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The mathematical concept of function is used in physics to represent different physical quantities. For example the air pressure variation with time and space is called an acoustic wave. We use a function to represent a charge distribution (or even electric field strength) in space and time.In gravitation we use it to represent a mass distribution (and momentum distribution) in space and time. In Quantum mechanics we use it to represent a Quantum state and hence the probability distribution of position and momentum of a particle in space and time. Most of the time we do a variety of mathematically well defined operations on these functions to make some conclusions and to answer some questions regarding a physical system.Even the fundamental laws of physics give the mathematical relation between functions representing different physical quantities. Now if i ask a question on the mathematical nature of a function that represents a physical quantity (say for example a mass distribution in space and time), would it be possible to find an answer ? Does every such question has a physical interpretation ?

EDIT 1

Here i give a sample question ?

Q. let a function $p(t)$ represent the acoustic pressure variation at a given point in space.Is the function well behaved ? Mathematically speaking "Is $p(t)$ a smooth function ?"

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Your question, if there is one in there, is impossible to answer. Please be more specific. –  dbrane Mar 2 '11 at 10:46
    
I'd like to know the explanation for the down votes ? –  Rajesh D Mar 2 '11 at 10:47
    
@dbrane : ok...i'll add a sample question for your convenience. –  Rajesh D Mar 2 '11 at 10:50
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I've no idea what your question is really asking, but it may be noted the mathematical concept is actually not of function, but of distribution aka functional, i.e., a mapping between functions and a field (real or complex); a functional corresponds to a function iff the functional is continuous. Charge distributions, quantum bra/kets--not always functions, but always functionals (e.g., Dirac deltas, plane wave states, etc.). –  Stan Liou Mar 2 '11 at 10:54
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Related Stack Question: physics.stackexchange.com/q/1324 –  Roy Simpson Mar 2 '11 at 13:57

5 Answers 5

up vote 2 down vote accepted

Dear Rajesh, first of all, different questions have different answers. So the answer below is purely for the actual question about the pressure that you stated and may be wrong for many other, "similar" questions that deal with other functions. You're very naive if you think that all questions about functions in physics have the same answer.

Pressure is a macroscopic concept because gases etc. are made out of atoms. So the number of atoms or molecules in a volume of space is an integer. If the volume is really small - if you really want to define the pressure at an "accurate locus" of space - then the integer-valuedness becomes really important and the pressure is behaving as a discrete, discontinuous variable.

However, if one averages the pressure over a large enough region of space so that the number of atoms in this region is much greater than one, then the discontinuities in the number of molecules - and their velocity - are suppressed by the statistically large number of the molecules, and the assumption that the pressure is approximately a continuous function of time becomes an acceptable approximation.

Classical field theory is a good model of the reality - at least some portions of it - and it is based on continuous (and differentiable) functions of space and time.

The discreteness above - that arises from the atomic structure of matter - isn't true for other things. If you ask what is the electric field at a given point, it is classically a totally smooth function and there is no disclaimed at all. However, quantum mechanically, the actual values of the quantities don't exist prior to the measurement. There are lots of subtleties.

Various layers of physics are more or less accurate and neglect various things. Correspondingly, some objects may be continuous, discontinuous, differentiable, or non-differentiable in different descriptions. And in quantum mechanics, there doesn't really exist any objective function that describes the state of the system at each moment at all. Even though the electron is point-like, at least with the accuracy of $10^{-18}$ meters, one can't say what was its trajectory through space, $x(t),y(t),z(t)$, at the same accuracy.

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I have several questions on your answer but i mention one of them here. You said that the pressure cannot be well behaved due to conceptual reasons (atoms molecules etc.)..I really want to setup an experiment to verify that...my question is what sort of equipment would i need, does a simple microphone and a computer with sound card (A/D convertor built in) would be sufficient ? –  Rajesh D Mar 2 '11 at 12:09
    
Do you have any comment on this. –  Rajesh D Mar 2 '11 at 13:05
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@Rajesh D : You seem to be asking Lubos here for an experiment to demonstrate the existence of atoms. Was that your real question all along? –  Roy Simpson Mar 2 '11 at 13:55
    
@Roy Simpson : i am sorry it turned out that way. I am actually laughing at myself after seeing your comment. –  Rajesh D Mar 2 '11 at 14:35
    
@Rajesh: Look up something on Brownian motion. That is precisely the experiment you're talking about. –  Jerry Schirmer Aug 30 '11 at 18:46

This reply is to the concept illustrated by your example. The answer is that not all functions used in physics are mathematically well behaved/smooth over all the variable field. For example some have singularities. Some are described by delta or theta functions. There are branch cut discontinuities in the imaginary plane for certain classes of functions used in physics. etc.

In general boundary conditions can lead to discontinuities in functional representations when solving physics problems.

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If your question is whether functions in physics don't necessarily have to be smooth, I can point out a few examples:

  1. surface charges on a conductor are discontinuous functions in 3-space
  2. the corresponding electric field undergoes a discontinuity when it crosses a surface charge
  3. first-order phase transitions have a discontinuity in the first derivative of some thermodynamic variable (say, in specific heat)
  4. second-order phase transitions have a discontinuity in the second derivative of some thermodynamic variable
  5. most of the paths which contribute to the Feynman path integral are not differentiable

Maybe others can add a few more.

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A lot of these happen when the continuum approximation to a matter field breaks down, and smoothness is restored when you start making a more detailed approximation--there are no true surface charges in microscopic electrondynamics, the thermodynamic limit doesn't work in the neghborhood of a phase transition, and a more careful partition function is continuous, etc. –  Jerry Schirmer Aug 30 '11 at 18:50

Physicists use discontinuous functions all the time, as pointed out by @anna v, and even sometimes use «generalised functions» or «distributions» which, although written as if they are functions, and inspired by functions, are not functions of any kind, as point out somewhere else by somebody else...

But, as pointed out by @Jerry Schirmer, sometimes, at least, we know that the reason the function is discontinuous is because we are working in an approximation and stretching it a little bit further than it is really valid.

As far as I know, we are not yet sure whether ultimately, in a fully accurate physical theory, all functions will be continuous and finite-valued everywhere. To get rid of the singularities in General Relativity is actually one of the motivations of quantum gravity...

The answer to your particular question, then, is, maybe: maybe all physical functions are continuous and well-defined. But for purposes of approximation and calculation, at least, we use discontinuous ones, with singularities, etc.

Your general question can be rephrased as follows: if $f$ is a physical function, i.e., a function which represents something physical, does every mathematical feature of $f$ have physical significance? The answer is yes. The only way to express a mathematical property of $f$ or, indeed, any mathematical object, is to use logical and linguistic combinations of the definition of $f$. If $f$ is physically significant, every logical combination of its properties will also be physically significant merely because logic also applies to physics.

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One can say a few things about every function arising in physics:

  • It is Lebesgue measurable
  • It does not encode the solution to the halting problem.

These (and minor modifications) are just about it.

These properties are a complete cheat. They are actually true of all functions which you can compute, and they should be thought of as just mathematically true of all functions which exist in the strong sense of the word. But human imagination being what it is, mathematicians have the idea that they can imagine more functions than this, and they have a philosophy of "if I think I can conceive it, then it must be true", for completely silly reasons.

In the case of the halting problem, it is useful to pretend that you can solve this with a given real number, so the mathematicians have the right convention. But in the case of measurability, it is absolutely useless to pretend that there are nonmeasurable functions, so here they chose the wrong convention.

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