90
$\begingroup$

I mean, why is $F=ma$? Why not $m^{0.123}$, $a^{1.43}$ or some random non-integers or irrational?

I hope you understand that my question isn't limited just to force, energy, velocity, etc.; it also extends to the area of a square, circle, etc. and all other formulas.

I think the whole thing starts with direct proportionality.

Most of them tell about the area of a circle, $A = πr^2$, where π is 3.14159..... an irrational number! It's not about the constant. I am talking about the power of a physical quantity.

I know why it has pi. It's because we chose the constant for area square to be one. If we chose it to be 1 for the circle then the area of a square will have a constant, 1/pi.

I've edited the question to 'rational exponents' since all are giving me examples of decimal non-integers.

$\endgroup$
  • 1
  • 1
    $\begingroup$ Possible duplicate (at least, I started to write the same answer): physics.stackexchange.com/q/112959/44126 $\endgroup$ – rob Feb 2 '15 at 8:18
  • 1
    $\begingroup$ @rob I wouldn't call it a duplicate since the other question is specifically about prefactors (although your same answer applies well here too!) $\endgroup$ – Kevin Lyons Feb 2 '15 at 9:36
  • 17
    $\begingroup$ Do not change the meaning of the question after it's been asked and answered. Make a new question. (For context, this used to say "integer exponents", but now it says "rational exponents") $\endgroup$ – immibis Feb 4 '15 at 21:52
  • $\begingroup$ There were a bunch of good comments here, but those sorts of discussions should be held in chat, so I've moved most of the comments to an associated chat room. I'm leaving comments regarding whether this question is a duplicate and the appropriateness of any edits; for anything else, please take it to the chat room. $\endgroup$ – David Z Feb 5 '15 at 20:23

18 Answers 18

71
$\begingroup$

There is a non-subjective and quite mathematical approach to this question.

First, we have the simple linear proportionalities that aren't really physical laws but just definitions of physical quantities. Why are different sensible measurable quantities usually in linear or power-law proportions will be further clarified later. An example is $F=ma$ (just defines what force is - it's convenient to define it like that) and all unit-conversion formulas (essentially, there is only one unit - time and space can be made equal by $x=ct$, energy and momentum as well, then you have $E=\hbar \omega$ from quantum mechanics and so on).

Linear relationship is not just a mathematical thing. Linearity means the principle of superposition holds: that a sum of causes creates a sum of effects. It's almost universal that when the effect is small, perturbation theory is valid and you have the first correction as a linear term. Imagine a Taylor expansion: it's a power series, no fractional exponents. This also means that a lot of the linear relations like that are approximation for weak perturbation. There's Ohm's law, heat conduction, hooke's law and so on. Even if you expand it further, it's still a power law. However, this may just be an approximation of some general result with some nonlinear function (could be exponential or something worse). But some of these relations are exact: in electrodynamics/vacuum optics, superposition principle is fundamental. But this brings us to the next point:

Natural laws are local (ok, they can be expressed variationally, but that's another discussion). Local means that relationships between quantities obey differential equations. And differential equations are linear and when operated on power laws, they just shift the exponent by one. They are also usually linear (superposition), because nonlinearity most likely has a physical interpretation of a system acting on itself by changing its environment. Linearity in differential equations does not necessarily yield power laws: all exponential and oscillatory phenomena are results of linear differential equations. Here, nonlinearity means something different: dependence of the phenomena on the amplitude. A linear differential law means that twice the cause has twice the effect. Nonlinear means that twice the cause can have a completely unrecognizable effect. For instance, a pendulum at small amplitudes has a constant frequency. But when amplitudes are too big, the nonlinearity kicks in and you can have quite interesting behaviour.

Fundamental laws are usually linear (Maxwell equations, for instance), and while there is an inherent question why the universe is so beautiful and elegant, the fact is, that if there are conserved quantities in a system, the relationship between them will be something simple.

With differential equations, we again see not only laws but also plain definitions... velocity as derivative of position, acceleration as derivative of velocity, that's all just our decision what to measure. There's also $dE=F\,dx$ to get the work (energy contribution) caused by the force, which leads to all quadratic energy laws (of course: if forces are linear, at least in approximation, then integration brings you to quadratic).

One very interesting point is, that fundamental laws of nature don't involve time derivatives higher than two (acceleration). That's somewhat related to the conservation of energy (Lagrangian functional) and tells you "how far a phenomenon can see" -- how much about history influences the now. But even with higher derivatives, we would still just the exponents by 1.

So all in all, you can't really define a sensible differential law that would give you constant, but noninteger exponents. You may get rational exponents if you express quantities which have different powers to each other (from $a^3=b^2$ you will get $a=\sqrt[3]{b^2}$), but that's just an algebraic development.

You do see weird exponents in empirical relationship: if there is no theoretical physical law behind it, but you measured some dependency and made up a function to draw a curve through the measurements, then a power function is something simple enough for people to try if it works. This is again an approximation and probably hides some more general theoretical result that is not a weird power law but a transcendental function or something that's just too complicated to write out algebraically. This is very common in material science: dependence of heat capacity, conductivity,... on temperature or current, are very strange functions. Transmission spectra are even worse. When things get complicated, a bunch of linear and nonlinear processes together yield a complex behaviour that's best just measured or at least simulated on a computer. However, the base laws and defining formulas of our chosen fundamental quantities, are mostly linear, or at least, something manageable. Superposition and proportionality are the most natural phenomena and even outside physics (economics, general statistics), this is just how things are.

$\endgroup$
  • 13
    $\begingroup$ Excellent answer. I'm not sure if this covers the feature that I was going to comment on -- that many of these laws describe spatial relationships, and space comes in countable dimensions. So this explains the relationship between one-dimensional and two-dimensional measures of regular shapes (circles, spheres, squares, cubes). I suspect that it also accounts for how the intensity of forces decreases with distance. $\endgroup$ – adam.r Feb 3 '15 at 15:55
  • 5
    $\begingroup$ Good point, geometry by itself also comes all in integer powers (rational, with small-powered roots if you play with norms). And indeed the inverse square law and its relatives come directly from that. $\endgroup$ – orion Feb 3 '15 at 17:36
  • 4
    $\begingroup$ This answer hits the nail on the head at the beginning: the reason is that the relationships in question are all linear. Why are they linear relationships? For the geometrical measurements, the answer is proportional scale. For the physical measurements, the answer is (usually) conservation of energy. $\endgroup$ – dotancohen Feb 5 '15 at 13:05
  • $\begingroup$ "An example is F=ma (just defines what force is - it's convenient to define it like that)" There is no universal agreement on this point of view, there are others. For example, in one view force is not defined by $a$ but by spring deformation or weight measurements. Also, in special relativity this equation is not used any more, not because the definition of force was changed, but because Newtonian mechanics is approximate to special relativistic mechanics. $\endgroup$ – Ján Lalinský Feb 7 '15 at 10:44
  • $\begingroup$ Sure, the more proper definition of force would be the derivative of a particular conserved value -- the linear momentum. But that doesn't change the fact it's linear and that it's not even strictly needed as a physical quantity. Physics works quite well without ever talking about forces specifically. $\endgroup$ – orion Feb 7 '15 at 11:23
36
$\begingroup$

I think a very easy-to-understand answer is that we humans rig the game to make things easier for us. For example, we choose to express the volume of a sphere as a function of its radius because a sphere's radius is EASY for us to measure. We are lazy creatures:

$$V_{sphere}=\frac{4}{3}\pi{R^3}$$

Now suppose over the course of human history, we instead decided to express a sphere's volume as a function of its surface area;

$$A_{sphere}={4}\pi{R^2}$$ $${R^2}=\frac{A_{sphere}}{{4}{\pi}}$$ $${R^3}=\frac{A_{sphere}^{3/2}}{{8}{\pi}^{3/2}}$$

Then we'd have;

$$V_{sphere}=\frac{1}{6}\pi^{-1/2}{A_{sphere}^{3/2}}$$

The exponent is now uglier because we chose a different property to routinely define our sphere. But nobody is going to want to physically measure its surface area in order to determine its volume. That would be silly.

$\endgroup$
  • 2
    $\begingroup$ Another example would be Kepler's third law which we might write $T^2 = K r^3$. We could use fractions, $r = K' T^{2/3}$, but that looks uglier. $\endgroup$ – Jeppe Stig Nielsen Feb 3 '15 at 0:34
  • 3
    $\begingroup$ A sphere's radius is not easy to measure. If you are given a solid steel ball or a marble, you would be unable to measure the radius directly. A more relevant reason for describing spheres in terms of the radius comes from the standard equation of a sphere: $x^2 + y^2 + z^2 = R^2$. Whether or not the radius of a sphere is easily accessible to measurement (often it is not), in mathematical formulas it is a convenient way to distinguish one sphere from another (with the same center). $\endgroup$ – KCd Feb 3 '15 at 4:11
  • 33
    $\begingroup$ @KCd: My calipers beg to differ. The diameter is trivial to measure, and radius has a natural and simple relationship to diameter, of course. $\endgroup$ – R.. Feb 3 '15 at 5:01
  • 2
    $\begingroup$ @R.. The diameter, yes, but not the radius (directly, as I wrote). Even though a diameter can be measured, nevertheless we still most often give mathematical formulas related to a sphere in terms of its radius rather than its diameter. $\endgroup$ – KCd Feb 3 '15 at 5:56
  • 1
    $\begingroup$ For a real world example of this, many view $2\pi$ as a bletcherous fossil from the past. $\endgroup$ – imallett Feb 5 '15 at 6:15
25
$\begingroup$

I'll bring in a less technical answer. Actually, even more dominant than integer exponents, is exponent 1, i.e., linear dependence between causes and effects. And often, exponent 2 comes from integrating this effect (i.e. kinetic energy), or multiplying two linear effects in which the same cause appears.

Why linearity? Well, this is as easy as 1 and 1 is 2, I'd say! This is simply the quite general principle that in the absence of a very specific interaction, doubling the intensity of a cause that has some effect will double the effect — e.g., twice as much inertia for twice as much mass.

Edit: the excellent and detailed answer by Orion is expanding on this.

$\endgroup$
  • 11
    $\begingroup$ I don't think your argument for linearity makes much sense. $\endgroup$ – Nit Feb 2 '15 at 9:10
  • $\begingroup$ @Nit: Could you develop? $\endgroup$ – Joce Feb 2 '15 at 9:47
  • 1
    $\begingroup$ You could expand the concept - a natural/obvious/default relation is a flat one, where something is constant. Say, the amount of energy required to remove a single atom out of a certain crystallic structure in process of melting it. Integrating this constant effect it over a variable results in an exponent of 1 - i.e. energy for melting as proportional to the number of atoms. For other relations you get to the second power, etc. $\endgroup$ – Peteris Feb 2 '15 at 9:55
  • $\begingroup$ @Peteris: Of course one can deepen the answer, but I think the answer matches the question, which is quite general and non-technical. You can ask another question for more detailed answer. $\endgroup$ – Joce Feb 3 '15 at 10:07
  • 2
    $\begingroup$ why is $1/r^2$ in the gravity formula instead of $1/r^{2.0000001}$ $\endgroup$ – chaohuang Feb 4 '15 at 2:09
17
$\begingroup$

A simpler answer would be: units. For example, in $F = ma$ the unit of measurement for force would match the units of measurement for mass and acceleration. By definition, a Newton of force is a Kilogram of mass accelerated by a Meter per Square Second. There is no need to clutter the equation by defining a different unit to measure force that requires a constant multiplier.

In the case of pi and other natural constants, these are not "chosen" numbers. These numbers occur in nature and have real significance in mathematics, physics, and other sciences.

$\endgroup$
  • 2
    $\begingroup$ +1 it seems obvious to me that dimensional consistency enforces integer (or simple ratio) exponents whenever the quantity being exponentiated has dimensions. $\endgroup$ – John Rennie Feb 3 '15 at 11:43
  • 2
    $\begingroup$ I don't see how the definition of a Newton is relevant. $F=ma$ and Newton's third law say that if two objects interact with (and only with) each other, the ratio of the accelerations is the ratio of the masses raised to the power of $n$. I think the question here is why this power $n$ is an integer. $\endgroup$ – JiK Feb 3 '15 at 15:14
  • 1
    $\begingroup$ @JiK but that takes $F = ma$ for granted; the whole point here is that $F = ma$ exactly only because of how the units are defined. $\endgroup$ – shadowtalker Feb 4 '15 at 17:24
  • $\begingroup$ A good illustration for this point would be to work out $F=ma$ for some non-metric units. $\endgroup$ – Phil Frost Feb 5 '15 at 11:11
  • $\begingroup$ @ssdecontrol No need to take $F=ma$ for granted. Experiments can show that if two objects interact with (and only with) each other, the ratio of the accelerations is the ratio of the masses raised to the power of $n$, where $n$ is an integer. The fact that $n$ is an integer seems to be a result of natural laws (not units) which is what this question is about. (Here I assume that mass and acceleration are too natural things to depend on the definition of units, because the margin is too small for a proof.) $\endgroup$ – JiK Feb 5 '15 at 12:28
11
$\begingroup$

You ask a very interesting question. The other answers here point at some fine examples and reasonable explanations. However, I think they only touch on the largest cause for your observation: the human factor.

Assumption

  • Most formulas in physics have integer exponents.

No other answer really challenges this hypothesis. To have an irrefutable proof of that hypothesis, we would need to fully enumerate all formulas in physics and divide them into those with integer exponents and those with non-integer exponents. We, as a collective seeker and store of knowledge, are discovering new physics all the time. Will there ever be a final equation? Will there be an end to what we mortal beings can know? I cannot answer these questions.

I will instead deflect your question with a true statement about myself:

  • Most formulas that I know have integer exponents.

Amongst the entire population of people alive today, I have an above average understanding of physics. Amongst registered users of this site, my understanding of physics is probably way below average. On an absolute scale of everything humans know about physics, my understanding is laughably incomplete. Regardless of what the absolute count integer vs. non-integer exponents of physical or natural laws comes out to be, I think we can formulate a sound explanation of your observation thus:

Propositions

  • Things that are easy to understand have simple representations.
  • Integer exponents are easier for humans to understand than fractional, irrational and complex exponents.
  • Formulas which have non-integer exponents will be related to or even derived from formulas which have integer exponents.
  • Human learning (individually, which necessarily scales to collective learning) begins with understanding the simple things.
  • Attempts to understand the complex without understanding of the basics are less successful.

Therfore

  • The first formulas humans discovered and recorded which accurately described the natural world used integer exponents.1
  • The first formulas one would learn in studying physics (as humans understand it) will have integer exponents.

And Q.E.D., most formulas an arbitrary human knows will have integer exponents.

1: Inquisitive points out an excellent example of this with the sphere. Applying these propositions to that example: 1. A (2 dimensional) circle is easier to understand than a (3 dimensional) sphere. 2. A (1 dimensional) line is easier to understand than a circle. 3. Beginning understanding with a point, then a line, then a circle and a sphere, the equations for each would build upon the most basic, shared elements of each: the center point and the radius. Therefore, more people know the formula for a circle than a sphere, because it's simpler, easier to understand, and has been a part of human knowledge for longer.

Now that opens the door to the question I think you're really looking to be answered: why? Or more properly stated: Why do the most basic laws of the physical world involve integer exponents?

That depends, I think, primarily on what you mean by basic. From a human perspective, the most basic laws are the ones we learned the first and understand the best, which puts my argument back in play. From the cosmic perspective of which laws are involved in most of the relationships that make up the universe, upon which the macroscopic view depends, the most basic laws are high energy particle physics and quantum mechanics. (Or so I suppose; I think I've already sufficiently declaimed that I know nothing in this field.)

I don't know what the equations in these fields look like. As Inquisitive demonstrates, you could probably find ways to express some or all of the equations in any field with non-integer exponents. However, if and when we arrive at 'the theory of everything,' I would wager the most popular form for the equations, the form that gets recorded in textbooks and taught in classrooms, will involve as many integer exponents as we can make fit.

A delicious example of this is Euler's formula, which describes trigonometric functions by means of an imaginary exponent: eix = cosx + isinx. Wikipedia summarizes it by saying "This formula can be interpreted as saying that the function eix is a unit complex number, i.e., traces out the unit circle in the complex plane as x ranges through the real numbers." In other words, while the simple, real circle is described with integer exponents, there is another circle describable with imaginary numbers. Who is to say which is more fundamental? Or which leads to more formulas or a broader description of physics and reality?

$\endgroup$
  • 2
    $\begingroup$ IOW, "selection bias". Very good re framing the question as an observation rather than underlying fact, as a general thing to keep in mind. $\endgroup$ – JDługosz Feb 4 '15 at 16:03
  • 1
    $\begingroup$ Formulas in compressible fluid dynamics are filled with non integer exponents. $\endgroup$ – Rick Feb 5 '15 at 13:26
10
$\begingroup$

In most cases, the answer is: Because the creator of the formula wanted to express it in a simple way.

E.g. in $F = ma$ we are defining the mass ($F/a$), as the property that matter has to offer resistance to acceleration when applied a particular force. In short, Newton chose to represent this model as simple as possible.

We could use a different value for $m$, but it would only make things more complex than they need to be.

Weird exponents start to come up when we combine previously assumed axioms, like this one. Suppose I make a new assumption based on these laws. I can no longer define what $m$ is, so my expressions will be more complex.

$\endgroup$
  • 16
    $\begingroup$ This could explain pre-factors, but not exponents. In the case of $m$ in $F=ma$, you could introduce a multiplicative constant, but how would an exponent different from 1 appear...? $\endgroup$ – Joce Feb 2 '15 at 16:47
  • 4
    $\begingroup$ This doesn't answer the question. Newton didn't choose to have $F=ma$. He found that $F=ma$ (up to multiplicative constants depending on the choice of units) was the formula that represents how those quantities relate to each other. The question is why such formulae overwhelmingly have integer exponents. $\endgroup$ – David Richerby Feb 2 '15 at 17:47
  • 2
    $\begingroup$ "so my expressions will be more complex." -> to me this sounds like $F=ima$ $\endgroup$ – Kyle Oman Feb 2 '15 at 18:02
  • $\begingroup$ It's just a simplification.. of course, some things are harder to describe, but -given the knowledge level of the question- a taylor series explanation seems a bit overkill.. sorry if complex is mistaken with complex numbers, I'm not a native speaker $\endgroup$ – Alvaro Feb 2 '15 at 21:07
  • $\begingroup$ @DavidRicherby he found out they were related. He could've written something completely different to represent the integral $\endgroup$ – Alvaro Feb 2 '15 at 21:08
8
$\begingroup$

The math we use to describe the behavior of the world around us has two types of quantities: values and units. "3.4" is a value. "Meter" is a unit. There are extra rules that value-unit pairs have to follow in order for the results to have any physical meaning:

  • Two value-unit pairs must have the same type of unit if they are added or subtracted, i.e., $\text{length} + \text{length}$, $\text{force} + \text{force}$, etc. $1 \text{meter} + 2 \text{feet}$ makes sense. $5 \text{gallons} + 4 \text{acres}$ does not.
  • Units multiply and divide just like values. $\text{length} * \text{length} = \text{length}^2$. $\text{length}/\text{time} = \frac{\text{length}}{\text{time}}$ (which we rename $\text{speed}$ for convenience).
  • Unit-value pairs that are compared must have the same type of units. $5 \text{seconds} < 1 \text{hour}$ works. $5 \text{seconds} < 1 \text{inch}$ does not (let's not bring relativity into this).
  • Values without units have the equivalent of $1$ for a unit, so $2*(4 \text{cm}) = 8 \text{cm}$.

These are the rules and also the only allowed operations on value-unit pairs. Now, every physical measurement you can make results in set of value-unit pairs. Is it possible, according to the above rules, to come up with a valid expression with non-integer exponents? No.

Roots, including the square root, sometimes work. The formula for the period of a pendulum is $$T = \frac{1}{2\pi}\sqrt{\frac{L}{g}}$$ where $T$ is the period, $L$ is the length of the pendulum, and $g$ is the acceleartion due to gravity. If you work out the units, then you're taking the square root of $\text{time}^2$, which is $\text{time}$ (note that $\pi$ is a circle's circumference divided by its diameter: $\frac{\text{length}}{\text{length}} = 1$). So, a partial rule:

  • Square, cube, or any other root is fine as long as the result is a product or quotient of integer powers of units.

No one has been able to attach meaning to $\sqrt{\text{2 miles}}$. By "meaning," I mean something like "come up with a way to measure that quantity." I can imagine how to measure a square mile of land or a cubic mile of water, but what thing in the universe could have a quantity of a root mile?

"But, wait," you say, "I see lots of other functions in science formulas like sines and logs. What about them?" You're right. For example, the amount of radioactive material left in a sample is given by $$N(t) = N_0 e^{-t/\lambda}$$ where $N(t)$ is the number of undecayed atoms left, $N_0$ is the original number at time $t = 0$, and $\lambda$ is a constant that indicates the average lifetime of an undecayed atom. Note that $t$ and $\lambda$ both have units of time, so the total exponent is unitless. In this equation, the exponent not only isn't an integer, it isn't even constant. It more negative with time.

This brings me to one more rule about units:

  • The input and output of transcendental functions (sin, cos, log, $e^x$, etc.) must be unitless. Edit: See note below.

We can see why using the Taylor series of the exponential function: $$e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \cdots$$ What would happen if $x$ had a unit $X$? The first term is unitless; the second term has unit $X$; the third term has unit $X^2$. These quantities can't be added together according to the first set of rules unless $x$ was unitless to start with (that is, $X = 1$).

In science, what we want to measure and study are aspects of the universe around us. That means we are interested in physical measurements--value-unit pairs. We humans invented math in order to work with value-unit pairs and came up with the rules because they were the ones that worked. We don't understand what $\text{length} + \text{area}$ could mean, so we make up rules to make that calculation impossible. Integer exponents on quantities with units is the result.


As pointed out by Eli Rose in the comments, it is not only transcendental functions that must have unitless inputs and outputs. Consider the Lorentz factor from relativity: $$\gamma = \frac{1}{\sqrt{1-\beta^2}}.$$ In order for this formula to make physical sense, $\beta^2$ (and thus $\beta$) must be unitless because $1$ is unitless. Indeed, in relativity, $$\beta = \frac{v}{c}$$ where $v$ is the velocity of an object and $c$ is the speed of light. This results in a unit of $1$. The full Lorentz formula is written: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$

$\endgroup$
  • $\begingroup$ This is really interesting, but I'm not sure I understand. The Taylor series for $\frac{1}{1 - x}$ is $1 + x + x^2 + x^3 + \dots$, but it isn't transcendental. $\endgroup$ – Eli Rose Feb 6 '15 at 21:04
  • $\begingroup$ @EliRose I forgot about that one. So, it's not just transcendental functions. However, look at the denominator. If $x$ has units, then it can't be a valid scientific formula because $1$ is unitless. $\endgroup$ – Mark H Feb 7 '15 at 1:52
  • $\begingroup$ Your argument for why the input of a transcendental functions must be dimensionless is not correct. The taylor expansion of a function is given by $$f(x)=f(0)+f'(0)x+\dfrac{f"(0)}{2}x^2+...$$ the first term has the dimensions of f(0) while the second term has also the same units since $f'(0)x=\dfrac{df}{dx}x$ where the dimensions of $dx$ cancel with $x$ and same goes for higher powers $\endgroup$ – Omar Nagib Jul 24 '17 at 6:39
  • $\begingroup$ @OmarNagib What you wrote only shows that the Taylor expansion preserves the units of the original function. It says nothing about whether the function itself can accept or return dimensioned values. Looking at the pure function $e^x$, we can conclude that $x$ must be dimensionless because the first two terms are $1+x$. This means that the entire Taylor expansion is dimensionless, hence the result must be dimensionless as well. $\endgroup$ – Mark H Jul 24 '17 at 9:29
  • $\begingroup$ Yes indeed what I wrote just shows that. But it's this very argument that you invoke to try to prove that $e^x$ Can only take dimensionless input(when you say what if $x$ had units $X$?). Your argument can at best prove that $e^x$ itself is dimensionless(because the 1st term is pure number). You gotta realize that the second term in the expansion is actually $\dfrac{d}{dx}[e^x]|_{x=0}x$ and it's dimensionless regardless of whether $x$ is dimensional or not. One can prove using other methods that indeed $x$ in this case must be dimensionless, but your method of proof here is flawed. $\endgroup$ – Omar Nagib Jul 24 '17 at 21:41
7
$\begingroup$

If you think of physics as the study of how certain operators act on certain state spaces (a translation into mathematics of "stuff happens in the World" - the operators are the "stuff happens" and the state space is the "World") and you make the further assumption that the operators are analytic in their actions (this is similar, but a stronger notion than "smooth"), i.e. that there is an $\epsilon>0$ such that for every $x$ within "distance" $\epsilon$ of some point $x_0$, we can describe the operator's action as a convergent Taylor series, then relationships between operators can be expressed as relationships given by differential equations, which contain only integer co-efficients of the differentiation operator, i.e. there are only terms of the form $D^n\,T$ where $D$ is differentiation and $T$ the operator in question. Therefore, only integer powers appear in the formulas.

This argument is somewhat circular insofar that assuming convergent Taylor series is just another way of ultimately saying that you only get integer powers. However, even noninteger powers of operators, when meaningful, can often be written in terms of infinite Taylor series, so one could argue that we could do physics by taking enough terms if the "real" relations were non integer powers, and thus get a finite series that would be experimentally indistinguishable from the "real" operator.

$\endgroup$
  • $\begingroup$ I agree with you, because it supports the most useful harmonic oscillator potential, as an approximation to any real potential. So in this sense, integers are useful in modeling the observations simply. $\endgroup$ – anna v Feb 7 '15 at 13:39
4
$\begingroup$

Another way of thinking about it is that people tend to break problems down into chunks that are easy to understand. For example, with Newton's second law there is no fundamental problem defining force using the weird exponents you mentioned, but derivations which follow from that definition will get messy very quickly. It is much more convenient to define force as $ma$ in the direction of the acceleration. We then use that as a simple building block for more complicated problems. The simple harmonic oscillator or driven damped oscillator for example are built upon $F=ma$ but have no simple exponent in their equations of motion.

As a second example, take Coulomb's Law. By taking a spherically symmetric problem (i.e. a point charge) as a first step, the resulting electric field is spherically symmetric and we get $r^2$ as the exponent. From that simple problem we can break tougher ones down into easier ones that make more sense to us. By adding or integrating over all of the point charges in a system, we can end up with much more complicated fields that don't resemble a simple exponent (see for example general multipole expansions).

For examples of ugly exponents in physics, quantum mechanics and optics are full of complex phase factors and Fourier transforms where you integrate over some $e^{ikx}$. Even those problems have the common thread of complicated problems built as a superposition of simple ones, just like classical mechanics and electrostatics mentioned above.

$\endgroup$
3
$\begingroup$

Let's ignore the uninteresting cases where fractional exponents arise from bad definitions of relevant physical quantities.

In general, the simplicity of expressions governing many phenomena has to do with existence of unique length/time scales. In general situations, where there are multiple length scales or when there are no length scales, you often find fractional exponents. You can find plenty of examples in fluid mechanics where there are often multiple scales. You find fractional exponents also when the system is scale invariant such as at second-order phase transitions.

$\endgroup$
2
$\begingroup$

The reason for this is not just 'simplicity', as others have pointed out. It is also due to the way these expressions come into existence.

Usually, we have a simple, linear relation, obtained intuitively or due to definitions; i.e., $$v=a*t$$

The velocity traveled is acceleration $v$ multiplied by time $t$. Now, if we we want to get the distance traveled due to acceleration, we integrate this expression w.r.t. time and obtain $$x=\frac{1}{2}a*t^2$$ which should look familiar; as should $E=\frac{1}{2}mv^2$ for example, which momentum $m*v$ integrated wrt time. This immediately explains all the silly $\frac{1}{2}$ you see in equations. So, essentially, due to integration, the linear (i.e., exponent 1) equation becomes quadratic.

The fun part is that many (virtually all) physics equations are related through each other by differentiation and integration, which is why physics students are all obliged to take boring calculus classes.

Note that in many cases, the integration or differentiation is presented in a simpler way, however (which is why you don't use calculus in high school physics); for example

$$E=F*s$$

(Energy equals force times path length) is actually a simpler way of describing that energy is a force integrated along its path.

$\endgroup$
2
$\begingroup$

The answer to your question is three-fold. First, a number of integer exponents come from the definition of "being to the power of something" itself. Second, fundamental physical laws are (at least effectively) regular and local which does not allow for non-integer exponents. And third, the non-irrationality of exponents is not true in non-linear dynamics, critical phenomena in thermodynamics, and certain formulations of quantum mechanics.


1) Tautological geometrical powers

In the twenty-first century we are spoiled by modern mathematics giving us the feeling that every operation we do is somehow backed by some basic axioms. But this was not the case for millennia and twentieth century mathematics only built a scaffolding of axioms around well understood mathematics which was already there.

So let me ask: "What does it mean to multiply two non-integer numbers?" How would a nineteenth century mathematician understand it? Well, he would say that a multiplication of two numbers $a$ and $b$ denoted $a \cdot b$ is defined by the operation of drawing a rectangle with sides $a$ and $b$ and then counting the number of unit squares which fit into it. Nothing more, nothing less.

Relationships such as $c^2=d^3$ then also have a well defined meaning: Take a cube with side $d$ and rearrange it into a rectangular cuboid with one side equal to $1$ (in given units) and if the other two sides are equal, then they are equal to $c$. By formally iterating such definitions you can reach any rational exponent.

Statements such as $A_{\rm{C}(r)} = \pi r^2$ do not necessarily talk about the numbers - they talk about the proportionality of two objects of the same kind: If I can fit a certain number $N$ of unit squares into a square of side $r$, what is going to be the ratio $M/N$ where $M$ is the number of unit squares I can fit into a circle of radius $r$?

When you say a circle has an area of $1.2 \rm m^2$, you are only talking about the number of "unit-meter-squares" that can be fit into it, but the "unit-square-meter" is only a conventional reference object.

You could as well decide to ask how many Wibbly-gaga-tons fit into the circle. Wibbly-gaga-tons may be fractals of irrational fractal dimension $1<\alpha<2$ and a single length parameter $a$ so that the area it covers is $$ A_{\rm{Wgt}(a)}=K a^\alpha$$ Your formula of proportionality between the area of a circle of radius $a$ and a Wibbly-gaga-ton of characteristic length $a$ is going to be $$A_{\rm{C}(a)} = \pi a^2 = \pi K^{-\frac{2}{\alpha}} A_{\rm{Wgt}(a)}^{2/\alpha}$$ Say we now establish a Wibbly-gaga-ton with $a=1 \rm m$ as a new referential object. Then we would express the area of a circle as for instance $1.2 \, \rm Wgt(1m)^{2/\alpha}$. That is, most of the integer exponents in geometry are conventional.


I was just in the process of writing my answer to this question when an answer got accepted, so I am just going to post the first part and will not finish 2) and 3).

$\endgroup$
1
$\begingroup$

I would in fact put the question on another level: Why do most formulas in mathematics&science have rational exponents?

The thing is – in a way all exponents are rational: the power operation is first only defined for integers (by iterated multiplication). When we then rearrange the equations, rationals arise naturally as roots. But you can never get to irrational exponents this way. Well... of course you can, by interpolation: you postulate that $x^y$ is a continuous function of $y$ (actually, you first need to proove that $y \mapsto x^y$ fulfills the $\epsilon$-$\delta$ criterium on $\mathbb{Q}$), and since the rationals are dense in $\mathbb{R}$ that can be used to fix the values for all $x \in \mathbb{R}$. But this is pretty much an ad-hoc thing to do, it's not like the definition of $\pi$ or $\sqrt{2}$ where it follows from some definition that a value must be irrational. Such formulas do turn up in some physics applications (where it is just assumed that some relationship is a power law, and the exponent is obtained by least-square fitting), but they don't really make for very elegant mathematical models.

A much more straightforward manner in which “irrational exponents” turn up, that you find all the time in science, is as arguments to the exponential function $\exp$. In that case, it is natural to use real numbers rather than just rationals, because the function itself is defined through calculus rather than algebra (as a solution to the differential equation $f'(x) = x$ with $f(0)=1$. If you will, $\exp$ doesn't even make sense as a function on the rationals; it just so happens that when evaluated on rationals it coincides with $e^x$ where $e = 2.718...$. But the exponential function is nevertheless not really a power function, it's just a particularly important elementary function. If this function is chosen for some physical model, then usually not so much because of its algebraic properties (power laws), but because some differential equation resembles its very definition, or some integral is easily solved when the kernel is e.g. Gauss-shaped (which has a lot to do with the differential properties).

$\endgroup$
1
$\begingroup$

Using a similar argument as @Jason, we should first concentrate on the nature of the observables we use and also on geometry and conservation laws.

Surely we use time, space, mass as references precisely because the laws of physics relate these quantities in a simple way. This is precisely for this reason that we have been able to discover relations at all.

Science has created a conceptual frame based on these units and the principle of measurement: We deal with measurable quantities and we express the results in a unit system. The unit system is by all means arbitrary. Here arbitrary means that it can be any system, as long as we agree on the definition of a few reference values. As consequence of this arbitraryness, we can perform dimensional analysis.

A few words concerning dimensions. Geometry imposes a few constraints to units and dimensional analysis. Conservation laws (mass conservation, energy conservation, etc.) are governed by geometry. The electric field created by a point charge varies as $r^{-2}$ because we are in three space dimensions (in a space of $d$ dimensions, it would decrease as $r^{-(d-1)}$ by Gauss's law). Many other examples can be found. Geometry and conservation laws will provide mostly integer exponents. More generally, geometry imposes simple scaling laws that are governed by integer exponents.

Suppose now we look for the result $A$ of a physical experiment involving some parameters $p$, $q$, $r$. If we suppose that $A$ is proportionnal to a combination of the form $p^xq^yr^z$ where $x$, $y$ and $z$ are unknowns exponents, dimensional analysis will reasonably often give a single solution. For instance, an object of mass $m$ falls from a height $h$. How long does the fall last. The solution is $t\propto m^xh^yg^z$ ($g$ is acceleration of gravity). We find that $x=0$ (it does not depend on mass) $y=1/2$ and $z=-1/2$. We have obtained rational numbers because we have solved a linear system with integer coefficients.

In problems where there are more parameters, dimensional analysis must be improved to relate dimensionless quantities. If we take into account friction during the fall in my example, the quantity $Q=(\frac{\alpha}{m})^2\frac hg$ (where $\alpha$ is the friction coefficient) is the parameter of a transcendantal equation for $u=\frac{\alpha}mt$ which is $\mathrm e^{-u}+u=1+Q$. The dimensionless quantities are formed with rational exponents for the same reasons as in the first case.

In situations containing parameters with no dimension, non-rational exponents can appear. For instance, consider a polymer formed by $N$ monomers of length $a$. When this polymer is in a good solvent, its extension is $R=aN^\nu$ where $\nu=0.588...$ is a non-rational number. Dimensional analysis cannot say anything about it. It only states that the exponent of $a$ must be $1$.

$\endgroup$
1
$\begingroup$

This answer started as a comment to orion's excellent answer. I'd like to expand on it. There are two different classes of measurement being addressed here: Geometric measurements and physical phenomenon measurements. I'll address them separately.

The geometric measurements such as $A = πr^2$ have rational exponents because the relationship between measurements of the same dimension is linear, and relationships varying by N dimensions is to the N+1 power.

That is, a square that has twice the side length of a unit square will have twice the width as well, since both length and width are of the same dimension. However that same square will have $2 ^ {(1+1)} = 4$ times the area of the unit square, because area is one dimension higher than length.

The physical phenomenon measurements such as $F=ma$ have rational exponents when conservation of energy is a constraint.

The force that a 1 KG book puts on the table comes from the total mass of the book's baryonic components, as acted upon by the curvature of spacetime as defined by gravity. Doubling the amount of baryonic particles will double the amount of energy being acted upon. When you see physical measurements that have exponents other than 1, such as $F = G \frac{m_1 m_2}{r^2}$, (exponent -2) you will find that the relationships of "where is the energy coming from" become quite complex. Here, you have two masses which both "tell spacetime how to bend", quite a complex relationship indeed.

$\endgroup$
0
$\begingroup$

Good question, and I don't have a good general answer.

However in the specific example you quote it's because the minimal dimension of space you can inscribe the circle is a surface of some kind ie a plane, or the surface of a sphere.

Similarly, with the formula for the volume of a sphere, the exponent is three - which refers to the fact a volume can only be inscribed in a three dimensional space.

$\endgroup$
0
$\begingroup$

I think what symanzik138 wrote is the right way to think about this. The laws of physics that we know about and use, should be considered to be effective laws that one could in principle derive from fundamental laws by integrating out the degrees of freedom that the effective laws do not describe. Whether the topic is quantum field theory, fluid dynamics or classical mechanics, this picture is always valid (except, of course, if one where to consider candidates for the theory of everything).

Now, given an effective theory that is supposed to arise from a more fundamental theory by integrating out degrees of freedom that exist on a length scale smaller than $\Lambda$, one should be able to integrate out the degrees of freedom between $\Lambda$ and $\Lambda(1+\epsilon)$ and then rescale the lengths by a factor of $1-\epsilon$, so that in terms of the new length variables the cut-off is restored to the old value. This then has the effect of rescaling the entire system. That rescaled system is just as good as the original system so it should also be described by the same model, except that the parameters that define it, will have changed. This leads to differential equations for the parameters of the model in terms of the lengths scale (changing the length scale is the same as keeping it the same but instead changing the parameters in a certain way).

These equations for the scale transforms are called "renormalization group transforms". In cases where there is no nontrivial integration over the degrees of freedom to be done, one ends up with simple scaling relations involving with integer or fractional exponents, but such exponents can also arise in nontrivial cases.

$\endgroup$
-2
$\begingroup$

I think everyone has missed the obvious answer: because the equations of physics simply use math to model the way the universe works. Put some fractional exponent in your F = ma, and the answers come out wrong.

Now if you're asking why the universe happens to be that way... Well, I don't know, but I think it's more of a question for philosophy than physics.

$\endgroup$
  • 3
    $\begingroup$ Actually, as orion points out, there are important mathematical reasons for linear equations, so the value of the exponent is not arbitrary $\endgroup$ – Phil H Feb 3 '15 at 13:13
  • $\begingroup$ But the mathematical equations used in physics are (as the OP points out) only a small subset of all possible equations. We use the ones with integer exponents precisely because they provide accurate models of real-world phenomena. $\endgroup$ – jamesqf Feb 3 '15 at 19:29
  • $\begingroup$ On reflection, I think I'd go even further, and suggest that for example the operations of integration and differentiation were defined as they are precisely because they accurately described physical relationships. The subset of math used in physics (or any science) is descriptive, not prescriptive. $\endgroup$ – jamesqf Feb 4 '15 at 6:05

protected by Qmechanic Feb 3 '15 at 0:10

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.