc^2$ and OK. However, the form of the expression still allows us to see that we made an illegitimate step because even "subexpressions" of expressions should have coherent units and $mc^2+U$ simply doesn't have one. The case of $\exp(a)$ is fully analogous; it doesn't have coherent units for a dimensionful $a$, either. – Luboš Motl May 25 '12 at 09:11
  • I agree that dimensional analysis cannot identify all errors, but that's exactly my question: can we extend it to identify more errors? You are viewing all of this firmly from inside the existing framework for dimensional analysis. Dimensional analysis is just a man-made formal tool with formal rules that allows us to easily identify some equations as incorrect. We can easily design different formal rules. Some of those rules are more powerful, like Siano's orientational analysis. – Jules May 25 '12 at 11:16
  • A system that assigns different units to $A-B$ than to $A$ and $B$ is definitely not meaningless. For example take units of position, $m^3$. We can identify the position of an object by some value with unit $m^3$. But to identify the position we need to define an origin. Clearly no physical law can depend on where we define the origin. So any physical law will take the form $p_1 - p_2$ where $p_{1,2}$ are positions. A simple way to catch such errors with units is the following. If $A$ is a unit, then define $\overline{A}$ as the unit corresponding to differences in $A$. – Jules May 25 '12 at 11:19
  • If $x,y$ have unit $A$ then $x-y$ has unit $\overline{A}$. Using this we can identify an equation $p_1 = p_2 - p_3$ as invalid, because the left hand side has unit $m^3$ and the right hand side has unit $\overline{m}^3$. And indeed, it is an invalid equation because its meaning depends on where you define the origin, and we know that physical laws do not depend on where you define your origin. Sure, this is just the very start of the development of a unit system that identifies translation invariance errors, hence my question about whether there already is a well understood theory about this. – Jules May 25 '12 at 11:24
  • It's meaningless to assign different units to $A-B$ than to $A,B$ because you would have to decide what is the unit of $C+D$ or $C+D/2$ as well and you would need infinitely many units that are not really units but just some nonsensical bookkeeping devices for some coefficients that may really be arbitrary in general. Also, $(A-B)+B$ must clearly have the units of $A$ because it is $A$. But if you assign "less general" (and no longer remembering the zero) units of a difference to $A-B$, you won't be able to "recall" that $C+B$ where $C=A-B$ should behave as $A$ again. – Luboš Motl May 25 '12 at 13:52
  • Lets apply the exponential homomorphism to what you wrote there: – Jules May 25 '12 at 16:01
  • It is meaningless to assign different units to $A/B$ than to $A,B$ because you would have to decide what is the unit of $C\cdot D$ or $C\cdot \sqrt{D}$ as well and you would need infinitely many units that are not really units but just some nonsensical bookkeeping devices for some powers that may really be arbitrary in general. Also, $A/B\cdot B$ must clearly have the units of $A$ because it is $A$. But if you assign "less general" (and no longer remembering the one) units of a division $A/B$, you won't be able to "recall" that $C\cdot B$ where $C = A/B$ should behave as $A$ again. – Jules May 25 '12 at 16:01
  • As you can see such a relative unit system is exactly isomorphic to the conventional unit system where you replace addition by multiplication and subtraction by division (i.e. do the transformation $exp(f(log(x)))$). You might argue that the conventional system is more useful, but then you'd be arguing that scale invariance is more useful than translation invariance. So I am still looking for an answer to the questions in the post. – Jules May 25 '12 at 16:09
  • Dear Jules, what is true for multiplication (or division) isn't true for subtraction (or addition). These are different operations. Your "argument" is equivalent to the argument that 2+2 cannot be 4 because 2+2 cannot be 5, either, and 4 and 5 are analogous numbers. The units are just some "universal factors". Take 2 meters plus 3 meters. It's $2m+3m=5m$. I used italics fonts because $m$ literally behaves as a mathematical variable. By the distributive law, $2m+3m=(2+3)m$ so the sum has the same units. But $2m/3m=2/3$ has no units - different than $m$ - because $m$ simply cancels. – Luboš Motl May 25 '12 at 16:43
  • If there is any natural reason to write a mathematical object in physics as $\exp(s)$ where $s$ is something, $s$ simply has to be dimensionless because of the dimensional analysis. That's what the dimensional analysis is all about. So if something gets scaled by $q=\exp(s)$, then both $q$ and $s$ are dimensionless numbers which means that there is nothing to adjust about their "units" and indeed, there can't exist any "system of units" that would be able to tell you anything about such $q$ or $s$. I still insist that I have answered all your questions; you just don't seem to like the truth. – Luboš Motl May 25 '12 at 16:47
  • Scalings are related to a multiplication by a dimensionless number such as $q=\exp(s)$ while translations are translations by a dimensionful quantity, the distance. So there is a huge difference between the units (and ability to use dimensional analysis) related to scaling and to translations. You seem to dislike this basic fact. But this basic fact is the very reason why we use units and dimensional analysis in the first place. In scaling, there are preferred objective scalings everyone agrees are universally important, like 1-times or $e$-times, but for translations, it's not the case. – Luboš Motl May 25 '12 at 16:49
  • For translations, you may have translations by 1 meter or 1 inch and there is no way to say that one of them is more universal or important than others. They're just translations by different lengths or different units of lengths. Which translations are viewed by someone as "more canonical" depends on his subjectively preferred units (for scaling, it's not the case: there are no units). In infinite space, the generator of translations is (just like the distance) dimensionful so there is no way how you could uniquely identify it with the generator of scalings. – Luboš Motl May 25 '12 at 16:52
  • I think it might be helpful to read what I wrote again. All the arguments that you can propose against the relative unit system also hold against the conventional unit system, but because you are too familiar with the latter you just see one side of the isomorphism. Anyway, this discussion is not going anywhere, so I'll wait until perhaps somebody else answers my question. Thanks for your time :) – Jules May 25 '12 at 16:57