Can an imminent laser strike ever be observable?

[davidjoe]

If you are in line with it, arrival is without notice, but what if it were fired to intersect with the point where earth “will be” say from the second closest star, several light years away, and during that period of travel, the bodies get further apart, in non parallel paths.

Is it possible to see it fired, not directly at you, then a few minutes or hours later, be hit by it, and what would you see if you trained the telescope on that beam from when you saw it fired, to impact?

 

[Vanadium 50]

I do not understand your scenario, but there is no scenario in which you get a single that the light is coming before the light does.

 

[Rive]

Nope. Light is going the fastest (straightest) between two points: anything sideways will be just slower.

You need something like a black hole or 'space-time anomaly' (=> technobabble) to have a suitable path to be observed.

 

[hutchphd]

This of course assumes that you not bounce the light off of a mirror or other nontrivial intervention in the primary ("light gets to it") path. I can concoct tricky ways but that is not the spirit of the question I think

 

[davidjoe]

I’m envisioning the “leading the target” scenario, but over years of travel. But it might be similar to envisioning a long transparent optical cable where I’m walking close to the source of a pulse of light emitted, that I can look over and see. The cable could loosely arc out in front of both of us, with the other end coming back over to me. If some light escaped as it entered the long optical cable, the direct path to me is shorter, and I can see that escaped pulse slightly before the pulse exiting the cable is visible.

 

[hutchphd]

That represents one of the tricky scenarios I think. You define the early light you see directly as "not that light" and so we are in a semantic muddle. I feared the unequivocal statements for this reason although they are correct.

 

[davidjoe]

That represents one of the tricky scenarios I think.

… just trying to better illustrate the question of if it could be possible to see the laser fired, then watch it come to you.

Earth’s travel further away from the source would be likened to the laser being fired out to the point where earth “would be” years later.

I would imagine that if you could see it fired before being hit in any kind of scenario, then, you could conceivably track it, but what you would see if you could eludes me.

But it appearing curve into a straight on aimed targeting of the earth, even though it followed a straight line, seems inevitable.

 

[hutchphd]

I think any complicated (fair) method you devise can be analyzed in terms of the simplest case and the triangle inequality will always prevail.

 

[davidjoe]

That represents one of the tricky scenarios I think. You define the early light you see directly as "not that light" and so we are in a semantic muddle. I feared the unequivocal statements for this reason although they are correct.

The early light counts, that would be the equivalent of seeing the laser fired, that is going to hit you, before it does.

 

[Ibix]

The only way you can see it coming is if there is also an available shorter path than that taken by the laser pulse. "Leading the target" is not doing that, because if you fail to lead your target you miss. If I spot a laser pulse passing by and want to fire off a message to warn Earth my signal also needs to lead its target by the same amount as the laser pulse or I miss, so the best it can do (if I react in zero time) is arrive at the same time as the pulse.

You need a fairly sharp corner in the light path, some reason why they took an indirect shot, and someone stationed before the corner and close enough to the beam path to detect it (and that probably means getting hit by it - laser beams in space won't glow) who has a direct shot at Earth. And you are still just getting warning, not seeing it coming.

Also, note that laser beams spread. At interstellar ranges the laser muzzle would need to be many kilometres across to deliver damaging energy densities to a target.

 

[davidjoe]

The only way you can see it coming is if there is also an available shorter path than that taken by the laser pulse. "Leading the target" is not doing that, because if you fail to lead your target you miss. If I spot a laser pulse passing by and want to fire off a message to warn Earth my signal also needs to lead its target by the same amount as the laser pulse or I miss, so the best it can do (if I react in zero time) is arrive at the same time as the pulse.

You need a fairly sharp corner in the light path, some reason why they took an indirect shot, and someone stationed before the corner and close enough to the beam path to detect it (and that probably means getting hit by it - laser beams in space won't glow) who has a direct shot at Earth. And you are still just getting warning, not seeing it coming.

Also, note that laser beams spread. At interstellar ranges the laser muzzle would need to be many kilometres across to deliver damaging energy densities to a target.

Supposing that the beam is visible in the way that we are used to seeing lasers, it doesn’t widen, it’s 100 feet long, doesn’t weaken, etc., and the difference between its speed and you, the target, is very minimal, say it’s gaining on you from behind and slightly to the right, at the rate of an inch a day, in a nearly parallel line to the one you are traveling, and it’s only a few feet to the right, and closing.

In the future, it’s going to intersect your course, and that is where the strike will occur, but prior to that happening, it is behind, then beside, and then in front of your for days or weeks. Does its light not emanate to its side such that you may be able to see it, at least when it ahead of you, and slightly to your right?

 

[Ibix]

Supposing that the beam is visible in the way that we are used to seeing lasers

For that to be possible the beam must be passing through a cloud of particles with density similar to smoke in a nightclub - the kind of place where you can only see about ten yards. First, I have no idea where you'd find that in space, and second, the beam will scatter into nothingness in a few meters.

the difference between its speed and you, the target, is very minimal,

Unfortunately, the universe is governed by relativity and light is always doing $3\times 10^8\mathrm{ms^{-1}}$ relative to everybody. So this situation is not possible.

Does its light not emanate to its side such that you may be able to see it, at least when it ahead of you, and slightly to your right?

You mean, can laser light can overtake itself? No.

 

[hutchphd]

Does its light not emanate to its side such that you may be able to see it, at least when it ahead of you, and slightly to your right?

Wha does this mean? (Emanate to the side???.) Light in free space travels sperically out from a source. So no because any piecewise (not straight) path will be longer than the direct path.

 

[davidjoe]

I follow on not seeing it beside or behind you, although it’s there and closing. If you and the beam collide just several inches behind its front, it has still been in front of you many hours, (you are just under C) where I would imagine it is casting light in all directions that you are seeing, and I would suppose is getting brighter as you get closer.

 

[Ibix]

you are just under C

There is no such thing. If you see me travelling near light speed then I see you travelling near light speed and both of us see light travelling at $c$. Time dilation, length contraction and the relativity of simultaneity make it so.

I would imagine it is casting light in all directions

...which is travelling at $c$, same as the laser beam. So it doesn't get to me before the beam does. Light cannot overtake light!

 

[DaveC426913]

Not sure if anyone mentioned it, but light that does not directly enter your eye/sensor is invisible - certainly in the vacuum of space a laser light beam cannot be seen at all, unless there is something in its path like a cloud of dust or gas (which will incidentally weaken its power).

If you go outside on a clear night with a laser pointer, you cannot see the beam, unless there is sufficient haze or fog to interfere with it, reflecting it to your eyes.

 

[davidjoe]

 

[davidjoe]

It’s like being on the highway beside a car nearly at your speed. So, with filters or sensors, or such, to make the laser visible, once it gets a little ahead of you casting light such that the photons don’t fall behind before they reach you, can you see the laser growing brighter and brighter until the strike?

 

[DaveC426913]

Unless the laser path (or at least a portion of it, if it has spread) directly intersects your location, you will see nothing.

Note that there is nothing you can do about it anyway, since no signals you send can reach the target before the laser light does.

 

[davidjoe]

Unless the laser path directly intersects your location, you will see nothing.

Yes, the premise is that it will cross your path. But prior to that, it’s going to get ahead of you by a little bit, because let’s say, it was timed that way to guarantee a hit because you present a small target and the beam’s length is being used to full effect. Once it has passed you, albeit, at a snail’s pace from your perspective, some of that light emitted is sent back your way. You haven’t hit it yet, that’s coming, but hasn’t occurred yet. You can see that light, right, before you collide?

 

[DaveC426913]

Yes, the premise is that it will cross your path. But prior to that, it’s going to get ahead of you by a little bit, because let’s say, it was timed that way to guarantee a hit because you present a small target and the beam’s length is being used to full effect. Once it has passed you, albeit, at a snail’s pace from your perspective, some of that light emitted is sent back your way. You haven’t hit it yet, that’s coming, but hasn’t occurred yet. You can see that light, right, before you collide?

Sorry, this is not what your diagram seems to show.

I don't know what you mean by 'get ahead of you a little bit' or how 'some of the light is sent back your way' or 'you haven't hit it yet'.

 

[Ibix]

It’s like being on the highway beside a car nearly at your speed.

...except the velocity addition formula you are using is only valid at low speeds. I still measure the light beam doing $c$ relative to me.

And the fact remains that light from the laser is travelling at light speed, as is any light scattered from it. The scattered light cannot overtake the beam! They are doing the same speed!

 

[DaveC426913]

Once it has passed you, albeit, at a snail’s pace

No. Light is always observed to be moving at c by any observer. Even if you are headed for the target at .9999999c, you will experience the light beam passing you at c relative to you.

It is NOT like two cars on a highway.
If you are doing 59.9999mph on the highway you will still clock the "light beam" car next to you passing you at what you measure as 60mph faster than you.

Light is measured to be moving at c in all reference frames, including yours, moving at .999999c.

That's relativity. It has no classical counterpart.

 

[davidjoe]

No. Light is always observed to be moving at c. Even if you are headed for the target at .9999999c, you will experience the light beam passing you at the speed of light relative to you.

It is NOT like two cars on a highway.
If you are doing 59.9999mph on the highway you will still clock the "light beam" car next to you passing you at what you measure as 60mph faster than you.

Light is measured to be moving at c in all reference frames, including yours, moving at .999999c.

That's relativity. It has no classical counterpart.

Hmm. I have seen that illustrated where the moving object projects a light from its own platform. The light picks up no net speed. You cannot ride alongside a photon?

If I see the laser pass me at C, the laser nevertheless takes a long time to catch up to me, as illustrated, right?

 

[Ibix]

If I see the laser pass me at C, the laser nevertheless takes a long time to catch up to me, as illustrated, right?

Depends who's doing the measuring. To me, watching you travelling along at 0.99c, yes. To you, seeing yourself at rest and me doing 0.99c, no.

 

[DaveC426913]

Hmm. I have seen that illustrated where the moving object projects a light from its own platform. The light picks up no net speed. You cannot ride alongside a photon?

No and no.
If you shine a flashlight as you pass Earth at .9999c, and someone on Earth shines a flashlight as well, both of you will measure both flashlight beams to be moving at c relative to yourselves.

That is, you will measure your beam and his beam both taking one second to travel a distance of one light second.

He will measure his beam and your beam taking one second to travel one light second.

If I see the laser pass me at C, the laser nevertheless takes a long time to catch up to me, as illustrated, right?

I don't know what this means. If it 'passes' you then it's not 'catching up' to you.


Let's put it this way. You are in your spaceship and just woke up so you have no idea, without looking at a star or planet, how fast you are going. You might be stationary for all you know.

You see a laser light beam played across your ship. Nothing* about that laser light beam will tell you how fast you are going (relative to some planet or star). It behaves the same way as far as you can tell, whether you are standing still or moving at .9999c.

*_{almost nothing}


There's more to why this is so, but you'll have to go down the relativity rabbit hole.

 

[davidjoe]

I pass the laser at a constant speed just under C… my path is observed and computed and the laser is fired to intersect (hit) me, and the laser is aimed and fired.

In this case the laser beam travels at C and I am just under that. The laser does not close in on me as if I am stationary, relative to it, right? It closes in on me at the rate of an inch a day, as illustrated… right? In other words if fired a second after I passed it, it does not hit me in a second, it hits me in more like the number of days equal to the number of inches light travels in a second, right?

 

[Ibix]

The laser does not close in on me as if I am stationary, relative to it, right?

As I said in my last post, it depends who is doing the measuring. You will measure it closing on you at $3\times 10^8\mathrm{ms^{-1}}$. Whoever it is who says you are doing "just under $c$" will say it closes the gap very slowly (and that your clocks are ticking very slowly, they aren't synchronised properly, and your rulers are length contracted, which is why he will not have a problem reconciling his measurements with yours).

 

[davidjoe]

As I said in my last post, it depends who is doing the measuring. You will measure it closing on you at $3\times 10^8\mathrm{ms^{-1}}$. Whoever it is who says you are doing "just under $c$" will say it closes the gap very slowly (and that your clocks are ticking very slowly, they aren't synchronised properly, and your rulers are length contracted, which is why he will not have a problem reconciling his measurements with yours).

I think it’s the proper answer but does it mean I get hit a second after I pass it?

Because if I perceive that I got hit a second after I passed it, which is when it was fired, then I really may as well be going 10 mph, right? If we assume that the laser targeting always needs a second to compute and get the shot off.

I think what you are saying is that I’m going to get hit roughly a second after the laser fires, as I measure it, no matter how fast or slow I’m going, up to and including an inch a day under C.

 

[Ibix]

I think it’s the proper answer but does it mean I get hit a second after I pass it?

A second after you pass what , according to who?

 

[davidjoe]

Let’s just say a second on the targeting system’s clock, tripping a sensor.

 

[davidjoe]

I realize that a second on my clock may be a long time to someone else at my speed, but it does not mean I can accomplish any more in that ship-second.

 

[davidjoe]

To reduce any ambiguity, inside my own ship, does the amount of time I have before being struck, vary at all in any meaningful way, as a result of the speed I am going, if no matter what that speed is, the laser is going to be fired at me after I pass a certain point? (Such that I realize inside the ship, no benefit of extended time, for traveling almost as fast as the laser is going)?

 

[DaveC426913]

You really need ri simplify your scenario. It is too complex to analyze.

If any scenario has any observer measuring any beam of light moving at any speed other than c, that scenario is wrong.

 

[davidjoe]

not to try to answer my own queries but I think the answer under relativity is that I don’t have more time inside the ship, as a result of how fast or slow I may have propelled it relative to the laser. It’s tough to wrap my mind around that because an inch a day slower than C, or a micron a millennium slower than C, are both less than C, it means I’d see the laser pass me at C as if I am not moving relative to it, at all. For that matter I think if I were to pass it in the opposite direction a micron a year under C, same result. This really wasn’t the question I was wondering about originally but it (SR GR) has a certain way about it, of becoming the analysis.

A cartooned steam train with a lantern by a pedestrian with a lantern is sure easier for me to take at face value.