Dimensional Analysis in Mathematics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:57:22Z http://mathoverflow.net/feeds/question/63749 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics Dimensional Analysis in Mathematics Alex R. 2011-05-02T23:02:13Z 2011-10-28T04:14:41Z <p>Is there a sensible and useful definition of units in mathematics? In other words, is there a theory of dimensional analysis for mathematics?</p> <p>In physics, an extremely useful tool is the <a href="http://en.wikipedia.org/wiki/Buckingham_Pi" rel="nofollow">Buckingham Pi theorem</a>. This allows for surprisingly accurate estimates that can predict on what parameters a quantity depends on. Examples are numerous and can be found in this short <a href="http://www.math.ntnu.no/~hanche/notes/buckingham/buckingham-a4.pdf" rel="nofollow">reference</a>. One such application (pages 6-7 of the last reference) can derive the dispersion relation exactly for short water ripples in deep water, in terms of surface tension, density and wave number. In this case an exact relation is derived, but in general one expects to be off by a constant. The point is that this gives quick insight into an otherwise complex problem.</p> <p>My question is: can similar techniques be used in mathematics? </p> <p>I envision that one application could be to derive asymptotic results for say, ode's and pde's under certain asymptotic assumptions for the coefficients involved. For any kind of Fourier analysis, dimensions naturally creep up from physics if we think about say, time and frequency. I find myself constantly using basic dimensional analysis just as a sanity check on whether a quantity makes sense in these areas.</p> <p>Otherwise, let's say I'm working on a problem involving some estimate on a number theoretic function. If I have a free parameter, can I quickly figure out the order of the quantity i'm interested in in terms of my other fixed parameters?</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63757#63757 Answer by Michael Hardy for Dimensional Analysis in Mathematics Michael Hardy 2011-05-03T00:41:57Z 2011-05-03T00:41:57Z <p>If you didn't know that the area of a circle of radius $r$ is $\pi r^2$, you would still know that it's some constant times $r^2$ because the area must be proportional to the squares of any distances involved. Similarly, if Heron's formula giving the area of a triangle as $$A = \frac14 \sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$ where $a,b,c$ are the side lengths, were not homogeneous of degree 2 in $(a,b,c)$, you would know there's a mistake because $a,b,c$ are distances and $A$ is an area. And if you find that some other expression in $a,b,c$ is homogeneous of degree $0$, you conclude it's a trigonometric function of the angles well in advance of knowing the specifics.</p> <p>Could those be considered instances of the same sort of reasoning?</p> <p>(One interesting thing about the proof of the Buckingham pi theorem is that it's an instance of the use of vector spaces over the field of rational numbers in physics. I'd otherwise have guessed that such a thing might never happen.)</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63761#63761 Answer by Brian Borchers for Dimensional Analysis in Mathematics Brian Borchers 2011-05-03T01:55:37Z 2011-05-03T01:55:37Z <p>You might want to look at the book "Street Fighting Mathematics" by Sanjoy Mahajan, which contains lots of examples of heuristic approaches of the sort you're thinking about. </p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63762#63762 Answer by Todd Trimble for Dimensional Analysis in Mathematics Todd Trimble 2011-05-03T02:07:11Z 2011-05-03T02:07:11Z <p>This may be somewhat obliquely along the lines you are asking about, but I think it's interesting enough that it deserves to be made public. </p> <p>My friend James Dolan has been developing with a number of other people a big program in which large portions of algebraic geometry are interpreted and explained in terms of concepts from categorical logic. A basic chapter in this program is one he explicitly identifies as "dimensional analysis", which in his rendering is another term for a general theory of line bundles or more general line objects in symmetric monoidal categories -- each line bundle can be considered a "dimension", and quantities of that dimension are sections of that bundle. Dimensions of course multiply (correspondingly, line bundles are tensored), and are the objects of a symmetric monoidal category enriched in a category of vector spaces, which he calls a <i>dimensional category</i>. Jim proposes to study objects in algebraic geometry (schemes, stacks, etc.) in terms of the dimensional categories that are attached to them, and representations of them in other symmetric monoidal categories. These often take the form "the dimensional category attached to (some specified important scheme studied by algebraic geometers) $X$ is the universal dimensional category such that...". </p> <p>Jim gave a number of very accessible and thought-provoking introductory lectures a few years ago at UC Riverside. Videos of those lectures (as well as a brief written description of his program) can be accessed <a href="http://ncatlab.org/jamesdolan/published/Algebraic+Geometry" rel="nofollow">here</a>. John Baez (one of Jim's collaborators) has also written on this program at the n-Category Caf&eacute;, as for example <a href="http://golem.ph.utexas.edu/category/2009/06/algebraic_geometry_for_categor.html" rel="nofollow">here</a> and I believe also in week 300 of This Week's Finds. </p> <p>Not that this program is <i>completely</i> novel, by any means. Right now Jim and I are discussing toric varieties, and I note that Vladimir Arnol'd once remarked on the spiritual kinship between dimensional analysis and toric varieties, in connection with a problem he solved as a young boy, in <a href="http://www.ams.org/notices/199704/arnold.pdf" rel="nofollow">this article</a>. There was a brief Math Overflow discussion about this, <a href="http://mathoverflow.net/questions/52585/on-a-schoolchild-puzzle-of-v-i-arnold-re-toric-varieties" rel="nofollow">here</a>. </p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63769#63769 Answer by jc for Dimensional Analysis in Mathematics jc 2011-05-03T03:33:02Z 2011-05-03T03:56:58Z <p>Regarding the use of dimensional analysis to derive asymptotic solutions to PDE, dimensional analysis techniques can be seen as just a small part of a certain well-developed theory for "self-similar" asymptotic solutions with lots written in e.g. the fluid mechanics literature. I learned most of this in "applied mathematics" courses, for what it's worth. If anyone knows where the stuff below is treated from a higher-tech point of view, I'd love to hear of it.</p> <p>For example, if one is trying to solve an equation of the form </p> <blockquote> <p>$\partial_t u(x,t)=F[u]$ </p> </blockquote> <p>where $F$ is some typically nonlinear differential operator, one often believes for physical reasons that a solution of the type</p> <blockquote> <p>$u(x,t)=(t-t_0)^\alpha H\left(\frac{x-x_0}{(t-t_0)^\beta}\right)$</p> </blockquote> <p>(called self-similar, as the function $H$ is fixed under certain simultaneous scalings of time and space) is likely. For instance if the system is approaching some kind of singularity at $x=x_0$, $t=t_0$, the length and time scales which are going to dominate will be just those describing the distance to the singularity in space-time.</p> <p>By using this ansatz of this type, one often finds that the PDE we originally had to solve is reduced to an ODE for the function $H$.</p> <p>It's important to note that naïve dimensional analysis can only get you $\alpha$ and $\beta$ in a small subset of these asymptotic solutions due to the existence of so-called self-similarities of the second kind!</p> <p>An idiosycratic but readable book along these lines that I enjoyed is G.I. Barenblatt's <em>Scaling, self-similarity, and intermediate asymptotics</em>. As far as I know, Barenblatt introduced and emphasized this division of "self-similar" intermediate asymptotic solutions into two types. </p> <p>In self-similarities of the first kind, naïve dimensional analysis works. Consider for instance the self-similar solution $u(x,t)=\frac{u_0}{\sqrt{4\pi t}} e^{-x^2/4Dt}$ to the diffusion equation $\partial_t u-D\partial^2_{xx}u=0$. We can derive this solution by assuming that the solution depends only on the dimensionless combination $x^2/4Dt$, then the PDE can be reduced to an ODE.</p> <p>In self-similarities of the second kind, non-trivial scalings appear, where the exponents in power laws turn out to be determined by nonlinear eigenvalue problems. Roughly speaking, one has a continuum of $\alpha$ and $\beta$ values which work, and imposing proper boundary conditions yields the ones which are of relevance. I believe you get irrational values of $\alpha$ and $\beta$ by using some "microscopic" length scales from short-distance / short-time cut-offs in your system. <strong>(Sorry, I'll fill in details / an example when I get my hands on my copy of the book)</strong>. </p> <p>For a recent review with plenty of fluid mechanics examples, see <a href="http://iopscience.iop.org/0951-7715/22/1/R01" rel="nofollow">this paper by Eggers and Fontelos</a>.</p> <p>Intriguingly, this seems to be analogous to dimensional analysis in quantum / statistical field theory, where whenever a phase transition is described by "mean field theory" (i.e. a Gaussian RG fixed point), all the critical exponents governing the behavior near the phase transition can be derived from the dimensional analysis arguments (I think some books say that the scaling dimensions are equal to the engineering dimensions). When the RG fixed point governing the phase transition is nontrivial, then there are anomalous dimensions which appear essentially by the same mechanism as above. Nigel Goldenfeld emphasizes this point of view in his book <em>Lectures on Phase Transitions and the Renormalization Group</em>. You might also find his papers connecting the two subjects interesting, e.g. <a href="http://pre.aps.org/abstract/PRE/v54/i1/p376_1" rel="nofollow">L.Y. Chen, N. D. Goldenfeld and Y. Oono. The renormalization group and singular perturbations: multiple-scales, boundary layers and reductive perturbation theory. Phys. Rev. E 54, 376-394 (1996)</a>.</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63771#63771 Answer by Deane Yang for Dimensional Analysis in Mathematics Deane Yang 2011-05-03T03:47:07Z 2011-05-03T03:47:07Z <p>I consider dimensional analysis to be an extremely useful and powerful tool in pure mathematics. A lot of mathematics, especially PDE's and differential geometry, comes from or is closely related to the real world, so that dimensional analysis is relevant shouldn't be so surprising. And yet it did come as a surprise to me, because as far as I can tell pure mathematicians almost never discuss it in a systematic manner. When you learn about PDE's, you do learn about scaling arguments for figuring out what exponents or norms are appropriate in a given setting (see article by Tao cited by Knutson). However, I find that by keeping track of "dimensions" and "units", I am able to spot errors in my formulas, calculations, and inequalities. Sometimes, this kind of analysis even tells you immediately what the correct formula or inequality is. I would like to see a systematic exposition of how to use "dimensional analysis" in areas of mathematics such as differential geometry.</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63781#63781 Answer by Denis Serre for Dimensional Analysis in Mathematics Denis Serre 2011-05-03T07:11:30Z 2011-05-07T20:24:05Z <p>Dimensional analysis tells you which norms of functional spaces must be compared. Often, this results into useful inequalities. A basic example is the Gagliardo-Nirenberg-Sobolev inequality</p> <p><code>$$\|f\|_{L^q}\leq C_{p,n}\|\nabla f\|_{L^p},\qquad\forall f\in\mathcal{D}(\mathbb{R}^n),$$</code></p> <p>which is valid only if $1\le p&lt; n$ and (dimensional analysis) $$\frac1q=\frac1p-\frac1n.$$</p> <p>A more complicated situation is that of Moser's inequalities, which involve more than two norms. Among them is the Ladyzhenskaia inequality</p> <p><code>$$\|f\|_{L^4}\leq C\|f\|_{L^2}^{1/2}\|\nabla f\|_{L^2}^{1/2},\qquad\forall f\in\mathcal{D}(\mathbb{R}^2).$$</code></p> <p>This one was fundamental in the proof by C. Foias and G. Prodi of the global regularity of the Navier-Stokes solutions in two space dimension. This led me to mention that a modern attempt to prove the same result in three space dimension (1M dollar problem) is to study the Navier-Stokes equation within functional spaces that are <em>scaling invariant</em>, again a concept that comes from dimensional analysis.</p> <p>Finally, an even more advanced situation arises when you deal with norms involving two variables of different nature, typically a time variable and space variables. Read about the <a href="http://wiki.math.toronto.edu/DispersiveWiki/index.php/Strichartz_estimates" rel="nofollow">Strichartz estimates</a>. Their validity depends upon an equality between the parameters that reflects again dimensional analysis. But then you assume that the Fourier transform of the functions is supported by a submanifold with non-zero curvature.</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/63786#63786 Answer by Stefan Waldmann for Dimensional Analysis in Mathematics Stefan Waldmann 2011-05-03T09:25:29Z 2011-10-26T21:56:57Z <p>Dimensional analysis can be viewed as the study of graded objects in algebra. The grading then corresponds to "counting the units" in a precise way. There are of course many examples and I believe that morally, every graded (say $\mathbb{Z}$-graded) algebra/vector space/etc can be viewed as collection of objects having an intrinsic dimension. If you have a graded algebra then this just means that the dimensions mutliply correctly. If you have a grading on a vector space and a bilinear map (product, Lie bracket...) which is still homogeneous but not respecting the grading additively then the bilinear map itself carries a fixed dimension:</p> <p>On example is the polynomial algebra $\mathbb{C}[x,y]$ with its usual $\mathbb{Z}$-grading by the total polynomial degree. Then the canonical Poisson bracket is determined by $\lbrace x, y \rbrace = 1$ and hence has dimension $-2$ times the units of the generators. Relabelling then into $q$ and $p$, you have the Poisson bracket of classical mechanics with its usual dimension being that of an "inverse action".</p> <p>More sophisticated examples can be found e.g. in differential geometry where you have zillions of graded algebras/spaces arising naturally. Staying in the realm of Poisson geometry, the canonical Poisson bracket on a cotangent bundle has "momentum degree $-1$": Mathematically, this means that one first has an Euler vector field $E = \sum_i p_i \frac{\partial}{\partial p_i}$ on $T^*M$ as on every vector bundle (where the $p_i$ are fiber coordinates). Heuristically, this vector field "counts" how many $p$'s you have. Then the Poisson tensor is something like $\pi = \sum_i \frac{\partial}{\partial q^i} \wedge \frac{\partial}{\partial p_i}$ which satisfies $\mathcal{L}_E \pi = - \pi$, making the above statement precise.</p> <p>In general, associated to a grading one can attach the grading "derivation" (or better just operator) which satisfies $deg = k id$ on the homogeneous components $V^k$ of the graded space $V$. In the above case, $deg$ is just the Euler vector field and the graded spaces are the tensor fields on $T^*M$...</p> <p>I hope that these examples have convinced you that a "dimensional analysis" happens quite often and naturally in many areas of mathematics. The associated grading operators usually play an important role and help a lot. The least is certainly a "self-correcting" aspect, that handling dimensions correctly avoids stupid errors.</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/64228#64228 Answer by Kevin O'Bryant for Dimensional Analysis in Mathematics Kevin O'Bryant 2011-05-07T19:27:13Z 2011-05-07T19:27:13Z <p>Here is a cute "dimensional analysis" proof of the Pythagorean theorem. Consider a right triangle with legs $a,b$, hypotenuse $c$, and angle $t$ between $a$ and $c$. Let $A(t,c)$ be the area of the triangle. </p> <p>Now, the quantity $A(t,c)/c^2$ is dimensionless, and so $$\frac{A(t,c)}{c^2}=\frac{A(t,b)}{b^2}=\frac{A(t,a)}{a^2}.$$ By dropping an altitude from the right-angle corner to the hypotenuse, we cut the triangle into two smaller similar triangles: $$A(t,c)=A(t,a)+A(t,b)=\frac{a^2}{c^2}A(t,c)+\frac{b^2}{c^2}A(t,c) ,$$ whence $$c^2=a^2+b^2.$$</p> <p>The moral here is that "dimensional analysis" is physics code for "exploiting symmetry across scales".</p> http://mathoverflow.net/questions/63749/dimensional-analysis-in-mathematics/79230#79230 Answer by Phil Isett for Dimensional Analysis in Mathematics Phil Isett 2011-10-27T04:08:39Z 2011-10-28T04:14:41Z <p>People have mentioned so far how dimensional analysis is fundamental in many inequalities in analysis, especially estimates (and even formulas) which come from partial differential equations. I want to elaborate on this point. For example, in ${\mathbb R}^{n + 1}$ ($n$ spatial dimensions), look at an equation like the wave equation $(-\partial_t^2 + \Delta) u = f$ -- for which we have the formal units $U/T^2 \sim U/X^2 \sim f$ where $T$ is the unit for the time variable and $X$ is the unit for the space variable. or the Schrodinger equation $(i \partial_t + \Delta)u = f$ (for which $U/T \sim U / X^2 \sim f$). The rigorous meaning of these units for Schrodinger is that for $\lambda \neq 0$, we can rescale a given solution to construct another solution $u(t/\lambda^2, x/\lambda)$ whose forcing term is $\lambda^{-2} f(t/\lambda^2, x/\lambda)$ (here we are imagining $\lambda$ has the units of $X$). (Similarly, scaling in the $x$ variable alone, we can change the equation to remove or reinsert the physical constants which usually appear in these equations.) Consider a large forcing term, e.g. $f$ belongs some $L^p$ space or mixed space-time $L^p$ space; you use dimensional analysis to figure out to what space we could possibly guarantee the solution $u$ lives in -- the goal here is to make rigorous the idea that $u$ depends continuously on $f$, and thanks to some abstract facts of functional analysis, proving this continuous dependence for a linear equation can essentially be achieved by no other means than by proving a (Strichartz) estimate such as </p> <p>$(\int |u|^r dt dx)^{1/r} \leq C (\int |f|^p dt dx)^{1/p}$ </p> <p>for some constant $C$ independent of $f$. Together with linearity, this estimate when applied to the difference of nearby $f$'s, says, if true, that as $f$ varies in $L^p$, $u$ varies continuously in $L^r$. </p> <p>The function space norms themselves have units'': e.g. $||u||_{L^2} = (\int |u|^2 dt dx)^{1/2}$ has units of $u$ times $T^{1/2}X^{n/2}$ where $T$ are the time units and $X$ are the spatial units ($T \sim X$ for the wave equation, but $T \sim X^2$ for Schrodinger). One can use these "units" to figure out what kind of estimates might possibly be possible. E.g. for Schrodinger, </p> <p>$(\int |u|^r dt dx)^{1/r} \leq C (\int |f|^p dt dx)^{1/p}$ </p> <p>would be impossible if the units were compatible, the left hand side is like </p> <p>$U (TX^n)^{1/r} \sim UX^{(n+2)/r}$ </p> <p>whereas the term with $f \sim (U/X^2)$ has dimensions $(U/X^2)\cdot(TX^n)^{1/p} = U(X^{-2+(n+2)/p})$. </p> <p>So dimensional analysis tells us that we have no chance of proving this kind of estimate unless $(n+2)/p - 2 = (n+2)/r$; otherwise one could rescale any solution to produce a counterexample. Similar considerations apply to the problem of determining in what sense $u$ can depend continuously on its Cauchy data.</p> <p>The dimensional analysis is very far from a proof of the estimate, which is not always true and requires a real understanding of, say, the dispersion properties of the specific equation at hand (you could have been looking at a different equation with the same scaling properties but which is qualitatively very different). Denis Serre mentioned how these dispersion properties can be read off from the curvatures of the parabola (for Schrodinger) and cone (for wave) in frequency space, although the proofs also bring in complex interpolation theory to also take into account the energy, and to get the most sharp results you need an interpolation scheme which treats individual frequency scales differently. You also need to look at norms which are different in the space and time variables, because the equation also distinguishes these variables. (The space of functions whose "energy" or "mass" is bounded in time gives a particularly natural spacetime norm.) Suffice to say, dimensional analysis is far from enough to justify a bound.</p> <p>For nonlinear problems with some kind of scaling symmetry, the units of the solution and the time or space dimensions can be tied together. E.g. consider the incompressible Euler equation without force for an unknown velocity field $\partial_t u + (u \cdot \nabla) u = - \nabla p$, $\nabla \cdot u = 0$. Then formally $U/T \sim U^2/X \sim P/X$ or $U \sim X/T$ (which is good for a velocity'' field), which means that given a solution $u(t,x)$ you can rescale to obtain a new solution $\lambda^{b-a} u(t/\lambda^a, x/\lambda^b)$ with pressure $\lambda^{2(b - a)} p(t/\lambda,x/\lambda)$. After all, if $\lambda^b \sim X$ and $\lambda^a \sim T$, then this rescaled $u$ looks like has units of $U$ and $P$ appears to have units of energy $U^2$. Any estimate for the Euler equations has to be consistent with this scaling, so it has consequences for studying the equation (of course, not so many a priori estimates exist, which is another problem...). On the other hand, like Deane Yang said, it's very good for dummy-checks.</p> <p>I guess all I've said is that the equations of physics often have a scaling symmetry, and due to this symmetry dimensional analysis is of use for rigorous mathematical treatment of the equation. (But that's no surprise since everyone's talking about the same equation.) But in mathematics the idea that inequalities must be consistent with a scaling symmetry goes a long way. For example, consider the Hölder inequality</p> <p>$\int f(x) g(x) d\mu(x) \leq ( \int |f(x)|^p d\mu(x))^{1/p} (\int |g(x)|^q d\mu(x))^{1/q}$</p> <p>This inequality is valid for any measure $\mu$; so you could be integrating over a surface or just taking a discrete weighted sum or something and it's still true. In particular, it is true for $\mu$ and true for the renormalized measure $\mu/\lambda$. If you compare "units", the left hand side is like $f g \mu$ and the right hand side is like $f g \mu^{1/p + 1/q}$. Thus, for Hölder's inequality to be true for arbitrary measure spaces, we see we need $1/p + 1/q = 1$. Here I am talking about abstract measure spaces -- something which has nothing to do with a physical problem or Euclidean spaces. Terry Tao has harmonic analysis notes where he uses this symmetry to renormalize $\mu$ and reduce the proof of Hölder's inequality to the case where $g$ is not even present, which allows one to view Hölder's inequality as an interpolation statement. The point of view I've expressed in great part derives from and is elaborated in those notes, which can be found <a href="http://www.math.ucla.edu/~tao/247a.1.06f/" rel="nofollow">here</a>.</p> <p>But from a broader mathematical viewpoint, dimensional analysis is probably only one example of paying attention to a group of symmetries (not just scaling symmetries). But at the moment, I cannot think of a particularly good example to illustrate this point. Probably one can also be found in the linked notes.</p>