9 ms·

I know nothing about this: what if the color did change, to be slightly redder?

ForOldHack 2y ago
Then we ( and Hubble ) may over estimate the speed of the expansion of the universe, an the galaxies that we have measured their red shift and estimated their relitive speed based on less of a red shift. It is certainly possible.
gus_massa 2y ago
I don't remember the details, but I took a look at Wikipedia to remember some details. The redshift of far away galaxies is calculated multiplying the frequencies by 1+z, so you get the same displacement if you draw the spectrum in the correct logarithmic scale https://en.wikipedia.org/wiki/Redshift https://en.wikipedia.org/wiki/Redshift The emission/absorption lines of Hydrogen are calculated using the Rydberg constant https://en.wikipedia.org/wiki/Rydberg_constant https://en.wikipedia.org/wiki/Rydberg_constant that use the mass of the electron. But it's not the actual mass of an isolated electron, you must use the "reduced mass" because the electron moves around the proton but the proton also moves a little [1]. So the reduced mass of the electron is reduced_mass_electron = 1 / ( 1 / actual_mass_electron + 1 / actual_mass_proton ) The proton is much heavier [2], so the difference of the reduced and the actual mass of the electron is only .1%. If you magically change the mass of the electron to be equal to the mass of the proton, then the reduced mass of the electron would be 1/(1/1+1/1)=1/2 of the actual mass of the magical electron, and all the lines in the spectrum of Hydrogen will change. But the spectrum astronomer get includes other elements and isotopes. For example Deuterium that has the same charge but the double of mass. In the real word, the reduced mass of the electron in Deuterium is also almost equal to the actual mass of the electron, the difference is like .05%. In the magical world where the mass of the electron to be equal to the mass of the proton, then the reduced mass of the electron would be 1/(1/1+1/2)=1/3 of the actual mass of the magical electron, and all the lines in the spectrum of Deuterium will change but in a different way. I think it's possible to see all these lines in a spectrum, but to be sure ask an expert. Anyway, a magical change to make the mass of protons and electrons equal will change the spectrum of all the other atoms in different and strange and weird and unexpected ways. [3] So it will be easy to spot. Smaller changes will change thing and be visible too, unless they are too small. [1] If you want to go down the Quantum Mechanics rabbit hole, they don't move and you must use orbitals. But the correction is the same than in the fake/simplified classical calculation. [2] * massive [3] A heavy [2] electron is very similar to a muon, and in Hydrogen molecules with muons both Hydrogen are very close so you get fusion reactions https://en.wikipedia.org/wiki/Muon-catalyzed_fusion https://en.wikipedia.org/wiki/Muon-catalyzed_fusion so the magical would be very strange.
- gus_massa 2y ago
  Toooo late to edit. Fixing it for posterity: The correct number is 1/(1/1+1/2)=2/3 I incorrectly wrote 1/3 instead.

Then we ( and Hubble ) may over estimate the speed of the expansion of the universe, an the galaxies that we have measured their red shift and estimated their relitive speed based on less of a red shift. It is certainly possible.

gus_massa 2y ago

I don't remember the details, but I took a look at Wikipedia to remember some details. The redshift of far away galaxies is calculated multiplying the frequencies by 1+z, so you get the same displacement if you draw the spectrum in the correct logarithmic scale https://en.wikipedia.org/wiki/Redshift https://en.wikipedia.org/wiki/Redshift The emission/absorption lines of Hydrogen are calculated using the Rydberg constant https://en.wikipedia.org/wiki/Rydberg_constant https://en.wikipedia.org/wiki/Rydberg_constant that use the mass of the electron. But it's not the actual mass of an isolated electron, you must use the "reduced mass" because the electron moves around the proton but the proton also moves a little [1]. So the reduced mass of the electron is reduced_mass_electron = 1 / ( 1 / actual_mass_electron + 1 / actual_mass_proton ) The proton is much heavier [2], so the difference of the reduced and the actual mass of the electron is only .1%. If you magically change the mass of the electron to be equal to the mass of the proton, then the reduced mass of the electron would be 1/(1/1+1/1)=1/2 of the actual mass of the magical electron, and all the lines in the spectrum of Hydrogen will change. But the spectrum astronomer get includes other elements and isotopes. For example Deuterium that has the same charge but the double of mass. In the real word, the reduced mass of the electron in Deuterium is also almost equal to the actual mass of the electron, the difference is like .05%. In the magical world where the mass of the electron to be equal to the mass of the proton, then the reduced mass of the electron would be 1/(1/1+1/2)=1/3 of the actual mass of the magical electron, and all the lines in the spectrum of Deuterium will change but in a different way. I think it's possible to see all these lines in a spectrum, but to be sure ask an expert. Anyway, a magical change to make the mass of protons and electrons equal will change the spectrum of all the other atoms in different and strange and weird and unexpected ways. [3] So it will be easy to spot. Smaller changes will change thing and be visible too, unless they are too small. [1] If you want to go down the Quantum Mechanics rabbit hole, they don't move and you must use orbitals. But the correction is the same than in the fake/simplified classical calculation. [2] * massive [3] A heavy [2] electron is very similar to a muon, and in Hydrogen molecules with muons both Hydrogen are very close so you get fusion reactions https://en.wikipedia.org/wiki/Muon-catalyzed_fusion https://en.wikipedia.org/wiki/Muon-catalyzed_fusion so the magical would be very strange.

gus_massa 2y ago
Toooo late to edit. Fixing it for posterity: The correct number is 1/(1/1+1/2)=2/3 I incorrectly wrote 1/3 instead.