Let's say we have a cloud of dust which is a lightyear across and someone shoots a beam of light from point A to B , why it is not possible for an observer far far away to see the light while it travels through the cloud at the speed of light?


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    $\begingroup$ Do you know it is not possible? $\endgroup$ – paparazzo Jul 29 '15 at 2:59
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    $\begingroup$ Here's the rub: compute the intensity needed for the original source when you are talking about length scales on the order of light years. Even with a highly (but not perfectly!) collimated original beam, the scattered light is projected into a lot of solid area and therefore falls off with the familiar dependence. So, design the right detector system and that's not a huge problem, but rely on, say, the Mk I eyeball and you have problems. $\endgroup$ – dmckee --- ex-moderator kitten Jul 29 '15 at 3:18
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    $\begingroup$ Whatever light enters your eye is not light that is traveling to B. $\endgroup$ – 200_success Jul 29 '15 at 5:09
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    $\begingroup$ I'm not sure whether this is a question about light echoes, or a fundamental misunderstanding of what it means to see something. $\endgroup$ – Lightness Races in Orbit Jul 29 '15 at 9:52
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    $\begingroup$ In most cases you'd call that 'seeing the dust as the light pulse travels through it' rather than 'seeing the light pulse as it travels through the dust'. Of course, what you actually interact with is neither, it's the light emitted from the dust or reflected from the dust as the light pulse encounters the dust. What you call it isn't really all that relevant, but may lead to a new perspective or understanding. $\endgroup$ – Martijn Jul 29 '15 at 12:14

Sometimes we do, and the phenomenon is called a light echo.

enter image description here

What you're looking at there is NOT moving gas. It's an "echo" exactly as you describe.

The problem is that you need a pulse of light. If you have a constant stream of light, the "light echos" will be exactly like what you see in fog on earth.

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    $\begingroup$ This is a fantastic link, but I think some deeper explanation is needed. At least say that the pulsing is owing to a supernova. $\endgroup$ – Selene Routley Jul 29 '15 at 2:55
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    $\begingroup$ I am not convinced that this answers OPs question, at least not directly. I think this could use quite a bit more detail on the how/why it does answer it. $\endgroup$ – Kyle Kanos Jul 29 '15 at 3:08
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    $\begingroup$ @CeesTimmerman then Point B includes the observer. $\endgroup$ – WBT Jul 29 '15 at 17:16
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    $\begingroup$ What is the amount of time represented in this gif? $\endgroup$ – duzzy Jul 29 '15 at 20:45
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    $\begingroup$ @duzzy - Based on the captions on the Wikipedia picture, it's a bit under 2 years: May 2012 to Feb 2014. $\endgroup$ – Johnny Jul 29 '15 at 23:27

It would be possible to see the progress of photons through space if the light pulse were exceedingly intense, and if the dust cloud from which they reflect were positioned and shaped to reflect the light toward us. Rather than shooting a beam from Point A to Point B, it would be better if the light source were between us and the dust cloud, as light reflected off the cloud would appear more intense to us and more likely to be seen by us. It's best if the pulse were instantaneous, as otherwise it would appear to be an expanding diffuse gas ball with no interior detail, as John Rennie points out in his comment.

All these conditions were met by the link which NeuroFuzzy provided, which appears to be time lapse photography of a pulse of light from V838 Monocerotis echoing off of a dust could behind it, the most spectacular light echo in the history of astronomy, according to the European Space Agency.

It is much brighter than even a supernova, but it is not exactly an explosion. The light from the initial pulse reached the earth in 2002. V838 Monocerotis didn't lose its outer skin. Instead, it expanded enormously in size, until its outer skin was not much hotter than a light bulb. This is very unusual behavior and may have been caused by one star cannibalizing another.

The animation of the light echo does NOT depict ejected debris. What you see is the light itself reflecting off of interstellar dust that is mostly behind the progenitor, not in front of it. The light echo forms an expanding ellipsoid with V838 Monocerotis at one focus and we observers at the other. It's concave toward us. A good description is in the Wikipedia article, which is the last link below.

Here is an account of the event, and an even better time lapse video, thanks to the Hubble telescope: http://www.theatlantic.com/technology/archive/2014/06/space-cannibalism-is-beautiful/373260/. It's fascinating to see the progress of photons themselves as they cover vast distances.

Here is a more complete history of the event, with a bibliography: http://www.phschool.com/science/science_news/articles/enigmatic_eruption_v838.html.

The wikipedia article also includes a good bibliography of references to this light pulse and the subsequent light echo: https://en.wikipedia.org/wiki/V838_Monocerotis.

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    $\begingroup$ I'm having a hard time trying to understand why it looks like an expanding cloud of dust. What would we see if instead of a flash it was a continuous illumination? Would it be the result of blending all the frames together? $\endgroup$ – Ruslan Jul 29 '15 at 11:05
  • $\begingroup$ @Ruslan: you'd see an expanding ball of glowing gas. The glow from the outer regions of the ball would tend to mask details from the interior. How much inner detail you'd see would depend on how dense the gas was. $\endgroup$ – John Rennie Jul 29 '15 at 11:35
  • $\begingroup$ @JohnRennie Glowing gas? Surely you mean reflecting dust. $\endgroup$ – Cees Timmerman Jul 29 '15 at 11:44
  • $\begingroup$ @CeesTimmerman: yes, I mean only that it wouild look as if it was glowing because it would be scattering the light. The dust itself is cold. $\endgroup$ – John Rennie Jul 29 '15 at 12:42
  • $\begingroup$ How is this an answer to the question? At best, this is an explication on Neuro's complete lack of details in his answer. $\endgroup$ – Kyle Kanos Jul 30 '15 at 3:46

If some of the light is reflected off the dust at such an angle that it is diverted to reach the observer, the observer will see that light. However, those specific photons reaching the observer will not reach B (unless they are reflected there by the observer). Similarly, unless the observer is at point B (which is not the case in the question as asked), or light is reflected from B to the observer, the light which reaches B will not reach the observer.

The observer may use past experience about the behavior of light etc. to infer that the source of the light is not the dust particles but A, and that any light not dissipated by dust will reach B. For example, in the following image, we can infer that the source of the light is the sun and that some of the light, not dissipated by particles in the air, will likely reach particular spots on the ground in that field. Here, we can also see that some of the light reaching those points on the ground is reflected back to the observer, confirming the inference.

shafts of sunlight over field poking through clouds

The observer, illustrated as an eye, is a light detector which only detects light reaching it, the observer, not light which doesn't reach it. (That's the short answer to the question.) The observer will also have to wait whatever time is necessary for the photons to go from A to whatever it's being reflected off of, to the observer.

By contrast, in this second photo, certain frequencies of light are reflected from B (points on the ground) to the observer, but the path the light is following is not as clearly shown because of the differing amount of particles on the path between those ground-points and the light source.

shafts of sunlight field


Using a camera that can capture "Motion at a Trillion Frames Per Second", this can be done at the laboratory scale. The technique used has been called femto-photography.

enter image description here

(Image credit to Ramesh Raskar, Associate Professor, MIT Media Lab)

Of course a camera that literally takes one trillion full frames per second is totally impossible with today's technology. Assuming 1000x1000pixel frame size and 3 bytes per pixel, such a camera would have to output a total raw data rate of at least $3*10^{18}$ bytes/second or 24 billion Gigabit/second!

Instead, the photographed scene must be repeatable (i.e. totally still and variations in ambient lighting eliminated). A laser is used to send many short pulses of light into the scene. As well as as illuminating the scene, the pulse of laser light triggers a streak tube, which effectively captures one scan line of image. With multiple repeated exposures, the scan lines can be built up into an image, and multiple images into a full-motion video. The trick is in careful synchronization of time and viewpoints.

More info from MIT here.

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    $\begingroup$ This is a nice example of "seeing light move" but it would be good to include a little more description of the experiment rather than just linking to an external site (which can go stale) $\endgroup$ – Floris Jul 29 '15 at 22:39
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    $\begingroup$ Your description is misleading. The animation you included does not show the actual propagation of a single light pulse. As far as I know no existing camera is capable of capturing such a frame rate in such a quality. ------ Instead they captured millions of images of millions of repeated discrete light pulse expositions: We use an indirect 'stroboscopic' method that records millions of repeated measurements by careful scanning in time and viewpoints. Then we rearrange the data to create a 'movie' of a nanosecond long event. $\endgroup$ – pabouk Jul 30 '15 at 9:46
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    $\begingroup$ I remember all the news articles at the time being equally misleading. Everybody said "omg a camera that is fast enough to show light actually moving", which is complete nonsense. $\endgroup$ – Lightness Races in Orbit Jul 31 '15 at 22:43
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    $\begingroup$ @LightnessRacesinOrbit I hope I clarified this a bit, though judging by the continued upvotes to pabouk's comment, perhaps I didn't do a great job of that. In other news: can you be observed while you race around in orbit? Perhaps that would be the best answer to this question ;-) $\endgroup$ – Digital Trauma Jul 31 '15 at 22:47
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    $\begingroup$ @DigitalTrauma: :P It is indeed a similar question! $\endgroup$ – Lightness Races in Orbit Jul 31 '15 at 22:52

Let's say you build a ping pong ball counter. It increments the count every time a ping pong ball hits the sensor.

You throw a ball, and it hits the sensor: Detected!

You throw a ball across the sensor from left to right... no detection, because you didn't hit the sensor.

Your eyeball is a light sensor, which creates pictures from the light that hits your optic nerve. You can't see light that doesn't enter your eye.

  • $\begingroup$ To extend this metaphor to the "light echo" in the other answer - you throw a trillion ping pong balls at the same time, and some small fraction of them bounce off of obstacles in the room and hit the sensor (which also knows what direction it was hit from) $\endgroup$ – Random832 Jul 29 '15 at 21:32

My masters project was on something like this (though with hydrogen alpha emission lines for gas clouds between galaxies rather than dust between stars) and the answer in that case (and almost certainly in this one too) is that you can't see it because it's just too dim, though using hundreds of telescope hours can get you close (maybe).

Again, this is not exactly the same thing but is a related concept: http://arxiv.org/pdf/0711.1354v1.pdf

(that wasn't me but was the paper I referred to most often.)


Under the conditions stipulated, the question is a false statement. We would, in fact, be able to "see" the light beam between points A & B. As the photons travel from point A to B, some photons will be deflected (by colliding with the dust particles) in our direction, allowing us to "see" the beam. Only in the absence of dust particles (or any other particles), we would not be able to see the light beam.
To avoid complications due to light intensity, big distances, and diffusion, the experiment is done in a lab, using a laser beam, and the distance between A & B equals a light-nanosecond.


If the photon you described is going to move from A to B, there's no reason you can see it at position C. Because if you were able to see the photon, it had moved from A to C, where you are, not B. Therefore, you are not able to see the path light travels through, unless, as others said, something like an atom deflects it. In all the answers above, something has reflected light so that it comes to the photographer's camera. Otherwise, it's impossible to see that photon.


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