I'll test bendy light theory

Quote from: parsec on March 21, 2010, 05:39:29 PM

Quote from: Lorddave on March 21, 2010, 05:37:50 PM

Quote from: parsec on March 21, 2010, 05:35:32 PM

Quote from: Lorddave on March 21, 2010, 05:35:01 PM

Quote from: parsec on March 21, 2010, 05:24:15 PM

Quote from: Lorddave on March 21, 2010, 05:22:12 PM

Quote from: parsec on March 21, 2010, 05:13:26 PM

Quote from: corleone on March 21, 2010, 05:06:40 PM

If you are asking about being perfectly vertical I'll tell you that id does not matter at all if we only want to see bendy light in action. If we want to take precise data it matters, and i'll do my best with the precision of the angles. But bendy light phenomena should be easy to spot without that precission.

ORLY? At a length of less than 1 m, you better have precise measurements because your confidence level will be less than 50%, i.e. you will be basically tossing a coin.

But since the light bounces between mirrors thousands of times, wouldn't that translate to a large distance for the light to bend?  Sure the net distance would be small, but the travel distance shouldn't be.

No, because BLT takes into account the difference in refraction index over large height differences. Your tube has pretty much a uniform refraction index and you do not get the same effect as a true propagation of light through the atmolayer.

Let me make sure I understand you're reasoning.
Light will only bends when the light passes through the whole atmosphere?

No.

So it bends based on the atmosphere's index of refraction for various heights?

I'm afraid that even if I said yes, you would not be able to comprehend what that meant since you obviously do not have a clue what a refraction index is.

>_>
The index of refraction is how a light beam will bend through a solid but translucent or transparent objects, such as water, and appear to distort the object.  This can easily be seen by putting a pencil in a cup of water.   The pencil appears to bend at the surface of the water.  

So yes, I know what an index of refraction is.
And yes, I know the atmosphere has a refraction index that changes based on density and temperature of the air.
What I'm trying to figure out is how that matters since we can measure the index refraction of air and calculate the true path of light in the vacuum.  Atmospheric refraction is one of the things astronomers try to limit by building large telescopes high up.  Less air, less matter for the light to bend.

I'm still trying to figure out what is different between bendy light and simple atmospheric refraction.

That copypasta looks delicious, but my point stands unshaken.

That copypasta looks delicious, but my point stands unshaken.

Copypasta'd from where exactly?

The only point you've made is "baaawww I won't tell!".

Quote from: Lorddave on March 21, 2010, 05:45:39 PM

Quote from: parsec on March 21, 2010, 05:39:29 PM

Quote from: Lorddave on March 21, 2010, 05:37:50 PM

Quote from: parsec on March 21, 2010, 05:35:32 PM

Quote from: Lorddave on March 21, 2010, 05:35:01 PM

Quote from: parsec on March 21, 2010, 05:24:15 PM

Quote from: Lorddave on March 21, 2010, 05:22:12 PM

Quote from: parsec on March 21, 2010, 05:13:26 PM

Quote from: corleone on March 21, 2010, 05:06:40 PM

If you are asking about being perfectly vertical I'll tell you that id does not matter at all if we only want to see bendy light in action. If we want to take precise data it matters, and i'll do my best with the precision of the angles. But bendy light phenomena should be easy to spot without that precission.

ORLY? At a length of less than 1 m, you better have precise measurements because your confidence level will be less than 50%, i.e. you will be basically tossing a coin.

But since the light bounces between mirrors thousands of times, wouldn't that translate to a large distance for the light to bend?  Sure the net distance would be small, but the travel distance shouldn't be.

No, because BLT takes into account the difference in refraction index over large height differences. Your tube has pretty much a uniform refraction index and you do not get the same effect as a true propagation of light through the atmolayer.

Let me make sure I understand you're reasoning.
Light will only bends when the light passes through the whole atmosphere?

No.

So it bends based on the atmosphere's index of refraction for various heights?

I'm afraid that even if I said yes, you would not be able to comprehend what that meant since you obviously do not have a clue what a refraction index is.

>_>
The index of refraction is how a light beam will bend through a solid but translucent or transparent objects, such as water, and appear to distort the object.  This can easily be seen by putting a pencil in a cup of water.   The pencil appears to bend at the surface of the water.  

So yes, I know what an index of refraction is.
And yes, I know the atmosphere has a refraction index that changes based on density and temperature of the air.
What I'm trying to figure out is how that matters since we can measure the index refraction of air and calculate the true path of light in the vacuum.  Atmospheric refraction is one of the things astronomers try to limit by building large telescopes high up.  Less air, less matter for the light to bend.

I'm still trying to figure out what is different between bendy light and simple atmospheric refraction.

That copypasta looks delicious, but my point stands unshaken.

So you're calling me a liar now eh?
Can you show me where I got that?  Or would any source that gives a definition of "the index of refraction" do?

Please parsec, answer my question. I'm higly interested.

Quote from: parsec on March 21, 2010, 05:41:40 PM

Quote from: corleone on March 21, 2010, 05:41:08 PM

I've got a simple question: does BL happen in vacuum?

Define vacuum.

In this context i would refeer to a space without air, therefore, without any changes on the refractive index.

Yes.

Quote from: corleone on March 21, 2010, 05:44:31 PM

Quote from: parsec on March 21, 2010, 05:41:40 PM

Quote from: corleone on March 21, 2010, 05:41:08 PM

I've got a simple question: does BL happen in vacuum?

Define vacuum.

In this context i would refeer to a space without air, therefore, without any changes on the refractive index.

Yes.

Then if bendy light happens in a medium without changes on the ref index it will surely happen inside our resonant cavity, right?

Quote from: parsec on March 21, 2010, 05:54:27 PM

Quote from: corleone on March 21, 2010, 05:44:31 PM

Quote from: parsec on March 21, 2010, 05:41:40 PM

Quote from: corleone on March 21, 2010, 05:41:08 PM

I've got a simple question: does BL happen in vacuum?

Define vacuum.

In this context i would refeer to a space without air, therefore, without any changes on the refractive index.

Yes.

Then if bendy light happens in a medium without changes on the ref index it will surely happen inside our resonant cavity, right?

Bendy light does not 'happen'. What happens is that light bends and not proportionally to the height difference but progressively increasing. How high is your apparatus?

Bendy light does not 'happen'. What happens is that light bends and...

Anyone else see the contradiction in this?
If it bends, it bends as a result of some variable.  If it bends in a vacuum then it doesn't need an index of refraction to bend, making it independent of the refraction of the atmosphere.

I think I got the idea of "progressively increasing", what happens is that I have a small english vocabulary. When I say "bendy light happens" i really mean "bendy light phenomena happens" refering to the phenomena itself.

Explain me:

You say that, i.e., if I fire a laser at sealevel and i get that laser bent up ten degrees in 100 meters height (i.e.) and I repeat the same experiment at 5km heigth I would get, say, 20 degrees of desviation at the height of 5100 meters?

EDIT: this would be pretty easy to understand if we had a proper BL model, but we don't so I have to ask questions in order to make my experiment.

You say that, i.e., if I fire a laser at sealevel and i get that laser bent up ten degrees in 100 meters height (i.e.) and I repeat the same experiment at 5km heigth I would get, say, 20 degrees of desviation at the height of 5100 meters?

Yes, that would be the general idea behind what I meant as "progressively increasing deflection", although the numbers are certainly not realistic.

Ok, now we are beggining to understand each other. My experiment would take place at sealevel. According to the BL theory would I be able to detect it? I understand that, the lower you get, the lesser light is bent, therefore I might think that at sealevel no BL phenomena happens. If it's false I should be able to detect BL phenomena with my experiment. If it's true there is a conflict with the ships dissapearing.

Is my experiment at sealevel able to detect BL phenomena?

Ok, now we are beggining to understand each other. My experiment would take place at sealevel. According to the BL theory would I be able to detect it? I understand that, the lower you get, the lesser light is bent, therefore I might think that at sealevel no BL phenomena happens. If it's false I should be able to detect BL phenomena with my experiment. If it's true there is a conflict with the ships dissapearing.

Is my experiment at sealevel able to detect BL phenomena?

I never said altitude above sea level is the single variable that determines the bent of the light ray. Come to think of it, we did not even agree what a measure of a bent actually is.

Ok, well, I think it would be faster if you tell us what do you know about BL, all of it, if you don't mind.

Ok, well, I think it would be faster if you tell us what do you know about BL, all of it, if you don't mind.

There are plenty of threads discussing Bendy Light Theory. Perhaps you should go through some of them using the search function of the forums before you embark on 'testing' it. Don't you think so?

There are plenty of threads discussing Bendy Light Theory. Perhaps you should go through some of them using the search function of the forums before you embark on 'testing' it. Don't you think so?

But non of them state anything conclusive. Most involve RoboSteveMcParsiFail dancing about.

If there's a thread that offers up anything which succinctly defines bendy light in a way that doesn't immediately fail then I've not seen it. And I thought bendy light was outlawed by TFES?

Of course I did, but I don't get any equation, or answer better than "unknown" or "we are working on it". Where is the thread that explains bendy ligt properly?

Quote from: parsec on March 21, 2010, 06:36:11 PM

There are plenty of threads discussing Bendy Light Theory. Perhaps you should go through some of them using the search function of the forums before you embark on 'testing' it. Don't you think so?

But non of them state anything conclusive. Most involve RoboSteveMcParsiFail dancing about.

If there's a thread that offers up anything which succinctly defines bendy light in a way that doesn't immediately fail then I've not seen it. And I thought bendy light was outlawed by TFES?

I tried making a thread but the people who answered my questions gave answers that are vague and contradictory to what parsec has said.

Of course I did, but I don't get any equation, or answer better than "unknown" or "we are working on it". Where is the thread that explains bendy ligt properly?

A light ray emitted at an angle α with the horizontal in the north-south direction from an altitude y₀ traces a trajectory given by (to first approximation):

y = y₀ + x*tan α + π*(1 + 2*tan² α)*x²/(4*L),

where:

L = 10⁷ m.

Funny to see how, at the first sign of someone actually having a smart and practical way to prove or disprove bendy light, there is immediately a response basically saying..

no no no.. wait, you gotta do the experiment in a way that bendy light will happen and that is not contradictory to FE model..if you do it like this, there won't be any bendy light and it will be because you're doing it wrong..

Pretty soon they'll suggest that the best way to prove it is to look at the horizon and see that ships sink beneath it..
Parsec, you'd think that you'd be thrilled to finally see someone about to prove bendy light theory. The setup is smart, practical and repeatable and all things are taken into consideration. If you are gonna try to delay it long enough, trying to find an explanation to why it won't bend int theexperiment.. we'll never prove bendy light for you in this lifetime.. That would be a shame, wouldn't it?

Quote from: corleone on March 21, 2010, 06:42:50 PM

Of course I did, but I don't get any equation, or answer better than "unknown" or "we are working on it". Where is the thread that explains bendy ligt properly?

A light ray emitted at an angle ? with the horizontal in the north-south direction from an altitude y₀ traces a trajectory given by (to first approximation):

y = y₀ + x*tan ? + ?*(1 + 2*tan² ?)*x²/(4*L),

where:

L = 10⁷ m.

At last! an equation! Well, I'm trying to plote it with mathematica, to see how it works. BUT I think that that equation cannot be right. See the dimensions:

[M]=metres

[M]=[M]+[M]*(non-dimensional)+(non-dimensional)*[M]^2/[M]^7

You got:

[M]=[M]+[M]+[M]^-5

Where it should be:

[M]=[M]+[M]+[M]

In order to get the answer in metres. Also, you can't add [M.]^-5 to [M].

Re-work your marths. I would appreciate a lot if pastifal come here and join the debate, because he seems to be the one who carries on the research about BL.

EDIT: LOL, I read L=10 m^7 instead of L=10^7 m so the dimensions are ok. Let me some time to plot the equation and do some thinking. Do you have more equations?

I see your quoted text has no Greek letters. Just to clarify, the angle with the horizontal is 'alpha', and the formula reads:

Somehow the quote didn't recognize greek letters. After looking to the equation I obtain these conclusions:

1-The light path relative to the source of light does not depend on height. The path is the same either at sealevel or 10km high, only "transported" 10km higher.

2-The change rate of the angle between light and the horizontal is constant. It's a parabollic equation, therefore its derivate is linear. When you said "in first aproximation" I suppose you refer to some Taylor-like development and you show me only the first two terms.

3-None of these assumptions prevent my experiment from working. Since it does not depend on height, at first aproximation, It doesn't matter how high is my resonant cavity.

4-Since I only have the model of the north-south firing I'll have to do the experiment in these direction.

Am I wrong?

I agree with all your conclusions, except number 3.

I agree with all your conclusions, except number 3.

Why?

I agree with all your conclusions, except number 3.

If I understand the proposed experiment correctly, then the light only needs to bend enough to not hit the photo-detector.  We are talking probably 1mm or less.

Quote from: parsec on March 22, 2010, 05:56:37 AM

I agree with all your conclusions, except number 3.

If I understand the proposed experiment correctly, then the light only needs to bend enough to not hit the photo-detector.  We are talking probably 1mm or less.

Notice that the first two terms in the formula just describe straight propagation of the light ray. So, the third term (quadratic in x) gives the deflection from straight line propagation. Taking alpha = 0, we get that term (Deltay) to be equal to 1 mm, if x is equal to:

x = Sqrt[4*L*Deltay/Pi] = Sqrt[4*10⁷ m*10^-3 m/3.142] = 1.27*10² m = 127 m

Quote from: markjo on March 22, 2010, 06:18:31 AM

Quote from: parsec on March 22, 2010, 05:56:37 AM

I agree with all your conclusions, except number 3.

If I understand the proposed experiment correctly, then the light only needs to bend enough to not hit the photo-detector.  We are talking probably 1mm or less.

Notice that the first two terms in the formula just describe straight propagation of the light ray. So, the third term (quadratic in x) gives the deflection from straight line propagation. Taking alpha = 0, we get that term (Deltay) to be equal to 1 mm, if x is equal to:

x = Sqrt[4*L*Deltay/Pi] = Sqrt[4*10⁷ m*10^-3 m/3.142] = 1.27*10² m = 127 m

I'll get this as your answer to my question ("WHY?")

-->Inside the resonant cavity a photon may bounce millions of times before exiting trough the semi-transparent mirror. Since the cavity legth is over some cm, light really travels far distances inside the cavity. If, according to the formula, every 127m light is 1mm higher, there would be noticeable loss of bright in my experiment since the mirrors only have a radius over 1 cm or even less (I don't know the numbers exactly, I'll look)

Also, markjo, this is not the way my experiment works. The more powerful is BL the less light exits the resonant cavity.

No, bouncing off the light ray has nothing to do with my calculation. We have been through that. My calculation refers to markjo's proposed scheme. It is not what your suggested. Your experiment won't work since the mirrors were aligned parallel by implicitly assuming straight line propagation (see Fabry-Perrot etalon). So, in essence, your experiment will look for misalignment of aligned mirrors.

What about shining a laser light from the bottom floor of a building and walking away so you can no longer see the bottom of the building?

If light bends below the ground to make the horizon, why would you still be able to see a laser light above your head?

Hey parsec if you're so wise about the whole bendy light thing, how it works and how it doesn't, then why don't you tell us how to test it?

... because someone did test it, right?