Question about relativistic acceleration:

Originally posted by sonhouse
Consider Alpha Centauri. There are no other suns within 4 light years except us and AC. The nearest other star is Sirius, at twice the distance and therefore 1/4th any gravitational effect from AC. For all intents and purposes, there is no other gravitational effect from AC so the wave front will come to us more or less un-molested by other gravitational ef ...[text shortened]... the sun, you could feel intuitively there would be no focus from the sun on those waves, right?

Sirius is about twice the mass of the sun, and 8 light years distant. So the Lagrange point is (off the top of my head) ~ 2.5 ly away. My argument relies on order of magnitudes, not precise calculations. It does not affect your argument provided you remove the rest of the universe, but it does matter for the extreme end of the focus. Incidentally, it's in favour of your argument for the line of foci ending, as near the Lagrange point there would be no deflection of rays.

Originally posted by DeepThought
Sirius is about twice the mass of the sun, and 8 light years distant. So the Lagrange point is (off the top of my head) ~ 2.5 ly away. My argument relies on order of magnitudes, not precise calculations. It does not affect your argument provided you remove the rest of the universe, but it does matter for the extreme end of the focus. Incidentally, it ...[text shortened]... nt for the line of foci ending, as near the Lagrange point there would be no deflection of rays.

Visualize the center line of the sun going through AC, then visualize that angle going up a bit. That is the flight path of the wave front in that direction. At some point as that angle gets greater, the divergence of those rays from the centerline becomes too great to be bent back inwards to a focal point because the the greater that angle the higher off the surface of the sun they fly by and so encounter a smaller and smaller angle of deflection by the sun.

At some angle the two angles are equal which means loss of focus because all the sun can do to that angle which flies by at some distance above the surface and so runs into a smaller deflection angle so it is only straightened out not bend back. That is the end of the focus because now those rays are traveling parallel to the rays on the center line. Get that part?

Originally posted by sonhouse
Visualize the center line of the sun going through AC, then visualize that angle going up a bit. That is the flight path of the wave front in that direction. At some point as that angle gets greater, the divergence of those rays from the centerline becomes too great to be bent back inwards to a focal point because the the greater that angle the higher off t ...[text shortened]... cus because now those rays are traveling parallel to the rays on the center line. Get that part?

Yes, I understand your picture. The problem with it is that the line of focii isn't finite. It still goes to infinity.

Originally posted by DeepThought
Yes, I understand your picture. The problem with it is that the line of focii isn't finite. It still goes to infinity.

Yes but the photons from each angle only undergo a brief change in direction, through lesser gravity bending to max then reducing again so if you follow that photon, after it passes by the sun the bending becomes so small for all intents and purposes the photon is on a new straight path. Of course there are other spacetime bends around of a lesser degree but that doesn't alter the flight path of the photon in question very much and can be pretty much ignored for the short interstellar distances we are talking about, a few light years, local stuff.

Originally posted by sonhouse
At some angle the two angles are equal which means loss of focus because all the sun can do to that angle which flies by at some distance above the surface and so runs into a smaller deflection angle so it is only straightened out not bend back. That is the end of the focus because now those rays are traveling parallel to the rays on the center line. Get that part?

I will have to draw a diagram when I am on my other computer, which has word on it.
But what you say above occurs when the focus reaches infinity. There is no sudden cutoff before infinity.

Originally posted by twhitehead
I will have to draw a diagram when I am on my other computer, which has word on it.
But what you say above occurs when the focus reaches infinity. There is no sudden cutoff before infinity.

I predict one of us will have egg on one's face🙂

OK here is a diagram. Tell me what is wrong, or what you want added.

http://www.copperain.com/qgt/img/gravitational_lensing.jpg

Originally posted by twhitehead
OK here is a diagram. Tell me what is wrong, or what you want added.

http://www.copperain.com/qgt/img/gravitational_lensing.jpg

I am not sure what you mean by "Increasing total energy with increasing distance
from the sun at a rate of r^2"...

Also, I want to mirror the whole thing so the light rays are going left to right. 😛

Originally posted by googlefudge
I am not sure what you mean by "Increasing total energy with increasing distance
from the sun at a rate of r^2"...

The light goes in circles around the sun. For each circle, the light bends at the same angle and focus' onto the line of focus. As the circles get bigger, there is more energy for any given circle. However, this competes with the fact that the light is spread out and thus the further from the sun, the less light arrives at any given point on the focus line from any given point on the circle.

I think I got the rate wrong. The total energy for a circle is proportional to the circumference (not the area) so expands proportional to the radius.

Originally posted by twhitehead
The light goes in circles around the sun. For each circle, the light bends at the same angle and focus' onto the line of focus. As the circles get bigger, there is more energy for any given circle. However, this competes with the fact that the light is spread out and thus the further from the sun, the less light arrives at any given point on the focus lin ...[text shortened]... ircle is proportional to the circumference (not the area) so expands proportional to the radius.

Ah, OK I get what you are talking about now.

Originally posted by googlefudge
Ah, OK I get what you are talking about now.

It's pretty easy to visualize, there is X amount of energy in any total wavefront, all 360 spherical degrees of it. So the energy density gets less but the total energy from that star is constant.

So what reaches Earth from Alpha centauri clocks in at roughly 10 nanowatts per square foot, or roughly 40 nanowatts per square meter.

So if you look at the ring of energy from AC that just skims by the surface, you have a ring of roughly the sun's circumference. So if you just limited that ring to a 1 km width you have what I calculated to be about 70 kilowatts of energy which would come to some sort of focus at 580 AU.

But that same 1 Km width 1R above the surface would now have 140 Kw of energy focused at 2320 odd AU. and so forth.

At that distance, the energy in one square kilometer of light from the sun clocks in at about 4 milliwatts per square meter. Times 1 million, = about 4000 watts per kilometer squared. So if you are in the focal line area, at first focus you get 17 times the energy you would get from the sun. I guess you would get both so if you had a 1 km^2 collector you would get almost 75 kilowatts of energy. The thing about that focal line is the energy doesn't keep reducing like the inverse square thing of the sun, so at 2320 AU you would get 1/16th the energy you had at 580 AU, now at about 250 watts but you would still be collecting many kilowatts in the energy stream from AC if you stay on that line. If you had 1000 square Km of collector you would get of course 70 odd MEGAwatts of energy, which starts to be serious. Of course you have to deal with a collector 33 km on a side....

Originally posted by sonhouse
But that same 1 Km width 1R above the surface would now have 140 Kw of energy focused at 2320 odd AU.

The thing is though, it is more spread out when it hits the focal line, so it may actually have significantly less energy at any given point along the line.

Originally posted by twhitehead
The thing is though, it is more spread out when it hits the focal line, so it may actually have significantly less energy at any given point along the line.

That is what I would dearly love to show, or refute. I would love to have good instrumentation in a manned craft, manned by me of course🙂 and do detailed measurements of just that. Just how focused is the energy, what is the shape of the wavefront there and how does a particular focal line defocus?

Could it be as if you are just drawing straight lines from the sun to the focal point or is something else going on that would cause the line to deviate?

My guess is once a photon leaves the main bang of gravitation from the sun and is only bent by the lesser field out a few million miles, that photon will be going in a straight line, or at least as straight as it can go in a curved universe.

Originally posted by sonhouse
That is what I would dearly love to show, or refute. I would love to have good instrumentation in a manned craft, manned by me of course🙂 and do detailed measurements of just that.

No need for a spacecraft. Not only can the problem be reliably solved with maths and physics, but we can do actual observations using other objects than the sun for our lens. Gravitational lensing can be observed for many stars at different distances to see what the differences are.

Could it be as if you are just drawing straight lines from the sun to the focal point or is something else going on that would cause the line to deviate?
Did you look at my diagram?
As the light passes the sun, photons further away from the sun are deflected less. So for example, a circle of 1cm thick will be much thicker by the time it reaches the focal line. Couple that with the fact that it is arriving at a very small angle results in significant spread along the focal line.

Do you happen to know the formula for how much light will be deflected at a given radius?

Originally posted by twhitehead
No need for a spacecraft. Not only can the problem be reliably solved with maths and physics, but we can do actual observations using other objects than the sun for our lens. Gravitational lensing can be observed for many stars at different distances to see what the differences are.

[b]Could it be as if you are just drawing straight lines from the sun ...[text shortened]... ne.

Do you happen to know the formula for how much light will be deflected at a given radius?

Yes, it is A (angle)= 4GM/C^2R.

Notice R is the radius but not R^2, which was a surprise. I saw this formula at the Smithsonian during an Einstein exhibit. I copied it down.