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August 8, 2005

Surprises

We physicists tend to use ordinary English words, like “energy” or “field” in unconventional, technical ways. Mostly — even to a layman — it is recognizable, from the context, when a word like “energy” is being used in its technical, rather than its colloquial sense.

But, as I was reminded recently, that’s not always true, and can lead to some confusion. In particular, the use of the adjective “naïve” differs significantly from its colloquial meaning. In physics, a “naïve calculation” is one that omits some important effect, or makes some unwarranted simplification that, when properly understood, may significantly change the answer.

Which, in turn, reminds me of one of my favourite book, Rudolf Peierls’s Surprises in Theoretical Physics (and its sequel, More Surprises…). It’s a compendium of examples from various branches of Physics, where the naïve calculation yields the wrong answer, followed by a lucid explanation of what went wrong, and how a correct calculation fixes the problem.

To give you a taste, here’s a summary of one of my favourite “surprises,” that I sometimes talk about when I teach E&M.

Why isn’t a free electron gas paramagnetic?

In a uniform magnetic field, the trajectories of electrons are helices, which project to circles, when projected to the plane perpendicular to the applied magnetic field. Let v v_\perp be the projected velocity. The radius of the circle is r=mv eB r = \frac{m v_\perp}{e B} and the magnetic moment should be ev re v_\perp r, or μ=mv 2B \mu = \frac{m v_\perp^2}{B} (If you want the magnetic susceptibility, you would multiply this by the number density of electrons.)

That was easy. But, after a moment’s reflection, you realize that it’s complete garbage. This answer is independent of the electron charge, and it’s inversely proportional to the strength of the magnetic field. So it diverges for B0B\to 0!

What went wrong?

The first answer, as understood by H.A. Lorentz, and by Niels Bohr, is that you really need to think about a finite region. If you pick some rectangle in the plane, those orbits that lie entirely outside the rectangle do not contribute to the magnetic susceptibility of this rectangular region. Those orbits that lie entirely inside the rectangle, contribute paramagnetically, as described.

However, we made a mistake by neglecting the orbits which cross the boundary of the rectangle. We should only consider those portions of these orbits which lie inside the rectangle. These contribute a net current in the opposite sense (clockwise, instead of counterclockwise) from the electrons, whose orbits lie entirely inside the rectangle. This is a small current, but it encloses a large area (nearly the whole rectangle). The upshot, if you do it carefully, is that this boundary current precisely cancels out the effect of the electrons whose orbits lie inside the rectangle.

This is a somewhat gnarly calculation, and the sensitive dependence on the boundary conditions makes you think that we’re going about it the wrong way.

There’s a more robust calculation that yields the same result. The susceptibility μ=FB \mu = - \frac{\partial F}{\partial B} where FF is the Helmholtz free energy. In Boltzmann Statistics, Z=e βF=e βH(r,p)d 3rd 3p Z = e^{-\beta F} = \int e^{-\beta H(\mathbf{r},\mathbf{p})} d^3 \mathbf{r}d^3\mathbf{p} where H=12m(peA) 2+V(r) H = \frac{1}{2m}(\mathbf{p}- e \mathbf{A})^2 + V(\mathbf{r}) and V(r)V(\mathbf{r}) is the potential energy (which, say, confines the electrons to this rectangular region). The point is that, whereas our previous calculation was highly sensitive to the boundary conditions, calculating the free energy first, and then differentiating with respect to BB is more robust, as the free energy is much less sensitive to the boundary conditions.

Indeed, just by changing variables of integration from p\mathbf{p} to q=peA\mathbf{q}=\mathbf{p}-e \mathbf{A}, we see, immediately, that free energy is independent of BB, and hence the susceptibility vanishes.

Peierls then goes on to discuss Landau’s famous extension of this story to the quantum case. Still neglecting electron spin, one finds (using either Boltzmann statistics, as above, or Fermi-Dirac statistics)

(1)
μ=β 2e 212m 2B=13βμ B 2B \mu = - \frac{\beta \hbar^2 e^2}{12 m^2} B = -\frac{1}{3} \beta \mu_B^2 B

per electron.

So, now we see that, neglecting spin, the free electron gas in the quantum case, is diamagnetic. The magnitude of the above result is 1/31/3 the contribution due to electron spin, which is paramagnetic. So, including spin, we expect the electron gas to be paramagnetic.

Hmmm. So that poses another puzzle. Why are some metals diamagnetic?

In a periodic potential, the effective mass of the electrons in the conduction band will, in general, not be equal to the free electron mass, mm. This shifts the energies of the Landau levels. But the contribution of the spin still depends on the intrinsic magnetic moment, which depends on mm. For m effmm_{\text{eff}}\ll m, as in Bismuth, the contribution computed by Landau above wins, and one finds a large diamagnetic susceptibility.

Not all of the “surprises” have as many twists and turns as this one, but it’s all great stuff.

Posted by distler at August 8, 2005 4:53 PM

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Re: Surprises

This is interesting, although I sense a cultural barrier (between math and physics) that prevents me from appreciating the paradox, and therefore also its resolution.

It does remind me of my own pet favorite E&M problem, which I invented on my own when I learned that the momentum in an electromagnetic field is E cross B. Suppose that you have coil with a constant B inside. Suppose further that you place between two capacitor plates that generate a perpendicular E. (Perhaps coil is even square so that the plates are flush with the surface of the coil.) Then inside the coil there is a constant E cross B, while outside of the coil B is negligible. Where did this momentum come from?

Posted by: Greg Kuperberg on August 8, 2005 11:20 PM | Permalink | Reply to this

Surprises versus Paradoxes

There is, indeed, a cultural barrier. In math, if a chain of reasoning leads to a false conclusion, you call it a paradox. And the resolution is always the either one of the assumptions was false, or that one of the steps in the reasoning was incorrect.

In physics, the situation is muddier. We are always making some approximation. The question is: when is that approximation justified?

  • When can we neglect boundary effects?
  • When is treating the system as classical valid?
  • How good is the free electron gas approximation?

and so on.

The first, “naïve,” calculation got the answer hopelessly wrong. The “surprise” was that some approximation we thought was valid … wasn’t. Each subsequent refinement brought us closer to the truth.

We learned something about what the “right” way to think about the problem is (Compute the free energy, first. Then differentiate with respect to BB. If you try to differentiate under the integral sign, you run into the same sensitive dependence on the boundary conditions that Lorentz and Bohr found.). And we got a better feel for what the important effects at play are.

That really gets to the heart of what theoretical physics is about. And it’s why I love Peierls’s book.

Posted by: Jacques Distler on August 9, 2005 12:33 AM | Permalink | PGP Sig | Reply to this

Re: Surprises

No, the cultural barrier that I have in mind is not as philosophically significant as what you describe. I have papers in packing and covering problems, so I am certainly used to the idea of ignoring boundary terms. After all, in the limit of infinite volume, a negligible boundary effect shouldn’t be an approximation; it should limit to exactitude.

So in the abstract, many of the same intuitive ideas appear over and over again in different areas of mathematics and physics. Certainly the idea of ignoring boundary effects does.

But in practice, there is a more complicated, less interesting, more annoying problem. Namely, the conventions and jargon for possibly equivalent ideas in mathematics and physics becomes incompatible. For example, Rena related to me how a physics audience struggled with basic jargon in a colloquium talk given by a mathematician. The speaker referred to edges of a graph; unfortunately most of the audience thought that “edge” meant boundary!

In the case of your free electron gas paradox, my instinct would be to treat the electron gas as a continuous current function. Since it is translationally and rotationally invariant, the average current at any spot is zero. So it doesn’t create any magnetic field. Where is the surprise in that? To a PDE analyst, your naive calculation might seem more like a fallacy than a paradox.

On the other hand the quantum result of Landau really is a surprise. Here we run into a different incompatibility of conventions. It is not that mathematicians are unwilling to think about quantum mechanics; of course many of them do. But mathematicians view switching from classical to quantum systems as a sweeping change of topic. In physics, the quantum model is just a more accurate description of the same system.

Posted by: Greg Kuperberg on August 9, 2005 1:42 AM | Permalink | Reply to this

Re: Surprises

I am certainly used to the idea of ignoring boundary terms. After all, in the limit of infinite volume, a negligible boundary effect shouldn’t be an approximation; it should limit to exactitude.

But, indeed, that’s the first “surprise.” Lorentz, and Bohr showed that the “flaw” in the naïve calculation is that the boundary terms are not neglible. Indeed, they precisely cancel the “bulk” effect.

To avoid the sensitivity to boundary effects, you need to shift to the second procedure, of calculating the free energy before differentiating with respect to BB.

In the case of your free electron gas paradox, my instinct would be to treat the electron gas as a continuous current function. Since it is translationally and rotationally invariant, the average current at any spot is zero.

This, “continuum approximation” is an excellent way to get at the classical answer (assuming the continuous current distribution is a good approximation to the gas of electrons).

But Landau’s quantum mechanical solution actually harks back to the helical electron orbits discussed at the beginning, and quantizes those. One finds a set of energy levels, E(n,p)=p 22m+(n+1/2)eBm,n=0,1,2, E(n,p) = \frac{p^2}{2m} + (n + 1/2) \frac{\hbar e B}{m},\qquad n=0,1,2,\dots where pp is the momentum parallel to the applied magnetic field. And the upshot is that, in the quantum theory, the free energy is not independent of BB.

Posted by: Jacques Distler on August 9, 2005 2:25 AM | Permalink | PGP Sig | Reply to this

Re: Surprises

Upon reflection I partly agree with your comment about approximations being the difference between math and physics. Yes, mathematics also has its approximations. However, there aren’t as many of them because they come with more mental baggage. I am used to seeing an approximation together with an argument that it really is an approximation, whence it is usually called a “bound”. Often the upper bound and the lower bound are very different assertions.

The idea of the “continuum approximation” for the current of a charged gas is a case in point. Is it really an approximation? Since a gas is a statistical ensemble, my mental definition was that the current function is a probabalistic expectation, or average. Since Maxwell’s equations are linear, taking the expectation commutes with them. Hence I thought of it as exact and not an approximation.

The difference between approximations in physics and math is like the difference between approximations in cooking and preparing chemotherapy. Each for its own purpose.

Actually now that I think about it, the continuous argument makes Landau’s answer even more interesting or surprising. An electron sampled from the ensemble must be in some mixed quantum state which is rotationally and translationally invariant, but which nonetheless induces a magnetic field on average. Hmm…

While I think about this, what about my puzzle about the Poynting vector?

Posted by: Greg Kuperberg on August 9, 2005 9:45 AM | Permalink | Reply to this

Poynting

While I think about this, what about my puzzle about the Poynting vector?

Again, a cultural barrier. I don’t see the paradox.

Imagine I take a charged particle from one plate of the capacitor to the other, so that the electric field decreases. If I integrate the Lorentz force law, I see that the momentum imparted to the particle, while traversing the region between the plates is exactly equal to the momentum “lost” by the electromagnetic field.

So the momentum stored in the crossed electric and magnetic fields is real. I can extract it by moving charged particles about.

Posted by: Jacques Distler on August 9, 2005 11:31 AM | Permalink | PGP Sig | Reply to this

Re: Poynting

Maybe there is no real cultural barrier for either of these puzzles; maybe we’re just not fooling each other. The missing momentum is indeed in the charge carriers in the coil. Some people who have not had the practice of teaching E&M have had trouble finding the missing momentum in my exercise. The same may be true of the cancelling flux in your exercise. (Well, I never taught E&M either, but when you got into math vs. physics, them were fighting words!)

Maybe I can catch you with a pet math puzzle instead. How many Lie groups are there that consist of two circles? The problem is more fun if you write down a two-minute answer before thinking it through.

Posted by: Greg Kuperberg on August 9, 2005 12:11 PM | Permalink | Reply to this

Re: Poynting

Ooops! Forgot about this.

How many Lie groups are there that consist of two circles? The problem is more fun if you write down a two-minute answer before thinking it through.

Well, my first question would be: how many smooth manifolds “consist of two circles”? Depending on the definition of “consist,” I can think of two.

  1. The disjoint union of two circles
  2. The Cartesian product of two circles

Each of these can be endowed with an obvious Lie group structure.

  1. SO(2)× 2SO(2)\times\mathbb{Z}_{2}
  2. SO(2)×SO(2)SO(2)\times SO(2)

Are there any other Lie group structures we can impose? I don’t see any others, but I may be missing something.

Posted by: Jacques Distler on August 9, 2005 4:15 PM | Permalink | PGP Sig | Reply to this

Re: Poynting

I meant the disjoint union of two circles. Yes, SO(2) × ℤ2 is such a Lie group. O(2) is another one. So the most common two-minute answer that I have heard is that there are two such Lie groups. Some quite prominent people, who shall remain nameless, have given that answer.

They shall remain nameless because there is a third one. Although I am not sure, I suspect that it is relevant to string theory or at least conformal field theory. So I will refrain from saying what it is.

On the other hand, there is only one Lie group with the topology of a torus, as you describe.

Posted by: Greg Kuperberg on August 9, 2005 4:53 PM | Permalink | Reply to this

Re: Poynting

The possible semidirect products of SO(2) by Z_2 are given by the homomorphisms from Z_2 into Aut(SO(2)) = O(2). There is the trivial homomorphism which corresponds to the direct product SO(2) x Z_2. The order two subgroups of O(2) consist of the group generated by -I and and an infinite family of groups consisting of the identity and a matrix of the form (cos t sin t, -sin t cos t), where t is an arbitrary parameter. All members in the second case are Lie groups isomorphic to O(2). The first case is Greg’s mystery manifold.

Posted by: Richard on August 10, 2005 4:25 AM | Permalink | Reply to this

Two circle Lie group

Aut(SO(2))= 2Aut(SO(2))=\mathbb{Z}_2!
Inn(SO(2))=Inn(SO(2))=the trivial group
Out(SO(2))= 2Out(SO(2))=\mathbb{Z}_2

:):):)

The third “mystery Lie group” isn’t a semidirect product :) :)

If NN is a normal subgroup of GG, GG doesn’t always have to be the semidirect product of NN with G/NG/N :) :)

I guess “up to isomorphism” is implicit :) :)

So far, that took less than two minutes.

OK, after more than two minutes…,

Call the two circles + and - respectively. Each circle is coordinatized by an angle θ\theta which goes from 0 to 2π2\pi mod 2π2\pi. (+,θ 1)(+,θ 2)= (+,θ 1+θ 2) (+,θ 1)(,θ 2)= (,θ 1+θ 2) (,θ 1)(+,θ 2)= (,θ 1θ 2) (,θ 1)(,θ 2)= (+,π+θ 1θ 2)\array{\arrayopts{\colalign{right left}} (+,\theta_1)(+,\theta_2)=&(+,\theta_1+\theta_2)\\ (+,\theta_1)(-,\theta_2)=&(-,\theta_1+\theta_2)\\ (-,\theta_1)(+,\theta_2)=&(-,\theta_1-\theta_2)\\ (-,\theta_1)(-,\theta_2)=&(+,\pi+\theta_1-\theta_2) }

Posted by: Jason on August 10, 2005 1:35 PM | Permalink | Reply to this

Re: Two circle Lie group

I left out mod 2pi on the right hand side, but oh well…

But this really gives me an idea. It’s well known that we can have discrete gauge groups nonperturbatively, like on a lattice, for example or if we can UV-complete the theory with some larger gauge group which is spontaneously broken. We can also have Lie groups which aren’t connected for the gauge group and if such a Lie group doesn’t necessarily have to be a semidirect product…

Posted by: Jason on August 10, 2005 1:50 PM | Permalink | Reply to this

Use MarkdownMML, itex2MMLpara, or whatever

Jason,

I took the liberty of converting your comment to use the automatic TeX→MathML filter. This should make it much more readable.

I hope you don’t mind.

Posted by: Jacques Distler on August 10, 2005 1:55 PM | Permalink | PGP Sig | Reply to this

Re: Two circle Lie group

Jason: You found it, although you did not start your two minutes from the beginning of the question.

Let’s accept that spin groups are important in physics :-). Now SO(n) is to Spin(n) as O(n) is to Pin(n). (This is an obscene pun in French supposedly due to Serre.) But there are two Pin groups, Pin+(n) and Pin(n), depending on whether the square of a reflection lifts to the identity or minus the identity. The two Pin groups lead to two different kinds of “pinors” on non-orientable manifolds. Which I would suppose is relevant to string theory, although I don’t really know.

In particular, Pin±(2) are the two non-abelian Lie groups that consist of two circles. Pin+(2) is just isomorphic to O(2). Pin(2) appears, for example, as the subgroup of Spin(3) = SU(2), the fermionic rotational group in non-relativistic quantum mechanics, that stabilizes an axis.

I remember springing my question on a very capable representation theorist, a few minutes after talking about normalizers of Cartan tori in compact Lie groups. He still overlooked Pin(2).

Posted by: Greg Kuperberg on August 10, 2005 2:39 PM | Permalink | Reply to this

Application to String Theory

Pin ±(2)Pin_\pm(2) are central extensions of O(2)O(2) by 2\mathbb{Z}_2.

In Jason’s notation the multiplication rule for O(2)O(2) is (+,θ 1)(+,θ 2)= (+,θ 1+θ 2) (+,θ 1)(,θ 2)= (,θ 1+θ 2) (,θ 1)(+,θ 2)= (,θ 1θ 2) (,θ 1)(,θ 2)= (+,θ 1θ 2)\array{\arrayopts{\colalign{right left}} (+,\theta_1)(+,\theta_2)=&(+,\theta_1+\theta_2)\\ (+,\theta_1)(-,\theta_2)=&(-,\theta_1+\theta_2)\\ (-,\theta_1)(+,\theta_2)=&(-,\theta_1-\theta_2)\\ (-,\theta_1)(-,\theta_2)=&(+,\theta_1-\theta_2) } The homomorphism from Pin ±(2)Pin_\pm(2) to O(2)O(2) is, in both cases, given by (+,θ)(+,2θ),(,θ)(,2θ) (+,\theta)\mapsto (+,2\theta),\qquad (-,\theta)\mapsto (-,2\theta) and the kernel is 2={(+,0),(+,π)}\mathbb{Z}_2=\{(+,0),(+,\pi)\}.

As Greg said, Pin +(2)Pin_+(2) is just isomorphic to O(2)O(2), and Pin (2)Pin_-(2) is the group identified by Jason.

The closed-string sector of Type I string theory involves fermions on unoriented, compact 2-manifolds. Care to guess which of these two double-covers of O(2)O(2) is the relevant pinor group for Type I strings?

Posted by: Jacques Distler on August 10, 2005 5:02 PM | Permalink | PGP Sig | Reply to this

Re: Application to String Theory

In the spirit of my question, I will guess that it’s the more interesting one, Pin(2). Later I will cheat by looking in Polchinski or elsewhere.

Posted by: Greg Kuperberg on August 10, 2005 6:15 PM | Permalink | Reply to this

Re: Application to String Theory

That’s correct. What you’ll find discussed in Polchinski is the special case of flat worldsheets, where the structure group reduces from O(2)O(2) to 2={(+,0),(,0)}\mathbb{Z}_2=\{ (+,0),(-,0)\}. This lifts to the four element group ΓPin(2)\Gamma\subset Pin(2), Γ={(+,0),(,π),(+,π),(,0)} \Gamma=\{(+,0),(-,\pi),(+,\pi),(-,0)\} which is either 2× 2\mathbb{Z}_2\times\mathbb{Z}_2 or 4\mathbb{Z}_4, depending on which Pin(2)Pin(2) group we are talking about.

You can find the action of Γ\Gamma on the fermions in Polchinski or Green, Schwarz and Witten.

Posted by: Jacques Distler on August 10, 2005 8:20 PM | Permalink | PGP Sig | Reply to this

Re: Application to String Theory

On a more fundamental level, you might ask, why String Theory picks out one or the other of the two Pin(2)Pin(2) groups for the worldsheet theory.

That has to do with the existence of a global obstruction to lifting an O(n)O(n) bundle to a Pin ±(n)Pin_\pm(n) bundle. For Pin +(n)Pin_+(n), the obstruction is the 2nd Stiefel-Whitney class, w 2w_2. For Pin (n)Pin_-(n), the obstruction is w 2w 1w 1w_2-w_1\cup w_1.

On P 2\mathbb{R}P^2, for example, the latter obstruction vanishes, while the former does not. This singles out Pin (2)Pin_-(2) as the “correct” PinPin group for the Type I string.

Posted by: Jacques Distler on August 11, 2005 2:05 AM | Permalink | PGP Sig | Reply to this

Re: Application to String Theory

I don’t quite understand why this settles which Pin group is relevant to Type I string theory. Why can’t you just sum over “pun” surfaces, so that there would be two Type I string theories? From the way that you summarize it, I infer that there must be some inconsistency if you use the wrong group.

Another answer that you might have given me is that the two Pin groups give you equivalent string theories.

Posted by: Greg Kuperberg on August 11, 2005 9:28 AM | Permalink | Reply to this

Re: Application to String Theory

The P 2\mathbb{R}P^2 amplitude enters at the same order in string perturbation theory as the disk. The disk amplitude doesn’t care about the choice of Pin ±(2)Pin_\pm(2), but it does depend on the choice of gauge group. It yields a dilaton tadpole whose magnitude is proportional to “NN” of SO(N)SO(N).

The P 2\mathbb{R}P^2 amplitude does care about the choice of Pin ±(2)Pin_\pm(2). For Pin (2)Pin_-(2), it also yields a dilaton tadpole whose sign and magnitude is such as to cancel the disk tadpole for SO(32)SO(32) — the same gauge group picked out by anomaly cancellation.

For Pin +(2)Pin_+(2), there’s an obstruction to defining pinors on P 2\mathbb{R}P^2. Rather than producing a dilaton tadpole which cancels the disk amplitude, the P 2\mathbb{R}P^2 amplitude produces a Ramond-Ramond tadpole. Spacetime supersymmetry is broken, and the theory is inconsistent.

Posted by: Jacques Distler on August 11, 2005 10:31 AM | Permalink | PGP Sig | Reply to this

CPT and Pin(3,1)

Does this mean that the CPT theorem has to be generalized? Most books which go over the CPT theorem automatically assume the “standard” double cover of O(3,1) (why?), but I think there are four different pin groups which are double covers of O(3,1). The differences will only involve the spinors, but most QFTs do include spinors.

Posted by: Jason on August 11, 2005 3:26 PM | Permalink | Reply to this

Re: CPT and Pin(3,1)

Cécile DeWitt-Morette had some thoughts on this. I don’t remember the upshot, though.

Posted by: Jacques Distler on August 12, 2005 11:44 AM | Permalink | PGP Sig | Reply to this

Re: Surprises


Nice, enjoyable article.

Speaking of paradoxes, I used to be endlessly confused about conservation of momentum. The textbooks (Siegel terms them history texbooks) would talk about mechanical momentum, electromagnetic momentum, etc and how various paradoxes arise when one speaks of conservation of mechanical momentum etc…

But later on I realized that the right way of understanding all of it is via the Lagrangian. From there, one clearly sees that it is the canonical momentum that is fundamental. Of course, symmetries and conservation laws are transparent in the Lagrangian formalism. The advanced way of doing things is really the simplest way.

On an unrelated note, does string theory go beyond the Lagragian (in terms of fundamental concept from which all follows), which is the starting point of physics upt to QFT?

Posted by: mfa on August 9, 2005 9:57 AM | Permalink | Reply to this

Lagrangians

On an unrelated note, does string theory go beyond the Lagragian (in terms of fundamental concept from which all follows), which is the starting point of physics upt to QFT?

If I really knew what String Theory was, I would be in a better position to answer (as well as rich and famous :-).

The attempt to come up with a Lagrangian formulation of String Theory is called String Field Theory.

Open String Field Theory has had some notable successes. Closed String Field Theory is a murkier subject. Barton Zwiebach is the world’s expert on the subject. And I don’t think he’s totally happy with it, though he did report some interesting progress at Strings 2005.

Posted by: Jacques Distler on August 9, 2005 12:25 PM | Permalink | PGP Sig | Reply to this

Re: Lagrangians

Open String Field Theory has had some notable successes. Closed String Field Theory is a murkier subject.

And this is just the bosonic string. Various flavors of Berkovits-like super(or heterotic)string field theories seem even murkier, even though there are some nice underlying principles.

I find this bothersome. Would have hoped that the superstring is in the end prettier than the bosonic string.

Posted by: Urs on August 10, 2005 1:57 AM | Permalink | Reply to this

Re: Surprises

One of my favorite abuses of language is the word ‘formal’.

When physicists ‘proceed formally’, it basically means we have an equation that is completely nonsensical mathematically, but we sweep those technicalities aside and go with our gut. For instance we might do an operation that is prohibited mathematically on some sort of operator equation, but typically ‘a miracle occurs’ and often the right answer comes out.

Posted by: Haelfix on August 9, 2005 11:08 AM | Permalink | Reply to this

Re: Surprises

Ironically, in physics a formal argument is less reliable, while in mathematics a formal argument is more reliable.

Posted by: Greg Kuperberg on August 9, 2005 4:54 PM | Permalink | Reply to this

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