4 ms·

I’ve been trying to learn all this stuff lately. One thing that bothers me though, that perhaps someone here can help with. In Minkowski diagrams you set thing

I’ve been trying to learn all this stuff lately. One thing that bothers me though, that perhaps someone here can help with. In Minkowski diagrams you set things up to have rays as diagonal lines. But this way of illustrating things suggests that light moves in time. But they should really not be interpreted as that since they essentially define the limit of the time dimension. So I was thinking that I would like to have another illustration of things where light moves in 90-degree angles instead to fix this. Are there some specific, established ways to do this?

archgoon 9y ago
I'm not entirely sure what your objection is; light does move in time.
- CuriousSkeptic 9y ago
  Still learning ;) But IIUC, could you send a clock along 45-degree angles in a minkowski diagram it would not measure any time. But perhaps there is time there relative to an observer at 0-degrees? But what does that even mean then? If that observer sent a photon towards a mirror, would the photon have aged when returned to the sender?
  - soVeryTired 9y ago
    You can make your statement more accurate by saying that a clock will measure vanishingly little time as you get sufficiently close to the 45 degree line. But the 45 degree line isn't achievable. (Note I've made the same inaccuracy further up the page. Sometimes it's just easier to drop the epsilons and speak informally.) Physicists are precise people :)
    - fjsolwmv 9y ago
      Mathematicians are precise people. Physicists just assume their series converge :-)
QAPereo 9y ago
I think you’re talking about null vectors/geodesics, which define light, which from the viewpoint of any external observer does move through spacetime. https://physics.stackexchange.com/questions/23058/why-is-light-described-by-a-null-geodesic https://physics.stackexchange.com/questions/23058/why-is-lig... https://physics.stackexchange.com/questions/188859/what-is-a-null-geodesic https://physics.stackexchange.com/questions/188859/what-is-a... Remember that it’s deceptive to imagine things like the POV of a photon, and better to think in the usual terms of external observation.
philipov 9y ago
If you were to imagine photons having an internal clock, that clock never ticks. That's not the same as light not moving through time. Light does not experience proper time. Any statement that combines the word "move" with the claim that it shows how light lacks time has already contradicted its goal. I think it is better to imagine light as line-like objects that extend through time, not point-like objects that move through time, and do away with speaking of light as something that moves, if you wish to express that it lacks a clock (proper time).
- fjsolwmv 9y ago
  Every particle has a life line (segment). Light particle's life line segment is perpendicular to the time axis, jumping across empty space (from birth/emission to death/absorption) at one constant time. If an atom emits an alpha particle (helium atom) and gamma particle (photon) at the same time, the alpha particle moves through space while its proper time changes (its "imaginary clock" ticks), whereas the gamma particle instantly arrives at its destination somewhere else in the Universe without experiencing any change in time, absorbed by another particle. Since light has no mass, it has no impact on spacetime geometry, so it only exists at emission and absorption. There is no way to detect or affect it between those to moments in its life, so it makes sense to consider them exactly one moment. Lights line segment is a "dotted line" the line segment (exclusive of endpoints) between emission and absorption does not really represent any part of our Universe; it's purely virtual. The only way to affect a photon is to plan ahead and set up an apparatus before it is born. Once the photon exists (relative to any observer), it's already gone (absorbed) by whatever first observed it. (And nothing else can directly observe it. All of us can bounce lots of stuff off the moon to observe it, but only one person can bounce only one thing off a photon , which kills it)
- CuriousSkeptic 9y ago
  But still, that line, in a minkowski diagram, represents a boundary, so it still seems more intuitive to me to eliminate the area beyond it completely from the diagram.
  - ajkjk 9y ago
soVeryTired 9y ago
In any reference frame, light moves one (light) second of distance in one second of time. So the worldline of light should always be at 45 degrees in a minkowski diagram (with appropriate units). Is your issue that, loosely speaking, if you travelled from X to Y at the speed of light, you wouldn't experience any time elapsing as you travel? If so, remember that the Minkowski diagram relates to the observer's reference frame, not the light's reference frame.
- jacobolus 9y ago
  Also note that the “light’s reference frame” is not a meaningful concept.
  - fjsolwmv 9y ago
    AIUI, "light has no reference frame" is the same as "light does not experience time" is the same as "light photon teleports from its birthplace (emission) to its death place (absorption), in its own reference frame".
    - jacobolus 9y ago
      “Reference frame” is a way of assigning a coordinate system so as to split spacetime into one dimension of “time” and three dimensions of “space”. Light doesn’t have a reference frame. Light doesn’t experience “space” either. Two points in spacetime along the path of the light are displaced by a “null vector”, i.e. the relativistic magnitude of the displacement is zero.

archgoon 9y ago

I'm not entirely sure what your objection is; light does move in time.

CuriousSkeptic 9y ago
Still learning ;) But IIUC, could you send a clock along 45-degree angles in a minkowski diagram it would not measure any time. But perhaps there is time there relative to an observer at 0-degrees? But what does that even mean then? If that observer sent a photon towards a mirror, would the photon have aged when returned to the sender?
- soVeryTired 9y ago
  You can make your statement more accurate by saying that a clock will measure vanishingly little time as you get sufficiently close to the 45 degree line. But the 45 degree line isn't achievable. (Note I've made the same inaccuracy further up the page. Sometimes it's just easier to drop the epsilons and speak informally.) Physicists are precise people :)
  - fjsolwmv 9y ago
    Mathematicians are precise people. Physicists just assume their series converge :-)

QAPereo 9y ago

I think you’re talking about null vectors/geodesics, which define light, which from the viewpoint of any external observer does move through spacetime. https://physics.stackexchange.com/questions/23058/why-is-light-described-by-a-null-geodesic https://physics.stackexchange.com/questions/23058/why-is-lig... https://physics.stackexchange.com/questions/188859/what-is-a-null-geodesic https://physics.stackexchange.com/questions/188859/what-is-a... Remember that it’s deceptive to imagine things like the POV of a photon, and better to think in the usual terms of external observation.

philipov 9y ago

If you were to imagine photons having an internal clock, that clock never ticks. That's not the same as light not moving through time. Light does not experience proper time. Any statement that combines the word "move" with the claim that it shows how light lacks time has already contradicted its goal. I think it is better to imagine light as line-like objects that extend through time, not point-like objects that move through time, and do away with speaking of light as something that moves, if you wish to express that it lacks a clock (proper time).

fjsolwmv 9y ago
Every particle has a life line (segment). Light particle's life line segment is perpendicular to the time axis, jumping across empty space (from birth/emission to death/absorption) at one constant time. If an atom emits an alpha particle (helium atom) and gamma particle (photon) at the same time, the alpha particle moves through space while its proper time changes (its "imaginary clock" ticks), whereas the gamma particle instantly arrives at its destination somewhere else in the Universe without experiencing any change in time, absorbed by another particle. Since light has no mass, it has no impact on spacetime geometry, so it only exists at emission and absorption. There is no way to detect or affect it between those to moments in its life, so it makes sense to consider them exactly one moment. Lights line segment is a "dotted line" the line segment (exclusive of endpoints) between emission and absorption does not really represent any part of our Universe; it's purely virtual. The only way to affect a photon is to plan ahead and set up an apparatus before it is born. Once the photon exists (relative to any observer), it's already gone (absorbed) by whatever first observed it. (And nothing else can directly observe it. All of us can bounce lots of stuff off the moon to observe it, but only one person can bounce only one thing off a photon , which kills it)
CuriousSkeptic 9y ago
But still, that line, in a minkowski diagram, represents a boundary, so it still seems more intuitive to me to eliminate the area beyond it completely from the diagram.
- ajkjk 9y ago

soVeryTired 9y ago

In any reference frame, light moves one (light) second of distance in one second of time. So the worldline of light should always be at 45 degrees in a minkowski diagram (with appropriate units). Is your issue that, loosely speaking, if you travelled from X to Y at the speed of light, you wouldn't experience any time elapsing as you travel? If so, remember that the Minkowski diagram relates to the observer's reference frame, not the light's reference frame.

jacobolus 9y ago
Also note that the “light’s reference frame” is not a meaningful concept.
- fjsolwmv 9y ago
  AIUI, "light has no reference frame" is the same as "light does not experience time" is the same as "light photon teleports from its birthplace (emission) to its death place (absorption), in its own reference frame".
  - jacobolus 9y ago
    “Reference frame” is a way of assigning a coordinate system so as to split spacetime into one dimension of “time” and three dimensions of “space”. Light doesn’t have a reference frame. Light doesn’t experience “space” either. Two points in spacetime along the path of the light are displaced by a “null vector”, i.e. the relativistic magnitude of the displacement is zero.