A Fun with Deltas and ##L^2##: Physicists vs Mathematicians

[George Jones]

It does have a derivative

This is one of the nice properties of distributions. From "Fourier Analysis and its Applications" by Folland

"It is possible to extend the operation of differentiation form functions to distributions in such a way that every distribution possesses derivatives of all orders that are also distributions."

 

[secur]

In some of your posts, you are trying to make up maths on the fly.

It may seem that way at times. In fact, I'm trying to remember math on the fly. I need to find my old textbooks; pretty sure they're in the attic somewhere. However no doubt my math is usually right. Much bigger problem, I have to adjust to your language, which is often different from the way we used to do it.

But the biggest problem I'm having is adjusting to the fact that, sometime in the last 40 years, all the physicists morphed into pure mathematicians! And, predictably, you take these details more seriously than real mathematicians do.

Here's an analogy. Real mathematicians are like professional soldiers who might casually chuck a hand grenade to each other, since they know precisely when it's safe to do so. Whereas you're like reserves who've trained only on duds. When you have to deal with a real one you put on bomb-handling equipment before even looking at it.

Thus when I'm in a situation where a function like our example "g" above can't possibly occur (like, physics), I might casually assume the additive identity is unique, knowing that if the square of a physical function integrates to 0 it has to be 0 everywhere. But a week-end warrior has to first take equivalence classes (under the SI norm) - then invoke the Axiom of Choice to select a representative member - then carefully use the result, praying it won't blow up in his face.

My opinion is: when a physicist starts worrying about AC he's seriously off track! But if I'm wrong I want to be set straight.

Where in all of physics do you meet with a non-physical function like our example g, because you need infinite precision? And, tell me a real physical situation where you'll get the wrong answer if you assume - or, don't assume - AC. (By the way in most of math we simply assume it.)

Suppose I boil water, stick in a thermometer, and see the temperature is 100. Now, suppose I realize - with a shock - that AC is not implied by Zermelo–Fraenkel axioms! Oh no - better check that temperature again. So I re-boil the water, and look at the thermometer. Will it still read 100? What if I hold the thermometer at precisely pi inches (with infinite precision) below the surface - will it make a difference?

Actually I know the answers to those questions: "no".

But there must be some physical situation where infinite precision and Axiom of Choice play a part, or else you wouldn't care about such minutiae. So, please tell me where that situation arises.

A physics book would use the constant of integration and move on, without worrying about equivalence classes.

Now you're making sense

L2L2L^2 spaces are part of every undergraduate education in physics.

I hope, in the midst of all this abstract math, you find time to teach those poor undergraduates a little physics as well. But I can't help wondering, where in all of physics do you use, as a norm, the 5th root of the 5th powers? Or L3, or anything other than L1, L2, and L-infinity?

By the way, I didn't notice the difference between regular L and script L, so I probably said something incomprehensible. Sorry about that.

Mathematically rigorour is indispensable in modern research in physics.

Sure it is - within reason. Dirac was a great mathematician, and he was just as rigorous as he had to be - no more, no less. Please tell me what physical situation requires infinite precision, or Axiom of Choice, to analyze. No details needed, just a general description of where, and why, you'll get the wrong physical answer if you don't handle these details right.

Here's the type of answer I'd like to see:

"Well, the number of multiple universes, considering Guth's inflation and Everett's work, is clearly Aleph-2. And, in order to show that QM probabilities really come out to the (complex) square of the wave function, we have to integrate over all the infinite universes in which the different experimental outcomes occur. To select those universes from this (very large) set, we need the Axiom of Choice. See how obvious it is?"

Or,

"we know exactly what conditions occur 10-33 meters from a Kerr BH ring singularity. But we need to see what happens at the singularity itself, whose coordinates are known with infinite precision (of course). If we don't exclude functions like g - we might get any answer at all, due to spurious functions in our solution space. We might calculate that the wormhole takes us back to August 12, 1066, at 10:30 in the morning. But when we actually go through the wormhole, it turns out to be 10:31! That's what can happen without rigorous mathematics."

Please provide a common-sense, real-world, 3-line example like these samples, with as much hand-waving as necessary. Please don't tell me I'll understand these things when I'm older, now eat those veggies.

 

[micromass]

OK secur, you clearly believe physicists should be able to say incorrect things and get away with it. Got it.

Why is it important to say that $L^2$ is a Hilbert space and $\mathcal{L}^2$ is not, etc? It's because being precise means easier communication between experts. Saying that $L^2$ is defined by equivalence classes isn't even heavy math! It's something every physicist can easily understand and grasp.
I agree we shouldn't get lost in mathematical details. But sometimes those details really do shine a nice light on certain stuff.

 

[samalkhaiat]

Mathematically rigorour is indispensable in modern research in physics.

Really? So, how come our most accurate and successful theory (QED) is built on mathematical “gibberish”? Why are there no room for improvement left concerning the purely electromagnetic phenomena?
The “wish” to understand the QED-miracle in terms of a rigorous mathematical structure was shared by many physicists in the early 1950’s, (so your view is rather old not modern). That was the motivation for a development subsumed under the name “axiomatic QFT”, merging into “constructive field theory” in the late 1960’s. In the last 70 years, an army of great mathematicians tried but failed to provide a rigorous reformulation of QED that can account for its miracles. That, I believe, was a waste of effort because the great and essential progress came ultimately from elsewhere, namely from applying the same mathematical “gibberish” of QED to non-abelian groups, and thus covering a wider range of high energy physics.
Physics, my friend, is an ill-defined mathematical structure, for one cannot associate a unique and well-posed mathematical problem for every experimental question. Rigorous mathematic is good, very good indeed. However, physicists should not expect too much of it all the times.

 

[secur]

OK secur, you clearly believe physicists should be able to say incorrect things and get away with it. Got it.

Great! All I, or any reasonable person, has to insist on is that his point be understood - not agreed to.

Why should physicists - and mathematicians, and any group of experts - be allowed to say incorrect things? Two conditions must be satisfied. One, it's simpler that way. Why be incorrect if it's more trouble? Two, we all know what we mean.

Why is it important to say that L2L2L^2 is a Hilbert space and L2L2\mathcal{L}^2 is not, etc? It's because being precise means easier communication between experts. Saying that L2L2L^2 is defined by equivalence classes isn't even heavy math!.

It's precisely among experts where precision is not necessary, because they (should) all be on the same page.

Anyway, we all see you can't allow something like the example g in Hilbert function space; it can't even be a linear vector space (under SI) because the kernel's non-trivial. Dirac and I simply assume those functions aren't there, because this is physics. Another approach is to assume they are there at first and get rid of them by factoring over the kernel. To me, the second approach seems pedantic. To you, the first seems sloppy. That's alright; we both wind up at the same place, with the same understanding, and can just move on.

Note there are cases where that's no good. If someone asks "why does the force of gravity cause clocks to run slow", you must point out that gravity's not a force (like EM) but curvature of spacetime. On the other hand, consider "If a ball falls for 3 seconds, and the force of graviy is 9.8 m/s/s, how fast will it be going?" It's ridiculous to insist on GR to answer this question.

It's something every physicist can easily understand and grasp..

I hope so!

I agree we shouldn't get lost in mathematical details. But sometimes those details really do shine a nice light on certain stuff.

I respect your attitude and agree with it maybe 60%. We should have little trouble communicating.

 

[micromass]

Anyway, we all see you can't allow something like the example g in Hilbert function space; it can't even be a linear vector space (under SI) because the kernel's non-trivial.

But the square integrable functions do form a linear vector space. No problem there. They just don't form a Hilbert space.

 

[strangerep]

Hmm. Since the OP seemed to disengage from this thread back at post #5, perhaps everything from post #6 onwards should be split off into a new thread. (The original thread was "I", but it has now become "A".)

 

[secur]

Since the OP seemed to disengage from this thread back at post #5, perhaps everything from post #6 onwards should be split off into a new thread.

That illustrates my main point perfectly. Post 5, Kevin McHugh asked stevendaryl how to get the limit of the test functions to show dirac(f) = f(0). stevendaryl had explained it in terms of distributions, which is more math than Dirac used, and not necessary for this elementary discussion; but OP seemed to be OK with it. Unfortunately stevendaryl had said it was defined on the square-integrable functions NOT the Test Function Space! micromass pointed out the mistake - stevendaryl made one last post to OP - rubi seconded micromass - and they were off and running about Test Function spaces and distributions for pages. That's when I tried to clear up the confusion, but (mistakenly, more-or-less) said square-integrable functions were a Hilbert Space, ignoring the equivalence classes. Everyone jumped on that essentially trivial "error". Etc.

Perfect illustration how physics, i.e. actually using dirac function, gets buried under a mass of nit-picky abstract math.

But the square integrable functions do form a linear vector space. No problem there. They just don't form a Hilbert space.

Ok, this is an illustration that math rigor is important (not as much as physics, but still important). Only when you actually know/remember the terminology are you entitled to casually blow by details. Not good enough to have known the terminology 40 years ago.

I meant square-integrable functions with SI norm aren't a normed vector space, because norm of functions like our example g is 0 - but g is not 0. Now, I'm hoping I remember correctly that normed vector space requires non-trivial kernel. ... If I can't find my old copy of Royden I'll buy one!

 

[vanhees71]

It's a little confusing. When people talk about L^2, they do mean functions modulo equivalence. But it seems to me that we can certainly talk about the set of square-integrable functions without taking equivalence relations.

Yes, and from this is clear that $f(0)$ is not well defined, because you can vary $f(0)$ arbitrarily without changing the Hilbert space vector in $L^2$. I can only stress it again, the $\delta$ distribution is defined on some dense subspace like the Schwartz space of $C^{\infty}$ functions that fall faster than any power. In QM the test-function space is the maximal domain of the position operator in the sense of an essentially self-adjoint operator, and $\delta$ is a distribution in the dual space of this "nuclear space". That becomes very clear with the modern formulation in terms of the rigged Hilbert space (see, e.g., Ballentine's textbook for a physicists' introduction or the one by Galindo an Pascual, which is mathematically more rigorous).