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October 13

Planck relation and time

The kinetic energy of an asteroid is independent of time and energy in general, otherwise we speak of power in Watts. In Planck's relation the radiant energy depends on a frequency which is temporal. How is it possible ? — Preceding unsigned comment added by Malypaet (talk • contribs) 08:40, 13 October 2022 (UTC)[reply]

Sorry, what? Radiant energy is a function of temperature. See blackbody radiation. It is unrelated to kinetic energy. The first is about the motion of individual atoms and molecules, the second is about the motion of the entire object. Also, it's vague and unclear what you mean by "depends on". Dimensional analysis will also tell you that all measurements of energy have time embedded in their units; a joule is a kilogram meter squared per second squared (kg m² s^-2), or if you want to think in terms of unitless measure, energy is mass times distance squared divided by time squared. So your first statement is nonsensical; time is embedded in the measurement of energy. A watt, by the way, is a kilogram meter squared per second cubed, or power is mass times distance squared divided by time cubed. --Jayron 32 10:54, 13 October 2022 (UTC)[reply]

Sorry, but 1 W is equal to 1 kg m² s⁻³. --Lambiam 12:42, 13 October 2022 (UTC)[reply]

So corrected. --Jayron 32 12:08, 14 October 2022 (UTC)[reply]

Effectively radiation energie (not radiant) and when you have "E=hv" you don't agree that "E" depends on "v" ? Malypaet (talk) 21:46, 13 October 2022 (UTC)[reply]

That's the energy of one photon. Your question is still unclear, which is why it is hard for people to respond meaningfully. Can you please explain again, using more details and in different words, what you are trying to understand? --Jayron 32 12:17, 14 October 2022 (UTC)[reply]

Perhaps you meant asteroids are not losing or gaining energy with time. That is true for KE + PE (potential energy), but not KE alone due to differences in orbital speeds. In fact, since Compton's experiments 99 years ago scientists have known that radiated photons are particles that carry energy that gain or lose energy during collisions with matter. Photons also have an EM quantum wave nature that interacts with matter so as to impart or acquire their characteristic frequencies such as with Doppler effects. Modocc (talk) 21:32, 13 October 2022 (UTC)[reply]

If we compare with a bullets flow from a machine gun with firing frequency "f" (I'm peaceful, it's just a thought experiment), so that we can write the kinetic energy equation for a group of bullets "Ec=k 1/2mv2 f" this group must have been emitted over 1s with the constant "k=1s". In this case we can also write "h=k 1/2mv2" and "v=f", we then find "Ec=hv". There it is coherent for me as it would also be if we replaced the bullets by the peaks of an elementary wave, whereas for the Planck relation I cannot manage to find coherence with the frequency. Malypaet (talk) 22:25, 13 October 2022 (UTC)[reply]

If a pulse of light (possibly a single photon) has a certain energy

E p

, then

n

such pulses have, together, the energy

nE p

. If these pulses are fired with a frequency

f

, then there are

f s

pulses per second, giving a power of

f E p

. This holds equally if we fire bullets instead of light pulses. The only difference is in the energy

E p

. The photon energy of a single photon equals

hν

, whereas the kinetic energy of a single bullet equals

½ mv 2

. There is no relation between the frequency

f

of firing, and the frequency

ν

of the photon. --Lambiam 06:16, 14 October 2022 (UTC)[reply]

Per Lambiam, and to clarify: The vibration of a light wave is unrelated to how often a photon hits a target. While both could be described as a "frequency", they are completely different and unrelated things. --Jayron 32 12:13, 14 October 2022 (UTC)[reply]