3 ms·

He's just describing Einstein's Theory of Special Relativity.

chrisweekly 2y ago
Huh. I'm admittedly no astrophysicist, but I don't recall encountering this "photons have no velocity in the time dimension" before. Maybe I'm one of today's lucky 10,000[1]. 1. https://xkcd.com/1053/ https://xkcd.com/1053/
- bee_rider 2y ago
  But, it is sort of common to observe the somewhat trippy fact that, unlike matter, photons don’t experience time passing, right? (Although, I will be honest, I could not ever really grok relativity like a physicist. However, I can enjoy the Lorentz factor going to infinity as v goes to c).
  - da_chicken 2y ago
    From my understanding, it's more correct to say that time is undefined for massless objects. Through special relativity, our understanding of mass, gravity, and spacetime are linked. If something has no mass, then special relativity can't describe how gravity affects it's spacetime reference. Remember, however, that this explanation is based on the mathematics that explain the observations we've made or theorized. Just as the map is not the territory, the math is not the universe.
    - deleted 2y ago
      [deleted]
- salviati 2y ago
  I enjoyed this video on this topic a lot. Maybe you will too: https://youtu.be/fB8eatgkOyM?si=D9s01MY8jByWREPX https://youtu.be/fB8eatgkOyM?si=D9s01MY8jByWREPX
- nh23423fefe 2y ago
  It's not true. It's a mashup of a few ideas that don't belong together. 1. an object at rest has a world line which points (has tangent vectors) in the time dimension 2. light has proper time (dtau) = 0, so clocks moving at the speed of light dont tick 3. the magnitude^2 of an objects 4-velocity is c^2 (objects move through spacetime at c) 4. light has no 4-velocity (because dtau=0, you cant divide by zero) You can't say (3) means objects have 4-velocity c in spacetime and light has 3-velocity c in space and so that means the time component of 4-velocity for light is zero. Because light has no 4-velocity.

Huh. I'm admittedly no astrophysicist, but I don't recall encountering this "photons have no velocity in the time dimension" before. Maybe I'm one of today's lucky 10,000[1]. 1. https://xkcd.com/1053/ https://xkcd.com/1053/

bee_rider 2y ago
But, it is sort of common to observe the somewhat trippy fact that, unlike matter, photons don’t experience time passing, right? (Although, I will be honest, I could not ever really grok relativity like a physicist. However, I can enjoy the Lorentz factor going to infinity as v goes to c).
- da_chicken 2y ago
  From my understanding, it's more correct to say that time is undefined for massless objects. Through special relativity, our understanding of mass, gravity, and spacetime are linked. If something has no mass, then special relativity can't describe how gravity affects it's spacetime reference. Remember, however, that this explanation is based on the mathematics that explain the observations we've made or theorized. Just as the map is not the territory, the math is not the universe.
  - deleted 2y ago
    [deleted]
salviati 2y ago
I enjoyed this video on this topic a lot. Maybe you will too: https://youtu.be/fB8eatgkOyM?si=D9s01MY8jByWREPX https://youtu.be/fB8eatgkOyM?si=D9s01MY8jByWREPX
nh23423fefe 2y ago
It's not true. It's a mashup of a few ideas that don't belong together. 1. an object at rest has a world line which points (has tangent vectors) in the time dimension 2. light has proper time (dtau) = 0, so clocks moving at the speed of light dont tick 3. the magnitude^2 of an objects 4-velocity is c^2 (objects move through spacetime at c) 4. light has no 4-velocity (because dtau=0, you cant divide by zero) You can't say (3) means objects have 4-velocity c in spacetime and light has 3-velocity c in space and so that means the time component of 4-velocity for light is zero. Because light has no 4-velocity.