Does Physics need non-analytic smooth functions? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:00:39Z http://mathoverflow.net/feeds/question/114555 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions Does Physics need non-analytic smooth functions? Qfwfq 2012-11-26T16:58:48Z 2013-04-25T06:54:05Z <p>Observing the behaviour of a few physicists "in nature", I had the impression that among the mathematical tools they use a lot (along with possibly much more sofisticated maths, of course), there is certainly Taylor expansion. They have a quantity (function) that they need to approximate: they expand it in Taylor series, keep the order of approximation that is useful for their purposes, and discard the irrelevant terms.</p> <p>Appearently, there is little preoccupation for mathematically justifying this procedure, even if the to-be-approximated quantity is not given by an explicit form which is clearly known to be analytic. As Physics clearly gets no problems from the above mathematical subtleties, this may just mean that the distinction between analytic and smooth functions is somehow irrelevant to the basic equations of physics, or rather to the approximations of their solutions that are empirically testable.</p> <blockquote> <p>If non-analytic smooth functions are irrelevant to Physics, why is it so?</p> <p>Are there equations of physical importance in which non-analytic smooth solutions actually are important and cannot be safely considered "as if they were analytic" for the approximation purposes? </p> </blockquote> <p>Remark: analogous questions may arise about Fourier series expansions.</p> <p>One possible way the practice goes might be:</p> <ol> <li>Consider a (differential or otherwise) equation $P(f)=0$ usually with analytic coefficients.</li> <li>Expand the coefficients in Taylor series around a point in the scale of physical interest.</li> <li>Discard higher order terms obtaining an approximated equation with polynomial coefficients $\tilde{P}(f)=0$.</li> <li>Make the <em>ansatz</em> that the solutions $f$ of interest must be analytic.</li> <li>Find the coefficients of $f$ by hand or by other means.</li> </ol> <p>This leaves open the question why the ansatz is mathematically justified, if the equation of interest was $P$ not $\tilde{P}$. Do <em>analytic</em> solutions of $\tilde{P}$ aptly approximate solutions of $P$? Edit: I understand now that these last two lines are not very well formulated. Perhaps, ignoring the $\tilde{P}$ thing, I should have just asked something like: </p> <blockquote> <p>Given any $\epsilon>0$, does knowing the <em>analytic</em> solutions (i.e. knowing their coefficients, possibly up to an arbitrarily large but finite number of digits) of $P$ give all the information about all solutions of $P$ up to $\epsilon$-approximation? Are there physically well known classes of equations $P$ in which this may not happen (perhaps even up to taking very regular approximations of the coefficients/parameters of $P$ itself)?</p> </blockquote> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114556#114556 Answer by paul garrett for Does Physics need non-analytic smooth functions? paul garrett 2012-11-26T17:06:00Z 2012-11-26T17:06:00Z <p>It may be that for many purposes <em>asymptotics</em> (in a precise sense) are needed, rather than exact formulas. The prototype for this is the fact that a <em>finite</em> Taylor expansion, with error term, correctly approximates a smooth function, whether or not the smooth function is actually analytic, whether or not the infinite Taylor series converges. A similar, more complicated, asymptotic idea was legitimized by Poincare, after he realized that certain series expansions "formally" solving problems in celestial mechanics were divergent, but their finite truncations provided good approximations.</p> <p>There is a different objection one can have to analyticity, namely, the "action at a distance" aspect implied by "the identity principle", namely, that sufficient (but, yes, infinite) information about the function near a single point perfectly determines its behavior everywhere. This seems to me more delicate and more dubious than asymptotics.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114561#114561 Answer by Rhys Davies for Does Physics need non-analytic smooth functions? Rhys Davies 2012-11-26T17:56:09Z 2012-11-26T17:56:09Z <p>Part of your question has already been answered: a finite number of terms in a Taylor expansion can give a very good approximation, and we don't care whether the true solution is analytic; in any case, we are almost certainly only solving equations which themselves constitute an approximation.</p> <p>But I also wanted to give an example where non-analytic smooth functions are important: in quantum field theory, 'instantons' give rise to terms proportional to $e^{-\frac{k}{g^2}}$, where $k$ is some numerical constant, and $g$ is a 'coupling constant'. Perturbation theory consists of a finite series expansion in $g$ (useful when $g \ll 1$), and so completely misses instanton effects. This is obviously of most interest when the 'perturbative' terms are all zero.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114562#114562 Answer by Alexandre Eremenko for Does Physics need non-analytic smooth functions? Alexandre Eremenko 2012-11-26T17:58:24Z 2012-11-26T17:58:24Z <p>Yes, physicists really like Taylor series. However in many situations the Taylor series diverges. This is the case for almost all non-trivial "perturbation series" in physics. In the simplest situation, we have a function on $[0,1]$ which is infinitely differentiable at $0$, however its Taylor series at $0$ diverges. And there is an enormous literature about how to connect this series to some properties of the function.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114563#114563 Answer by Jeff Harvey for Does Physics need non-analytic smooth functions? Jeff Harvey 2012-11-26T18:04:36Z 2012-11-28T04:14:43Z <p>As a physicist "in nature" perhaps I can give a few examples that illustrate how non-analytic functions can appear in physics and counter the idea that physicists do not worry about the justification of these procedures.</p> <p>Example 1 involves one of the most precise comparisons between experiment and theory known to physics, namely the g factor of the electron. The quantity g is a proportionality factor between the spin of the electron and its magnetic moment. Perturbation theory in QED gives a formula $$g-2= c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 + \cdots$$ where the coefficients $c_i$ can be computed from i-loop Feynman diagrams and $\alpha=e^2/\hbar c \simeq 1/137$ is the fine structure constant. Including up to four loop diagrams gives an expression for $g$ which agrees to one part in $10^{8}$ with experiment. Yet it is known that that this perturbative series has zero radius of convergence. This is true quite generally in quantum field theory. Physicists do not ignore this, rather they regard it as evidence that QFT's are not defined by their perturbation series but must also include non-perturbative effects, generally of the form $e^{-c/g^2}$ with $g$ a dimensionless coupling constant. Much effort has gone into understanding these non-perturbative effects in a variety of quantum field theories. Instanton effects in non-Abelian gauge theory are an important example of non-perturbative phenomena.</p> <p>Example 2 involves the Hydrogen atom in an electric field of magnitude $E$, aka the Stark effect. One can compute the shift in the energy eigenvalues of the Hydrogen atom Hamiltonian due to the applied electric field as a power series in $E$ using perturbation theory and again one finds excellent agreement with experiment. One can also prove that this series has zero radius of convergence. In fact, the Hamiltonian is not bounded from below and does not have any normalizable energy eigenstates. The physics of this situation explains what is going on. The electron can tunnel through the potential barrier and escape from being bound to the nucleus of the Hydrogen atom, but for reasonable size electric fields the lifetime of these states exceeds the age of the universe. The perturbation theory does not converge because there are no energy eigenstates to converge to, but it still provides an excellent approximation to the energy eigenstates measured experimentally because the experiments are done on a time scale which is very short compared to the lifetime of the metastable state.</p> <p>So I would say that at least in these examples there is a very nice interplay between the physics and the mathematics. The lack of analyticity has a clear physical interpretation and this is something that is understood by physicists. Of course I'm sure there are other example where such approximations are made without a clear physical justification, but this just means that one should understand the physics better.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114568#114568 Answer by unknown (google) for Does Physics need non-analytic smooth functions? unknown (google) 2012-11-26T18:33:00Z 2012-11-26T18:33:00Z <p>In the field of mathematical general relativity, certain uniqueness results for black holes are known for real-analytic space-time but are open for smooth space-time. For example, the "No hair conjecture" was proven by Stephen Hawking for analytic space-time but is open in general. For more details, see slides by Klainerman:</p> <p>www.ihes.fr/~vanhove/Slides/Klainerman-ihes-fev2011.pdf</p> <p>(page 24 and on)</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114572#114572 Answer by Piero D'Ancona for Does Physics need non-analytic smooth functions? Piero D'Ancona 2012-11-26T19:25:14Z 2012-11-26T19:25:14Z <p>Let me add to the several excellent answers. From a physicist' point of view, a concept like 'the value of a field at a point' does not have much meaning, or at least it can certainly not be measured. One can measure averages (norms), over several experiments, and over regions of space. Thus it makes more physical sense to work in the framework of distribution theory, and Sobolev spaces like $L^2$ or $H^1$, rather than of continuous functions. For instance, particles are represented by probability distributions, and their mass or energy is represented by suitable norms of such distributions. I would venture to say: if a result is only true for (say) $C^2$ but not rougher quantities, then it is unlikely to have much physical meaning.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114575#114575 Answer by Deane Yang for Does Physics need non-analytic smooth functions? Deane Yang 2012-11-26T19:45:55Z 2012-11-27T12:30:30Z <p>It is worth noting that it is impossible to solve the initial value problem for the standard heat equation in the real analytic category. Here, there are asymptotic expansions available but no Taylor series.</p> <p>ADDED: It should also be noted that the directionality of time, as exhibited in the heat and diffusion processes, is a phenomenon lives outside the real analytic category. I believe that any PDE in the real analytic category that is well-posed as an initial value problem can be solved in both positive and time directions. That's not true for the heat equation in the smooth category. So the need for going outside the real analytic category appears already in fundamental classical physics.</p> <p>This can be avoided, I suppose, by working purely with discrete models, but that for some of us is a cure worse than the original "problem".</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114581#114581 Answer by Liviu Nicolaescu for Does Physics need non-analytic smooth functions? Liviu Nicolaescu 2012-11-26T20:35:19Z 2012-11-27T14:36:52Z <p>Suppose that $P$ is a partial differential operator with constant coefficients. An old result of Petrowski shows that the following statements are equivalent.</p> <ol> <li>All the classical solutions $u$ of the differential equation $Pu=g$ are real analytic if $g$ is real analytic. </li> <li>The operator $P$ is elliptic.</li> </ol> <p>If a physicist is interested in a <strong>well posed</strong> evolution equation, then, according to J. Hadamard, that differential equation <strong>cannot be elliptic</strong>. Thus by the above result of Petrowski you must allow non analytic solutions for analytic data. If moreover, the operator $P$ is hypoelliptic (e.g. $P$ is the operator $\partial_t -\Delta$) then any solution $u$ of $Pu=g$ is smooth once $g$ is smooth.</p> <p>To conclude, yes, physicists do need to consider smooth, nonanalytic functions.</p> <p><strong>Addendum</strong>: J. Hadamard in his classic <em>Lectures on Cauchy problem in linear partial differential equations</em> discusses the role of smooth nonanalytic functions. I'll let the great master speak for himself.</p> <p><em>"I have often maintained, against different geometers, the importance of this distinction. Some of them indeed argued that you may always consider any function as analytic [...] as they can be approximated with arbitrary precision by analytic ones. But, in my opinion, this objection would not apply, the question not being whether such an approximation would alter the data very little, but whether it would alter the solution very little."</em> </p> <p>In this paragraph he hints at the well posedness of the initial value problem. He then proceeds to show that the initial value problem for the Laplace operator $\partial_t^2+\partial_x^2$ is ill posed. </p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114586#114586 Answer by Ben Crowell for Does Physics need non-analytic smooth functions? Ben Crowell 2012-11-26T21:34:54Z 2012-11-26T21:40:36Z <p>The answer to this question depends on what you mean by "need." In some sense, all physical measurements are integers, and usually small ones, so physicists don't even need the real number system -- in fact, physics (Galileo and Newton) predates the real number system (ca. 1870).</p> <p>Some of the other answers have involved a lot of extremely exotic examples, but it isn't necessary to reach that far. Suppose a particle is released near the bottom of a bowl whose shape is given by a function $y=f(x)$, which has a minimum at $x=0$. Typically we would expect that the particle would oscillate around this minimum with a frequency given approximately by $\omega=\sqrt{gf''}$, where $g$ is the acceleration of gravity and the second derivative is evaluated at $x=0$. The result is approximately valid for small amplitudes, and for those small amplitudes it predicts correctly that the frequency is approximately independent of the amplitude.</p> <p>But the bowl could be defined by some function like $y=|x|$, in which case the second derivative is undefined, and there is no amplitude small enough that the above analysis is a good approximation.</p> <p>Does this mean that physicists "need" non-analytic functions? No. In reality, there are reasons why the infinitely sharp kink in the function $y=|x|$ can't be infinitely sharp. We can also come up with examples in which $y=g(x)$ is some <em>analytic</em> function very nearly equal to $f(x)=|x|$, and the oscillations will have <em>exactly</em> the same behavior for $g$ as for $f$. This will happen whenever the curvature at $x=0$ occurs on such a small scale that the particle's finite size prevents it from feeling any effect from the different behaviors of $f$ and $g$.</p> <p>The issue here is not whether the function is analytic or non-analytic. The issue is whether a certain approximation is good or bad.</p> <p>Often physicists find it <em>convenient</em> to use all kinds of badly behaved functions, such as Dirac delta functions. Convenience isn't the same as necessity. We could do quantum field theory using Egyptian fractions, if we had to --- it would just be extremely inconvenient.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114587#114587 Answer by Bazin for Does Physics need non-analytic smooth functions? Bazin 2012-11-26T21:55:36Z 2013-02-07T20:27:52Z <p>A strong argument is given above on the heat equation; let me be more specific. The heat equation, one of the most basic in PDE and mathematical physics, already known to Fourier, is $$L=\frac{\partial }{\partial t}-\Delta_x,\quad t\in\mathbb R,\quad x\in\mathbb R^n,$$ has the fundamental solution $$E= H(t)(4\pi t)^{-n/2}e^{-\frac{\vert x\vert^2}{4t}},$$ i.e. $LE=\delta(x)\otimes\delta(t)$ (here the Heaviside function $H$ is the indicatrix of $\mathbb R_+$). It is easy to see that the $C^\infty$ singular support of $E$ is reduced to $0_{\mathbb R^{1+n}}$ whereas the analytic singular support is the hyperplane $t=0$. Since the function $E$ is $C^\infty$ except at $x=0,t=0$, one can see that it is indeed a flat function at $t=0,x=x_0\not=0$, i.e. all derivatives vanish at such a point. It is thus impossible to understand one of the simplest PDE using only analytic functions.</p> <p>A more refined -yet classical- fact is related to the notion of well-posedness as defined by Jacques Hadamard. Loosely speaking, a PDE problem is well-posed whenever the solution can be controlled by the data or the sources via suitable inequalities. A typical example of a well-posed problem: the Cauchy problem with respect to a spacelike hypersurface (e.g. $t=0$) for the wave equation. A typical example of an ill-posed problem: the Cauchy problem for the Laplace equation. Although the latter has uniqueness properties, the analytic solutions given for instance by the Cauchy-Kovalewski Theorem are extremely unstable: you have $$\partial_x^2 u+\partial_y^2 u=0,\quad u=e^{\lambda(x+iy)}, u(0,y)=e^{i\lambda y}.$$ The Cauchy data at $x=0$ are bounded by 1, whatever is $\lambda >0$, whereas the solution increases exponentially with $x>0$: no control of $u$ by its Cauchy datum could be expected. However the solutions are analytic and uniquely determined by the Cauchy datum. The analytic method given by the CK theorem provides analytic solutions which are unstable. The CK theorem fails to deliver stable solutions in that case. No understanding of stability phenomena (a very interesting physical property) for PDE is possible within the class of analytic functions and one should use much larger classes of functional spaces in which inequalities of well-posedness could be proven.</p> <p>I could have mentioned another effect, for instance for the Cauchy problem for the Laplace equation: take an analytic Cauchy datum $\phi_0$, then CK provides an analytic solution. Now, perturb $\phi_0$ by a smooth non-analytic function $w$ and take as a datum say $\phi_0+\epsilon w$. Then there is no solution to the Cauchy problem since the very existence of a (say continuous) solution is forcing the data to be analytic. It is not difficult to prove that by Fourier transformation: the analyticity will be forced by the fact that you have to compensate the exponential increase by some exponential decay of the data, triggering analyticity for this data.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114600#114600 Answer by jjcale for Does Physics need non-analytic smooth functions? jjcale 2012-11-26T22:53:50Z 2012-11-26T22:53:50Z <p>In the BCS-theory of superconductivity the energy gap that seperates the ground state from the excited states is a non-analytic function of the exchange energy.</p> <p>Here one doesn't use a taylor expansion of the interacting Hamiltonian around the non-interacting Hamiltonian. Instead one uses a Bogoliubov transformation.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114617#114617 Answer by Ben Wieland for Does Physics need non-analytic smooth functions? Ben Wieland 2012-11-27T01:06:34Z 2012-11-27T01:06:34Z <p>Here's an example that doesn't really fit the motivation. There is a setting where physicists consider discontinuous functions, continuous but not differentiable functions, and so on, all the way up to smooth but not analytic functions. Given that context, I wouldn't expect them to study Taylor series at the singularity.</p> <p>Phase transitions are typically not smooth. For example, the energy as a function of temperature is discontinuous at freezing or boiling. More exotic are second-order phase transitions, which are continuous but not differentiable; and $n$-th order, which have several derivatives. But there are a few phase transitions best modeled as being the infinitely differentiable meeting of two real analytic functions. </p> <p>I doubt that there are practical physical consequences distinguishing infinite order transitions from, maybe, tenth order ones. In particular, if the model is of a bulk substance made of smaller particles, then for sufficiently large derivatives (probably much larger than ten) the model breaks down. But the model is good in its domain and solving it yields smooth but not analytic functions.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114706#114706 Answer by Andreas Rüdinger for Does Physics need non-analytic smooth functions? Andreas Rüdinger 2012-11-27T21:29:36Z 2012-11-28T09:09:20Z <p>One example seems for me important and obvious, so I'm wondering, why no one has posted it so far: The time evolution of the state vector in quantum mechanics (Schrödinger's equation) is $|\psi(t)\rangle = \exp (-i \frac{H t}{\hbar}) \quad |\psi(0) \rangle$ and thus non-analytic in Plancks constant $\hbar$. This has important consequences in semiclassics ($\hbar \to 0$). </p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/120789#120789 Answer by Sönke Hansen for Does Physics need non-analytic smooth functions? Sönke Hansen 2013-02-04T17:45:45Z 2013-02-04T17:45:45Z <p>Here is a differential equation, a simple prototype of the equations of gas dynamics, which cannot be solved with analytic functions although coefficients and data are analytic: The <code>$C^1$</code> solutions of Burgers' equation <code>$u_t+uu_x=0$</code> are constant along characteristics <code>$\dot x=u$</code>. If the initial data satisfy <code>$u(0,a)&gt;u(0,b)$</code> for some <code>$a&lt;b$</code>, then the characteristics through these points which will intersect, and the solution will, after finite time, cease to be even a <code>$C^1$</code>-solution. Trying a Taylor series ansatz is not an adequate approach to solve Cauchy problems with analytic data for Burgers' equation (or other hyperbolic conservation laws).</p> <p>The lack of well-posedness of Cauchy problems in real analytic settings has already been pointed out in previous answers. It means that one should not confine one's attention to analytic solutions. </p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/128686#128686 Answer by ACL for Does Physics need non-analytic smooth functions? ACL 2013-04-25T06:54:05Z 2013-04-25T06:54:05Z <p>A beautiful <a href="http://en.wikipedia.org/wiki/Bertrand%27s_theorem" rel="nofollow">19th century theorem of mechanics</a> due to Joseph Bertrand says the following. <em>Consider the motion of a particle which is driven by a central force potential $V$</em>, that is, the potential $V$ is a function from the distance of the particle to some fixed origin and the force exerted on the particle is given by $\vec F=-\mathop{\rm grad}(V)$. <em>If (almost) all trajectories are periodic, then either $V(r)\propto 1/r$ (celestial mechanics), or $V(r)\propto r^2$ (harmonic oscillator).</em></p> <p>At some point of the proof one knows that all periods of all trajectories are rational multiples of a common period, and one needs to conclude that there is a common period. That part of the proof is always incorrect in the litterature I know. (In particular, Wikipedia's argument is not complete.)</p> <p>It is in fact easy to construct smooth real families of periodic functions with non-constant rational period. The only way I can correct the proof assumes that the potential is real analytic.</p>