# Teach a physicist (me) about FET

• 90 Replies
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#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #30 on: June 04, 2017, 04:12:42 PM »
There is no such thing as a flat earth model. To call any flat earth ideas a model is a travesty and misuse of the word. The best the could hope for would be a con-jackture.
There are plenty of models, they just don't match reality.

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #31 on: June 04, 2017, 04:23:02 PM »
DET doesn't use any math at all. JDs model is the only one as far as I know.

How did the ridiculously long equation come into being?
Pretty simply.
You map the surface of a sphere onto a flat circle, invert that mapping and then figure out the distance between them.

One of the big issues I have with it is it is just 2D, not 3D, there is no z.

The cool part is looking at the distance between (1,0) and (-1,0):
It boils down to
acos(sin(atan(0))*sin(atan(0))+cos(atan(0))*cos(atan(0))*cos|atan(0)-atan(0)|)
=acos(sin(0)*sin(0)+cos(0)*cos(0)*cos(0))
=acos(0+1)
=acos(1)
=0

So 2 points on the opposite side of this flat Earth, are actually the same point.

Of course, if you try it with (0,1) and (0,-1), you get a bunch of divide by 0 and it doesn't work any more.

#### Pezevenk

• 15331
• Militant aporfyrodrakonist
##### Re: Teach a physicist (me) about FET
« Reply #32 on: June 05, 2017, 01:53:35 AM »
Is it not a bit silly to do all that just to get a flat earth that looks exactly like a round earth?
Member of the BOTD for Anti Fascism and Racism

It is not a scientific fact, it is a scientific fuck!
-Intikam

Read a bit psicology and stick your imo to where it comes from
-Intikam (again)

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #33 on: June 05, 2017, 02:42:08 AM »
DET doesn't use any math at all. JDs model is the only one as far as I know.

How did the ridiculously long equation come into being?
Pretty simply.
You map the surface of a sphere onto a flat circle, invert that mapping and then figure out the distance between them.

One of the big issues I have with it is it is just 2D, not 3D, there is no z.

The cool part is looking at the distance between (1,0) and (-1,0):
It boils down to
acos(sin(atan(0))*sin(atan(0))+cos(atan(0))*cos(atan(0))*cos|atan(0)-atan(0)|)
=acos(sin(0)*sin(0)+cos(0)*cos(0)*cos(0))
=acos(0+1)
=acos(1)
=0

So 2 points on the opposite side of this flat Earth, are actually the same point.

Of course, if you try it with (0,1) and (0,-1), you get a bunch of divide by 0 and it doesn't work any more.
Actually it works fine, you'd be taking the arctan of infinity which is defined. If you want to be rigorous, take a limit, it's all good.
But, yep, it's just 2-D, it was a bored afternoon with simplified geometry. Theoretically you could to a 3-D mapping as well, but this isn't my area, involving z would make it so, so much worse. If you want the eigenvalues of the Laplacian, call me, this much is a bit beyond me.

Is it not a bit silly to do all that just to get a flat earth that looks exactly like a round earth?
It's just an illustration of how the distances of a round world can exist on a flat. It wouldn't look exactly the same though, you'd still need to be able to explain a whole host of mechanisms once you get at all above or below the Earth's surface.

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #34 on: June 05, 2017, 03:57:00 AM »
Actually it works fine, you'd be taking the arctan of infinity which is defined. If you want to be rigorous, take a limit, it's all good.
No, it doesn't.
You end up with 1/0, not infinity.
The arctan of 1/0 is not defined.
If you tried taking limits you would realise why, there is not a single value for the limit of 1/x as x approaches 0, that is because it will vary depending upon which side you come from.

Going from a positive value of x, you get 1/x approaches +infinity and atan(1/x) approaches pi/2.
Going from a negative value of x you get 1/x approaches -infinity and atan(1/x) approaches -pi/2.

Is it not a bit silly to do all that just to get a flat earth that looks exactly like a round earth?
It's just an illustration of how the distances of a round world can exist on a flat. It wouldn't look exactly the same though, you'd still need to be able to explain a whole host of mechanisms once you get at all above or below the Earth's surface.
I think the biggest issue is how light behaves, which is fundamentally appealing to 3D.
For a spherical universe, light should travel along the surface, effectively bending with it as that is what a "straight" line would be.
But in reality, it doesn't. So not even this model matches reality.
« Last Edit: June 05, 2017, 04:01:46 AM by JackBlack »

#### rabinoz

• 26528
• Real Earth Believer
##### Re: Teach a physicist (me) about FET
« Reply #35 on: June 05, 2017, 05:36:22 AM »
I think the biggest issue is how light behaves, which is fundamentally appealing to 3D.
For a spherical universe, light should travel along the surface, effectively bending with it as that is what a "straight" line would be.
But in reality, it doesn't. So not even this model matches reality.
Reality, reality! Is that all you can think of? Jane's not bound by such trivia.
Come on play fair! That's physics and the (amost) real world. Jane only does mathematics, where you can say "Let us postulate!"
I'm stuck in engineering where things really have to work! Poor me, so limited by reality like this .

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #36 on: June 05, 2017, 06:34:37 AM »
No, it doesn't.
You end up with 1/0, not infinity.
The arctan of 1/0 is not defined.
If you tried taking limits you would realise why, there is not a single value for the limit of 1/x as x approaches 0, that is because it will vary depending upon which side you come from.

Going from a positive value of x, you get 1/x approaches +infinity and atan(1/x) approaches pi/2.
Going from a negative value of x you get 1/x approaches -infinity and atan(1/x) approaches -pi/2.

if you want to get technical there's no unique answer to any arctan. We take pi/2 by convention, and taking both left and right limits pretty clearly both give that, just look at the graph. What you're saying is just wrong.

Quote
I think the biggest issue is how light behaves, which is fundamentally appealing to 3D.
For a spherical universe, light should travel along the surface, effectively bending with it as that is what a "straight" line would be.
But in reality, it doesn't. So not even this model matches reality.
Yep, that's one of the mechanisms. Theoretically you could answer that with non-Euclidean space just as easily, the shortest possible distance between two points, which light takes, could get weird then.

Reality, reality! Is that all you can think of? Jane's not bound by such trivia.
Come on play fair! That's physics and the (amost) real world. Jane only does mathematics, where you can say "Let us postulate!"
I'm stuck in engineering where things really have to work! Poor me, so limited by reality like this .
Someone should really make t-shirts saying "Welcome to FET!"

?

#### Lonegranger

• 4083
##### Re: Teach a physicist (me) about FET
« Reply #37 on: June 05, 2017, 07:23:37 AM »
No, it doesn't.
You end up with 1/0, not infinity.
The arctan of 1/0 is not defined.
If you tried taking limits you would realise why, there is not a single value for the limit of 1/x as x approaches 0, that is because it will vary depending upon which side you come from.

Going from a positive value of x, you get 1/x approaches +infinity and atan(1/x) approaches pi/2.
Going from a negative value of x you get 1/x approaches -infinity and atan(1/x) approaches -pi/2.

if you want to get technical there's no unique answer to any arctan. We take pi/2 by convention, and taking both left and right limits pretty clearly both give that, just look at the graph. What you're saying is just wrong.

Quote
I think the biggest issue is how light behaves, which is fundamentally appealing to 3D.
For a spherical universe, light should travel along the surface, effectively bending with it as that is what a "straight" line would be.
But in reality, it doesn't. So not even this model matches reality.
Yep, that's one of the mechanisms. Theoretically you could answer that with non-Euclidean space just as easily, the shortest possible distance between two points, which light takes, could get weird then.

Reality, reality! Is that all you can think of? Jane's not bound by such trivia.
Come on play fair! That's physics and the (amost) real world. Jane only does mathematics, where you can say "Let us postulate!"
I'm stuck in engineering where things really have to work! Poor me, so limited by reality like this .
Someone should really make t-shirts saying "Welcome to FET!"

Using mathematics to prove a point.....but I though you said it wasn't that type of forum.....sorry I'm confused, all these mixed messages.

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #38 on: June 05, 2017, 07:43:00 AM »
Using mathematics to prove a point.....but I though you said it wasn't that type of forum.....sorry I'm confused, all these mixed messages.
Sometimes it's like there's more than one kind of conversation that can go on. Such as conversations with you, which are humouring a troll. This specific conversation with jackblack, which is correcting a misconception. Conversations with JRowe, which are seeing what kind of thing you can get him to say next...

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#### Lonegranger

• 4083
##### Re: Teach a physicist (me) about FET
« Reply #39 on: June 05, 2017, 12:38:16 PM »
Using mathematics to prove a point.....but I though you said it wasn't that type of forum.....sorry I'm confused, all these mixed messages.
Sometimes it's like there's more than one kind of conversation that can go on. Such as conversations with you, which are humouring a troll. This specific conversation with jackblack, which is correcting a misconception. Conversations with JRowe, which are seeing what kind of thing you can get him to say next...

I think when it comes to indulging in trollish behaviour you win the race.
Resorting to attacking the person rather than the points they attempt to make is the true sign of a Troll. Why not try addressing valid points rather than indulging in mud slinging. But again mud slinging is a far easier approach rather than trying to put together a well constructed response.

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #40 on: June 05, 2017, 12:41:45 PM »
I think when it comes to indulging in trollish behaviour you win the race.
Resorting to attacking the person rather than the points they attempt to make is the true sign of a Troll. Why not try addressing valid points rather than indulging in mud slinging. But again mud slinging is a far easier approach rather than trying to put together a well constructed response.
I honestly cannot express to you how little I care about your opinion. You're one of those users that doesn't engage in any kind of actual discussion, so even the most valid points in the world would be wasted when delivered to you. You just suck all the possible enjoyment out of this site. I'm not wasting energy on a troll, so this is all you get. When someone's having an interesting discussion, like jackblack is here or tomato, then I'll say more. I just really don't see the point in putting thought in to a reply to you.
Have fun.

#### rabinoz

• 26528
• Real Earth Believer
##### Re: Teach a physicist (me) about FET
« Reply #41 on: June 05, 2017, 01:27:18 PM »
Someone should really make t-shirts saying "Welcome to FET!"
Maybe at the portals signs should be erected saying:
Warning:
and other hazardous items
in the provided receptical,
You are now leaving reality,

Welcome to FET!

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #42 on: June 05, 2017, 02:41:01 PM »

if you want to get technical there's no unique answer to any arctan. We take pi/2 by convention, and taking both left and right limits pretty clearly both give that, just look at the graph. What you're saying is just wrong.
No. If you want to get technical, there is a single unique answer for every arctan.
That is how the function is defined.
This is a limitation of inverse functions, sometimes they will lose values.

Just because you have y=f(x), where 2 values of x can give the same y DOES NOT MEAN if you have y=f-1(x) that a single value of x will give 2 (or more) values of y.

Inverse functions lose a lot of the potential mapping between functions, and this applies to lots of things.

(note: ignoring complex numbers)
For y=tan(x), the domain is the real numbers, and the range is the real numbers.
But for y=atan(x), the domain is the real numbers and the range is -pi/2 to pi/2, non-inclusive.

Similarly, for cos(x), the domain is the real numbers and the range is -1 to 1.
But for acos(x), the domain is -1 to 1 and the range is 0 to pi.
For sin(x), the domain is the real numbers and the range is -1 to 1.
For asin(x), the domain is -1 to 1 and the range is again -pi/2 to pi/2.

The same happens with other (non trig) functions as well.
For x^2, the domain is real and the range is >=0.
But for the inverse, sqrt(x), the domain and range is >=0.

So no, just because tan(x) allows 2 values of x to give the same y value, that doesn't mean that atan(x) can give to different y values.

They are rigorously defined functions.

Quote
I think the biggest issue is how light behaves, which is fundamentally appealing to 3D.
For a spherical universe, light should travel along the surface, effectively bending with it as that is what a "straight" line would be.
But in reality, it doesn't. So not even this model matches reality.
Yep, that's one of the mechanisms. Theoretically you could answer that with non-Euclidean space just as easily, the shortest possible distance between two points, which light takes, could get weird then.
No, you can't. That is kind of the point and a massive problem with this model.

If you make your Earth flat by using non-Euclidean space, such that a straight line goes straight along the surface of Earth, then light will follow that straight line.

That would mean if you are at the south pole, and projected a light due north (so basically any direction), the light would travel along a straight line, going straight to the north pole, at which point it starts heading back south and then goes back to the south pole.

Effectively, light shone parallel to the surface of Earth would make a great circle.

That is one way we can know that Earth is not flat in non-flat space.

Light doesn't travel the shortest possible distance between 2 points. It travels in a "straight" line, which for spherical geometry is a great circle. That can allow it to travel some of the longest distances (without turning) between 2 points, such as going pretty much all the way around the sphere.

But don't worry, there are other consequences of it as well.

For example, the intensity of various things as a function of distance, which appears in lots of laws and has been experimentally verified.
These follow inverse square laws, where I=k/r^2, where k is a bunch of other things, like GMm.
These laws only hold in flat space (or basically flat space).
This is because of their derivation.

The flux, which is I*A is constant, regardless of r.
So that means I0*A0=I1*A1.
And thus I1=I0*A0/A1.
And the area, for a spherically symetric flux (the surface of a sphere), is given by:
A=(4)*pi*r^2.

Thus I1=I0*(4)*pi*r0^2/((4)*pi*r1^2)
=I0*r0^2/r1^2

Where I0*r0^2 becomes that constant k.

Note that this is a result of flat space, that the area of a sphere, in flat space, is 4*pi*r^2

In non-flat space, this no longer holds.

But as we are restricted to 2D, we need to use a formula for a circularly symmetric one, which is I=k/r.

Again, this holds experimentally, but in our non-flat geometry, it wouldn't.
If you take a light which is shone out equally in a circle and put it on the south pole, it would fade with distance, until the equator at which point it increases in intensity.
Also, it would not follow a 1/r law.

Instead it would be based upon the area of a ring on this non-flat geometry.

This spherical geometry has a radius of R, and thus the length from a point to some other point (l), which subtends an angle alpha would be given by:
R*alpha=l.
The radius of the ring (r) would be given by sin(alpha)=r/R
And thus the circumference of the ring would be given by 2*pi*r=2*pi*sin(alpha)*R=2*pi*sin(l/R)*R

And thus instead of getting a 1/r law, you will get a 1/sin(l/k).

Reality, reality! Is that all you can think of? Jane's not bound by such trivia.
Come on play fair! That's physics and the (amost) real world. Jane only does mathematics, where you can say "Let us postulate!"
I'm stuck in engineering where things really have to work! Poor me, so limited by reality like this .
Someone should really make t-shirts saying "Welcome to FET!"
It's not a theory.

?

#### Aries

• 76
• And evolved monkeys can laugh!
##### Re: Teach a physicist (me) about FET
« Reply #43 on: June 05, 2017, 02:51:58 PM »
Dear Sir,

As a physicist I'm am convinced that you can answer some of my questions. Please take in consideration the fact that I'm not a physicist and English is not my first language so I might make some mistake but I would greatly appreciate if you could ignore these mistakes and take in consideration the essence of my queries.

My question:

1. Why in all the pictures presented by NASA and other spatial agencies the Earth is perfectly round but in geography books the Earth is an oblate spheroid? I have personally watch Apollo mission on the moon from NASA website and even in the video presented the Earth looks perfectly round and about the size of the moon, should'n be about 4 times bigger than the moon? Also I've watch another video presented by NASA from Hubble Telescope and in the video is showing the moon crossing the face of the Earth, but the question here is why when viewed from the moon(384,400 km) the earth has moon's size but when viewed from Hubble (a million miles away) the Earth is bigger and of course perfectly round?

3. '' To make one complete rotation in 24 hours, a point near the equator of the Earth must move at close to 1000 miles per hour (1600 km/hr). The speed gets less as you move north, but it's still a good clip throughout the United States. Because gravity holds us tight to the surface of our planet, we move with the Earth and don't notice its rotation in everyday life.'' Source: https://astrosociety.org/edu/publications/tnl/71/howfast.html.
What I understand from here is that  if I place two cities, A in the North and B at equator on the imaginary latitude the city in the north A will move slower comparing with the city B at equator? How is this possible? City A moving slower means city B will be much ahead? If the speed is decreasing does it mean that at the poles the speed is 0? When I look at a bicycle's wheel the center of the wheel appears to move faster than the tire. I think I've lost it here!

4. The Earth is surrounded by an immense vacuum and other stars, planets etc. Our planet is surrounded by atmosphere (air, gas) and gases are expandable which means they will fill any available space. We also see the smoke from furnace going up, so the smoke goes against gravity. My question is why the vacuum stopped absorbing us? Are we loosing air?

Any help would be much appreciated, thank you!

Flori Mltr

#### Junker

• 3796
##### Re: Teach a physicist (me) about FET
« Reply #44 on: June 05, 2017, 03:12:28 PM »
I think when it comes to indulging in trollish behaviour you win the race.
Resorting to attacking the person rather than the points they attempt to make is the true sign of a Troll. Why not try addressing valid points rather than indulging in mud slinging. But again mud slinging is a far easier approach rather than trying to put together a well constructed response.
I honestly cannot express to you how little I care about your opinion. You're one of those users that doesn't engage in any kind of actual discussion, so even the most valid points in the world would be wasted when delivered to you. You just suck all the possible enjoyment out of this site. I'm not wasting energy on a troll, so this is all you get. When someone's having an interesting discussion, like jackblack is here or tomato, then I'll say more. I just really don't see the point in putting thought in to a reply to you.
Have fun.

Nice to see Longdanger still trolling and not contributing in the slightest. He un-ironically accused me of actually being you not that long ago...

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #45 on: June 05, 2017, 03:50:34 PM »
They are rigorously defined functions.
I have an MSc, I really don't need you to explain maths to me. Like I said, by convention we take pi/2 because we usually just work on set intervals (either -pi/2 to pi/2, or 0 to pi, generally). All you're doing is repeating what I said to you, with different phrasing, and none of it's important because all I was doing was pointing out that your claim that arctan(1/x) taking the limit as x tends to zero was undefined, was false.
You just get onto discussions of shorthand, partial inverses, practicalities in various disciplines... None of which matters to what's actually under discussion.

Quote
Light doesn't travel the shortest possible distance between 2 points. It travels in a "straight" line, which for spherical geometry is a great circle. That can allow it to travel some of the longest distances (without turning) between 2 points, such as going pretty much all the way around the sphere.
No, it's the shortest distance between two points. The reason for that gets into quantum physics, my basic understanding is essentially photons take every possible path between two points, but the light waves can interfere with themselves; essentially there's a light wave that 'cancels' another wave, and they can all be paired off that way except for the light that takes the unique shortest path. That's probably an extreme simplification, but it's how I learnt it.
Quantum physics is weird. But, yep, in Euclidean space the shortest path will always be a straight line. The moment you get non-Euclidean that changes.
If you shine a light, it takes the shortest path. This would not necessarily be a straight line, it just is in all but the most unintuitive of spaces. Don't rely on instinct. It's the same principle that makes light curve around large masses; we don't live in Euclidean space, but rather Minkowski space. The geodesic, the shortest distance between two points, would be that curve.
The whole point of non-Euclidean space is that you can basically make it do whatever you want. In this case, you'd just do something to make distances shorter at lower altitudes. Multiply the metric through by some increasing function of z (or z2), say.

Quote
It's not a theory.
You must be fun at parties.

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #46 on: June 05, 2017, 04:00:30 PM »
1. Why in all the pictures presented by NASA and other spatial agencies the Earth is perfectly round but in geography books the Earth is an oblate spheroid? I have personally watch Apollo mission on the moon from NASA website and even in the video presented the Earth looks perfectly round and about the size of the moon, should'n be about 4 times bigger than the moon? Also I've watch another video presented by NASA from Hubble Telescope and in the video is showing the moon crossing the face of the Earth, but the question here is why when viewed from the moon(384,400 km) the earth has moon's size but when viewed from Hubble (a million miles away) the Earth is bigger and of course perfectly round?
Firstly, it isn't.
The equatorial radius (half the "width" of Earth) is roughly 6378.1 km.
The polar radius (half the "height" of Earth), is roughly 6356.8 km.

That means Earth's "width" is roughly 100.34% of its height.

This is a very small difference, so without actually measuring, it will look round.
This image shows just how close it is to a perfect circle:

How does this compare to actual pictures?
Well take this one on wikipedia from Apollo 17

Again, hard to tell, but fortunately there is an objective measurement, the number of pixels (however it can vary depending on the camera lens. It is not always equal in x and y).

So in this picture, the width of Earth is 2676 pixels (although it is a bit blurry at the edge).
The height is 2690 pixels.
So while it looks like a sphere (or circle), it actually isn't, however, as I said, the edge is blurry so there is some significant error here.
This is mainly due to not having all of Earth illuminated by the sun.

Using more modern pictures, like those of EPIC, such as this one (this one is in L1 so basically all of Earth that it can see is illuminated):
https://epic.gsfc.nasa.gov/archive/natural/2017/05/30/png/epic_1b_20170530000830_02.png
you get a width of 1642 pixels and a height of 1638 pixels.
That makes the width 100.24% of the height.(each pixel gives around 0.05%.

Also, I'm pretty sure you didn't see it from Hubble. You likely saw it from EPIC (on DSCOVR).

This all depends on the distance, and they aren't all the same size.

All you can compare is the relative apparent size.

For Apollo, they were on the moon. This makes the moon appear very big.
Meanwhile, Earth was roughly 400 000 km away. This makes Earth appear quite small, however it would appear larger than the moon would appear from Earth.

For EPIC, that is in L1, 1.5 million km from Earth.
The moon is thus roughly 1.1 million km from EPIC.

As they are now both quite some distance away, they will appear closer to the relative sizes, however the moon will still be slightly larger than the simple ratio.

If you view it on the far side, the moon will be slightly smaller.

The plane is flying through the air. Moving through the air will result in drag. This will attempt to have the plane become stationary with respect to the air.
The engines provide thrust and combat this drag.
The maximum speed of the plane is thus a speed relative to the air.

As the air (for the most part) is moving with Earth, that means the trip will take roughly the same time in each direction.

However, note that this is not always the case.
There are various Jet-streams around the world.
One is one which blows east from Perth to Sydney.

The plane still flies with a speed relative to the air, however now the air is moving relative to Earth.
This results in trips from Perth to Sydney taking much less time (roughly 4 hour 10 minutes), than the trips from Sydney to Perth (roughly 5 hours and 5 minutes).
This is akin to trying to swim in a current.

3. '' To make one complete rotation in 24 hours, a point near the equator of the Earth must move at close to 1000 miles per hour (1600 km/hr). The speed gets less as you move north, but it's still a good clip throughout the United States. Because gravity holds us tight to the surface of our planet, we move with the Earth and don't notice its rotation in everyday life.'' Source: https://astrosociety.org/edu/publications/tnl/71/howfast.html.
What I understand from here is that  if I place two cities, A in the North and B at equator on the imaginary latitude the city in the north A will move slower comparing with the city B at equator? How is this possible? City A moving slower means city B will be much ahead? If the speed is decreasing does it mean that at the poles the speed is 0? When I look at a bicycle's wheel the center of the wheel appears to move faster than the tire. I think I've lost it here!
The rotational speed remains the same.
But the linear speed does not.

To try this yourself, spin a bike wheel slowly so you can clearly observe its motion. Perhaps put some tape on it (radially) so you can watch it better rather than having a bunch of spokes).

It is a simple consequence of the constant angular speed and circumference.
If the wheel turns once in a day, that means the centre basically just spins, while the rim will have to move the entire length of the circumference of the wheel.

You can also try this with a friend. Have a friend walk in a small circle and you walk around outside them. You will need to walk much faster to keep up.

This is also why in races (at the Olympics for example), if you are confined to lanes and there is a curve, the people on the inside track start back further to make it so they travel the same distance.

4. The Earth is surrounded by an immense vacuum and other stars, planets etc. Our planet is surrounded by atmosphere (air, gas) and gases are expandable which means they will fill any available space. We also see the smoke from furnace going up, so the smoke goes against gravity. My question is why the vacuum stopped absorbing us? Are we loosing air?
It is somewhat the other way around.
Smoke isn't rising against gravity.
The denser air around it is being pulled down to gravity and displacing the lighter air in the process, just like when you fall you displace the less dense air or if you drop a steel ball in water, it displaces the less dense water.

As for why the gas doesn't escape, this is due to what causes the air pressure in the first place and how vacuums work.
Vacuums don't actually suck at all.
Instead it is the air pressure which pushes things into the vacuum.

The gas would happily act as a bunch of particles flying through space, and that is what they do in space, or close to it.

However, when near a planet (such as Earth), the gravity of the planet pulls them down.
If you just had a little bit of gas, it would just bounce of the planet and keep flying following parabolic arcs.
That is similar to what the gas quite high up does.
But instead of bouncing off Earth, it hits other gas below it and the other gas below it supports its weight.
Thus the gas below it has its parabola cut short and starts falling not only with its momentum but also that transferred from the gas above it.
This gas also hits more gas below. This gas below is holding up not only this layer of gas, but all that above it.

This process continues until you get to Earth where finally the surface of Earth (or whatever is in the way) is holding up all the gas above it.

In each layer you have a balance of forces.
You have gravity pulling the gas down. You have the gas above it weighing down on it forcing it down and you have the pressure below forcing it up.
In each layer, these forces are balanced to produce no net force.

So effectively, gravity is holding the atmosphere to the surface of Earth.

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #47 on: June 05, 2017, 04:10:21 PM »
They are rigorously defined functions.
I have an MSc, I really don't need you to explain maths to me. Like I said, by convention we take pi/2 because we usually just work on set intervals (either -pi/2 to pi/2, or 0 to pi, generally). All you're doing is repeating what I said to you, with different phrasing, and none of it's important because all I was doing was pointing out that your claim that arctan(1/x) taking the limit as x tends to zero was undefined, was false.
No. I am saying something quite different to you, specifically that your claim that my claim was false, is in fact false due to how arc tan is defined.
Arctan is not simply solving y=tan(x) for a given value of x.

By definition of the arctan function, it is undefined.

If you don't want me to explain math to you then stop spouting such crap about it.

Quote
Light doesn't travel the shortest possible distance between 2 points. It travels in a "straight" line, which for spherical geometry is a great circle. That can allow it to travel some of the longest distances (without turning) between 2 points, such as going pretty much all the way around the sphere.
No, it's the shortest distance between two points. The reason for that gets into quantum physics, my basic understanding is essentially photons take every possible path between two points, but the light waves can interfere with themselves; essentially there's a light wave that 'cancels' another wave, and they can all be paired off that way except for the light that takes the unique shortest path. That's probably an extreme simplification, but it's how I learnt it.
Again, complete crap. There is nothing special about the shortest path, except that in flat space, the shortest path is a straight line.
Instead what you would have is the various non-straight paths have multiple possible pathways to reach them and thus various phase changes which result in deconstruction interference cancelling out the wave function there.

So it isn't the shortest path that is special here, it is the straight line.

But even that doesn't hold as you can have multiple path ways to get to the straight line.

If you wish to claim such stuff, make sure you can back it up.

But, yep, in Euclidean space the shortest path will always be a straight line.
Again, other way around. A straight line will always be the shortest path in Euclidean space.

The moment you get non-Euclidean that changes.
Yes, because you have multiple straight lines between 2 points, and some can be the same length and result in the same, shortest, path.

If you shine a light, it takes the shortest path. This would not necessarily be a straight line
In spherical geometry, the shortest path between 2 points will be a great circle. Again, it is more complicated when it gets to 3D.

It's the same principle that makes light curve around large masses; we don't live in Euclidean space, but rather Minkowski space.
It is a similar principle. Space is curved, light follows a straight path in this curved space, making it appear to curve from a flat space perspective.

If it was curved to such a great extent that Earth was really flat in this curved space, then light would follow a path which followed the curve of Earth.

The whole point of non-Euclidean space is that you can basically make it do whatever you want. In this case, you'd just do something to make distances shorter at lower altitudes. Multiply the metric through by some increasing function of z (or z2), say.
No. You can't.
You can get it to do a lot of the things you want, but not necessarily everything.

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #48 on: June 05, 2017, 04:44:10 PM »
No. I am saying something quite different to you, specifically that your claim that my claim was false, is in fact false due to how arc tan is defined.
Arctan is not simply solving y=tan(x) for a given value of x.

By definition of the arctan function, it is undefined.

If you don't want me to explain math to you then stop spouting such crap about it.
Again, just take a limit. You said left and right limits were different, but that's just not true. The graph shows that easily.
The limit of tan(x) as x tends to pi/2 does vary when you come at it from the right or the left, but the same cannot be said for arctan at 1/0. If you want to focus on the definition of the arctan function, its codomain is the interval (-pi/2,pi/2] and so saying it's even possible to get -pi/2 is wrong.
I'm not 'spouting crap,' I know what I'm talking about, you don't need to get so hostile.

Quote
Again, complete crap. There is nothing special about the shortest path, except that in flat space, the shortest path is a straight line.
Instead what you would have is the various non-straight paths have multiple possible pathways to reach them and thus various phase changes which result in deconstruction interference cancelling out the wave function there.

So it isn't the shortest path that is special here, it is the straight line.

But even that doesn't hold as you can have multiple path ways to get to the straight line.

If you wish to claim such stuff, make sure you can back it up.
I'd love to see you draw multiple straight lines between two points. Here's a couple of dots for you:
.        .
I did back it up in my post. I explained why the shortest distance is what matters. There is only one straight line between these two points: a unique minimizer.

Quote

But, yep, in Euclidean space the shortest path will always be a straight line.
Again, other way around. A straight line will always be the shortest path in Euclidean space.
Not the other way around, it's an if-and-only-if. A line in Euclidean space is the shortest distance between two points if and only if it is a straight line. The implication goes both ways.
Straight lines are just a specific instance of a geodesic, geodesics are what really matter here. Straight lines are barely even defined in non-Euclidean space, except by means of geodesic.

Quote
In spherical geometry, the shortest path between 2 points will be a great circle. Again, it is more complicated when it gets to 3D.
Not that complicated. That's true, if you're limited to the surface of the sphere, and that's 2-D, you just need two parameters (longitude and latitude) to define it. In 3-D, the shortest distance is still the straight line, you just might need to burrow through the sphere itself.

Quote
It's the same principle that makes light curve around large masses; we don't live in Euclidean space, but rather Minkowski space.
It is a similar principle. Space is curved, light follows a straight path in this curved space, making it appear to curve from a flat space perspective.

If it was curved to such a great extent that Earth was really flat in this curved space, then light would follow a path which followed the curve of Earth.
Yes, light follows a 'straight' path in the curved space. But, of course, how do you define straight once you leave Euclidean space? It's just a geodesic.
Curved space is only a very simple non-Euclidean metric (it just throws a -1 into the Euclidean metric for the time dimension), there are a lot of other ways it could go. In those cases light would just follow the geodesic. Pick a clever metric and you'd easily set up geodesics to head into the ground; that'd be like light curving in Minkowski space. There is no reason for it to follow the curve of the Earth if you're smart with the metric.

#### disputeone

• Ranters
• 20375
• Or should I?
##### Re: Teach a physicist (me) about FET
« Reply #49 on: June 05, 2017, 07:46:46 PM »
It's one of Jane's affections: she likes to call the collections of unevidenced ad-hoc responses a "model".

>I resort to ad-hominems when a poster is much brighter than me and I can't debate them.

We know jim, we know, let it all out buddy.
There there.
« Last Edit: June 05, 2017, 07:49:57 PM by disputeone »
BOTD member

For all the law is fulfilled in one word, even in this.

The reason I am consistently personally attacked here.
https://www.theflatearthsociety.org/forum/index.php?topic=69306.msg1960160#msg1960160

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#### Lonegranger

• 4083
##### Re: Teach a physicist (me) about FET
« Reply #50 on: June 06, 2017, 12:11:29 AM »
No. I am saying something quite different to you, specifically that your claim that my claim was false, is in fact false due to how arc tan is defined.
Arctan is not simply solving y=tan(x) for a given value of x.

By definition of the arctan function, it is undefined.

If you don't want me to explain math to you then stop spouting such crap about it.
Again, just take a limit. You said left and right limits were different, but that's just not true. The graph shows that easily.
The limit of tan(x) as x tends to pi/2 does vary when you come at it from the right or the left, but the same cannot be said for arctan at 1/0. If you want to focus on the definition of the arctan function, its codomain is the interval (-pi/2,pi/2] and so saying it's even possible to get -pi/2 is wrong.
I'm not 'spouting crap,' I know what I'm talking about, you don't need to get so hostile.

Quote
Again, complete crap. There is nothing special about the shortest path, except that in flat space, the shortest path is a straight line.
Instead what you would have is the various non-straight paths have multiple possible pathways to reach them and thus various phase changes which result in deconstruction interference cancelling out the wave function there.

So it isn't the shortest path that is special here, it is the straight line.

But even that doesn't hold as you can have multiple path ways to get to the straight line.

If you wish to claim such stuff, make sure you can back it up.
I'd love to see you draw multiple straight lines between two points. Here's a couple of dots for you:
.        .
I did back it up in my post. I explained why the shortest distance is what matters. There is only one straight line between these two points: a unique minimizer.

Quote

But, yep, in Euclidean space the shortest path will always be a straight line.
Again, other way around. A straight line will always be the shortest path in Euclidean space.
Not the other way around, it's an if-and-only-if. A line in Euclidean space is the shortest distance between two points if and only if it is a straight line. The implication goes both ways.
Straight lines are just a specific instance of a geodesic, geodesics are what really matter here. Straight lines are barely even defined in non-Euclidean space, except by means of geodesic.

Quote
In spherical geometry, the shortest path between 2 points will be a great circle. Again, it is more complicated when it gets to 3D.
Not that complicated. That's true, if you're limited to the surface of the sphere, and that's 2-D, you just need two parameters (longitude and latitude) to define it. In 3-D, the shortest distance is still the straight line, you just might need to burrow through the sphere itself.

Quote
It's the same principle that makes light curve around large masses; we don't live in Euclidean space, but rather Minkowski space.
It is a similar principle. Space is curved, light follows a straight path in this curved space, making it appear to curve from a flat space perspective.

If it was curved to such a great extent that Earth was really flat in this curved space, then light would follow a path which followed the curve of Earth.
Yes, light follows a 'straight' path in the curved space. But, of course, how do you define straight once you leave Euclidean space? It's just a geodesic.
Curved space is only a very simple non-Euclidean metric (it just throws a -1 into the Euclidean metric for the time dimension), there are a lot of other ways it could go. In those cases light would just follow the geodesic. Pick a clever metric and you'd easily set up geodesics to head into the ground; that'd be like light curving in Minkowski space. There is no reason for it to follow the curve of the Earth if you're smart with the metric.

That you using science again...tsk tsk....you know full well that sort of thing could get you into trouble involving kicking and out and other thngs......or so one 'nasty person' by their own admission once told me!

#### JimmyTheCrab

• 10340
##### Re: Teach a physicist (me) about FET
« Reply #51 on: June 06, 2017, 01:54:19 AM »
It's one of Jane's affections: she likes to call the collections of unevidenced ad-hoc responses a "model".

>I resort to ad-hominems when a poster is much brighter than me and I can't debate them.

We know jim, we know, let it all out buddy.
There there.
It's not an ad hominem.  Before you start flinging Latin about you should be clear about what it means.
Quote from: mikeman7918
a single photon can pass through two sluts

Quote from: Chicken Fried Clucker
if Donald Trump stuck his penis in me after trying on clothes I would have that date and time burned in my head.

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #52 on: June 06, 2017, 04:48:44 AM »
Again, just take a limit. You said left and right limits were different, but that's just not true. The graph shows that easily.
The limit of tan(x)
WE ARE NOT DISCUSSING TAN(X)!!
We are discussing atan(x).

As such, your graph of tan(x) is quite useless.

If you were to plot atan(x), or even better, atan(1/x) you will find just like I said, it goes to 2 different values. There is a positive limit, as x approaches 0 from a positive number, giving pi/2 and a negative limit, as x approaches 0 from a negative value, giving -pi/2.

And yes, the graph shows it easily.

If you want to focus on the definition of the arctan function, its codomain is the interval (-pi/2,pi/2] and so saying it's even possible to get -pi/2 is wrong.
Close, but not quite.
It's range (or co-domain)  is (-pi/2,pi/2), you can't actually get either of them

I'm not 'spouting crap,' I know what I'm talking about, you don't need to get so hostile.
No, you are spouting crap.
I will stop getting so "hostile" when you stop treating me like a moron for pointing out your mistake.

I'd love to see you draw multiple straight lines between two points. Here's a couple of dots for you:
In flat space or non-flat space? If the former, I never said you could.

I did back it up in my post. I explained why the shortest distance is what matters. There is only one straight line between these two points: a unique minimizer.
No you didn't. You baselessly asserted that it is special without providing any rational justification.

Straight lines are just a specific instance of a geodesic, geodesics are what really matter here. Straight lines are barely even defined in non-Euclidean space, except by means of geodesic.
Again, you claim that, but have provided no reason to accept it.
Yes straight lines are quite ill defined in flat-space, due to them no longer being straight.
Yes, straight lines are a specific instance of a geodesic, but that is not the only way to consider them.

You need to show it is the shortest path which is important, rather than the path which merely follows the curvature of the space.

In 3-D, the shortest distance is still the straight line, you just might need to burrow through the sphere itself.
The issue is that it is no longer straight except when viewing the geometry from the perspective of flat geometry.

Yes, light follows a 'straight' path in the curved space. But, of course, how do you define straight once you leave Euclidean space? It's just a geodesic.
The simplest way is to have it merely curve with the space.

In those cases light would just follow the geodesic.
WHY? You are yet to justify this?
Why wouldn't it simply follow the curvature of the space?

Pick a clever metric and you'd easily set up geodesics to head into the ground; that'd be like light curving in Minkowski space. There is no reason for it to follow the curve of the Earth if you're smart with the metric.
There is if it follows the curve of the space rather than magically seeking out the shortest path.

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #53 on: June 06, 2017, 06:34:46 AM »
Ok, this is getting out of hand, so to simply:

Arctan. You're right, if you want to be absurdly technical, that the arctan function is limited. However, any mathematician will tell you that context is always going to be important when dealing with inverse trig functions. Some might prefer to work with the codomain 0 to pi, for example, it depends entirely on what you're doing. There is a reason mathematicians always define their variables, convention varies between people. Some use arctan as a partial inverse, some use tan-1 as the normal inverse.
You're approaching this from a simplified and useless perspective. You cannot work with inverse tan without working with tan, especially when we're taking limits beyond those the simplified inverse is meant to be applied to. For example; saying it could not attain pi/2. Normally that is true, but we're basically working with arctan(1/x). Some mathematicians balk at taking a limit and saying it's attained, others have to. What you're saying is all well and good at certain levels or in specific fields, but just isn't generally true.

Light. To repeat myself, as you're insisting I said none of this, due to the properties of quantum particles and all the probabilistic weirdness that entails, photons take every possible path when going from A to B. When light waves are out of phase however, they cancel each other out and you see nothing. All of the probable-lines are cancelled out, except for the only one that takes the unique, shortest path; the geodesic. If it takes any other path, it's essentially cancelled by another one of the probabilities. Like I said, very simplified description, but it's the basic gist.
Light does not travel in 'straight lines' because there is no concept of 'straight' that could have any impact on it. It takes the shortest path, for the reason outlined above. Light travels along geodesics (technically null-geodesics; neither time-like nor space-like) which is a well-established fact you are more than welcome to look up and I don't understand why you're denying it. In Euclidean space, this is a straight line. In non-Euclidean space, this geodesic depends entirely on the metric.
I imagine the fact that in non-Euclidean space geodesics seem 'curved' is what you mean by 'follow the curvature of the space,' but that's just not a good way to think about it because it relies too much on intuition than on what would be mathematically possible, and geodesics by definition are the equivalent of a straight line. It gives you an automatic bias for things to behave a certain way, which just isn't going to be the case.

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#### Lonegranger

• 4083
##### Re: Teach a physicist (me) about FET
« Reply #54 on: June 06, 2017, 07:01:21 AM »
Ok, this is getting out of hand, so to simply:

Arctan. You're right, if you want to be absurdly technical, that the arctan function is limited. However, any mathematician will tell you that context is always going to be important when dealing with inverse trig functions. Some might prefer to work with the codomain 0 to pi, for example, it depends entirely on what you're doing. There is a reason mathematicians always define their variables, convention varies between people. Some use arctan as a partial inverse, some use tan-1 as the normal inverse.
You're approaching this from a simplified and useless perspective. You cannot work with inverse tan without working with tan, especially when we're taking limits beyond those the simplified inverse is meant to be applied to. For example; saying it could not attain pi/2. Normally that is true, but we're basically working with arctan(1/x). Some mathematicians balk at taking a limit and saying it's attained, others have to. What you're saying is all well and good at certain levels or in specific fields, but just isn't generally true.

Light. To repeat myself, as you're insisting I said none of this, due to the properties of quantum particles and all the probabilistic weirdness that entails, photons take every possible path when going from A to B. When light waves are out of phase however, they cancel each other out and you see nothing. All of the probable-lines are cancelled out, except for the only one that takes the unique, shortest path; the geodesic. If it takes any other path, it's essentially cancelled by another one of the probabilities. Like I said, very simplified description, but it's the basic gist.
Light does not travel in 'straight lines' because there is no concept of 'straight' that could have any impact on it. It takes the shortest path, for the reason outlined above. Light travels along geodesics (technically null-geodesics; neither time-like nor space-like) which is a well-established fact you are more than welcome to look up and I don't understand why you're denying it. In Euclidean space, this is a straight line. In non-Euclidean space, this geodesic depends entirely on the metric.
I imagine the fact that in non-Euclidean space geodesics seem 'curved' is what you mean by 'follow the curvature of the space,' but that's just not a good way to think about it because it relies too much on intuition than on what would be mathematically possible, and geodesics by definition are the equivalent of a straight line. It gives you an automatic bias for things to behave a certain way, which just isn't going to be the case.

Are you going to provide a translation for all our lay readers and those who are mathematically inept?

#### JackBlack

• 17277
##### Re: Teach a physicist (me) about FET
« Reply #55 on: June 06, 2017, 03:44:15 PM »
Arctan. You're right, if you want to be absurdly technical, that the arctan function is limited.
Yes, just like I said from the start, yet you felt the need to "correct" me.

However, any mathematician will tell you that context is always going to be important
Yes, context is important, and seeing it in context makes this much simpler.
You don't actually have a flat surface here, what you have is the surface of a sphere, so the point (0,1) is the same point as (0,-1) (and (1,0) and a bunch more).
No fancy formulas needed.

This also shows a limitation of this mapping.
It isn't a 1 to 1 map.
Assuming it is a north pole centred map, the south pole, rather than being a single point is shown as infinitely many points.

You're approaching this from a simplified and useless perspective.
I'm pointing out the problems with it, that isn't simplified nor is it useless.
Typically going to more technical aspects is not simplified.
The simplified way would be treating y=atan(x) as equivalent to x=tan(y).

For example; saying it could not attain pi/2. Normally that is true, but we're basically working with arctan(1/x).
And the exact same issue applies with that, it becomes undefined at x=0 and it has 2 different limits.

Again, the solution wouldn't be to pretend it has some value.
The solution would be to recognise what it is doing and that the 2 points are actually the same as you are going half way around the sphere.

Light. To repeat myself, as you're insisting I said none of this, due to the properties of quantum particles and all the probabilistic weirdness that entails, photons take every possible path when going from A to B. When light waves are out of phase however, they cancel each other out and you see nothing. All of the probable-lines are cancelled out, except for the only one that takes the unique, shortest path; the geodesic. If it takes any other path, it's essentially cancelled by another one of the probabilities. Like I said, very simplified description, but it's the basic gist.
No, I never said that you didn't say that.
I said you never justified it.
With it taking all possible routes, including non-straight ones, that means it can deviate from straight then go back to the straight path with a slight phase difference and thus cancel out the straight path.
For example, say instead of going straight out, it goes at a right angle for 1/4 of a wavelength, turns back around and goes back for 1/4 of a wavelength, then goes back along the straight path.
It is now 1/2 a wavelength out of phase with the light along the straight line and thus it will result in deconstructive interference and thus no light is obtained.

So do you have a justification, or just hand wavy bullshit?
If you don't have a justification, I have no reason to accept it.

You also need to show that what makes it special is that it is the shortest path, rather than it being straight/not curving relative to the space.

Light does not travel in 'straight lines' because there is no concept of 'straight' that could have any impact on it.
Sure there is, not curving away from space.
There is no concept of "shortest path" that could have any impact on it.

It takes the shortest path, for the reason outlined above.
Except you didn't really outline reasons, you just baselessly asserted them, pretty much effectively saying that it does it because it does.

Light travels along geodesics
PROVE IT!

which is a well-established fact
No, it is your baseless claim.

you are more than welcome to look up
It is your claim, the burden is on you to back it up.
Don't expect me to do your homework.

I don't understand why you're denying it.
Because you are yet to justify it in any rational way and it spits in the face of the typical propagation of light.

In Euclidean space, this is a straight line. In non-Euclidean space, this geodesic depends entirely on the metric.
Again, why not the other way around?

I imagine the fact that in non-Euclidean space geodesics seem 'curved' is what you mean by 'follow the curvature of the space,'
No, because in more complex spaces the geodesics do not follow the curve, and instead can completely ignore it and act like a straight line in flat space.

geodesics by definition are the equivalent of a straight line. It gives you an automatic bias for things to behave a certain way, which just isn't going to be the case.
No, they aren't. They are equivalent to some aspects.
A straight line is a geodesic in flat space. But that does not mean geodesics are the equivalent of a straight line.
A geodesic is the shortest distance between 2 points, that is not what a straight line is defined as.
A straight line is of the form ax+by+cz+d=0. Going into non-flat space, you would use an equivalent of this.

#### Slemon

• Flat Earth Researcher
• 11690
##### Re: Teach a physicist (me) about FET
« Reply #56 on: June 06, 2017, 04:04:02 PM »
Not going into the arctan stuff because I don't want to just repeat myself all over again.

Quote
No, I never said that you didn't say that.
I said you never justified it.
With it taking all possible routes, including non-straight ones, that means it can deviate from straight then go back to the straight path with a slight phase difference and thus cancel out the straight path.
For example, say instead of going straight out, it goes at a right angle for 1/4 of a wavelength, turns back around and goes back for 1/4 of a wavelength, then goes back along the straight path.
It is now 1/2 a wavelength out of phase with the light along the straight line and thus it will result in deconstructive interference and thus no light is obtained.

So do you have a justification, or just hand wavy bullshit?
If you don't have a justification, I have no reason to accept it.
And if you'd made that argument, I would have been able to respond. Yes, light can easily deviate from the straight path and go back in an effort to cancel it, but then there'd be a light wave that in turn cancels out that. The shortest path, being unique, is the only one that basically can't be paired off. And yes, like I said, this is handwaved because it's quantum physics and I'm a mathematician, but I'm pretty confident in the basic principle.

Quote
You also need to show that what makes it special is that it is the shortest path, rather than it being straight/not curving relative to the space.
What do you think being straight is? Literally the definition of a straight line, or of a geodesic, is the shortest distance between two points. They are equivalent statements.

Quote
Light travels along geodesics
PROVE IT!

which is a well-established fact
No, it is your baseless claim.

you are more than welcome to look up
It is your claim, the burden is on you to back it up.
Don't expect me to do your homework.
http://siegelsoft.com/phy303/ch12.html
"In space, light travels along geodesics. This explains why light is bent in gravitational fields. Because gravitational fields bend space and time, the geodesics through those regions of space are also bent."
https://infinityplusonemath.wordpress.com/2017/04/29/a-mathematical-intro-to-general-relativity-part-2/
"Light travels along null geodesics."
http://faculty.washington.edu/boynton/114AWinter08/LectureNotes/Le34.pdf
"Light travels along a geodesic path—the “straightest”* line between points"
"Light travels along geodesics."

Are we done yet? Hardly homework to type 'light travels along geodesics' into google. Sure, my explanation of the underlying principle wasn't great, because I though this much was common knowledge. Geodesics are just a generalisation of a straight line; if you believe light travels in a straight line, then it travels along geodesics. The only notion of straightness that you get in a 'curved' space is the geodesic.

Quote
In Euclidean space, this is a straight line. In non-Euclidean space, this geodesic depends entirely on the metric.
Again, why not the other way around?
What the other way around? I've already explained how straight line/geodesic on Euclidean space are equivalent statements so there is no 'other way around,' and I don't see how you could want a metric to depend on the geodesic. I mean if you wanted, sure, you could hypothetically find a metric to give you a specific geodesic, but given humans haven't yet been able to warp the fundamentals of spacetime I'd say we ought to keep calculating geodesics for now.

Quote
geodesics by definition are the equivalent of a straight line. It gives you an automatic bias for things to behave a certain way, which just isn't going to be the case.
No, they aren't. They are equivalent to some aspects.
A straight line is a geodesic in flat space. But that does not mean geodesics are the equivalent of a straight line.
A geodesic is the shortest distance between 2 points, that is not what a straight line is defined as.
A straight line is of the form ax+by+cz+d=0. Going into non-flat space, you would use an equivalent of this.
That is exactly what a straight line is defined as. The reason a straight line is of that form is because they crunched the numbers for calculating a geodesic in Euclidean space and the answer was 'anything that follows that equation.'
The only 'equivalent' of that in non-flat space would be a geodesic.

#### sokarul

• 18932
• Extra Racist
##### Re: Teach a physicist (me) about FET
« Reply #57 on: June 06, 2017, 05:38:52 PM »
Not going into the arctan stuff because I don't want to just repeat myself all over again.

Quote
No, I never said that you didn't say that.
I said you never justified it.
With it taking all possible routes, including non-straight ones, that means it can deviate from straight then go back to the straight path with a slight phase difference and thus cancel out the straight path.
For example, say instead of going straight out, it goes at a right angle for 1/4 of a wavelength, turns back around and goes back for 1/4 of a wavelength, then goes back along the straight path.
It is now 1/2 a wavelength out of phase with the light along the straight line and thus it will result in deconstructive interference and thus no light is obtained.

So do you have a justification, or just hand wavy bullshit?
If you don't have a justification, I have no reason to accept it.
And if you'd made that argument, I would have been able to respond. Yes, light can easily deviate from the straight path and go back in an effort to cancel it, but then there'd be a light wave that in turn cancels out that. The shortest path, being unique, is the only one that basically can't be paired off. And yes, like I said, this is handwaved because it's quantum physics and I'm a mathematician, but I'm pretty confident in the basic principle.
I don't know if I believe that.
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You also need to show that what makes it special is that it is the shortest path, rather than it being straight/not curving relative to the space.
What do you think being straight is? Literally the definition of a straight line, or of a geodesic, is the shortest distance between two points. They are equivalent statements.
You are thinking of a line segment. A line is infinite.
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Light travels along geodesics
PROVE IT!

which is a well-established fact
No, it is your baseless claim.

you are more than welcome to look up
It is your claim, the burden is on you to back it up.
Don't expect me to do your homework.
http://siegelsoft.com/phy303/ch12.html
"In space, light travels along geodesics. This explains why light is bent in gravitational fields. Because gravitational fields bend space and time, the geodesics through those regions of space are also bent."
https://infinityplusonemath.wordpress.com/2017/04/29/a-mathematical-intro-to-general-relativity-part-2/
"Light travels along null geodesics."
http://faculty.washington.edu/boynton/114AWinter08/LectureNotes/Le34.pdf
"Light travels along a geodesic path—the “straightest”* line between points"
"Light travels along geodesics."
Yes, GR.

ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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#### fliggs

• 567
##### Re: Teach a physicist (me) about FET
« Reply #58 on: June 06, 2017, 06:44:48 PM »
Hi, I'm a self-taught physicist - obviously raised around RET - but I'm open to anything. Actually, I'd like to contribute some mathematically rigorous and testable arguments for FET that could hold water in a scientific light.

Are there any FET topics, say, that have readily verifiable results, but no one has done the math or the physics involved in them yet?

'self taught physicist'.  You couldnt discredit yourself more than by using that descriptor. A real physicist has a PhD from an real university. Someone who has attended physics courses (like me) is just someone who has a bit of knowledge.  one of the big try-hard debunkers of the Moon Landing was a 'self taught engineer' and he demonstrated what that really means - total ignorance of the discipline.

And that is before you led with your request for 'verifiable FET' models etc. The one thing that sets FET apart from any hypothesis in the entire history of the world is that it has zero verifiable modesl and absolutely no evidence of any kind.

My guess is that you took High School science and now think you are a cosmologist.

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#### fliggs

• 567
##### Re: Teach a physicist (me) about FET
« Reply #59 on: June 06, 2017, 06:55:55 PM »
Trying to make what we know already fit the flat earth concept will never fit, just ask the size 24 lady.......so why try?
Who's trying that? The blank slate's the fun part. Most models I've seen, like DET, do start anew. It's just things like the basic rules of maths that are going to have to be constant, even if they're applied to something different.

Well yes actually....as the very act of asking it is a bit of a giveaway...
I think we've stumbled into Catch-22 territory.

STOP!
You cant label the nonesense put out by JRowe as a model. It does not meet the specification required to be considered as one. You say you are a mathematician. I imagine you have read mathematical papers, have you read the proof for Fermat's Last Theorem ?
I met Sir Andrew John Wiles, to give him his full and proper title, very briefly at a conference I was luckily enough to attend in 1999, though I understood very very little of what went on, what I did understand was the meticulous work and effort that man put into proving The Theorem. The same goes for any other scientific model, it has to be robust and packed with both evidence and references to allow others that follow to replicate the work.
The rubbish put forward by JRowe should not be labelled a model as it's pure unadulterated popycock of the highest quality with not a picogram  of sense or science contained within its badly written lines.

There is no such thing as a flat earth model. To call any flat earth ideas a model is a travesty and misuse of the word. The best the could hope for would be a con-jackture.

I totally agree. It is one thing to take the time to consider an alternate theory, but to simply ascribe the term 'model' to DET or Denspressure is ridiculous. Models of all types have a form of internal consistency. They hold together and they are at least at first look, credible. None of these concepts make any sense at all. They are mere examples of rampant fantasy with no effort at logic, nevermind science. And the absolute lack of any kind of evidence - even anecdotal - is stunning. We do them - and science - a disservice by describing this utter nonsense as anything other than that - nonsense.