Why does the moon "bend away" in case it is a disk/cylinder?

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #30 on: November 19, 2008, 02:39:51 AM »
Even if the Moon isn't a perfect reflector, the geometry doesn't allow light reaching the Moon to be reflected back to a viewer on Earth.

That's like saying it's not possible for a lamp shining straight down on a piece of paper to illuminate it to an observer looking at it from a 45-degree angle.



An interferometer doesn't show any bending of light down to accuracies of nanometers.  Since "bendy light" requires much more significant bending than this, there is evidence that light isn't bent in the manner required.

Light doesn't bend on small scales, only large ones. This is due to quantum uncertainties in the positions of the photons; when the uncertainty is greater than the bending effect, no bending is observed.

See told you he just used the formula for the curvature of the Earth rather than some sort of data to derive it.  Of course, the fact that experiments about the nature of light show that it doesn't bend to that extent would make it impossible to derive that equation from an observation of light.

Actually, that equation could quite easily be derived from observations of the Sinking Ship effect. But, since the radius of the Earth is known to agree with the Sinking Ship effect, I used existing data derived from such observations to derive my equation for bendy light.
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Earthquakesdontbend

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #31 on: November 19, 2008, 02:45:52 AM »
Could you please demonstrate this with a perfect, mathematical equation?

I am eager to learn.

- Earthquakes don't bend
I was thinking of putting up the "top ten shapes of the earth". I've got Pyramid Earth and Cubic Earth so far...

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #32 on: November 19, 2008, 03:06:42 AM »
Could you please demonstrate this with a perfect, mathematical equation?

I am eager to learn.

- Earthquakes don't bend

y = ax2

Where a = 7.84 * 10-8 m-1, and x and y are horizontal and vertical coordinates, respectively.
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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #33 on: November 19, 2008, 03:29:02 AM »
That's like saying it's not possible for a lamp shining straight down on a piece of paper to illuminate it to an observer looking at it from a 45-degree angle.

Except I don't have "bendy light" to account for with light reflecting off of the sheet of paper.  The light reflects off of the paper and proceeds in a straight line to my eyes.

With your "bendy light" the light has curved from the source to the surface of the Moon by missing the Earth.  There is no geometry that will allow that reflected light to reach observers standing on the surface of the Earth.


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Light doesn't bend on small scales, only large ones. This is due to quantum uncertainties in the positions of the photons; when the uncertainty is greater than the bending effect, no bending is observed.

I wouldn't describe distance of meters as "small scale" when we are talking about quantum effects.  What distance would you say is required to see a noticeable effect of your "electromagnetic acceleration?"  I don't see any part of your formula that implies that there is a minimum size for the effect.
« Last Edit: November 19, 2008, 07:06:54 AM by Rig Navigator »

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #34 on: November 19, 2008, 06:29:02 AM »
The Moon is not a retroreflector.
Actually, according to RET, you get a reflection from the Earth to the moon and then back, we call it Earth shine - and you can even see it with the naked eye (but you need a dark night to do so - so you will have to leave the cities).

If you look at the moon when it is nearly new (crescent), you will still be able to see the part that is in shadow. This is because of the light reflected first form the Earth and then back to the Earth from the Moon.

How does FE explain "Earth Shine" if the Moon is not a Retro Reflector?
Everyday household experimentation.

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britishgent

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #35 on: November 19, 2008, 06:31:46 AM »
the conspiracy: "they" put tippex on your retinas
Global warming: Liberal hoax
The earth is not getting warmer after all; the effect is really just the prevalence of air conditioning. It just seems warmer when we go outside.

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #36 on: November 19, 2008, 09:55:19 PM »
Except I don't have "bendy light" to account for with light reflecting off of the sheet of paper.  The light reflects off of the paper and proceeds in a straight line to my eyes.

With your "bendy light" the light has curved from the source to the surface of the Moon by missing the Earth.  There is no geometry that will allow that reflected light to reach observers standing on the surface of the Earth.

Apparently I'm going to need another analogy. Okay then, here goes:

Consider the "sinking ship" effect. According to FET+EA, light bends up to allow light with an initial downward trajectory reflecting off the ship to reach the eye of the observer. Thus, bendy light has caused light to bend away from the Earth. Are you telling me that there is no orientation at which a mirror may be held that will reflect this light to meet the ground?

I wouldn't describe distance of meters as "small scale" when we are talking about quantum effects.  What distance would you say is required to see a noticeable effect of your "electromagnetic acceleration?"  I don't see any part of your formula that implies that there is a minimum size for the effect.

The distance of metres isn't the small scale, what is small scale is the change in position and momentum of the photon perpendicular to its initial momentum vector. That equation only applies over large distances; I will see if I can come up with some mathematics based on the uncertainty principle to illustrate why the uncertainty in position and momentum is always greater than the effect of bendy light - and therefore unobservable - over short distances.

Again, your pulling figures and equations out of your ass.

Bravo.

Except... hold on... light could bend in any direction in 3d space. Which direction does it bend in and why? Does the Wonko Force have anything to do with it? (Hint: Yes)

Dark Energy always acts upward.

Also, the equation you gave contradicts your statement that

Light doesn't bend on small scales, only large ones.

Using the equation given would produce detectable changes over a small distance (few metres)

It would, if that equation were valid over short distances. In much the same way that General Relativity becomes useless over short distances and Quantum Mechanics over large ones, that equation is only valid when the quantum uncertainty in the position and momentum of photons is significantly less than the change in position and momentum caused by the bending of light.

I can see how you've simply picked the distance the the horizon, the average height of an adult, and pulled out a square law (cos it kinda fits, right?). This is classic Rowbotham style failure.

No, I chose a square law because a parabola is the simplest symmetrical function that is not a straight line, and things in nature tend to be as simple as possible.

Oh and ...

In this thread you promised Matrix an experiment to quantify the effects of bendy light. Got anything yet?

No. The recent emergence of the quantum effects in the bendy light theory have made it substantially more difficult to come up with something.

How does FE explain "Earth Shine" if the Moon is not a Retro Reflector?

The Moon is a retroreflector in RET now? Can you please provide a source?
« Last Edit: November 20, 2008, 12:56:04 AM by Robosteve »
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The Terror

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #37 on: November 19, 2008, 11:36:02 PM »
The Moon is not a retroreflector.
Actually, according to RET, you get a reflection from the Earth to the moon and then back, we call it Earth shine - and you can even see it with the naked eye (but you need a dark night to do so - so you will have to leave the cities).

If you look at the moon when it is nearly new (crescent), you will still be able to see the part that is in shadow. This is because of the light reflected first form the Earth and then back to the Earth from the Moon.

How does FE explain "Earth Shine" if the Moon is not a Retro Reflector?

Well, under FE theory the Earth Shine effect should be much greater, given the shape of the Earth and the moon being a lot closer to us.

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #38 on: November 19, 2008, 11:39:17 PM »
Apparently I'm going to need another analogy. Okay then, here goes:

Consider the "sinking ship" effect. According to FET+EA, light bends up to allow light with an initial downward trajectory reflecting off the ship to reach the eye of the observer. Thus, bendy light has caused light to bend away from the Earth. Are you telling me that there is no orientation at which a mirror may be held that will reflect this light to meet the ground?

Yes.  I am saying that if the light from the sinking ship would not be able to light the bottom or far side of an object.  If it could, because of the bend in the light, then it would not be reflected to an observer on the ground.  The angles just don't work out.



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The distance of metres isn't the small scale, what is small scale is the change in position and momentum of the photon perpendicular to its initial momentum vector. That equation only applies over large distances; I will see if I can come up with some mathematics based on the uncertainty principle to illustrate why the uncertainty in position and momentum is always greater than the effect of bendy light - and therefore unobservable - over short distances.

What is the definition of large vs short distances for this discussion.  This is an effect that should manifest itself over a distance of more than a meter which is smaller than the travel path for the light in most interferometers, and many are significantly larger than that.  In none of them has any observation of "bendy light" been made.

I don't think that Heisenberg's Uncertainty Principle applies to observations made on objects larger than atoms.  Even if we give you centimeters of leeway in your measurement, there are still no observations that show "bendy light" or EA.

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It would, if that equation were valid over large distances.

From your above...

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That equation only applies over large distances

Please make up your mind.


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In much the same way that General Relativity becomes useless over short distances and Quantum Mechanics over large ones, that equation is only valid when the quantum uncertainty in the position and momentum of photons is significantly less than the change in position and momentum caused by the bending of light.

I don't think that there have been any problems with GR in any measurements over the course of a meter.


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No. The recent emergence of the quantum effects in the bendy light theory have made it substantially more difficult to come up with something.

Then lets stick with observations of light as it travels at least one meter.  That should simplify things significantly.  There are no observations that show "bendy light" at greater than one meter, so that needs a lot of work before you start trying to describe anything on the quantum level.  Let's start with the simple, observable characteristics of light before you start moving toward extreme complexity.  We can worry about the quantum level once you figure out the macro level.

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #39 on: November 20, 2008, 01:02:23 AM »
Yes.  I am saying that if the light from the sinking ship would not be able to light the bottom or far side of an object.  If it could, because of the bend in the light, then it would not be reflected to an observer on the ground.  The angles just don't work out.

You aren't making any sense. Why would I not be able to reflect an almost horizontal light ray a metre or two above the surface of the Earth so that it goes straight down to meet the surface of the Earth?

What is the definition of large vs short distances for this discussion.  This is an effect that should manifest itself over a distance of more than a meter which is smaller than the travel path for the light in most interferometers, and many are significantly larger than that.  In none of them has any observation of "bendy light" been made.

Probably ten metres or so at least would be required to measure a curve in the path of a light ray. I don't know exactly; I haven't yet done the calculations regarding the quantum effect.

I don't think that Heisenberg's Uncertainty Principle applies to observations made on objects larger than atoms.  Even if we give you centimeters of leeway in your measurement, there are still no observations that show "bendy light" or EA.

Heisenberg's Uncertainty Principle always applies. The product of the uncertainties in position and momentum for any object must always be greater than or equal to Planck's constant divided by 2π.

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It would, if that equation were valid over large distances.

From your above...

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That equation only applies over large distances

Please make up your mind.

Sorry, my mistake. It is only valid for large distances.

I don't think that there have been any problems with GR in any measurements over the course of a meter.

I didn't say there were.

Then lets stick with observations of light as it travels at least one meter.  That should simplify things significantly.  There are no observations that show "bendy light" at greater than one meter, so that needs a lot of work before you start trying to describe anything on the quantum level.  Let's start with the simple, observable characteristics of light before you start moving toward extreme complexity.  We can worry about the quantum level once you figure out the macro level.

You fail to realise that because the bendy light effect over a few metres is very tiny, light travelling across a distance of a few metres is still subject to quantum effects. We are talking about the uncertainty in the photons' positions perpendicular to the direction in which they are travelling.
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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #40 on: November 20, 2008, 02:24:25 AM »
You aren't making any sense. Why would I not be able to reflect an almost horizontal light ray a metre or two above the surface of the Earth so that it goes straight down to meet the surface of the Earth?

Except we aren't talking about a meter off of the ground.  We are talking the Moon.

As for your example, I am saying that it is impossible to have a light source from a ship on the horizon illuminate the bottom of a sphere.


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Probably ten metres or so at least would be required to measure a curve in the path of a light ray. I don't know exactly; I haven't yet done the calculations regarding the quantum effect.

Why would this quantum effect be so large as to effect light moving less that ten meters and still not have a measurable effect on the light beyond this distance? 

There are interferometers that are significantly larger than this "upper limit of quantum effect" that still do not show the results that you are looking for.

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Heisenberg's Uncertainty Principle always applies. The product of the uncertainties in position and momentum for any object must always be greater than or equal to Planck's constant divided by 2π.

and for a measurement of meters this would mean that the uncertainty is what?


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Sorry, my mistake. It is only valid for large distances.



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I didn't say there were.

From two posts ago...

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In much the same way that General Relativity becomes useless over short distances...

You are now saying that anything less than 10 meters is subject to this "quantum effect" and thus, short distances.


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You fail to realise that because the bendy light effect over a few metres is very tiny, light travelling across a distance of a few metres is still subject to quantum effects. We are talking about the uncertainty in the photons' positions perpendicular to the direction in which they are travelling.

Tiny but measurable.  I am not concerned about measuring the position of each individual photon.  I am talking about comparing two beams of light that have traveled the same distance following two paths.  If these two beams of light don't travel the same distance because of EA, or whatever causes "bendy light", then there will be measurable differences.  These are not present.

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #41 on: November 20, 2008, 02:49:17 AM »
Except we aren't talking about a meter off of the ground.  We are talking the Moon.

The principle is the same.

As for your example, I am saying that it is impossible to have a light source from a ship on the horizon illuminate the bottom of a sphere.

Incorrect.

Why would this quantum effect be so large as to effect light moving less that ten meters and still not have a measurable effect on the light beyond this distance?

There are interferometers that are significantly larger than this "upper limit of quantum effect" that still do not show the results that you are looking for.

Please name one.

You are now saying that anything less than 10 meters is subject to this "quantum effect" and thus, short distances.

General Relativity and - as you REers like to call it, "bendy light" - are different theories.

Tiny but measurable.  I am not concerned about measuring the position of each individual photon.  I am talking about comparing two beams of light that have traveled the same distance following two paths.  If these two beams of light don't travel the same distance because of EA, or whatever causes "bendy light", then there will be measurable differences.  These are not present.

They are not present because the overall beam of light is the sum of many individual photons, each of which will not show measurable bending when its uncertainty in position is greater than the effect of the EA.
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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #42 on: November 20, 2008, 03:17:14 AM »
The principle is the same.

Not really.

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Incorrect.

Evidence?

Take light, shine it at a ball from a distance above, and light up the bottom.  I don't care if the light is 16 meters away or 16 miles away, you aren't going to be able to do it.


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Please name one.

Laser Interferometer Gravitational-Wave Observatory (http://en.wikipedia.org/wiki/LIGO)

or its 40 meter prototype (http://www.ligo.caltech.edu/~ajw/40m_upgrade.html)


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General Relativity and - as you REers like to call it, "bendy light" - are different theories.

You are the one that brought GR into the discussion.


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They are not present because the overall beam of light is the sum of many individual photons, each of which will not show measurable bending when its uncertainty in position is greater than the effect of the EA.

??!?
« Last Edit: November 20, 2008, 03:19:31 AM by Rig Navigator »

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #43 on: November 20, 2008, 04:00:17 AM »
Evidence?

Take light, shine it at a ball from a distance above, and light up the bottom.  I don't care if the light is 16 meters away or 16 miles away, you aren't going to be able to do it.

Over distances greater than ten megametres or so, it would be very easily possible to do just that.

Laser Interferometer Gravitational-Wave Observatory (http://en.wikipedia.org/wiki/LIGO)

or its 40 meter prototype (http://www.ligo.caltech.edu/~ajw/40m_upgrade.html)

I'm sorry, how does this provide any evidence for or against the EA?

You are the one that brought GR into the discussion.

Yes, to illustrate that large-scale rules are sometimes different from small-scale ones.

Wow really?! Holy crap this would finally give us proof of the presence of "Dark Energy".

Get thee to an academic institution now!

 ???

I really don't know why you're babbling on about Heisenberg. Maybe it somehow lends credence to yuor guff. Heisenberg has absolutely nothing to do with the straight line path of a photon (in this debate).

That's where you're wrong. Look up "uncertainty principle".

Once again, you seem to have great difficulty understanding the fact that large distances are made of many small ones. If light fails to bend over small distances, how can it bend over large ones?

An electron cannot arbitrarily change to any energy level in an atom, but rather must jump between discrete levels. Again, small-scale rules are not always the same as large-scale ones.

Exactly. You say "chose" I say "pulled out of ass". By your own admission, your equation breaks at low values. Now lets see what happens with high values. Let's see how far we can see from the top of everest for example.

RE says 335,764m
RoboSteve wonko light says 335,842m

Yikes! That's a difference of 78m!

No it isn't. I only gave the value of the parameter a in my equation to three significant figures. In other words, my theory agrees perfectly with RET at the precision to which it can be tested.

Working the other way, and using scientific equipment, would produce hidious differences between your equation and observation.

Wrong.

So anyway, I'm guessing you put a load of values into a scientific calculator and then pressed "find formula" then took the first term as an approximate. It doesn't work like that I'm afraid.

Nope.

Also, your equation (and proposition) has difficulty accounting for objects seen over the horizon, for example viewing one mountain from the summit of another.

It does? ???

Lets recap, your equation breaks down for small values, and it breaks down for large values. So what good is it?

It works perfectly for large values. You have yet to show that it does not.

Is that the sound of your rickety bike backpedalling out of the room? Quantum effects have nothing to do with this, that's just horsesh*t. I can see how you're desperate for your brand of manure to remain unproven, it's all just a bit sad really.

Bendy light can be tested very simply over any distance. It just doesn't happen the way you want it too.

I think you'll find it does. Search for "sinking ship effect", you'll find all the evidence you need.
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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #44 on: November 20, 2008, 05:02:09 AM »
Over distances greater than ten megametres or so, it would be very easily possible to do just that.

But you were talking about one or two meters in your thought experiment, not millions of meters. 

In your "small scale" experiment where you are reflecting light from a sinking ship type light source you can't reflect the light straight down or at an observer looking toward the light source from the other side of the mirror as is needed to model the full Moon.



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I'm sorry, how does this provide any evidence for or against the EA?

You were asking for examples of large interferometers that did not show evidence of "bendy light."


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Yes, to illustrate that large-scale rules are sometimes different from small-scale ones.

My retort is that your "small scale" is thousands of times bigger than small scale in GR.  Well beyond what is considered normal for quantum effects to be noticeable.


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That's where you're wrong. Look up "uncertainty principle".

Where we are concerned with measurements of millimeters or tenths of millimeters, we can probably not have to worry too much about the uncertainty of our measurements.  We can sacrifice the accuracy of our velocity measurement to gain extra precision in our position measurement if we need to.


Working the other way, and using scientific equipment, would produce hidious differences between your equation and observation.

Wrong.[/quote]

Except apparently at small distances (less than 10 meters).


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I think you'll find it does. Search for "sinking ship effect", you'll find all the evidence you need.

Most of the posts that I found were about how it didn't exist and any observations showing it were mistaken.

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #45 on: November 20, 2008, 05:55:08 AM »
Wow, this debate is still going?

The bendy light thing wouldn't light the moon properly, as the lack of shadows on the moon during a full moon suggests that it's being lit head on, not from the side.

Observations of a curved surface of the planet may suggest that the FE and bendy light hypothesis is possible, but it also suggests that the RE model is possible, and the RE model is simpler than a FE model with an extra unknown and otherwise undetectable force, and quantum effects at a random value large enough to mask the evidence against it.

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #46 on: November 20, 2008, 08:19:06 AM »
How does FE explain "Earth Shine" if the Moon is not a Retro Reflector?

The Moon is a retroreflector in RET now? Can you please provide a source?
Yes. Look at the Moon when it is a crescent and you are not near city lights. You can actually still see the "dark" part of the moon and this light you can see is reflected first from the Earth and then onto the Moon and back to Earth.

So my source is... /me points at the crescent moon...
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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #47 on: November 20, 2008, 09:43:06 PM »
Look post your revised equation in a new thread. I'll pick it apart again then.

I'll do that when I've figured out the mathematics of the quantum mechanics of it.

But you were talking about one or two meters in your thought experiment, not millions of meters. 

Uh... the distance between the Sun and the Moon during Full Moon is not one or two metres in either RET or FET. I don't know where you're getting that idea from.

In your "small scale" experiment where you are reflecting light from a sinking ship type light source you can't reflect the light straight down or at an observer looking toward the light source from the other side of the mirror as is needed to model the full Moon.

So you can't reflect almost horizontal light straight down? Interesting.

You were asking for examples of large interferometers that did not show evidence of "bendy light."

Scientific experiments that cannot distinguish between bendy light and straight light usually don't show any evidence for the former.

My retort is that your "small scale" is thousands of times bigger than small scale in GR.  Well beyond what is considered normal for quantum effects to be noticeable.

No it isn't.

Where we are concerned with measurements of millimeters or tenths of millimeters, we can probably not have to worry too much about the uncertainty of our measurements.  We can sacrifice the accuracy of our velocity measurement to gain extra precision in our position measurement if we need to.

No we can't. The position and momentum of a photon under the influence of the EA are intimately linked. If we know the initial and final positions of the photon, and that it bends according to a square law, we can calculate its momentum, and vice versa. We either know both its position and its momentum to a certain accuracy, or we don't know either one to that accuracy.

Except apparently at small distances (less than 10 meters).

Yes, because of quantum effects. Good to see you're keeping up.

Most of the posts that I found were about how it didn't exist and any observations showing it were mistaken.

So you disagree that ships disappear from the bottom up as they sail away over the horizon?

The bendy light thing wouldn't light the moon properly, as the lack of shadows on the moon during a full moon suggests that it's being lit head on, not from the side.

The light bends to approach it from a different angle. Hence, bendy light.

Observations of a curved surface of the planet may suggest that the FE and bendy light hypothesis is possible, but it also suggests that the RE model is possible, and the RE model is simpler than a FE model with an extra unknown and otherwise undetectable force, and quantum effects at a random value large enough to mask the evidence against it.

Yes, because RE doesn't have any forces that scientists can't explain. ::)

Yes. Look at the Moon when it is a crescent and you are not near city lights. You can actually still see the "dark" part of the moon and this light you can see is reflected first from the Earth and then onto the Moon and back to Earth.

So my source is... /me points at the crescent moon...

I understand the effect you are talking about. Why does the Moon need to be a retroreflector for this effect to be observed?
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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #48 on: November 21, 2008, 03:38:44 AM »
I'll save you the effort.

Quantum mechanics works on an atomic and subatomic level. Naturally the distances involved are not just tens of metres, as you seem keen to imply, but miniscule of the order pico-metres.

I don't think I ever claimed that.

The Heisenberg uncertainty principle, that the position or momentum of a particle can never both be known, since an observer will disturb one or the other, has no relevence here, other than another attempt at obfuscation.

If you like, you can calculate the probability of all the paths a photon could take. You'll find a very large percentage of them are in a straight line. Then you can calculate the probability paths of billions upon billions of photons.

Then you can devise a simple test to see if your calculations match observations.

Or I could just perform a few calculations using ΔyΔpy ≥ ℏ.
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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #49 on: November 21, 2008, 04:02:38 AM »
Shoo shoo little troll.

According to the EA theory, light does not follow a one-dimensional path. We can speak of quantum effects in the y dimension as being independent of those in the x dimension.
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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #50 on: November 21, 2008, 05:13:52 AM »
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The light bends to approach it from a different angle. Hence, bendy light.

I don't think it would bend so much that the light would hit the moon head on. If the light is approaching the Moon perpendicular to the ground to hit the Moon head on from our viewpoint, then it must have been emitted from the sun perpendicular to the ground. But then, what caused it to bend off to the side towards the Moon?

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Yes, because RE doesn't have any forces that scientists can't explain.

I never said that it didn't. Any theory or hypothesis, after delving deep enough into it is going to have something scientists can't explain. That's a moot point. The question is which has the least unknowns: The RE model with light going in straight lines, or the FE model with the light being bent up by something?

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #51 on: November 21, 2008, 05:28:07 AM »
I don't think it would bend so much that the light would hit the moon head on. If the light is approaching the Moon perpendicular to the ground to hit the Moon head on from our viewpoint, then it must have been emitted from the sun perpendicular to the ground. But then, what caused it to bend off to the side towards the Moon?

It doesn't need to be perpendicular, just close enough to it that we don't notice the difference.

I never said that it didn't. Any theory or hypothesis, after delving deep enough into it is going to have something scientists can't explain. That's a moot point. The question is which has the least unknowns: The RE model with light going in straight lines, or the FE model with the light being bent up by something?

How does the behaviour of light relate to the number of unknowns?
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Earthquakesdontbend

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #52 on: November 21, 2008, 06:15:19 AM »
The only thing I want to ask now is,

is the moon a disc or a sphere according to FE theory?
I was thinking of putting up the "top ten shapes of the earth". I've got Pyramid Earth and Cubic Earth so far...

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #53 on: November 21, 2008, 06:28:19 AM »
So you can't reflect almost horizontal light straight down? Interesting.

Nope.  Take a mirror, shine the light at the edge and see if it will be reflected toward the ground.  If you angle the mirror so it will reflect the light, it will no longer be reflected straight toward the ground, but will hit the ground at an angle.


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Scientific experiments that cannot distinguish between bendy light and straight light usually don't show any evidence for the former.

An experiment which relies on light moving in straight lines would not show the correct data if the light bent.


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So you disagree that ships disappear from the bottom up as they sail away over the horizon?

No, you suggested that I search the forum for posts relating to the sinking ship phenomenon and I reported what I found.

Personally, I think that my observations of ships shows that the Earth is round.  My observations of light show that it moves in straight lines.

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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #54 on: November 21, 2008, 06:52:11 AM »
The only thing I want to ask now is,

is the moon a disc or a sphere according to FE theory?

A sphere.

Nope.  Take a mirror, shine the light at the edge and see if it will be reflected toward the ground.  If you angle the mirror so it will reflect the light, it will no longer be reflected straight toward the ground, but will hit the ground at an angle.

What would that angle be if the light and the ground were perfectly horizontal, and the mirror it was reflecting off was angled at exactly 45 degrees to the horizontal?

An experiment which relies on light moving in straight lines would not show the correct data if the light bent.

I agree. Would you care to link me to information on such an experiment, or are you just going to claim that they exist without backing up your statement?

No, you suggested that I search the forum for posts relating to the sinking ship phenomenon and I reported what I found.

Personally, I think that my observations of ships shows that the Earth is round.  My observations of light show that it moves in straight lines.

Why assume that the Earth is curved and not the light?
I'm going to side with the white supremacists.

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #55 on: November 21, 2008, 08:16:34 AM »
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It doesn't need to be perpendicular, just close enough to it that we don't notice the difference.

Interesting, but if the light coming from the Sun that lights the Moon is close to perpendicular to the Planet's surface, then the light that lights the Planet's surface would have to be even closer. Assuming the Sun is a sphere outputting equally in all directions as it appears, then we are only receiving a tiny portion of it's total output. There's only so much energy a 50 kilometre wide ball can output.

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How does the behaviour of light relate to the number of unknowns?

Well, how many unknown forces or effects are required to make light go in a straight line, and how many are required to make it bend upwards in a giant parabola?

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #56 on: November 21, 2008, 09:08:06 AM »
What would that angle be if the light and the ground were perfectly horizontal, and the mirror it was reflecting off was angled at exactly 45 degrees to the horizontal?

45?. 

That isn't straight to the ground is it?

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I agree. Would you care to link me to information on such an experiment, or are you just going to claim that they exist without backing up your statement?

How many times do I have to send something to you about interferometers?


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Why assume that the Earth is curved and not the light?

Because I haven't seen observed the bending of light in interferometers, but I have observed with celestial observations the curved nature of the surface of the Earth.

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #57 on: November 21, 2008, 11:05:45 AM »
All experiments regarding optics rely, however implicitly, on the rectilinear propagation of light. The LHC for example catapults photons many times round a 27km ring, if there was a Dark Energy (Field?) tampering with photons don't you think they would notice?

Sorry to nit pick Goldsein, but the LHC is a proton accelerator, not a photon accelerator (photons already move at c).
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Parsifal

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Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #58 on: November 21, 2008, 09:59:50 PM »
Michelson Morley?

That only measures the path of light over fairly short distances, which means that quantum uncertainties are much larger than the effect of the EA and we wouldn't expect to observe any bending.

Interesting, but if the light coming from the Sun that lights the Moon is close to perpendicular to the Planet's surface, then the light that lights the Planet's surface would have to be even closer. Assuming the Sun is a sphere outputting equally in all directions as it appears, then we are only receiving a tiny portion of it's total output. There's only so much energy a 50 kilometre wide ball can output.

Not necessarily. The craters on the Moon are much shallower than they appear, so it doesn't need to be extremely close to perpendicular to eliminate shadows, only close enough to it that we don't notice any shadows.

Well, how many unknown forces or effects are required to make light go in a straight line, and how many are required to make it bend upwards in a giant parabola?

So because FE has one unknown force that RE doesn't, it automatically has more unknowns? ???

45?. 

That isn't straight to the ground is it?

No, it isn't. Unfortunately, it doesn't agree with the theory of optics you REers seem to be so keen to support, either. I'd take another look at how mirrors work, and then try to answer that question again. Alternatively, I look forward to reading your theory on the reflection of light.

How many times do I have to send something to you about interferometers?

Not every interferometer can distinguish between straight and bendy light.

Because I haven't seen observed the bending of light in interferometers, but I have observed with celestial observations the curved nature of the surface of the Earth.

I am not aware of any experiment that has ever been able to detect the difference between straight and bendy light. So, that point is invalid.
I'm going to side with the white supremacists.

Re: Why does the moon "bend away" in case it is a disk/cylinder?
« Reply #59 on: November 21, 2008, 11:43:58 PM »
That only measures the path of light over fairly short distances, which means that quantum uncertainties are much larger than the effect of the EA and we wouldn't expect to observe any bending.

I await you math then so I can see why distances of 1+ meters are short distances in quantum effects.  I have a feeling that this is going to be a pointless wait, because like other parts of EA that we have been waiting months for, this explanation will never come out.


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No, it isn't. Unfortunately, it doesn't agree with the theory of optics you REers seem to be so keen to support, either. I'd take another look at how mirrors work, and then try to answer that question again. Alternatively, I look forward to reading your theory on the reflection of light.



Now here is a diagram that shows how mirrors work on a RE.  Now if we use the FE theory of "bendy light" with this, it wouldn't look the same.  You couldn't reflect the light straight down to the ground.


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Not every interferometer can distinguish between straight and bendy light.

Why not?


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I am not aware of any experiment that has ever been able to detect the difference between straight and bendy light. So, that point is invalid.

You haven't explained why interferometers wouldn't detect "bendy light" yet.  So far, I would say that none of them has detected "bendy light" yet, and if it was present, it would be noticeable.