# Tag Info

44

The answer is simple: Yes, stars really do produce that many photons. This calculation is a solid (though very rough) approximation that a star the size of the sun might emit about $10^{45}$ visible photons per second (1 followed by 45 zeros, a billion billion billion billion billion photons). You can do the calculation: If you're 10 light-years away from ...

21

Although I agree with all three of the above answers let me present a slightly different perspective on the problem. It's tempting to think of the light from the star as a flood of photons that behave like little bullets. However this is oversimplified because a photon is a localised object i.e. we observe a photon when something interacts with the light ...

17

Photons don't have a rest frame, since in all inertial frames they must go at the speed of light. So the following statement: By that logic, photons don't age in a vacuum state as, to us, the time stops for them. is meaningless because one really can't talk about proper time for a photon. However, in a medium, their speed decreases, Nope. The ...

10

The only stars you can reliably see are ones that are spewing enough photons at your eyeballs to appear stable. Any star which is so dim that photons entering your eye can literally be counted one by one, simply will not register in your vision, because your eye's retina is not sensitive enough. So your question is basically embroiled in observer bias; it ...

7

I will turn my comment into an answer, because the question in the header: Do photons age? is very anthropomorphic , and physics is a discipline that discourages interpreting data by use of the anthropic principle. The photon is an elementary particle. Aging is not a verb to be used with elementary particles in general because a) they have no ...

7

Allow me to channel something akin to the anthropic principle here. You can only see the stars that have a lot of photons reaching your eye. If a star were so far away that photons were reaching your eyes only occasionally then the star would be too dim for you to see in in the first place. Even if you could see the photons, the star would appear to ...

6

Your confusion arises because the light is not a photon and it's not a wave. It's a quantum field, specifically the photon field, and this quantum field can interact with other matter in particle like or wave like ways. My preferred way of thinking about this is that the photon is the unit of interaction of the quantum field with something else, so the ...

5

The field lines in your drawing are not the trajectories of photons. The field lines show the direction of the force on a test magnetic dipole. The force, and therefore its direction, is mediated by virtual photons (or can be described that way) but those photons will travel in straight lines just like ordinary photons.

5

A star radiates in all directions. You would still see the star regardless of the number of steps you take to any side, just not the same photons. A laser radiates in only one direction (or in a very small cone). If you took a large enough step to the side (larger than the angular size of the emitted beam) so as to exit this cone, then you would no longer ...

2

We can interpret any solution of Maxwell's equation as the quantum state of one photon (I discuss how this arises in this answer to the question "How can we interpret polarization and frequency when we are dealing with one single photon?"). Given this fact, the HUP can't be applied to light for position-momentum because there are problems defining a ...

2

Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared. Also, you should realize that just because a ...

1

If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation: $$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$

1

The best account for photon emission when an electron drops to a lower eigenstate is the Wigner-Weisskopf Model for spontaneous emission, see this paper from the Photonics group at ETH Zürich and the co-efficients for this model can be calculated by standard quantum electrodynamics. I explain this model further and give references in my answer here. A ...

1

I wonder if you're getting mixed up with propagation of waves in a physical medium like a string. If you have a wave travelling on a string then it has a velocity along the string, but the string is also oscillating normal to its length. So if you stretched the string along the $z$ axis, as the wave travelled along the string (i.e. the $z$ axis) the string ...

1

In the specific case of slowing light with a Bose-Einstein condensate there will be a limit because the slowing of the light is due to an interaction of the light with the BEC to form a polariton. If you put too much energy in you'll destroy the BEC and it will stop slowing the light. Offhand I don't know what the limit is, but it will be a very small amount ...

1

Having determined the momentum precisely is equivalent to having an absolutely monochromatic wave, which is never true. There's always uncertainty in the frequency of a photon which shows up as a width of the peak in the spectrum of any signal. And don't forget that we never measure a single photon's momentum, but we always work with statistical averages, ...

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