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  1. When a ray of light is projected, (say) from the surface of Earth to outside in space. The condition is that, there is no obstruction to it till infinity (it travels only in vaccum). My question is that how far can that ray of light go?

  2. Also, instead of a ray of light, if I consider a beam of laser with same conditions, then how far can a beam of laser go?

    Compare both the situations.

    And does the light(ray of light and beam of laser) stops after traveling some distance or it has no end?

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    $\begingroup$ Not quite sure why there's a bounty. The answer by iantresman answers the question quite well. $\endgroup$ – HDE 226868 Dec 14 '14 at 18:13
  • $\begingroup$ I agree w/ @HDE226868 $\endgroup$ – TanMath Dec 14 '14 at 22:18
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/18555/2451 , physics.stackexchange.com/q/105980/2451 and links therein. $\endgroup$ – Qmechanic Dec 15 '14 at 15:23
  • $\begingroup$ Is your first question essentially a slight variant of this: What percentage of light from a star situated at the center of the universe directly reach the edge of the universe ? $\endgroup$ – Gaurav Dec 16 '14 at 11:28
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    $\begingroup$ After my answer including this cartoon has been upvoted, downvoted, flagged and deleted, I would like to share this with you as a comment. Merry christmas to all of you. $\endgroup$ – Aziraphale Dec 17 '14 at 7:09

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Theoretically, the photon (or the beam of photons, there really isn't a difference) can go an infinite distance, traveling all the while at a speed $c$.

Since photons contain energy, $E=h\nu$, then energy conservation requires the photon to only be destroyed via interaction (e.g., absorption in an atom). There is nothing that could make the photon simply stop after some distance, it can only be stopped via an interaction of some sort.

Note that some of the light we are seeing from very distant galaxies are some billions of years old and traveled many yottameters to get here. Had they not been absorbed by Hubble space telescope, for instance, they would have continued on their way through our galaxy (until something else stopped it).

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    $\begingroup$ That's such a fascinating thought, by the way... That all those things traveled for such a long time to reach us... (don't even start spoiling these mythical musings with talk about absorption and reemission) $\endgroup$ – Danu Dec 15 '14 at 22:36
  • $\begingroup$ Yeah, but the point is that then the one we see may not have traveled all that far :P $\endgroup$ – Danu Dec 15 '14 at 23:41
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    $\begingroup$ Could the downvoter explain what they think is wrong with my post? $\endgroup$ – Kyle Kanos Dec 16 '14 at 13:35
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  1. A photon will travel "at the speed of light" until obstructed. From the speed, and elapsed time, you can calculate how far the light will travel.

  2. Laser light consists of more than one photon "in phase", which has exactly the same property in this respect, as a solitary photon.

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  • $\begingroup$ Does light has a STOP after some distance? $\endgroup$ – Sushant23 Dec 7 '14 at 13:53
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    $\begingroup$ Only if it interacts with something else, ie. gets absorbed after hitting an atom, or another particle. Otherwise there is is no reason for it to just vanish. $\endgroup$ – iantresman Dec 7 '14 at 14:43
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    $\begingroup$ There's not only no reason to just vanish, it even cannot just vanish because doing so would violate conservation of energy. $\endgroup$ – celtschk Dec 14 '14 at 21:58
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    $\begingroup$ It may be worth noting that space is not a vacuum, but a thin plasma, which may interact with the photon, preventing it from travelling to infinity. $\endgroup$ – iantresman Dec 15 '14 at 10:37
  • $\begingroup$ @iantresman Please elaborate or provide a source for considering space to be a thin plasma. I am aware of the quantum foam nature of the vacuum state, is that what you're referring to? Celtschk- the photon can vanish as long as it does so very briefly ;) $\endgroup$ – electronpusher Apr 27 '17 at 2:09
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Note that it is correct that a photon can travel an infinite distance in an infinite time, but it can not reach any desired point in the universe.

This is caused by the expansion of the universe, which also leads to the fact that we can not receive information outside of the observable universe.

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    $\begingroup$ I've heard this called the "communication horizon"; the Wikipedia article currently calls it a future horizon. $\endgroup$ – rob Dec 17 '14 at 7:23
  • $\begingroup$ Probably there are several options how to call it; I am non-native so I don't know for sure $\endgroup$ – Christian Dec 17 '14 at 18:08
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One small addition to the other answers: While it is indeed true that the light will never stop if it doesn't hit anything, it will however get red shifted, and thus become less energetic, due to the expansion of the universe. For example, the cosmic microwave background consists of photons which were emitted back when the atoms formed. However, back then the temperature of the universe was about $3000\,\rm K$ (about the melting point of iron) while today the cosmic microwave background has a temperature of merely $2.7\,\rm K$. So the photons we see in the CMB have travelled for more than 13 billion years without vanishing, however they have shifted in frequency from visible light down to microwaves.

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  • $\begingroup$ Does this mean that individual photons lose energy? Or is it rather the decreasing number of photons that reach us, resulting in less energetic radiation? $\endgroup$ – Aziraphale Dec 15 '14 at 13:52
  • $\begingroup$ @Aziraphale the individual photons lose energy (see a lot of related questions/answers on this site on how this ties together with energy conservation etc) $\endgroup$ – BjornW Dec 16 '14 at 14:33
  • $\begingroup$ The photons don't lose energy or 'become red shifted'. In the rest frame they were emitted they still have their original energy. If we measure their absorption in a different rest frame (e.g by 'seeing'), we will perceive them to be red shifted. $\endgroup$ – Julian Dec 24 '17 at 10:30
  • $\begingroup$ @Julian: You are confusing two things: Red shift due to relative movement is something different than red shift due to cosmic expansion. That is also why at the cosmic horizon the "relative speed" of massive objects can reach the speed of light without violating relativity: It's not a relative speed between objects in the special-relativistic sense, but an expansion of space itself. $\endgroup$ – celtschk Dec 24 '17 at 10:39
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    $\begingroup$ @celtschk - interesting. I don't think expansion red shift is really different to normal red shift. I think the photon still has its original wavelength, momentum and energy 'as far as it is concerned'. But if the photon was emitted by a proton which has no momentum then that proton would 'see' the expansion red shift. So my original rest frame doesn't really exist any more. $\endgroup$ – Julian Dec 30 '17 at 11:06
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Provided that there is nothing for the photon to interact with (i.e. we look at it in vacuum), the mean free path will be infinite; that is, it will continue travelling forever in a given direction. There's nothing which will stop the photon's path. Hence, it will go arbitrarily far. Whether you have a single photon or a laser, the answer won't change.

The fact that photon lines will never end manifests itself in another relevant fact. If you look at the intensity $I$ of light on a sphere of radius $r$ away from a point source, the intensity drops off as $1/r^2$. More specifically, if $P$ is the power of that source, then $I(r) = \frac{P}{4 \pi r^2}.$ The $4 \pi r^2$ in the denominator is just the surface area of the sphere.

You may think this is relatively trivial, but in fact, it's actually a pretty deep fact. We know from 20th century work that there are particles similar to photons but with some differences. One of these is the $Z$-boson. Unlike the massless photon, the $Z$ boson is massive. Its mass is around $91 GeV/c^2$, which is about 97 times as massive as a proton. If you did the corresponding analysis for $Z$-bosons, you'd find that they decay, and the decay length is on the order of $10^{-18}m$. A $Z$ boson will on average only travel about that far in vacuum. This leads to a different functional form for the above intensity, which will have an exponential dampening. In fact, this mass is essentially equivalent to studying photons in a medium which provides dissipation (e.g. inside a superconductor).

The fact that the photon doesn't suffer this same fate is really a consequence of its masslessness. There are many possible bounds on the photon mass. Of course, just the fact that we see photons from very long distances away provides a (rather strong) upper bound on the photon mass, though it is perhaps a bit deceptive as there are certain unusual models which avoid this strong bound. The most robust, model-independent bounds we have to date are about $10^{-14} eV/c^2$, that is, a factor of about $10^{23}$ lower than the proton mass.

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A ray of light or a laser beam will not stop until it reaches an obstruction.

If there is no obstruction, light will NEVER stop. It has no end.

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Whether it be a beam or ray of light, photons will keep traveling until they are absorbed. Photons can't stop because they travel at a constant velocity, the speed of light, i.e., they can't accelerate or decelerate. However, their wavelengths change over time due do the expansion of the universe, i.e, their wavelengths get larger and loose energy as such because $E_{\gamma}$ and $\lambda$ are inversely proportional,

$E_{\gamma} = \frac{hc}{\lambda}$.

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A "ray of light" must be respelled as "photon" because here we are talking physics.

Between a single photon and a laser beam, in this case, there is no difference. Every photon will continue his travel until stopped, every single photon is "indistinguible" from others (in the sense that they are no different intrinsecally). The photons of a laser beam are only at the same energy level and travel in the same direction (assuming a perfect laser) but this is of no importance for the question.

A photon can be stopped only by interacting with it with enough energy. If the interaction is of lower energy or is a gravitational field the photon will be deviated but continue "moving".

And does the light(ray of light and beam of laser) stops after traveling some distance or it has no end?

I think that you want to know if a photon can travel outside the Universe. If a photon reach the limit of the Universe it will continue his travel, extending the Universe itself!

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Newton's first law states a particle will have constant velocity unless an external force acts upon it. The photon has no mass, but nonetheless the first law still holds true in the case of light.

  1. When a ray of light is projected, (say) from the surface of Earth to outside in space. The condition is that, there is no obstruction to it till infinity (it travels only in vaccum). My question is that how far can that ray of light go?

$$x=vt$$

In this case $c=v$ where $c$ is the speed of light travelling in a vacuum (a constant) and $t$ seems to $\rightarrow \infty$ seconds based on the information given in your question.

The distance the light travels depends on the time it travels for because $c$ is constant in a vacuum which implies:

$$x\rightarrow\infty$$

  1. Also, instead of a ray of light, if I consider a beam of laser with same conditions, then how far can a beam of laser go?

Same as with 1.

Compare both the situations.

One is a ray of light travelling infinitely in a vacuum and the other is several rays of coherent light travelling infinitely in a vacuum.

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The distance that a particle can travel is partly set by its mass.

If the particle has a mass less than something like 7 eV, then it could cross the universe without attenuation.

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    $\begingroup$ Care to elaborate on how you came to this conclusion? $\endgroup$ – Kyle Kanos Dec 15 '14 at 13:45
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    $\begingroup$ what is "crossing" the universe ? $\endgroup$ – Gowtham Dec 15 '14 at 15:38
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    $\begingroup$ And, how do you "attenuate" a particle? $\endgroup$ – hdhondt Jul 17 '17 at 7:33

protected by Qmechanic Dec 15 '14 at 15:21

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