# How quickily does a photon reach c?

1. Nov 28, 2012

### Neghentropia

Hi,
I have a very dumb question in my head: when a photon gets emitted by a source, how long does it take to reach c?
the fact it has no mass makes me thinking the acceleration is instantaneous (t=0) but this is from a classical physics point of view (and I also know that around speed of light few things go as it would be intuitive to think them going)

thanks :)

2. Nov 28, 2012

### Staff: Mentor

It does not accelerate at all. It is produced as particle travelling with c, and it travels at c all the time.

3. Nov 28, 2012

### dkotschessaa

That's actually a great question. You should keep asking questions like that. I never even thought to ask it.

Here's an interesting Q&A on the subject:

http://van.physics.illinois.edu/qa/listing.php?id=2026

-Dave K

4. Nov 28, 2012

### Neandethal00

That is an excellent question.
or most people simply gave up.

But my suspicion is, the photon emerges with constant velocity c
from whatever system it is trapped into. In other words, the acceleration
of the photon takes place inside the atom as it builds up energy to
leave the atom (I am here considering Bohr's atom).

5. Nov 28, 2012

### Neghentropia

and in those means where the photons travel slower than c? (i.e. everything except the void)

6. Nov 28, 2012

### Staff: Mentor

It isn't a dumb question, but "acceleration is instantaneous" is gibberish... Informative gibberish, though: As you correctly noted, instantaneous means t=0. That's not acceleration, it is just time. Acceleration is v/t, so if t=0, acceleration is undefined.

7. Nov 28, 2012

### Staff: Mentor

The same thing applies to any other elementary particle produced in a decay or interaction process. A muon decays into an electron and two neutrinos, each of which has a certain energy and velocity at the moment of decay. They don't accelerate to those velocities.

Afterwards, of course, the decay products may accelerate or decelerate depending on the conditions. When a nucleus decays via beta-decay, the emitted electron starts out with a certain energy and velocity, but then slows down a bit as it recedes from the nucleus because of the Coulomb attraction.

8. Nov 28, 2012

### dkotschessaa

According to the link I posted above "Those photons didn't really exist before. There aren't a fixed number of photons. They can be destroyed and created. The energy, momentum, and angular momentum they have were there beforehand, but not in the form of photons."

I think this is probably important to understand. C isn't just the "speed of light," but it is the maximum speed of other things as well, like the electrons that are running through the light bulb.

Do I have this about right?

-Dave K

9. Nov 28, 2012

### superdave

Now, when a photon is transitioning through a boundry, it must accelerate right?

10. Nov 28, 2012

### Michael Redei

This mental image may explain why some here think that there's some acceleration happening. Imagine someone, call him John, throwing a ball. First the ball in John's hand is at rest, and later it travels through the air with some speed v. Somehow it must have accelerated from zero to v, and so there's an acceleration involved.

When a photon is emitted by another particle, this is not like John who held the ball before he emitted it. The ball existed before the throw, and so it had to "be somewhere". But the photon came into existence, and so it couldn't have had a speed before, since there was no "before" from its point of view.

I think many people find this thought to be quite a hurdle. How can something just come into being if it wasn't there before? If the photon was emitted, it must have been hiding somewhere. Even a conjurer who pulls a rabbit out of a hat had that rabbit hidden somewhere, it didn't just appear from nothing.

Photons aren't rabbits though, and they don't behave like rabbits.

11. Nov 28, 2012

### Neghentropia

ok. that's actually pretty straightforward once you deal with the fact that photons are "generated"

but the questions I posted above remains. what happens to those photons that are generated into a medium (i.e air, water, glass) where the light is slower than in the vacuum?
are they decelerated?

if yes, what happens to a photon crossing a sheet of glass into the vaccum? it slows down and than accelerates again?

12. Nov 28, 2012

### dkotschessaa

Sorry to sound redundant, and I know you probably want a discussion - but the link I posted above pretty much answers all these questions.

(to repeat, it's here: http://van.physics.illinois.edu/qa/listing.php?id=2026)

-Dave K

13. Nov 28, 2012

### derek101

When photons travel through a medium they travel at the speed of light,but their rate of propagation is slowed by being absorbed and re-emitted by atoms.

14. Nov 28, 2012

### Neghentropia

sorry, I didn't see your previous post!

15. Nov 28, 2012

### Staff: Mentor

It depends on the interpretation - see derek101's for one possible model.
You can describe light propagation with "effective" photons in matter as well - when photons enter a medium, they just have a different velocity inside (but no "acceleration process"!) than outside.

Note that my description is a simplified model and not the actual quantum mechanics behind that.

16. Nov 28, 2012

### Neghentropia

it explains why the speed of a beam of photons travels at v<c, without the need to slow the actual photons down

here ( http://van.physics.illinois.edu/qa/listing.php?id=2026 ) I read:

"...As photons go into materials, they become something a little different, dressed in excitations of electrons. Assuming no reflection, the net energy E of the whole packet stays essentially the same, since the frequency f doesn't change and the relation E=hf is universal. However, the wavelength λ becomes shorter, so the momentum p of the packet goes up, since p=h/λ is also universal. (The momentum conservation is maintained by a slight recoil of the whole material.)

So since p goes up and v goes down, if you define inertial mass as p/v, it goes up by a factor (c/v)2=n2, where n is the index of refraction. In a typical material with electric but not magnetic susceptibility at optical frequencies, this is just a factor of ε, the electric susceptibility..."

I assume the v that goes down is the v of the beam of light, not the v of the actual photons, right?

17. Nov 28, 2012

### Staff: Mentor

From the Frequently Asked Physics Questions section at the top of this forum:

Last edited by a moderator: May 6, 2017
18. Nov 28, 2012

### Neghentropia

thanks

19. Nov 28, 2012

### derek101

Question.So does the model I described work ok for gas and liquids?

A photon entering a solid is no longer a particle until it is re-emitted/re-created.

20. Nov 28, 2012

### Staff: Mentor

Liquids are very similar to solids, and gases can behave similar, too (if the wavelength is long compared to the typical distances between atoms).

21. Nov 28, 2012

### Neghentropia

I would imagine it is like you describe. Although I can't quite grasp the meaning of a photon being absorbed by a phonon, given the latter is a quasiparticle.

anyway, from this topic as a whole, I think I got that photons, no matter the medium they are travellig through are always going to speed = c. Aristotele would say it's their natural state :)

at this point I'm wondering: what happens to those photons that enter a black hole??
I'm tempted to believe a black hole core (where the photons are dragged by gravity) has no vibrational state, so there are no phonons to absorb the photons.
Maybe I'm asking something beyond our current knowledge?

22. Nov 28, 2012

i would like to discuss this question from the wave hypothesis of light.
an analogy. if a photon is a wave like a sound wave, then think of the
creation of a photon as the plucking of a string. as the string is plucked
a train of waves, sound waves are created. and then the question becomes
how quickly do the sound waves build up? anybody familiar with music
will recognize this as the attack time.
so now the acceleration question becomes a question about how quickly
the wave of a photon builds up. and i think that this may have real
significance,

23. Nov 28, 2012

### ZapperZ

Staff Emeritus
But is this really a valid analogy for light?

For sound, you require that string to be there. Light requires no such medium.

Secondly, there is "build up" required to generate photons. After all, we have ample single-photon sources here. Are there "single photon" sources for sound? Have you ever heard of single-phonon sources for vibrational modes?

The wave picture breaks down rather easily here, and such analogy doesn't help.

Zz.

24. Nov 28, 2012

### Staff: Mentor

I think this will need a theory of quantum gravity.
Photons do not need phonons to get absorbed in a medium, but the central singularity (if it is one) cannot be described with current physics.

25. Nov 29, 2012

### Kazz

Although I very much doubt it... It could be C(m/s)/Planck Time(seconds)=5.5609185948589E+51 M/S^2
But, like I said. I doubt it.