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Of course, the space was not made explicit. With R the set of real numbers, the usual assumption in the physics is that the math is done in R^3 with the usual

Of course, the space was not made explicit. With R the set of real numbers, the usual assumption in the physics is that the math is done in R^3 with the usual inner product, norm, metric, and topology.

kgwgk 4y ago
The usual assumption in physics is that the observable universe is bounded. https://www.feynmanlectures.caltech.edu/I_05.html#Ch5-S6 https://www.feynmanlectures.caltech.edu/I_05.html#Ch5-S6 Edit: you may be talking about the paragraph "For an atom at rest," in https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 The section starts saying that "We want now to talk a little bit about the behavior of probability amplitudes in time. [...] We are always in the difficulty that we can either treat something in a logically rigorous but quite abstract way, or we can do something which is not at all rigorous but which gives us some idea of a real situation—postponing until later a more careful treatment. With regard to energy dependence, we are going to take the second course. We will make a number of statements. We will not try to be rigorous—but will just be telling you things that have been found out, to give you some feeling for the behavior of amplitudes as a function of time. "
- graycat 4y ago
  I put in a qualification: > ... the usual assumption in the physics is that the math ... Soooo, you got into the physics. I avoided getting into the physics. E.g., when Newton wrote out his second law or Maxwell wrote out his equations, it was just math and the assumption was as I mentioned R^3. Stokes theorem, the Navier-Stokes equations were implicitly in R^3.
  - kgwgk 4y ago
    > Soooo, you got into the physics. I avoided getting into the physics. What did you think that Feynman's Lectures on Physics were about?
    - graycat 4y ago
      When physicists talk math concepts, e.g., probability densities, differentiability, continuity, probability, inner products, Fourier transforms, solutions to differential equations, unitary transformations, etc. they are talking math. They should get the math correct. So, roughly the paradigm might be -- as a student I assumed it was -- start with a physics problem, convert to a math problem, see what the math says, e.g., solution to a differential equation, Stokes theorem, the weak law of large numbers, then convert back to physics to see what those math results say about the physics. The idea of a finite universe, say, with a boundary, is popular in popular science and maybe science fiction but is missing from nearly all serious, accepted physics as taught today. So, really, the idea of a finite universe is no good as a patch up of what Feynman wrote. A simple approach for some of this issue is just to say, "I have a lab. A neutron is wandering around in it. I don't know where it is. So, I assume the probability density of its location is uniform in the lab." That's fine, that is, in the math. Notice, I never used rigorous. When physics profs used that word, it meant that it was time for me, no delay, don't wait for the end of the class period, just to stand, say nothing, walk out, drop the course, and return to the math department.
      - kgwgk 4y ago
        Fair enough. If you find the idea that "I have a universe." could be implicit in "A particle is somewhere." so unpalatable physics may not be for you. Note that (most) physicists don't accept infinities as a physical thing. They are mathematically convenient but only make sense as a limiting form. If you were indeed refering to https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 I agree it could be qualified sloppy. But it comes with a warning! ("not at all rigorous but [...] gives us some idea of a real situation") "If the uncertainty in momentum is zero, the uncertainty relation, ΔpΔx=ℏ, tells us that the uncertainty in the position must be infinite" doesn't make sense except when zero and infinity are understood as limits, for example.
        graycat 4y ago
        Congratulations on typing ΔpΔx=ℏ, Whenever I see that in a physics video on YouTube, one more such outrage and it's back to a Marilyn Monroe movie. Same for a book on quantum mechanics. If they have an application of Parseval's theorem, then be clear about it and trot it out. But usually the Δp and Δx are not defined clearly or defined at all. I just can't take such sloppiness seriously.

kgwgk 4y ago

The usual assumption in physics is that the observable universe is bounded. https://www.feynmanlectures.caltech.edu/I_05.html#Ch5-S6 https://www.feynmanlectures.caltech.edu/I_05.html#Ch5-S6 Edit: you may be talking about the paragraph "For an atom at rest," in https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 The section starts saying that "We want now to talk a little bit about the behavior of probability amplitudes in time. [...] We are always in the difficulty that we can either treat something in a logically rigorous but quite abstract way, or we can do something which is not at all rigorous but which gives us some idea of a real situation—postponing until later a more careful treatment. With regard to energy dependence, we are going to take the second course. We will make a number of statements. We will not try to be rigorous—but will just be telling you things that have been found out, to give you some feeling for the behavior of amplitudes as a function of time. "

graycat 4y ago
I put in a qualification: > ... the usual assumption in the physics is that the math ... Soooo, you got into the physics. I avoided getting into the physics. E.g., when Newton wrote out his second law or Maxwell wrote out his equations, it was just math and the assumption was as I mentioned R^3. Stokes theorem, the Navier-Stokes equations were implicitly in R^3.
- kgwgk 4y ago
  > Soooo, you got into the physics. I avoided getting into the physics. What did you think that Feynman's Lectures on Physics were about?
  - graycat 4y ago
    When physicists talk math concepts, e.g., probability densities, differentiability, continuity, probability, inner products, Fourier transforms, solutions to differential equations, unitary transformations, etc. they are talking math. They should get the math correct. So, roughly the paradigm might be -- as a student I assumed it was -- start with a physics problem, convert to a math problem, see what the math says, e.g., solution to a differential equation, Stokes theorem, the weak law of large numbers, then convert back to physics to see what those math results say about the physics. The idea of a finite universe, say, with a boundary, is popular in popular science and maybe science fiction but is missing from nearly all serious, accepted physics as taught today. So, really, the idea of a finite universe is no good as a patch up of what Feynman wrote. A simple approach for some of this issue is just to say, "I have a lab. A neutron is wandering around in it. I don't know where it is. So, I assume the probability density of its location is uniform in the lab." That's fine, that is, in the math. Notice, I never used rigorous. When physics profs used that word, it meant that it was time for me, no delay, don't wait for the end of the class period, just to stand, say nothing, walk out, drop the course, and return to the math department.
    - kgwgk 4y ago
      Fair enough. If you find the idea that "I have a universe." could be implicit in "A particle is somewhere." so unpalatable physics may not be for you. Note that (most) physicists don't accept infinities as a physical thing. They are mathematically convenient but only make sense as a limiting form. If you were indeed refering to https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 https://www.feynmanlectures.caltech.edu/III_07.html#Ch7-S1 I agree it could be qualified sloppy. But it comes with a warning! ("not at all rigorous but [...] gives us some idea of a real situation") "If the uncertainty in momentum is zero, the uncertainty relation, ΔpΔx=ℏ, tells us that the uncertainty in the position must be infinite" doesn't make sense except when zero and infinity are understood as limits, for example.
      - graycat 4y ago
        Congratulations on typing ΔpΔx=ℏ, Whenever I see that in a physics video on YouTube, one more such outrage and it's back to a Marilyn Monroe movie. Same for a book on quantum mechanics. If they have an application of Parseval's theorem, then be clear about it and trot it out. But usually the Δp and Δx are not defined clearly or defined at all. I just can't take such sloppiness seriously.