Talk:Photon energy

Latest comment: 22 days ago by 155.190.18.44 in topic Formulas

Energy of visual light edit

It might be helpful for the casual reader to include the energy of photons in the visual spectrum, and to compare their energy to that of other processes. I added one such example: the energy of a photon absorbed for photosynthesis in plants. Benjamin.friedrich (talk)

speed of light.... edit

c in this formula is the speed of light in the medium in question and not in vacuum or is it? Because if it would be speed of light in vacuum, i.e. a constant, that would mean the energy of a given ray of light would directly depend on wavenlenght (instead of on frequency). That would mean its energy would change, sometimes increase, when it crosses into a different medium, wouldnt it?

If somebody who knows these things would correct this, or tell me why i am wrong, i would be much obliged. Best regards. — Preceding unsigned comment added by 160.45.190.123 (talk) 13:01, 5 January 2017 (UTC)Reply

Photon mass edit

Hi, Adrian.

In the introduction to the Photon Energy page, you state: "Photons being massless, the notion of "photon energy" is not related to mass through the equivalence E = mc2."

This is not entirely true... there is a circumstance under which photons can acquire mass. When the electromagnetic fundamental force undergoes symmetry breaking (the underlying phenomenon which causes superconductivity), photons in the superconductor trade the magnetic component of their EM energy for invariant mass. Thus it becomes energetically disadvantageous for magnetism to exist. This is known as the Meissner Effect.

The hypothesis that there is a "cosmic superconductivity" led to the further hypothesis that there must exist a field which causes this symmetry breaking. This is known as the Higgs field. Subsequent experimentation led to the confirmation of the Higgs field, and high-energy collisions confirmed the existence of the Higgs boson in 2012. The Higgs boson doesn't freely exist anymore, since universal energy density has fallen too far, so it's solely a laboratory beastie, but is presumed to exist under energetic cosmic event conditions.

Under the circumstances described above, photon energy directly correlates to its resultant invariant mass (E2=p2c2 + m2c4). Once temperature goes above the critical temperature, the photon trades its invariant mass for the magnetic component of its EM energy, and superconductivity ceases.

Thus superconductivity is a condensed-matter analog to the Higgs mechanism.

References: http://web.mit.edu/8.701/www/Lecture%20Notes/8.701originsOfMass04FA13.pdf

As an aside, I hypothesize that this "cosmic superconductivity", combined with the 2.725 K temperature of space, causes photons to acquire mass in much the same manner as described above, which might explain the "missing mass" problem. Thus photons acquiring mass could very well be that long-sought "dark matter".

In much the same vein, the quantum vacuum (that universal pool of entropied EM energy) being a cold plasma, combined with the well-known phenomenon of EM frequencies above the plasma frequency experiencing a negative index of refraction, causing their wavelengths to lengthen while their frequencies remain unchanged, causing an effect which allows those EM frequencies above the plasma frequency to travel faster than the speed of light through that medium. Since the quantum vacuum is a universe-pervading plasma, it is the medium which sets c, the speed of light. Thus, the field radiation pressure of those EM frequencies traveling faster than c could explain universal expansion >c. This would make the quantum vacuum that long-sought "dark energy".

— Preceding unsigned comment added by 71.135.36.177 (talk) 05:54, 4 March 2017 (UTC)Reply

Additionally, photon being massless refers to its rest mass. Photons are never at rest and therefore this argument doesn't applies. — Preceding unsigned comment added by 46.42.34.17 (talk) 20:15, 26 May 2017 (UTC)Reply

The Article is in need of fixing edit

It is very concerning that references for this article contain links to known pseudoscientific sources and the text of the article shows influences from said sources. For example a reference to the website of one Miles Mathis. This person is not a scientist nir does his website contain any valde scientific information and I strongle suspect he may have been the one who wrecked this article. — Preceding unsigned comment added by Mtw416 (talkcontribs) 02:55, 6 October 2017 (UTC)Reply

Colors on the energy graphic are backwards edit

Red is lower energy, purple is higher energy. The colors are backwards on the chart. 2001:420:28C:1320:7CE0:480B:805B:5540 (talk) 22:46, 15 October 2018 (UTC)Reply

Photosynthesis arithmetic edit

"1 red photon = 3 * 10−19 J ... A minimum of 48 photons is needed for the synthesis of a single glucose molecule from CO2 and water (chemical potential difference 5 x 10−17 J) with a maximal energy conversion efficiency of 35%."

5 * 10^-17 / (48 * 3 * 10^-19) = 350%, not 35%. Is there a factor of 10 error somewhere? Septimus.stevens (talk) 12:52, 9 June 2019 (UTC)Reply

Dear Septimus, thank you so much for being attentive. I just checked the math again: indeed, I missed an order of magnitude in the chemical potential of glucose. I corrected the article. Benjamin.friedrich (talk) —Preceding undated comment added 10:20, 19 June 2019 (UTC)Reply

450 TeV gamma rays detected edit

  • Starr, Michelle (2019-07-08). "Highest-Energy Photons Ever Detected Have Been Traced to an Enigmatic Source". Science alert. Archived from the original on 2019-07-08.

Quote: Previously, the highest energy photon ever detected was just 75 trillion electron volts, or teraelectronvolts (TeV). This marks the first time astronomers have detected such high energy photons - known as gamma rays - over 100 TeV.

Quote, slightly reformatted: The research has been accepted into the journal Physical Review Letters archive, and is available on arXiv archive.

davidwr/(talk)/(contribs) 21:26, 8 July 2019 (UTC)Reply

Photon amplitude edit

A photon's amplitude (or width) η is always equal to its wavelength λ divided by pi (π).

However, I can't find a reliable source for this, so I have removed it from the article. — Preceding unsigned comment added by 96.227.223.203 (talk) 17:43, 4 September 2023 (UTC)Reply

Formulas edit

The last line of the Formulas sections states, "The photon energy at 1 Hz is equal to 6.62607015×10−34 J, which is equal to 14.135667697×10−15 eV." I believe there is an arithmetic error here and the value for electronvolts should be 4.135667697×10−15 eV (without the leading "1"). 155.190.18.44 (talk) 14:51, 15 April 2024 (UTC)Reply