Bendy light: The maths

Ok FEers, when your ready, post it all here, no distractions please everyone, just the formulas for EA.  You guys get it done, get it agreed on, then we can do an experiment to see if it's consistent with observations. 

Now compute! 

There is already a thread for this. It is locked temporarily because it was being trolled. You are officially FES's most redundant citizen. To summarise: in b4 teh lock!

Robosteve hasn't really figured out how the EA works, so atm doing any proper maths is impossible.

Here are some speculative calculations:

First, I calculated that y = x² / 2r if it is to approximate the effect of a round Earth at short distances (r being the radius of the RE). This means that dy / dx = x / r and therefore dy / dt = (x / r) * (dx / dt). Substituting x = f(t) and y = g(t), you get g'(t) = f(t) * f'(t) / r.

Next, if the speed is to remain constant, then by Pythagoras' theorem f'(t)² + g'(t)² = c². Squaring the above equation, c² - f'(t)² = f(t)² * f'(t)² / r². Rearranging this gives:

And the solution:

f² (df/dt)² = r² (c² - (df/dt)²)

f² = (rc)² (dt/df)² - r²

(f² + r²)^0.5 = +/- rc (dt/df)

Integrate w.r.t. f to get the equation in blue.

f * (f²-r²)^0.5 - log[ f + (f²-r²)^0.5] = +/- 2 r c t

Using a different set of postulates, it works out like this:

I used x(t) and y(t) instead of f(t) and g(t), just because it is more descriptive.

We know (if I understand EAT correctly):
y'(t)² + x'(t)² = c²                                       (speed of light is constant)
y''(t) = a                                                    (light accelerates upwards constantly with acceleration a)
y(0) = 0,  x(0) = 0,  y'(0) = 0                        (initial conditions)

y''(t) = a  =>  y'(t) = at  =>  y(t) = at²/2  This is assuming the EA is constant and directed upwards.

So, by pythagoras: (at)² + x'(t)² = c²
Re-arrange for x'(t) to get a difficult integral, type into the online integrator to get:

x(t) = (c²/2a) arctan (at/{c²-(at)²}^0.5) + (t/2)(c²-(at)²)^0.5

There is already a thread for this. It is locked temporarily because it was being trolled. You are officially FES's most redundant citizen. To summarise: in b4 teh lock!

I see no maths...

Robosteve hasn't really figured out how the EA works, so atm doing any proper maths is impossible.

Here are some speculative calculations:

Quote from: Robosteve on August 18, 2008, 10:20:42 AM

First, I calculated that y = x² / 2r if it is to approximate the effect of a round Earth at short distances (r being the radius of the RE). This means that dy / dx = x / r and therefore dy / dt = (x / r) * (dx / dt). Substituting x = f(t) and y = g(t), you get g'(t) = f(t) * f'(t) / r.

Next, if the speed is to remain constant, then by Pythagoras' theorem f'(t)² + g'(t)² = c². Squaring the above equation, c² - f'(t)² = f(t)² * f'(t)² / r². Rearranging this gives:

And the solution:

Quote from: ghazwozza on August 19, 2008, 04:59:22 AM

f² (df/dt)² = r² (c² - (df/dt)²)

f² = (rc)² (dt/df)² - r²

(f² + r²)^0.5 = +/- rc (dt/df)

Integrate w.r.t. f to get the equation in blue.

f * (f²-r²)^0.5 - log[ f + (f²-r²)^0.5] = +/- 2 r c t

Using a different set of postulates, it works out like this:

Quote from: ghazwozza on August 18, 2008, 10:50:14 AM

I used x(t) and y(t) instead of f(t) and g(t), just because it is more descriptive.

We know (if I understand EAT correctly):
y'(t)² + x'(t)² = c²                                       (speed of light is constant)
y''(t) = a                                                    (light accelerates upwards constantly with acceleration a)
y(0) = 0,  x(0) = 0,  y'(0) = 0                        (initial conditions)

y''(t) = a  =>  y'(t) = at  =>  y(t) = at²/2  This is assuming the EA is constant and directed upwards.

So, by pythagoras: (at)² + x'(t)² = c²
Re-arrange for x'(t) to get a difficult integral, type into the online integrator to get:

x(t) = (c²/2a) arctan (at/{c²-(at)²}^0.5) + (t/2)(c²-(at)²)^0.5

The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?

The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?

First off, I don't believe FE or EAT, I just find this an interesting theory.

You say the mineshaft was straight. How was this straightness measured? Spirit level? A beam of light? Looking down the length of it and seeing it is straight?

Over 1 mile light wouldn't deviate enough to be significant.

I see no maths...

When there is maths to post, it will be posted in that thread. Again, redundant thread is redundant.

The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?

Wait wait, let everyone finish doing it, they will just squeal conspiracy on that experiment.  Once it's done, there will be a simple experiment to prove or disprove it.  


And hopefully provide some lolz. 

Quote from: lolz at trollz on August 19, 2008, 07:18:36 AM

I see no maths...

When there is maths to post, it will be posted in that thread. Again, redundant thread is redundant.

No theory boy has no theory. 

There is already a thread for this. It is locked temporarily because it was being trolled. You are officially FES's most redundant citizen. To summarise: in b4 teh lock!

It's locked because you have no idea what you are talking about. 

Quote from: AmatureAstronomer on August 19, 2008, 07:26:29 AM

The speed of light was derived by sending a focused beam of light along a 1 mile long perfectly straight mineshaft. This beam did not bend. Does this fit with your calculations?

First off, I don't believe FE or EAT, I just find this an interesting theory.

You say the mineshaft was straight. How was this straightness measured? Spirit level? A beam of light? Looking down the length of it and seeing it is straight?

Over 1 mile light wouldn't deviate enough to be significant.

Do your own research... That's just a soundbite I remembered from high school.

OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.



A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).

Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.

FE+EAT predicts A will arrive first. RET predicts B will arrive first.

Over 1 mile light wouldn't deviate enough to be significant.

According to the EAT, the light should bend away from the surface of the Earth over 1 foot.  That is a significant difference if you are expecting something to be accurate within fractions of an inch.

OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.



A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).

Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.

FE+EAT predicts A will arrive first. RET predicts B will arrive first.

I don't think this will work. If FE+EAT is correct, then the light in A will keep getting bent upwards and reflected back down away from the edge of the optic fibre. This will result in a path that zigzags through the fibre optic. Of course, the same effect will be observed if RET is true and the light tries to follow a straight line through a curved optic fibre. This zigzag effect will probably be of greater significance to the time it takes the light to go from one end to the other than will the shape of the Earth.

OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.

A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).

Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.

FE+EAT predicts A will arrive first. RET predicts B will arrive first.

Of course, just setting up the experiment would give you the answer about which one was longer.  Just measure the amount of cable on each "leg" of the experiment as you string it up.

Of course, just setting up the experiment would give you the answer about which one was longer.  Just measure the amount of cable on each "leg" of the experiment as you string it up.

True, if you make it large enough. I was thinking that my experiment could be performed in a lab, maybe over just a few metres.

I'm not sure what would be cheaper and easier -- getting hold of a laser and inferometer, or a few miles of cable.

Here's another approach regarding the OP.  Why don't you just Lorentz transform the straight-line path of light in an inertial frame to an accelerating frame?

Here's another approach regarding the OP.  Why don't you just Lorentz transform the straight-line path of light in an inertial frame to an accelerating frame?

I do believe you may have simplified the necessary calculations quite considerably, good sir!

Quote from: jdoe on August 19, 2008, 01:11:06 PM

Here's another approach regarding the OP.  Why don't you just Lorentz transform the straight-line path of light in an inertial frame to an accelerating frame?

I do believe you may have simplified the necessary calculations quite considerably, good sir!

Honestly, the calculations haven't simplified much at all.  I've been messing around with the equations, and I obtained an expression for the path of light in an accelerated reference frame.  The path is not parabolic and the expression is very complicated. 

Using the conditions of the height of the sun, ~4800km, and the radius of the spotlight, ~10000km, I'm trying to solve for the required acceleration.  It's going to require some numerical approximations, and I don't know if I get a precise value or not.  So far it, just by playing with numbers, it seems that an acceleration of around 10^10 m/s^2 will be required.

Using the conditions of the height of the sun, ~4800km

Am I missing something, since when did the sun get that close to the earth? If it was that close, the heat would be so intense that the earth could not support life.

Quote from: ghazwozza on August 19, 2008, 08:09:55 AM

OK, I've come up with a way to test EAT theory, but it requires a fair bit of equipment.



A and B are fibre optic cables. A is arranged so that it would be straight (but appear curved) on a FE with EAT. B is arranged so that it would be straight on a RE (and appear straight on a FE + EAT, but not actually be straight).

Simply split laser light to travel down both fibres. At the other end, the beams are combined in an inferometer. This allows you to see which beam arrived first.

FE+EAT predicts A will arrive first. RET predicts B will arrive first.

I don't think this will work. If FE+EAT is correct, then the light in A will keep getting bent upwards and reflected back down away from the edge of the optic fibre. This will result in a path that zigzags through the fibre optic. Of course, the same effect will be observed if RET is true and the light tries to follow a straight line through a curved optic fibre. This zigzag effect will probably be of greater significance to the time it takes the light to go from one end to the other than will the shape of the Earth.

Umm...the concept of fiberoptic cables is that the difference in refractive index between the cable and the medium is such that light will bounce off the edges of the inside of the cable as it travels down. All light in fiberoptic cables travels in a zigzag fashion. In fact, if light were to bend, long-distance fiberoptic cables could not work as eventually the light would be too angled to continue reflecting. It would reach a point where its vector relative to the edge of the fiber is so direct that it would not reflect. Thank goodness light doesn't bend. Otherwise we wouldn't have fiberoptic technology.

Umm...the concept of fiberoptic cables is that the difference in refractive index between the cable and the medium is such that light will bounce off the edges of the inside of the cable as it travels down. All light in fiberoptic cables travels in a zigzag fashion. In fact, if light were to bend, long-distance fiberoptic cables could not work as eventually the light would be too angled to continue reflecting. It would reach a point where its vector relative to the edge of the fiber is so direct that it would not reflect. Thank goodness light doesn't bend. Otherwise we wouldn't have fiberoptic technology.

You misunderstand the EA, and you fail to acknowledge that in a perfectly straight optical fibre light (if it travels in a straight line) would not need to zigzag to get to the other end.

Honestly, the calculations haven't simplified much at all.  I've been messing around with the equations, and I obtained an expression for the path of light in an accelerated reference frame.  The path is not parabolic and the expression is very complicated. 

Using the conditions of the height of the sun, ~4800km, and the radius of the spotlight, ~10000km, I'm trying to solve for the required acceleration.  It's going to require some numerical approximations, and I don't know if I get a precise value or not.  So far it, just by playing with numbers, it seems that an acceleration of around 10^10 m/s^2 will be required.

I was using a different method of trying to figure it out. I worked out the concavity of the secant curve formed by plotting the height of a light ray tangential to the surface of the Earth at the point where it meets the Earth, and attempted to construct a parabolic arc with equal concavity to the secant curve at that point, such that it would approximate the effect of a round Earth over short distances.

Quote from: Mohnzh on August 19, 2008, 01:42:18 PM

Umm...the concept of fiberoptic cables is that the difference in refractive index between the cable and the medium is such that light will bounce off the edges of the inside of the cable as it travels down. All light in fiberoptic cables travels in a zigzag fashion. In fact, if light were to bend, long-distance fiberoptic cables could not work as eventually the light would be too angled to continue reflecting. It would reach a point where its vector relative to the edge of the fiber is so direct that it would not reflect. Thank goodness light doesn't bend. Otherwise we wouldn't have fiberoptic technology.

You misunderstand the EA, and you fail to acknowledge that in a perfectly straight optical fibre light (if it travels in a straight line) would not need to zigzag to get to the other end.

It is only theoretically possible for a photon to enter a fiberoptic cable in perfect parallel to the cable (or perfectly orthogonal to the beginning of the cable). Because it is not practically possible, all light will zigzag inside of a fiberoptic cable - even a perfectly straight one. Suppose, however, that you do manage to get a photon to do this. It would be undetectable in relation to the numerous photons that will follow the "zigzag" pattern. The next argument would be LASER light, as all the photons should propagate parallel to each other. However, we do not have the technology to ensure perfectly ideal orthogonality. As such, even the LASER photons would be refracted, even if only slightly, upon entering the optic cable. Even supposing you managed a test where you did get perfect orthogonality, it would be virtually impossible to repeat. You simply cannot achieve such perfect conditions that this could work. Even vibrations of the earth or a single dust particle or even a handful of large molecules would cause light scattering so that perfect orthogonality could not be achieved. This would occur even if the LASER was in contact with the cable inside of the best vacuum achievable with our technology. Not to mention the variance in the synchronization of the LASER or tolerance level for non-simultaneous emission.

It is only theoretically possible for a photon to enter a fiberoptic cable in perfect parallel to the cable (or perfectly orthogonal to the beginning of the cable). Because it is not practically possible, all light will zigzag inside of a fiberoptic cable - even a perfectly straight one. Suppose, however, that you do manage to get a photon to do this. It would be undetectable in relation to the numerous photons that will follow the "zigzag" pattern. The next argument would be LASER light, as all the photons should propagate parallel to each other. However, we do not have the technology to ensure perfectly ideal orthogonality. As such, even the LASER photons would be refracted, even if only slightly, upon entering the optic cable. Even supposing you managed a test where you did get perfect orthogonality, it would be virtually impossible to repeat. You simply cannot achieve such perfect conditions that this could work. Even vibrations of the earth or a single dust particle or even a handful of large molecules would cause light scattering so that perfect orthogonality could not be achieved. This would occur even if the LASER was in contact with the cable inside of the best vacuum achievable with our technology. Not to mention the variance in the synchronization of the LASER or tolerance level for non-simultaneous emission.

Okay, fair point. Still, the zigzag effect would be more pronounced in a curved optical fibre than a straight one.

Quote from: Mohnzh on August 19, 2008, 01:58:43 PM

It is only theoretically possible for a photon to enter a fiberoptic cable in perfect parallel to the cable (or perfectly orthogonal to the beginning of the cable). Because it is not practically possible, all light will zigzag inside of a fiberoptic cable - even a perfectly straight one. Suppose, however, that you do manage to get a photon to do this. It would be undetectable in relation to the numerous photons that will follow the "zigzag" pattern. The next argument would be LASER light, as all the photons should propagate parallel to each other. However, we do not have the technology to ensure perfectly ideal orthogonality. As such, even the LASER photons would be refracted, even if only slightly, upon entering the optic cable. Even supposing you managed a test where you did get perfect orthogonality, it would be virtually impossible to repeat. You simply cannot achieve such perfect conditions that this could work. Even vibrations of the earth or a single dust particle or even a handful of large molecules would cause light scattering so that perfect orthogonality could not be achieved. This would occur even if the LASER was in contact with the cable inside of the best vacuum achievable with our technology. Not to mention the variance in the synchronization of the LASER or tolerance level for non-simultaneous emission.

Okay, fair point. Still, the zigzag effect would be more pronounced in a curved optical fibre than a straight one.

Congratulations on your first concession (at least that I have ever witnessed). As to the more pronounced effect, that seems intuitively correct, but I seem to recall that one of the nifty features of a fiberoptic cable is that it would actually average out so that the amount of time it takes light to travel a severely coiled fiberoptic cable (or any random pattern) is exactly the same as light traveling another cable of the exact same length (worked out mathematically, not experimentally). I cannot remember the details, as I am nearly certain the phrae "averaging out" conveys the wrong concept. I did this research about 4 years ago while preparing a patent. I was considering the use of fiberoptic cable in the design of an instrument. The design was abandoned for something that put the light source and detector in nearly direct contact with the sample, eliminating the need for fiberoptics, so my study of fiberoptics was only tangential to my work and not done with vigor.

Congratulations on your first concession (at least that I have ever witnessed). As to the more pronounced effect, that seems intuitively correct, but I seem to recall that one of the nifty features of a fiberoptic cable is that it would actually average out so that the amount of time it takes light to travel a severely coiled fiberoptic cable (or any random pattern) is exactly the same as light traveling another cable of the exact same length (worked out mathematically, not experimentally). I cannot remember the details, as I am nearly certain the phrae "averaging out" conveys the wrong concept. I did this research about 4 years ago while preparing a patent. I was considering the use of fiberoptic cable in the design of an instrument. The design was abandoned for something that put the light source and detector in nearly direct contact with the sample, eliminating the need for fiberoptics, so my study of fiberoptics was only tangential to my work and not done with vigor.

I see. Well, if this is true, then the experiment would certainly work. Is there anyone here with more experience dealing with optic fibre?

Quote from: Mohnzh on August 19, 2008, 02:21:19 PM

Congratulations on your first concession (at least that I have ever witnessed). As to the more pronounced effect, that seems intuitively correct, but I seem to recall that one of the nifty features of a fiberoptic cable is that it would actually average out so that the amount of time it takes light to travel a severely coiled fiberoptic cable (or any random pattern) is exactly the same as light traveling another cable of the exact same length (worked out mathematically, not experimentally). I cannot remember the details, as I am nearly certain the phrae "averaging out" conveys the wrong concept. I did this research about 4 years ago while preparing a patent. I was considering the use of fiberoptic cable in the design of an instrument. The design was abandoned for something that put the light source and detector in nearly direct contact with the sample, eliminating the need for fiberoptics, so my study of fiberoptics was only tangential to my work and not done with vigor.

I see. Well, if this is true, then the experiment would certainly work. Is there anyone here with more experience dealing with optic fibre?

I second that call. I am by no means an expert and only have understanding of some of the foundational elements of the technology.

The reason I proposed using two fibre optic cables (rather than one cable and one laser beam) is that any slow-down due to the ligt zig-zagging in the cable would be present in both beams (A and B), so a camparison of arrival times would still be valid.

In a few days I'm going to see someone who is quite knowledgeable about fibre optics, so I'll put your questions to him.

So I have a bending light math question.  In the FE, light makes it from the sun to the earth in as little as .01 seconds to around .04 seconds.  How do we not notice the EA as it would have to be super strong?