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Please note the following sentence before you answer the question: you may assume the light itself is not affected by the observer perceiving the light.

Imagine the following schematic:

enter image description here

  • A = Source
  • B = Observer
  • C = Destination

Light wants to travel from point A to point C. Imagine an observer is standing in between A and C. You may assume the light itself is not affected by the observer perceiving the light.

The observers task is to stop the light from point A getting to point C, however, it can only undertake action (e.g. put up a barrier) once it has perceived the light.

Can the observer B, stop light from travelling from A to C after it has perceived the light? Not a single ray of light should ever reach point C.

My own guess is no as nothing is faster than light, so there is no way I can interfere with the light after it has passed


Edit: something I thought of myself is the gravitational deflection of light, I'm not an expert on this topic so i'm curious if there is an answer using this.

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    $\begingroup$ First issue: How can B even "perceive" the light without absorbing it? $\endgroup$ – The Photon May 15 '17 at 16:18
  • $\begingroup$ @ThePhoton ironic name :), 'the light itself is not affected by the observer perceiving the light.' But if you really need an answer: A is not emitting one straight ray, it's just like a bulb. So observer B might be 0.0000000000001m from the straight line between A and C. But for this question, it is neglectible. $\endgroup$ – Thomas W May 15 '17 at 16:23
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    $\begingroup$ Why is a question based on a completely false physical premise (which the poster insist on enforcing) being given "protected" status ? Surely it should be closed as off topic or poorly researched. $\endgroup$ – StephenG May 15 '17 at 23:25
  • $\begingroup$ @StephenG why is everyone being so salty? I never say something exist that can detect photons without absorbing them. Im just stating that for the sake of the question, B is not absorbing the light - this does not affect the answer. And if you have read my other comments, i said that if people are really so stubborn about making a statement that isnt possible but makes the question easier (which happens more on this site btw), you can assume observer B is 0.00000000001m away from the straight line between A and C. $\endgroup$ – Thomas W May 16 '17 at 5:59
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    $\begingroup$ It's the idea that a single photon can be observed by two observers without being altered (in practice seeing a photon means absorbing it - i.e. destroying it). In physics you might ignore something that you expect to have a small effect on the result (e.g. a small force component) but you don't ignore major effects, like this. That's why we're getting "salty". You might think of it as a point of principle or honor with us. $\endgroup$ – StephenG May 16 '17 at 9:33
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The answer is really "physics doesn't work the way you claim it does." You observe light by absorbing it. If you do not absorb energy from the light, you are blissfully unaware of its existence.

But let's permit this. Let's create Thomas's Daemon which can sense a light beam as it passes without affecting the beam, but is otherwise bound by the laws of physics.

If everything is being done in a vacuum, your guess is correct. By the time your Daemon has observed the light, it's too late to stop it. The signal from the Daemon to block the light couldn't travel any faster than light, and any process which can close some door or something to block the light must be slower than light.

However, what if the Daemon is allowed to be smarter? Our smarter Daemon might put something to slow light down. Any medium slows light to below its ultimate speed. Some are particularly good at this. Wikipedia's Slow Light page claims that some physicists have created particularly novel mediums which can slow light down to a crawl. The current record is a scant 17m/s (down from 299792458 m/s)!

If our Daemon is clever enough to get such a medium in place before the test begins, they can observe the light beam (using the Daemon's special skills) before it hits the medium. They can then transmit the "close the door" signal to a door on the other side of the medium. If they transmit it around the medium, and not through it, then the signal can indeed out-race the light beam that has already passed!

Of course, the easiest solution involves no Daemon at all. Simply put up a wall before the start of the experiment at B. It does nothing until the beam is detected, at which point it is absorbed and the observer can announce "I blocked it!"

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You guess right but you are mistaken. Once light passes a point, nothing can stop that light from reaching it's destination. That you got right. The part you are mistaken about is observing light. We can't observe light, either directly or indirectly. Photons are massless, colorless and travel very, very fast. Our retina can detect when a photon has struck it, but in the process the photon is destroyed.

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  • $\begingroup$ Please notice: you may assume the light itself is not affected by the observer perceiving the light. $\endgroup$ – Thomas W May 15 '17 at 18:52
  • $\begingroup$ In that case, 4 mirrors are all you need. Between B and C, place one mirror to reflect the light to the moon. Place 2 mirrors on the moon to reflect the light back to earth, parallel to the first light beam. Place the last mirror so it reflects the light from the moon towards C. You now have time to stop the light. $\endgroup$ – zane scheepers May 15 '17 at 19:06
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Let me give a concise answer:

Let the destination C be fitted with a photodetector to detect incoming photons.

If the observer at B is larger in size than the photodetector at C, then it'll create a shadow area over the photodetector. If the observer perceives light, then he'll absorb the light (irrespective of the fact that light travels at 299,792,458 m/s), and there will be a shadow over the photodetector, minimising the number of photons that reach it. If the observer perceives light, and the photodetector has been covered in a box with a narrow slit, then it'll not sound at all in ideal conditions.

However, if the photodetector is not in a box, it may get photons from the light rays that are not absorbed by the observer, and might show positive results in such cases.

Regarding Gravitational deflection:

You must know that the mass of photons cannot be estimated as in QED, they are simply the intermediate particles, just like fluorspar in strong force and the W and Z particles in weak interaction. So, they are least affected by Gravitational fields.

However, light is affected by Gravitational waves, but that too over a very long astronomical distance.

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  • $\begingroup$ I like your answer, but I stated 'the light itself is not affected by the observer perceiving the light.' in the question $\endgroup$ – Thomas W May 15 '17 at 18:46
  • $\begingroup$ I added something wrt your updated question. $\endgroup$ – Wrichik Basu May 15 '17 at 18:53

protected by Qmechanic May 15 '17 at 18:51

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