3 ms·
Not from the photon's point of view. It is emitted and absorbed immediately. The distance traveled (in 4d spacetime) of a photon is always 0, no matter (pun int
by ithinkso 5y ago
Not from the photon's point of view. It is emitted and absorbed immediately. The distance traveled (in 4d spacetime) of a photon is always 0, no matter (pun intended) what's in between or what you observe.
- sahil50 5y agoI don't think it is true to say that "it is emitted and absorbed immediately", because there is a very real difference from a photon going from the Sun to the Earth, to a photon going from the Sun to Jupiter. I know the theory the view you're saying comes from, and the weight it has in academia, but it always struck me as useless. And that whoever came up with it was lost in notation.
- ithinkso 5y ago>because there is a very real difference from a photon going from the Sun to the Earth, to a photon going from the Sun to Jupiter. Not from the photon's point of view. This is very crucial and it is not because of notation or maths. The real difference is from your point of view. I know it's counter-intuitive but that's different from being useless. Sun->Earth or Sun->Jupiter is a small beer. It is the same for photons from the CMB being recorded. For them, they have been emitted and absorbed at the same instant and experienced no passage of time or distance traveled
- sahil50 5y ago"Sun->Earth or Sun->Jupiter is a small beer"? How do you mean? This is a real physical example with real physical consequences for example for how much dimmer the light is per area (by 1/r^2) and by how much redder it is (by 1/r). What are the real physical consequences of thinking that the photon is "emitted and absorbed at the same instant"?