Timeline for Why Does Planck's Relation $E=hf$ Imply a Linear Relationship Only for Sinusoidal Frequency Bases?
Current License: CC BY-SA 4.0
26 events
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| Mar 13 at 0:50 | comment | added | DanielSank | @CuriousMind Look up Noether's Theorem and note that the functions $\exp(i \omega t)$ form an orthonormal basis of functions with time translation invariance (which is why the Fourier transform is so useful). | |
| Mar 10 at 15:12 | vote | accept | CuriousMind | ||
| Mar 9 at 23:15 | answer | added | Sturrum | timeline score: 5 | |
| Mar 9 at 22:37 | comment | added | CuriousMind | @DanielSank could you expand what you say about time translation invariance or link some resource where I could learn more about that? | |
| Mar 9 at 22:37 | answer | added | Agnius Vasiliauskas | timeline score: 0 | |
| Mar 9 at 22:36 | comment | added | CuriousMind | @DanielSank frequency is the number of oscillations per unit of time. If a sine oscillates with a frequency of 3Hz, that means that it makes 3 full oscilations in a second. The same goes for a periodic triangular or square signal, it would repeat the same pattern 3 times each second. In the world there are more frequencies than those of sinusoidal signals. | |
| Mar 9 at 19:55 | comment | added | DanielSank | I think everyone is missing the point here. The relationship $E = h f$ happens essentially because energy is the conserved quantity associated with time translation invariance and $\exp(i 2\pi f t)$ is precisely the function that is invariant (up to a scalar prefactor) under time translations. | |
| Mar 9 at 19:53 | history | edited | DanielSank | CC BY-SA 4.0 |
deleted 34 characters in body
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| Mar 9 at 19:52 | comment | added | DanielSank | What does "frequency" mean to you? In particular, what does it mean for non-sinusoidal functions? | |
| Mar 9 at 19:46 | answer | added | Noct | timeline score: 0 | |
| Mar 9 at 19:19 | answer | added | Thomas | timeline score: 1 | |
| Mar 9 at 18:58 | comment | added | Dale | I don’t think that matters specifically for the linearity. It is just a basis. In linear algebra a vector space is linear in every basis. This holds in quantum mechanics also. Hilbert spaces are vector spaces. They should be linear in all bases. Anyway, all I can tell you is that I personally am highly skeptical that the premise of your question is correct. Maybe your confusion comes from a false premise | |
| Mar 9 at 18:50 | comment | added | CuriousMind | @Dale No, it will be linear if you are thinking of it in terms of the sinusoidal frequencies of which the triangular wave is composed. But if you take into account the frequencies of the triangular wave (in a triangular wave basis, not as sinusoidal frequencies), they have a different non linear shape. | |
| Mar 9 at 18:48 | comment | added | Dale | I think it will still be linear. As you say, a triangular is simply a bunch of sinusoids. If you double the frequency of the triangular wave you will double the frequencies of all those sinusoids, and therefore the energy should be doubled. It will still be linear. | |
| Mar 9 at 18:16 | comment | added | CuriousMind | @Dale If you used a different basis, taking into account that a triangular wave is composed of lots of sinusoids of different frequencies, that means the frequency must have a different relation with the sinuoidal frequency (one which isn't linear) I think. | |
| Mar 9 at 18:00 | comment | added | Dale | Is the premise of this question true? If you used a different basis wouldn’t the relationship between energy and frequency still be linear? I mean, if you doubled the frequency of a triangular wave, I think it should still have double the energy. Other things might be weird, but I don’t think this is one of them | |
| Mar 9 at 17:32 | answer | added | Ванек Огонек | timeline score: 0 | |
| Mar 9 at 17:08 | review | Close votes | |||
| Mar 10 at 13:09 | |||||
| Mar 9 at 17:04 | comment | added | CuriousMind | But this relation is about photons, not electrons. Also, that would mean Planck relation is only an approximation? | |
| Mar 9 at 16:58 | comment | added | Kyle Kanos | Well square and triangular waves aren't harmonic, so that kind of prohibits the whole idea that electrons oscillating can be treated as masses on (massless) springs. | |
| S Mar 9 at 16:36 | history | suggested | Sancol. | CC BY-SA 4.0 |
Correcting formulaes
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| Mar 9 at 16:33 | comment | added | CuriousMind | @KyleKanos it means than in the equation E=h*f, f denotes the sinusoidal frequency components of the light wave. That is, the decomposition of the light wave in cosines with the Fourier Transform. Why Planck relation uses that decomposition of light instead of any other basis instead of cosines, like triangular or square waves. And why is it linear when choosing the cosines basis. | |
| Mar 9 at 16:14 | comment | added | Kyle Kanos | What does "sinusoidal bases for frequencies" mean? | |
| Mar 9 at 16:08 | review | Suggested edits | |||
| S Mar 9 at 16:36 | |||||
| S Mar 9 at 16:06 | review | First questions | |||
| Mar 9 at 17:44 | |||||
| S Mar 9 at 16:06 | history | asked | CuriousMind | CC BY-SA 4.0 |