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KJW

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Everything posted by KJW

  1. To complete the connection between Fourier transforms and differential operators with regards to conjugate variables that I started earlier in this topic: [math]\text{Let }F(\xi) = \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx \ \ \ \ ;\ \ \ \ f(\pm \infty ) = 0[/math] [math]\text{Then }\displaystyle \int_{-\infty}^{\infty} \dfrac{d}{dx}f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math] [math]= -\displaystyle \int_{-\infty}^{\infty} f(x)\ \dfrac{d}{dx}\exp(-2\pi i\ \xi x)\ dx[/math] [math]2\pi i\ \xi\ \displaystyle \int_{-\infty}^{\infty} f(x)\ \exp(-2\pi i\ \xi x)\ dx[/math] [math]= 2\pi i\ \xi\ F(\xi)[/math] [math]\text{Therefore }\dfrac{d}{dx} \equiv 2\pi i\ \xi[/math] [Please refresh browser window if the above LaTex doesn't render]
  2. The Heisenberg uncertainty principle relates the standard deviations of conjugate pairs of variables. But a variable having finite standard deviation does not necessarily have compact support (eg Gaussian function). So the Heisenberg uncertainty principle doesn't really say anything about the support of conjugate pairs of variables. However, I have indicated that at least one of a conjugate pair of variables must be without compact support. How this impacts on the physics is unclear since it is reasonable to assume that physical variables are bounded in value.
  3. "Compact support" means that the function is zero everywhere outside of some finite domain.
  4. There is an inverse relationship between the standard deviation width of a function and the standard deviation width of its Fourier transform. The value of the product of the two standard deviations depends on the function (hence the inequality) with the minimum value (where the inequality becomes the equality) achieved by the Gaussian function whose Fourier transform is also the Gaussian function. It can be proven that the Fourier transform of a function with compact support does not have compact support. To prove this, note that any function with compact support is the product of some function with the rectangular function. Therefore, the Fourier transform of any function with compact support is the convolution of some function (with or without compact support) with the sinc function, and therefore does not have compact support. The converse is not necessarily true. For example, both the Gaussian function and its Fourier transform (also the Gaussian function) are functions without compact support.
  5. This is a common misunderstanding of refraction. Refraction is easiest to understand in terms of classical electromagnetic waves. This can then be translated to the quantum picture provided the important aspects of the classical picture are maintained. When an electromagnetic wave passes through a medium, it exerts a force on the charges and charge dipoles of the medium. Depending on how easily the charges and charge dipoles of the medium can move in response to this force, the motion of the charges and charge dipoles of the medium creates an electromagnetic wave that combines with the original electromagnetic wave to produce a total electromagnetic wave that is delayed with respect to the original electromagnetic wave and therefore travels through the medium at a slower speed. Thus, the refractive index of the medium depends on how readily the charges and charge dipoles of the medium can respond to the passing electromagnetic wave. This depends on the frequency of the passing electromagnetic wave. Higher frequencies exert a greater force, but larger bulkier charges and charge dipoles respond more to lower frequencies. At visible frequencies, only electrons can significantly respond to the passing electromagnetic wave, and in this case, the refractive index depends on the polarisability of the electron orbitals of the medium and increases with frequency due to the increasing energy of the photons.
  6. Thanks. +1 I have an interest in supramolecular chemistry, in particular, host-guest chemistry.
  7. Why are you posting an image of melamine cyanurate?
  8. Interesting - substitution, also called replacement. Of course that is the function of the klein 4-group I was illustrating. But I'm not sure I would call group theory the most basic. I was referring to the process of doing mathematics rather than any subject within mathematics. Nevertheless, as far the subject matter of mathematics is concerned, I do consider group theory to be near the foundations of mathematics, and more fundamental than arithmetic. For example, if one considers the group of the integers under addition, then multiplications are endomorphisms of this group. Compare the distributive law with the definition of a morphism. Also note that under any morphism, the identity maps to the identity. By mapping the endomorphisms to the integers, one has created a ring. That is, whereas a group has one operation, the group of integers under addition naturally admits a second operation by considering its endomorphisms (though in general, the endomorphisms of a group are categorically different to the elements of a group). As a "formalist", I regard doing mathematics as a purely mechanical process, and require mathematics that is rigorous to be performed by a machine (or by a human emulating a machine) (but not AI). Human intuition and cleverness can still be used as a guide, but in the end, all the individual steps of a derivation or proof must be indistinguishable from the product of a purely mechanical process.
  9. Rowan Atkinson: Free Speech In particular, "the freedom to be inoffensive is no freedom at all".
  10. As I see it, the most basic operation or activity in mathematics is substitution. To transform one statement to another statement, one performs substitutions of expressions within the initial statement with other expressions.
  11. Useful shorthand, but not at all rigorous. I feel kinda dirty using them, like I need a shower. May I answer this question? I will wait to give @AVJolorumAV the opportunity to answer.
  12. I don't know if this is significant or just a strange coincidence, but if you consider the series: [math]1+2+4+8+16+32+64\>+\>...[/math] as a binary number, then that binary number would be represented as an infinite string of "1"s: [math]...1111111111111111[/math] In the two's complement representation of signed binary numbers, this corresponds to [math]-1[/math].
  13. That's why one doesn't attempt to sum the series... because the result depends on how it is done. That's why one uses a formula that gives the correct answer. In the case of the series [math]1-1+1-1+1-1+1-1\>+\>...\>[/math], this corresponds to: [math]1-1+1-1+1-1+1-1\>+\>...\\=1+x+x^2+x^3+x^4+x^5\>+\>...\>=\dfrac{1}{1-x}\\=\dfrac{1}{2}\>\>\text{for }\>\>x=-1[/math] Interestingly, it also corresponds to: [math]1-1+1-1+1-1+1-1\>+\>...\\=\eta(x) = 1-\dfrac{1}{2^x}+\dfrac{1}{3^x}-\dfrac{1}{4^x}+\dfrac{1}{5^x}-\>...\\\text{for }\>\>x=0[/math] Now lets consider the series [math]1+1+1+1+1+1+1+1\>+\>...\>[/math]. This corresponds to: [math]1+1+1+1+1+1+1+1\>+\>...\\=1+x+x^2+x^3+x^4+x^5\>+\>...\>=\dfrac{1}{1-x}\\=\>???\>\>\text{for }\>\>x=1[/math] That is, even the formula is undefined for [math]x=1[/math] But: [math]1+1+1+1+1+1+1+1\>+\>...\\=\zeta(x) = 1+\dfrac{1}{2^x}+\dfrac{1}{3^x}+\dfrac{1}{4^x}+\dfrac{1}{5^x}+\>...\>=\dfrac{\eta(x)}{1 - 2^{1-x}}\\=-\eta(0)=-\dfrac{1}{2}\>\>\text{for }\>\>x=0[/math]
  14. It appears that the LaTex code between the "math" tags (I only use "math" tags) has to be all on the same line. This appears to be different to the original forum which iirc allowed code to be split across multiple lines (between a single pair of tags). Usually, I use a pair of tags for each line, but some code (eg matrices) requires everything to be between a single pair of tags. However, the use of "\\" does allow rendered text to be on multiple lines.
  15. KJW replied to KJW's topic in The Sandbox
    [math]\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}[/math] [math]\begin{pmatrix} a & b\\c & d \end{pmatrix}[/math]
  16. Did you look at the entire video? Towards the end of the video, he explains how the identity: [math]1\ +\ 2\ +\ 3\ +\ 4\ +\ ...\ = -\dfrac{1}{12}[/math] arises from analytic continuation of the Riemann zeta function. If you look at the old thread I referenced, you'll see how I evaluated it, including the use of the relationship between the zeta and eta functions mentioned in the video: [math]\zeta(z) = \dfrac{1}{1 - 2^{1-z}}\>\eta(z)[/math] The important thing to note, both in the old thread I referenced, and in the simpler non-convergent series that I challenged you with, is that no attempt is made to sum the series or form the limit of a partial sum of the series. Instead, the entire series is assumed to be an entity in its own right and have a value, and by exploiting the self-similarity properties of this entity, algebraically obtain this value. In the old thread I referenced, you'll see that I derived the formula for the general infinite geometric series, and that the same technique was used for the non-convergent infinite geometric series in this thread, thus resulting in the same formula being implicitly used for the non-convergent series. The formula for the general infinite geometric series is valid in the domain that the series converges, but while the series does not converge outside this domain, the formula still provides a value. The general formula is: [math]1 + x + x^2 + x^3 + x^4 + x^5\>+\>...\>=\>\dfrac{1}{1 - x}[/math] For [math]x = 2[/math], . . . [math]\dfrac{1}{1 - x} = -1[/math]
  17. In case you were thinking that such series have no relevance to physical reality, the derivation of the Casimir effect makes use of the identity: [math]1\ +\ 8\ +\ 27\ +\ 64\ +\ 125\ +\ 216\ +\ ...\ =\ \dfrac{1}{120}[/math] [Please refresh browser window if the above LaTex doesn't render]
  18. A while ago in the thread titled "problem with cantor diagonal argument" on page 6, as part of a general discussion about infinity, I mentioned the identity: [math]1\ +\ 2\ +\ 3\ +\ 4\ +\ ...\ = -\dfrac{1}{12}[/math] which led me to evaluate various non-convergent infinite series, including a proof of the identity above using simpler maths than analytic continuation of the Riemann zeta function. Let's consider something simpler: [math]\text{Let }X\ =\ 1\ +\ 2\ +\ 4\ +\ 8\ +\ 16\ +\ ...[/math] [math]X\ -\ 1\ =\ 2\ +\ 4\ +\ 8\ +\ 16\ +\ 32\ +\ ...\ =\ 2\ X[/math] [math]X\ =\ -1[/math] [math]\text{Therefore: }1\ +\ 2\ +\ 4\ +\ 8\ +\ 16\ +\ ...\ =\ -1[/math] [Please refresh browser window if the above LaTex doesn't render] I challenge you to find the error in the above.
  19. I said that one can define the Riemann tensor this way. I didn't say that one has to define the Riemann tensor this way. I prefer to view GR analytically rather than geometrically, and in this view, the Riemann tensor is defined in terms of the integrability conditions of the coordinate transformation equation of the connection.
  20. KJW replied to m_m's topic in The Lounge
    I haven't eaten McDonald's food for a few years now. I quite like it, but around the time of COVID-19, the shop at the shopping centre I normally go to closed their business, and the only McDonald's restaurants in my area would require me to make a special journey to McDonald's, which lacks the spur-of-the-moment convenience of the shopping centre. Actually, I haven't even been to Hungry Jack's (Burger King) much since COVID-19, and they are quite conveniently located for me. Why this is I can't really explain, although I did go there about four times in a row over a period of about a year to buy a thick shake, and each time, the machine was out of service (I did wonder if they ever had thick shakes during this period).
  21. Displayed in the bottom-right image is the first four amino acid residues of the various endogenous opioid peptides: Tyr-Gly-Gly-Phe-... These four amino acid residues at the N-terminus of the peptide appear to be universal for opioid receptor activity. The fifth residue (not shown) appears to be either Met or Leu in the various endogenous opioid peptides. A careful look that the bottom-right image reveals some errors that indicate that one can't always trust artists with scientific knowledge.
  22. The stars in galaxies are not glued to the expanding space. They are quite free to move independently of any expansion of space. Thus, if a star exhibits some tendency to move away from the central core of its galaxy due to the expansion of space, then the gravity from the central core will pull the star back into the fold. If a star orbits its galaxy at some particular distance from the centre of the galaxy, then even after the space has expanded, the laws of physics will maintain the distance at which the star orbits, noting that the laws of physics are based on local scale, not cosmological scale.
  23. No, it's not what i think. You say that, but then you say things like: The notion of a variable c might seem reasonable from a superficial perspective, but when one considers it more deeply, it becomes clear that it is really quite meaningless, in the "how long is a piece of string" territory. No, that's not what I meant. Suppose that in the past, before the standard of length was defined in terms of the speed of light in a vacuum, one performs a measurement of the speed of light in a vacuum. One can do this by measuring the time it takes for light to travel in a vacuum the distance of the standard of length and return. From the measured time and the known distance, the speed of light in a vacuum is determined. Now consider that currently, with the standard of length defined in terms of the speed of light in a vacuum, one performs a measurement of a given length. One can do this by measuring the time it takes for light to travel in a vacuum the distance of the given length and return. From the measured time and the known speed of light in a vacuum, the given length is determined. But note that the same experiment is performed in both cases. That is, in both cases, one is measuring the time it takes light to travel some distance in a vacuum... a measurement of the speed of light in a vacuum. To make this more explicit, consider in the second case that one can define the given length as a temporary standard of length with the unit of "tmpstd". Then one is measuring the speed of light in a vacuum in units of tmpstds per second. Dividing the defined speed of light in a vacuum in units of metres per second by the measured speed of light in a vacuum in units of tmpstds per second gives a value in units of metres per tmpstd corresponding to the length in metres for the given length. I agree that the speed of light in a vacuum is conceptually distinct from c. The speed of light in a vacuum is a property of light or electromagnetism in general, whereas c is a value that relates units of length and units of time with regards to the equivalence of length and time in spacetime. Whereas one can use the same ruler to measure lengths in the east-west, north-south, and up-down directions by simply rotating the ruler, one can't rotate the ruler to the past-future direction. But relativistic effects do provide a way to determine the equivalence between length and time in spacetime. For example, one can measure c by measuring the velocity of light in water at rest and in water moving at velocity v. From the three velocities, the value of c can be obtained by rearranging the relativistic velocity-addition formula. Note that the length standard used to measure the three velocities is based on the speed of light in a vacuum, not c, so the measurement of c is a truly independent measurement. However, it is unfortunate that the length standard is based on the speed of light in a vacuum, and not on c, given that it is c that is about the equivalence of length and time in spacetime. In general relativity, the general metric is expressed succinctly as: (ds)2 = gpq xp xq How would a variable c even fit into this expression? And what do you think "ds" means in this expression? Bear in mind that regardless of what is used to define the standard of length, it will never be possible to determine the "true" value of that length. A length can be measured in terms of another length, but this ultimately leads to an infinite regress. And if a length is measured in terms of itself, it will have a definite value, but that value has no meaning.
  24. All the same they simplify the math and leads to the correct answer. That depends on the connection. For a Riemannian connection, normal coordinates exist which simplify the proofs of identities. But for a general connection, for example if the torsion tensor is non-zero, then normal coordinates do not exist in general. Also, if one is using normal coordinates to simplify the proofs of Riemann tensor identities, then one also needs to include the proof of the existence of the normal coordinates themselves. Thus, in the end it may actually be simpler to prove the Riemann tensor identities directly without invoking normal coordinates. And while one is at it, one may as well prove the Riemann tensor identities for the general connection.

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