this post was submitted on 10 Jul 2024
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[–] Brokkr@lemmy.world 26 points 4 months ago (2 children)

As long as they do not interact with any other particles then yes.

Remember, in the photon's frame of reference (i.e. It's point of view), time does not progress. So it is created and destroyed in the same moment. Any distance traveled for any amount of time in our reference frame, happens instantaneously for the photon.

[–] niktemadur@lemmy.world 6 points 4 months ago (1 children)

in the photon's frame of reference, time does not progress

Couldn't the same be said about black holes/singularities?
Yet they will evaporate via Hawking Radiation, over the course of eons upon eons of time.

[–] count_of_monte_carlo@lemmy.world 13 points 4 months ago (1 children)

in the photon's frame of reference

There are no valid inertial frames for an object moving at the speed of light. The idea that “a photon doesn’t experience time” is a common, but misleadingly incorrect statement, since we can’t define a reference frame for it. Sometimes this misconception can be useful for conveying some qualitative ideas (photons don’t decay), but often it leads to contradictions like your question about Hawking Radiation for black holes.

[–] niktemadur@lemmy.world 2 points 3 months ago* (last edited 3 months ago)

There are no valid inertial frames for an object moving at the speed of light.

Man, that is one weird concept to wrap one's head around. In fact, I'm not even sure what it means, how to visualize it, my mind trying to "make out the gears that make the contraption work", how do I make it let go of the classical physics it clings to?

[–] AbouBenAdhem@lemmy.world 11 points 4 months ago* (last edited 4 months ago) (1 children)

The wavelength of a photon isn’t intrinsic to the photon itself—it depends on the observer’s inertial frame. The Hubble redshift occurs because expansion affects the velocity of observers relative to the photons’ original frame, not because it affects photons directly.

[–] niktemadur@lemmy.world 1 points 3 months ago* (last edited 3 months ago) (1 children)

So I might perceive a stream of photons as radio waves from our current inertial frame, but if I were on a ship approaching the speed of light head-on towards that same stream of photons, I might perceive them as visible light? Or ultraviolet or gamma ray.

Wait... that doesn't sound right, for some reason.
It took billions of years of universe-stretching for the CMB to redshift to microwave, I don't think me pushing pedal to the metal for a few seconds on a rocket ship is going to counteract all those years.

Then also those CMB photons are more diffuse, spread out.
Doesn't Coloumb play into this?

[–] AbouBenAdhem@lemmy.world 4 points 3 months ago (1 children)

It took billions of years of universe-stretching for the CMB to redshift to microwave, I don’t think me pushing pedal to the metal for a few seconds on a rocket ship is going to counteract all those years.

We do see shifts in the CMBR due to local velocity changes though—for instance, we can tell that the sun is moving at about 370 km/s relative to the CMBR frame due to its radial movement through the galaxy and the motion of the galaxy itself through space.

[–] niktemadur@lemmy.world 1 points 3 months ago

Sure, that I do get, it's just that I'm guessing my local movement at the speed of light for a moment can blueshift a radiowave or microwave photon only a fraction of what is needed to get it into the visible light spectrum, surely never all the way to ultraviolet or gamma rays.

[–] count_of_monte_carlo@lemmy.world 10 points 4 months ago

Yes, the wavelength of photons will be preserved if they travel through non-expanding space. If the photon is emitted by a source that’s in motion with respect to a detector, there could still be redshift or blueshift from the relativistic Doppler effect. This would only depend on the relative velocity between the emitter and observer, and not on the distance the photon traveled between them.