4 ms·

Wouldn't c have to be infinite for "no time to pass", since light is still traversing space at a finite speed?

ben_w 8y ago
No, because the rules of Lorenz invariant field theories mean that the infinity you’re (presumably) expecting is mapped, via the joy of hyperbolic geometry, onto the propagation speed of that field. Or, to put it another way, the speed of light is where time dilation and length contraction run into each other and it all goes zero-divided-by-zero.
Yajirobe 8y ago
No, outside observer will measure clocks of photon run slower by a factor of γ (Lorentz factor), which (in scalar form) is equal to 1/sqrt(1 - v^2 / c^2 ). It diverges at v = c.
- pdonis 8y ago
  > outside observer will measure clocks of photon run slower by a factor of γ This is not correct since "clocks of photon" is a meaningless concept (at least if photons have zero mass, which they do according to our best current models).
  - Yajirobe 8y ago
    Obviously, I did not mean to imply that you can attach a clock on a photon. I thought this goes without saying.
    - pdonis 8y ago
      > I did not mean to imply that you can attach a clock on a photon. The issue isn't that you can't attach a clock to a photon in a practical sense. The issue is that even in principle, the concept of "the clock of a photon" is not well-defined. For it to be well-defined, there would need to be an inertial frame in which a photon was at rest. But that is impossible.
Florin_Andrei 8y ago
Speed of light is for-all-practical-purposes infinite in the reference frame of the moving object (i.e. photon), thanks to special relativity.
- pdonis 8y ago
  > Speed of light is for-all-practical-purposes infinite in the reference frame of the moving object (i.e. photon), thanks to special relativity. This is not correct, because there is no "reference frame" for a photon moving at the speed of light. The concept of "reference frame" is not well-defined for such objects.
pdonis 8y ago
> Wouldn't c have to be infinite for "no time to pass" You're using a different notion of "time passing". Yes, we, observing photons, certainly observe them to take a finite time to cover a finite distance. But that is not the same as the concept of "time passing" for the photons themselves, for example according to a clock that the photons carried along with them; that is the concept that has no meaning.
- omarchowdhury 8y ago
  If the photons were a conscious subject, would that mean, from their point of view, they are everywhere (everywhere being defined as the entire path the photon will take from its genesis to its final destruction or absorption), all at once? Because if no time is passing for them, doesn't that mean no space is being traversed either?
  - etaoins 8y ago
    The flip side of time dilation is Lorentz contraction. As you approach the speed of light objects in your direction of motion will become shortened from your frame of reference. For example, there are particles from cosmic rays that should not be able to make it to the surface without decaying. However, they're detected all the time. Two valid ways to think of this are: 1. From the Earth's frame of reference time moves more slowly for the particle. This slows down the process of decaying. 2. From the particle's frame of reference the Earth's atmosphere is considerably shorter so it doesn't need to travel as far. Things get a bit hairy to talk about once you actually reach the speed of light. One way to think of it might be from the photon's frame of reference its entire path has become infinitely short so it had no distance to travel at all.
    - omarchowdhury 8y ago
      > One way to think of it might be from the photon's frame of reference its entire path has become infinitely short so it had no distance to travel at all. From the photons frame of reference, then, they do not move at all? And the environment that photon "experiences", being the path in the universe that it traverses from our point of view; is the past, present, and future (from our point of view) all in instant simultaneity for the photon?

No, because the rules of Lorenz invariant field theories mean that the infinity you’re (presumably) expecting is mapped, via the joy of hyperbolic geometry, onto the propagation speed of that field. Or, to put it another way, the speed of light is where time dilation and length contraction run into each other and it all goes zero-divided-by-zero.

Yajirobe 8y ago

No, outside observer will measure clocks of photon run slower by a factor of γ (Lorentz factor), which (in scalar form) is equal to 1/sqrt(1 - v^2 / c^2 ). It diverges at v = c.

pdonis 8y ago
> outside observer will measure clocks of photon run slower by a factor of γ This is not correct since "clocks of photon" is a meaningless concept (at least if photons have zero mass, which they do according to our best current models).
- Yajirobe 8y ago
  Obviously, I did not mean to imply that you can attach a clock on a photon. I thought this goes without saying.
  - pdonis 8y ago
    > I did not mean to imply that you can attach a clock on a photon. The issue isn't that you can't attach a clock to a photon in a practical sense. The issue is that even in principle, the concept of "the clock of a photon" is not well-defined. For it to be well-defined, there would need to be an inertial frame in which a photon was at rest. But that is impossible.

Florin_Andrei 8y ago

Speed of light is for-all-practical-purposes infinite in the reference frame of the moving object (i.e. photon), thanks to special relativity.

pdonis 8y ago
> Speed of light is for-all-practical-purposes infinite in the reference frame of the moving object (i.e. photon), thanks to special relativity. This is not correct, because there is no "reference frame" for a photon moving at the speed of light. The concept of "reference frame" is not well-defined for such objects.

pdonis 8y ago

> Wouldn't c have to be infinite for "no time to pass" You're using a different notion of "time passing". Yes, we, observing photons, certainly observe them to take a finite time to cover a finite distance. But that is not the same as the concept of "time passing" for the photons themselves, for example according to a clock that the photons carried along with them; that is the concept that has no meaning.

omarchowdhury 8y ago
If the photons were a conscious subject, would that mean, from their point of view, they are everywhere (everywhere being defined as the entire path the photon will take from its genesis to its final destruction or absorption), all at once? Because if no time is passing for them, doesn't that mean no space is being traversed either?
- etaoins 8y ago
  The flip side of time dilation is Lorentz contraction. As you approach the speed of light objects in your direction of motion will become shortened from your frame of reference. For example, there are particles from cosmic rays that should not be able to make it to the surface without decaying. However, they're detected all the time. Two valid ways to think of this are: 1. From the Earth's frame of reference time moves more slowly for the particle. This slows down the process of decaying. 2. From the particle's frame of reference the Earth's atmosphere is considerably shorter so it doesn't need to travel as far. Things get a bit hairy to talk about once you actually reach the speed of light. One way to think of it might be from the photon's frame of reference its entire path has become infinitely short so it had no distance to travel at all.
  - omarchowdhury 8y ago
    > One way to think of it might be from the photon's frame of reference its entire path has become infinitely short so it had no distance to travel at all. From the photons frame of reference, then, they do not move at all? And the environment that photon "experiences", being the path in the universe that it traverses from our point of view; is the past, present, and future (from our point of view) all in instant simultaneity for the photon?