4 ms·

photons don't freeze in time in their own reference frame, and one doesnt get to priviledge any particular reference frame including those that are different fr

by platz 9y ago

photons don't freeze in time in their own reference frame, and one doesnt get to priviledge any particular reference frame including those that are different from the photon's

platz 9y ago
actually i'm reading now that photons do not even have their own reference frame simply by definition/axiom. Interesting..
cygx 9y ago
The (proper) time interval between any two points along a null geodesic is zero. I see nothing wrong with calling photons frozen in time.
- platz 9y ago
  no,the "Spacetime Interval" along a null geodesic is 0, but a null geodesic "does not have a Proper Time associated with it". Undefined is not the same as 0. "For lightlike paths, there exists no concept of proper time and it is undefined as the spacetime interval is identically zero. "
  - cygx 9y ago
    Tomayto, tomahto. You can't parametrize a null geodesic by proper time, sure, but I don't see anything particularly wrong with calling the arc 'length'/spacetime interval between events along any particle trajectory a proper time, even when it's 0.
    - platz 9y ago
      https://www.reddit.com/r/askscience/wiki/physics/light_frame#wiki_null_spacetime_interval https://www.reddit.com/r/askscience/wiki/physics/light_frame... (Δs)2 = 0 It is wrong to then equate this to some sort of proper time Δτ, find that Δτ = 0 for a photon, and thus conclude that photons always "experience" zero proper time. No. Remember that Δτ is defined as the time difference between two events as measured by an observer (i.e., an inertial frame) that actually travels between the events. Photons have no reference frames! So the definition of Δτ doesn't apply to null paths.
      - cygx 9y ago
        I disagree: While you cannot have a clock travelling alongside the photon, you can have a family of clocks travelling along light-like trajectories that have the photon trajectory as its limit. But as I was alluding to with my 'tomayto, tomahto' remark, this is a question of semantics, and we're not really arguing about physics, but labels.
        platz 9y ago
        fair enough

by platz 9y ago

photons don't freeze in time in their own reference frame, and one doesnt get to priviledge any particular reference frame including those that are different from the photon's

platz 9y ago

actually i'm reading now that photons do not even have their own reference frame simply by definition/axiom. Interesting..

cygx 9y ago

The (proper) time interval between any two points along a null geodesic is zero. I see nothing wrong with calling photons frozen in time.

platz 9y ago
no,the "Spacetime Interval" along a null geodesic is 0, but a null geodesic "does not have a Proper Time associated with it". Undefined is not the same as 0. "For lightlike paths, there exists no concept of proper time and it is undefined as the spacetime interval is identically zero. "
- cygx 9y ago
  Tomayto, tomahto. You can't parametrize a null geodesic by proper time, sure, but I don't see anything particularly wrong with calling the arc 'length'/spacetime interval between events along any particle trajectory a proper time, even when it's 0.
  - platz 9y ago
    https://www.reddit.com/r/askscience/wiki/physics/light_frame#wiki_null_spacetime_interval https://www.reddit.com/r/askscience/wiki/physics/light_frame... (Δs)2 = 0 It is wrong to then equate this to some sort of proper time Δτ, find that Δτ = 0 for a photon, and thus conclude that photons always "experience" zero proper time. No. Remember that Δτ is defined as the time difference between two events as measured by an observer (i.e., an inertial frame) that actually travels between the events. Photons have no reference frames! So the definition of Δτ doesn't apply to null paths.
    - cygx 9y ago
      I disagree: While you cannot have a clock travelling alongside the photon, you can have a family of clocks travelling along light-like trajectories that have the photon trajectory as its limit. But as I was alluding to with my 'tomayto, tomahto' remark, this is a question of semantics, and we're not really arguing about physics, but labels.
      - platz 9y ago
        fair enough