Probably not.

After some more consideration, I'm less confident that the proposal could work. There are two reasons: No guarantee of focusing at any one point, and an inability to control the parameters of the lens.

1. A gravitational lens has no focal point.

The analogy to a traditional lens does eventually break down. One reason is that gravitational lenses do not have focal points! They behave differently than the lenses you might be used to dealing with in optics; instead, gravitational lenses have focus lines. This means that if there is deviation in the positions of your lasers, you are definitely going to have problems.

The big issue is that the effects of lensing are not linear. A light ray twice as far from the lens will not experience twice the lensing. In optics, you do see linear dependence for lensing, which is what allows you to construct lenses with focal points. That's what makes things like glasses, cameras and telescopes possible and effective.

If a laser is offset the wrong distance from the central axis (the large black line in the figure below), its beam will reach the focus line at a different point than that of a laser positioned correctly. This means that your lasers will not, in fact, combine constructively. There's even the possibility for some destructive interference.

2. You need just the right setup for your lens.

I put together a diagram of a typical gravitational lensing scenario, assuming that the lensing object is a point mass:

I'm using some standard gravitational lensing notation:

• $\eta$: The initial distance of the source from some line connecting the source and the lens.
• $\xi$: The distance where the beam of light passes closest to the lens. Normally, $\xi\neq\eta$.
• $D_L$: The distance from the target to the lens.
• $D_{LS}$: The distance from the lens to the source.

The remaining three angles ($\alpha$, $\beta$, $\theta$) should be apparent. The path of the light is in yellow.

Your question features a ring of lasers spaced at some common distance from the center (I assume) with some radius $a$. I'm also assuming that the line connecting the ring and the target is perpendicular to the plane of the ring, meaning the initial path of the light is parallel to the line. Therefore, we have a special case where $\xi=\eta$.

One important quantity is $\tilde{\alpha}$, given by $$\tilde{\alpha}=\frac{4GM(\xi)}{c^2\xi}\tag{1}$$ where $M(\xi)$ is the mass contained within $\xi$. If we assume that $a=\xi$ is much greater than the radius of this object, then $M(\xi)=M$, the mass of the object.

The angle $\alpha$ itself is $$\alpha=\left(\frac{D_{LS}}{D_S}\right)\tilde{\alpha}\tag{2}$$ where $D_S=D_L+D_{LS}$, the distance to the target, and $\theta$ is $$\theta=\beta+\alpha\tag{3}$$ Trigonometry means that $$\tan(\beta)=\frac{\xi}{D_S},\quad\tan(\theta)=\frac{\eta}{D_L}=\frac{\xi}{D_L}$$ We can assume that $\beta$ and $\theta$ is small because $D_S\gg\xi$ and $D_L\gg\xi$, and so, by the small-angle approximation, $\tan(x)\approx x$: $$\beta\approx\frac{\xi}{D_S},\quad\theta\approx\frac{\xi}{D_L}\tag{4}$$ Inserting all of this into $\text{(3)}$ gives us, for our case where $\xi=\eta=a$, $$\boxed{\frac{a}{D_L}=\frac{a}{D_S}+\frac{D_{LS}}{D_S}\frac{4GM}{c^2a}}\tag{5}$$ You just need to find an object with approximately the right parameters.

I'd advise finding something relatively small. Galaxies are not good because we've assumed that all of these lasers point in the same direction (and don't spread out too much - though that's not likely). If a galaxy was the lensing object, you'd need $2a>d$, where $d$ is the diameter of the galaxy, and that's not really feasible in your scenario! If the lasers could be pointed independently, then maybe you'd have something, but it doesn't seem like that's the case - and anyway, if they could, you could then simply point them all at the target without using any lensing at all. However, I believe that's not the case with the sort of beams you're describing.

You're pretty limited by this method because you need three collinear points: The center of your array of lasers, the target, and the lens itself. We all know that Space is Big; more to the point, it's pretty empty when it comes to objects like stars. If you're using gravitational lensing, you really need to pick your targets wisely. $\text{(5)}$ is incredibly important (and, I think, accurate), because it must be satisfied for this to work.

Addendum

I'm a little concerned with the idea of giant space lasers, especially when used on the scale you're talking about. Let's assume that the beams are Gaussian (see also here), that is, the "radius" of the beam is given by $$w(z)=w_0\sqrt{1+\left(\frac{z}{z_R}\right)^2}\tag{6}$$ where $z_R$ is a scale length called the Rayleigh range, calculated in terms of wavelength of the beam, $\lambda$, by $$z_R=\frac{\pi w_0^2}{\lambda}\tag{7}$$ I have no idea what $w_0$ and $\lambda$ should be. Let's make each laser really big and say that $w_0=10^3\text{ m}$ and $\lambda=500\text{ nm}$. This is a gigantic green space laser, kind of like the Death Star's. This means that $z_R\simeq4.49\times10^{12}\text{ m}$, roughly $30\text{ AU}$.

Intergalactic space is really big, as are galaxies. Let's say that on the first test run, the civilization just wants to shoot the laser from one end of their galaxy to the other. If the galaxy is roughly the size of the Milky Way, then it's maybe $100,000$ light-years long, or about $9.461\times10^{20}\text{ m}$. Plugging this all into $\text{(6)}$, I get $$w\left(\text{Other side of galaxy}\right)=10^3\sqrt{1+\left(\frac{9.461\times10^{20}\text{ m}}{4.49\times10^{12}\text{ m}}\right)^2}\text{ m}=2.11\times10^{11}\text{ meters}$$ which is about $3\text{ AU}$ - actually a lot smaller than I thought, though still very big.

At distances this large, by the way, the $1$ inside the square root is negligible, and, roughly $w(z)\propto z$. Therefore, multiplying $z$ by, say, $42$ should multiply $w(z)$ by $42$. The nearest major galaxy, Andromeda, is about 2.5 million light-years away, on the order of $10$ times the Milky Way's diameter. Therefore, if the beam was sent from the Milky Way to Andromeda, it should have a $w\left(\text{Andromeda}\right)$ of perhaps $30\text{ AU}$. Strange.

It would seem, then, that this sort of beam could not grow very large - at least, not to the size of the galaxy, using these parameters. That's fortunate, because if it grew that large, the intensity would likely be so small that it wouldn't do much good at all!

Basically, only use your Death-Star-green-Gaussian-Nicoll-Dyson-beam-space-lasers on smaller targets - like maybe planets or other stars. Or Alderaan.

References:

• Notes by Kochanek (2004)

posted almost 7 years ago

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