# Bendy light: The maths

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#### ghazwozza

• 942
##### Re: Bendy light: The maths
« Reply #60 on: August 21, 2008, 10:27:26 AM »
y = x2 / 2r

How was this derived?

By calculating the concavity of the expected secant curve traced out by a light ray tangential to the surface of the round Earth at the point where it meets the Earth, and then using that to calculate the equation for the parabolic arc traced out by a light ray in FET with the EA, such that it will have the same concavity when horizontal. r is the radius of the round Earth.

You're assuming light travels in parabolic arcs. I see no reason that this should be true.

#### Parsifal

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##### Re: Bendy light: The maths
« Reply #61 on: August 21, 2008, 10:53:44 AM »
You're assuming light travels in parabolic arcs. I see no reason that this should be true.

I don't see a reason why it has to be true either. But I thought I'd pick a shape and run with it, and see what came of it. Perhaps the complexity of the calculations is a sign that it's a false assumption.
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#### ghazwozza

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##### Re: Bendy light: The maths
« Reply #62 on: August 21, 2008, 12:42:26 PM »
What shape would a ray of light trace out in a uniform gravitational field in GR? I think this would be the most logical shape to assume. Maybe it is a parabola.

#### sokarul

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##### Re: Bendy light: The maths
« Reply #63 on: August 21, 2008, 01:21:43 PM »
The distance to the horizon would depend on the angle of the light.

As it does in RET.

All this arguing is completely pointless, because Robosteve still fails to deliver.

I mean, he's saying all this stuff about what observations his EA can account for, but he still hasn't shown how it can even work.

Untill some sort of equation is presented, then everything he says with regard to the EA is just meaning less fluff.

Here's an equation:

y = x2 / 2r

And another:

(dx/dt)2 + (dy/dt)2 = c2

You are still not getting it.
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#### Parsifal

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##### Re: Bendy light: The maths
« Reply #64 on: August 21, 2008, 01:47:47 PM »
What shape would a ray of light trace out in a uniform gravitational field in GR? I think this would be the most logical shape to assume. Maybe it is a parabola.

My educated guess would be some sort of conic section, depending on various factors. Of course, one possible conic section is indeed a parabola. I don't know enough about General Relativity to say whether this is true for light, although I'm almost certain that it's true (bar energy loss through gravitational radiation) for material objects.

You are still not getting it.

Oh.
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#### lolz at trollz

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##### Re: Bendy light: The maths
« Reply #65 on: August 21, 2008, 02:14:11 PM »

Here's an equation:

y = x2 / 2r

And another:

(dx/dt)2 + (dy/dt)2 = c2

Ok, let's go with it.

Now for an experiment to show if it's correct or not.
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#### Parsifal

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##### Re: Bendy light: The maths
« Reply #66 on: August 21, 2008, 02:26:06 PM »
Here's an equation:

y = x2 / 2r

And another:

(dx/dt)2 + (dy/dt)2 = c2

Ok, let's go with it.

Now for an experiment to show if it's correct or not.

I think the second one has been proven to be correct many times. Not so certain of the first.
I'm going to side with the white supremacists.

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#### ghazwozza

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##### Re: Bendy light: The maths
« Reply #67 on: August 21, 2008, 02:43:36 PM »
My educated guess would be some sort of conic section, depending on various factors. Of course, one possible conic section is indeed a parabola.

The more I think about it, the more a parabola makes sense, but it's by no means a dead cert.

#### Parsifal

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##### Re: Bendy light: The maths
« Reply #68 on: August 21, 2008, 02:52:51 PM »
The more I think about it, the more a parabola makes sense, but it's by no means a dead cert.

That's pretty much my take on it too. A parabola seems to be the most likely shape, so I ran with that idea aware that it may not be entirely correct.
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#### jdoe

• 388
##### Re: Bendy light: The maths
« Reply #69 on: August 21, 2008, 04:33:50 PM »
OK, let me give you guys my approach to the problem; it's long so be patient.

First, I assume a uniform gravitational field.  Uniform gravitational fields have the property that the force is directly proportional to the mass of an object by a constant a.  (This is a property of all fictitious forces.)

F=ma

We also have that

F=dp/dt=d(γmv)/dt=γ3mdv/dt.
(by the chain rule)

Mass cancels so we have the differential equation

a=γ3dv/dt.

Solving this differential equation for v with the initial condition of v(0)=0, we have

v(t)=at/√(1 + a2t2/c2).

With our uniform gravitational field, we are working in a non-inertial frame.  We want to work in an inertial frame because that's where light travels in straight lines.  So let's move to the frame where objects fall freely, that is, the frame that moves with speed v(t) with respect to our non-inertial rest frame.  To do this we need to perform a Lorentz transformation.

First we need to find the position of the non-inertial frame in our inertial frame as a function of time, x(t).  To find this we simply integrate -v(t) and obtain

x(t)= -c2/a (√(1 + a2t2/c2) - 1)

with the initial condition that x(0)=0.  We can now apply the Lorentz transform on the x coordinate.

x'=γ(x - x(t))

x is the position in the inertial frame, and x' is the position in our non-inertial rest frame.  In addition,

γ=1/√(1 - v(t)2/c2)=√(1+a2t2/c2),

so we have

x'=√(1 + a2t2/c2)(x + c2/a(√(1 + a2t2/c2) - 1))),

which simplifies to

x'=√(1 + a2t2/c2)(x - c2/a) + at2 + c2/a
.

Now consider a light beam which travels in a straight line from a height of h to the ground at some angle θ from the vertical.  Its x coordinate is given by

x=h - cos(θ)ct

and y coordinate is given by

y=sin(θ)ct.

We notice that the y component is perpendicular to the direction of motion so it is unaffected by Lorentz contraction and y=y'.  We solve this relationship for t,

t=y'/(sin(θ)c),

and plug in to the Lorentz transformation equation along with the previous equation for x.  We now have an equation for the path of light in terms of x' (the vertical height), y' (the horizontal distance), θ (the angle from which the light was emitted from the sun), h (the height of the sun) and a (the constant acceleration).

x'=√(1+a2y'2/(c4sin2(θ)))(h - y'/tan(θ) - c2/a) + ay'2/(c2sin2(θ)) + c2/a

In the more familiar xy coordinates where y is vertical and x is horizontal this is

y=√(1+a2x2/(c4sin2(θ)))(h - x/tan(θ) - c2/a) + ax2/(c2sin2(θ)) + c2/a
.

Clearly, this path is not parabolic, but is nearly so, as can be seen if one graphs it.  It only remains to choose the free parameters a, h, and θ to match observation.
« Last Edit: August 21, 2008, 04:51:48 PM by jdoe »
Mars or Bust

#### Parsifal

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##### Re: Bendy light: The maths
« Reply #70 on: August 21, 2008, 04:45:56 PM »
Very interesting approach. Thanks for that. I'll have a look at differentiating that function, that I may calculate acceleration as a function of velocity.
I'm going to side with the white supremacists.

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#### Rig Navigator

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##### Re: Bendy light: The maths
« Reply #71 on: August 21, 2008, 11:35:45 PM »
OK, let me give you guys my approach to the problem; it's long so be patient...

and the answer is... four?

#### WardoggKC130FE

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##### Re: Bendy light: The maths
« Reply #72 on: August 21, 2008, 11:44:30 PM »
Yup....that's what I got.  Definitely 4.

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#### jdoe

• 388
##### Re: Bendy light: The maths
« Reply #73 on: August 22, 2008, 09:23:06 AM »
OK, let me give you guys my approach to the problem; it's long so be patient...

and the answer is... four?
Yup....that's what I got.  Definitely 4.

LOL.

But anyway, I don't like the idea of light being accelerated.  The only way the path of photons can be bent is through gravitation or interaction with a medium.  I've done the calculations for gravitation, and the field would have to be enormous.  We'd all be dead.

I have another idea that I think is more elegant.  I propose the existence of a quintessence permeating all space.  This quintessence has negative equation of state, P = wϵ, where w < -1/3.  (This is the same property of dark energy in RET)  This means that for any energy density of quintessence, there will be negative pressure.  This negative pressure is what makes dark energy so interesting.

Now I assume that there is a vertical gradient in the quintessence.  The energy density of the quintessence increases as one moves upwards.  By the equation above, there is also a pressure gradient.

Consider the earth trapped in this pressure gradient.  It feels a strong negative pressure pulling it upwards from the top.  On the bottom, the negative pressure is slightly less.  The difference between the two pressures gives rise to a buoyant force which pushes the earth upwards at 9.8m/s2.

Now, here is the interesting part.  The density of the quintessence increases with height, so presumably so does its index of refraction towards light, possibly in a nonlinear way.  The increase of refractive index with height is exactly what is needed to bend light to cause sunrises/sunsets, sinking ship effect, etc.   Keep in mind it was the negative pressure property of dark energy that made this possible.

Thoughts, criticisms?
Mars or Bust

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#### oldsoldier

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##### Re: Bendy light: The maths
« Reply #74 on: August 22, 2008, 09:45:09 AM »

Now, here is the interesting part.  The density of the quintessence increases with height, so presumably so does its index of refraction towards light, possibly in a nonlinear way.  The increase of refractive index with height is exactly what is needed to bend light to cause sunrises/sunsets, sinking ship effect, etc.   Keep in mind it was the negative pressure property of dark energy that made this possible.

Thoughts, criticisms?

jdoe... perhaps you can do this math... here's my issue with bendy light. It's a gut feeling and I'm kinda waiting on robosteve's equations, but in the mean time perhaps you can shed some light.

Consider the bendy light pictures shown here so far. All the rays appear to hit emanate from the sun at what I'm calling an "90 degree" angle. Furthermore all those curves have no inflection point. (This will be important.)

Now, consider what happens to a beam of light emanating from the sun at "89 degrees" or... lower angles if necessary.
Either all light that hits the earth from the sun comes from that "90 degree corridor" and consequently no light at even 89.9999 degrees hits the earth, or some light emanating at less than 90 degrees hits the earth.

I strongly believe that the models you and robosteve are kicking around imply that light emanating from the sun at something other than 90 degrees takes a path that does have an inflection point in it. And that just seems crazy to me. The only way that can happen is if the sun exerts its own component of EA and where the solar EA and the terrestrial EA are "equal" in strength is where this inflection point occurs.

And if you do the math there... explain the moon too as it will similarly need its own EA for similar reasons.

And if you do that... explain how we don't see any wobble in stellar positions when the moon occults them.

« Last Edit: August 22, 2008, 09:46:42 AM by oldsoldier »

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#### Rig Navigator

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##### Re: Bendy light: The maths
« Reply #75 on: August 23, 2008, 11:43:59 PM »
Consider the bendy light pictures shown here so far. All the rays appear to hit emanate from the sun at what I'm calling an "90 degree" angle. Furthermore all those curves have no inflection point. (This will be important.)

Now, consider what happens to a beam of light emanating from the sun at "89 degrees" or... lower angles if necessary.
Either all light that hits the earth from the sun comes from that "90 degree corridor" and consequently no light at even 89.9999 degrees hits the earth, or some light emanating at less than 90 degrees hits the earth.

I strongly believe that the models you and robosteve are kicking around imply that light emanating from the sun at something other than 90 degrees takes a path that does have an inflection point in it. And that just seems crazy to me. The only way that can happen is if the sun exerts its own component of EA and where the solar EA and the terrestrial EA are "equal" in strength is where this inflection point occurs.

This is where the "spotlight" model of the Sun comes in.  They need the Sun to only shine light straight at the ground in order for this model to be viable.

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#### Penispoop agogo

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##### Re: Bendy light: The maths
« Reply #76 on: August 23, 2008, 11:53:44 PM »
So, there currently is no math to explain it, therefore you make it up because it doesn't actually exist/happen right?

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#### Rig Navigator

• 808
##### Re: Bendy light: The maths
« Reply #77 on: August 24, 2008, 12:38:40 AM »
So, there currently is no math to explain it, therefore you make it up because it doesn't actually exist/happen right?

No, they have math to explain it, but no physics or other mechanism.

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#### trig

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##### Re: Bendy light: The maths
« Reply #78 on: August 24, 2008, 07:02:01 AM »

Now, here is the interesting part.  The density of the quintessence increases with height, so presumably so does its index of refraction towards light, possibly in a nonlinear way.  The increase of refractive index with height is exactly what is needed to bend light to cause sunrises/sunsets, sinking ship effect, etc.   Keep in mind it was the negative pressure property of dark energy that made this possible.

The more intriguing properties of the quintessence (or whatever) are:
• whatever property makes the light bend, it has to be anisotropic, otherwise the upper atmosphere would have to be solid.
• while some as yet unknown effect bends the light, it cannot be optical refraction, since optical refraction produces what in lenses is called chromatic aberration; you can also call it prism effect. This means that if a large part of the bending is caused differing refractive indexes at different atmospheric heights, you would see all the celestial objects with a big blue halo on top and a big red halo just below (to say the least).
• it has to magnify the intensity of the light that reaches the observer almost horizontally, but must not magnify the intensity of the light that comes directly down
• alternately, it has to dim down the light that comes directly down, but not the light that is bent almost 90 degrees and reaches the observer horizontally
« Last Edit: August 24, 2008, 07:11:43 AM by trig »

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#### ghazwozza

• 942
##### Re: Bendy light: The maths
« Reply #79 on: August 24, 2008, 03:14:34 PM »
^ Can we get this guy (Candleja) banned please?

#### General Douchebag

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##### Re: Bendy light: The maths
« Reply #80 on: August 24, 2008, 03:19:14 PM »
And you. It's an IP ban, all of your alts will die.
No but I'm guess your what? 90? Cause you just so darn mature </sarcasm>

#### Daniel

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##### Re: Bendy light: The maths
« Reply #81 on: August 24, 2008, 04:57:27 PM »
^ Can we get this guy (Candleja) banned please?

Done.  All of his posts are gone too.

#### General Douchebag

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##### Re: Bendy light: The maths
« Reply #82 on: August 24, 2008, 04:57:48 PM »
Yay!
No but I'm guess your what? 90? Cause you just so darn mature </sarcasm>

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#### dyno

• 562
##### Re: Bendy light: The maths
« Reply #83 on: August 25, 2008, 04:56:26 AM »
Does bendy light affect all things EM?

Magnetic fields?
X-rays?
Gamma rays?

What about particles?

Neutron radiation?
Electrons?

What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight

#### Parsifal

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##### Re: Bendy light: The maths
« Reply #84 on: August 25, 2008, 05:03:00 AM »
Does bendy light affect all things EM?

Only electromagnetic radiation.

Magnetic fields?

Only in the special case where they form half of a photon, along with an electric field.

X-rays?

Yes.

Gamma rays?

Yes.

What about particles?

Only photons.

Neutron radiation?

No.

Electrons?

No.

What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight

These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!
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#### Sir_Drainsalot

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##### Re: Bendy light: The maths
« Reply #85 on: August 25, 2008, 05:15:52 AM »
These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!

So you mean everyone has been worrying about the LHC creating a black hole and destroying the universe for nothing? Damn conspiracy.

#### sokarul

• 19303
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##### Re: Bendy light: The maths
« Reply #86 on: August 25, 2008, 05:18:16 AM »

These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!

I could comment on the rest of your shit but I will just comment on this.

You have no idea what you are talking about.  There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider.  It is real so stop making shit up.
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#### Parsifal

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##### Re: Bendy light: The maths
« Reply #87 on: August 25, 2008, 05:28:40 AM »
You have no idea what you are talking about.  There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider.  It is real so stop making shit up.

So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?
I'm going to side with the white supremacists.

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#### ghazwozza

• 942
##### Re: Bendy light: The maths
« Reply #88 on: August 25, 2008, 08:34:25 AM »
What about particle colliders? These things are built flat right? With the LHC does that mean we can't see the end because the light has curved up? By end I mean before it curves around. Apparently it is supposed to appear straight

These don't exist. They are part of the conspiracy. Imagine how much money people could make from funding intended to build a particle accelerator!

Hold that thought until the LHC is producing 1.8 GB/s, then tell me the government can produce this much hard scientific data that will stand up to rigorous scientific scrutiny.

Also, the 8000 scientists from all nations working at CERN must be part of the conspiracy. When people (like journalists) are given tours of the LHC, what are they being shown?

#### markjo

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##### Re: Bendy light: The maths
« Reply #89 on: August 25, 2008, 10:01:20 AM »
You have no idea what you are talking about.  There is A guy I have know for a while, his sister is over in France right now working on The Large Hadron Collider.  It is real so stop making shit up.

So a guy you have know [sic] for a while having a sister who is working on something lends it credibility, does it? How about you show me some primary evidence for its existence instead?

I have some pictures of the particle accelerator at Cornell University that I personally took.  If I were to post them, would that be credible evidence that particle accelerators exist?
« Last Edit: August 25, 2008, 10:02:52 AM by markjo »
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