Let's see what our gravitational lensing image is going to look like.
If there was nothing in the foreground, what we might see in our telescope field of view,
say there's our telescope field of view, is we'd see the background star.
But, because of the lens, instead we should see two images or maybe even a complete ring,
a distance and angular space, Alpha, from where this single image is going to be.
So, if Alpha isn't too small, we can see these things. So, what's the geometry here?
Well, let's say here we have the background star,
here we have the Earth and we've got the rays of light going either way around
and for the sake of simplicity we'll put our dark thing, whatever it is, exactly half way.
There's the distance to the background star and we have R here and R there, this middle then has a mass.
We know from the lensing equation that this deflection angle here, Theta, is given by the equation Theta equals four GM over RC squared.
So, from that we want to measure this half here which is this angle in there.
Okay, well, from symmetry, that's going to be Theta as well here, that's going to be Theta and all the way across here is going to be Theta.
So, we know that Alpha equals half Theta.
We also know that this right angle triangle here, tan Theta, equals opposite over hypotenuse which is half D,
but the angle Alpha and Theta are going to be very, very small...
Sorry, that should be tan Theta over two than tan Alpha.
As the angles are very, very small, as long as we can measure everything in radians, we can get rid of the tan,
so what we get is that Theta over two equals R, D over two, so R equals Theta D over four.
So, what we can do is substitute this one in here,
we've got two unknowns, R and Theta, and two equations so that should be solvable.
So, substitute this one into here and we end up with Theta equals four GM times four all over D Theta C squared, so four times four is 16.
You want to take the Theta up here, so that gives us the Theta squared equals 16 GM over DC squared.
Take the square roots of both sides and substitute, we know that Alpha is half of Theta,
so we end up that Alpha equals the square root four GM over DC squared.
This angle Alpha is so widely used, it's got a particular name, it's called the Einstein Radius.
This is the equation for when the lens is exactly half way between,
this is a highly more complicated equation for the Einstein Radius with the lens not being midway between.
This is where half way between is where it has the most lensing influence. So, what is this number?
Let's assume you've got a Jupiter mass planet half way to the middle of our galaxy,
in that case D is the distance from the sun to the bulge, which is about 25,000 lightyears.
M is the mass of Jupiter which is about 1.9 by 10 to the 27 kilograms, C is the speed of light, G the gravitational constant,
stick this into here and you end up with Alpha equals 1.5 by 10 to the minus 10 radians.
For you to convert that into some unit we can understand, so a radian is 180 over pi degrees, so multiply 180 over pi.
That will give us degrees, but the angle is going to be much smaller than the degree,
so then we have to times 60 for arc minutes times another 60 to get it to arc seconds,
because an arc minute is a 60th of a degree and an arc second is a 60th of an arc minute,
so multiply by 180 over pi times 60 times 60 gives you 31 times 10 to the minus six arc seconds, so 31 micro arc seconds.
Now, bear in mind, a typical telescope has it's images fuzzed out by about an arc second,
so we're talking a hundred thousand times smaller than that,
even the Hubble Space Telescope only gets about 0.1 arc seconds,
maybe 0.03 at very good times, they're still vastly smaller than that.
So, this is an incredibly small angle which means that we're basically never going to be able to see these multiple images.