7 ms·

It’s intuitively a reach, because photons don’t move. From a four dimensional perspective—which is the correct one, let’s not forget—the distance between a pho

It’s intuitively a reach, because photons don’t move. From a four dimensional perspective—which is the correct one, let’s not forget—the distance between a photon’s emission and absorption event is always zero. “Tired light” would require the photon to spontaneously lose energy based on the Manhattan distance of the path it takes. That distance can be arbitrarily large, sure—a photon can cover the entire universe before absorption—but it isn’t the path the photon actually takes, so it’s difficult to imagine why it would matter.

hvidgaard 5y ago
> From a four dimensional perspective—which is the correct one, let’s not forget—the distance between a photon’s emission and absorption event is always zero. That would require the emission and absorption events location in the space to be the same would it not? Or do you mean that a photon does not experience time, so relative to the photon it simply travel instantaneously from emitting coordinate to absorption coordinate?
- gizmo686 5y ago
  The distance metric for flat spacetime is given by: ds^2 = dx^2 + dy^s + dz^2 - cdt^2 For a particle traveling at the speed of light, this gives ds=0.
- creata 5y ago
  > That would require the emission and absorption events location in the space to be the same would it not? I think they're referring to the distance in Minkowski space,[0] where the time component and space component cancel out for light. [0]: https://en.m.wikipedia.org/wiki/Spacetime#Spacetime_interval https://en.m.wikipedia.org/wiki/Spacetime#Spacetime_interval
sahil50 5y ago
Where does your claim that photons don't move come from? Very simply, the Sun is 8 minutes from the Earth. The photon is most definitely moving from the Sun to the Earth.
- krastanov 5y ago
  They are talking about "intervals" in 3+1 dimensional space time. To explain it rigorously, one needs to define a few terms like "metric" or "light cone". One defining property of the photon is that its space-time interval between any two events in its history is zero. You can think of it as the dual of the fact that a clock moving at the speed of light would be completely stopped.
  - Filligree 5y ago
    Intervals are the square of the distance, so not quite, but… yes. There’s a tendency to treat spacetime intervals as something abstract and possibly metaphysical, because we can’t see or touch them, and they work counterintuitively. They’re actually the opposite. Our normal experience of 3D space is what is unphysical; the (square root of the) spacetime interval is simply distance, measured in meters. Or seconds, of course. The literal, physical distance between an absorption and emission event is always zero. Reality is non-Euclidean. The Manhattan distance is larger than the actual distance, which is usually true anyway. Well, I suspect I just spent a lot of words telling you something you already knew, so I’ll leave off with a book recommendation. Dichronauts, by Greg Egan. Between that and the Clockwork Rocket trilogy, I don’t know a better way to inculcate an intuition for special relativity. One is set in a 2+2 spacetime; the other in 4+0. The differences are instructive.
    - krastanov 5y ago
      This made me laugh given that, actually, Greg Egan is by far my favorite author. But I also think you are going way too far by saying that our normal 3D space is what is unphysical. The fact that our everyday reality is the limit of a more interesting reality when some parameter becomes small, does not make it any less real, for any reasonable definition of "real".
      - Filligree 5y ago
        Well, it's certainly consistent. And for the most part you're right; it's just light that gets hard to understand from this perspective. Unfortunately photon behaviour is fairly important...

hvidgaard 5y ago

> From a four dimensional perspective—which is the correct one, let’s not forget—the distance between a photon’s emission and absorption event is always zero. That would require the emission and absorption events location in the space to be the same would it not? Or do you mean that a photon does not experience time, so relative to the photon it simply travel instantaneously from emitting coordinate to absorption coordinate?

gizmo686 5y ago
The distance metric for flat spacetime is given by: ds^2 = dx^2 + dy^s + dz^2 - cdt^2 For a particle traveling at the speed of light, this gives ds=0.
creata 5y ago
> That would require the emission and absorption events location in the space to be the same would it not? I think they're referring to the distance in Minkowski space,[0] where the time component and space component cancel out for light. [0]: https://en.m.wikipedia.org/wiki/Spacetime#Spacetime_interval https://en.m.wikipedia.org/wiki/Spacetime#Spacetime_interval

sahil50 5y ago

Where does your claim that photons don't move come from? Very simply, the Sun is 8 minutes from the Earth. The photon is most definitely moving from the Sun to the Earth.

krastanov 5y ago
They are talking about "intervals" in 3+1 dimensional space time. To explain it rigorously, one needs to define a few terms like "metric" or "light cone". One defining property of the photon is that its space-time interval between any two events in its history is zero. You can think of it as the dual of the fact that a clock moving at the speed of light would be completely stopped.
- Filligree 5y ago
  Intervals are the square of the distance, so not quite, but… yes. There’s a tendency to treat spacetime intervals as something abstract and possibly metaphysical, because we can’t see or touch them, and they work counterintuitively. They’re actually the opposite. Our normal experience of 3D space is what is unphysical; the (square root of the) spacetime interval is simply distance, measured in meters. Or seconds, of course. The literal, physical distance between an absorption and emission event is always zero. Reality is non-Euclidean. The Manhattan distance is larger than the actual distance, which is usually true anyway. Well, I suspect I just spent a lot of words telling you something you already knew, so I’ll leave off with a book recommendation. Dichronauts, by Greg Egan. Between that and the Clockwork Rocket trilogy, I don’t know a better way to inculcate an intuition for special relativity. One is set in a 2+2 spacetime; the other in 4+0. The differences are instructive.
  - krastanov 5y ago
    This made me laugh given that, actually, Greg Egan is by far my favorite author. But I also think you are going way too far by saying that our normal 3D space is what is unphysical. The fact that our everyday reality is the limit of a more interesting reality when some parameter becomes small, does not make it any less real, for any reasonable definition of "real".
    - Filligree 5y ago
      Well, it's certainly consistent. And for the most part you're right; it's just light that gets hard to understand from this perspective. Unfortunately photon behaviour is fairly important...