1

Flat Earth Theory / Re: How do FE meteors work

« on: January 14, 2023, 05:09:25 AM »

If I may, here's 2 supplementary questions for JPJ.  I'm not a rocket scientist, but I am an aircraft engineer.  Aircraft actually do have a spirit level in the cockpit; its part of an instrument called the Turn and Slip Indicator.  It doesn't have any electronics, its just a plain and simple spirit level, and the pilot checks it to ensure that the aircraft is balanced about its longitudinal axis.

1.   Where do you think the bubble will be if the aircraft is banked 60 deg left in a level balanced turn?

2.   Where do you think the bubble will be if the aircraft is perfectly level in flight with the rudder hard over to the left?

(Hint; if you have access to any kind of computer flight sim you can try this yourself).

I don't know enough about aerodynamics to answer exactly, but in an uncoordinated turn, the bubble wouldn’t be in the center.  In a coordinated turn, where all the forces are balanced, it would.  Because...all gravitational and inertial forces are balanced. There is no lateral acceleration  relative to the planes center of gravity’ like you would feel in a car making a tight curve.

Relative to the plane, the indicator is stationary, so it responds to the same forces the plane, the pilot and passengers experience. Drinks and peanuts don’t go flying off tray tables and people don’t fall sideways out of their seats because within it’s own frame of reference, the plane is level.

A spirit level aligns itself to an equipotential surface.  Maybe the “spirit” means magic and the bubble and fluid just magically change positions.  It doesn’t really matter, because how it aligns isn’t the problem for FET.  Its the fact that a spirit level demonstrates differences in gravitational potential that according to FET shouldn’t exist.

Nah. We're waiting on your justification for assumption that the meteor is stationary relative to an inertial observer. Would you like to try again?

I have justified the relative velocity I used multiple times.  The RV of the meteor wrt to the stationary observer doesn’t have to be zero.  It could be anything from zero to the limit of -c. Either way, the relative velocity between the earth and the meteor is "ludicrous".   That is based on the meteor and the earth moving towards one another and relative velocity calculated Vac=Vab+Vbc.

If the earth’s velocity relative to b (stationary observer) is .77c and the meteor’s velocity relative to the stationary observer is 0, the relative velocity between the earth and meteor is .77c.  If the RV between the meteor and observer is .-.88c, the RV between the earth and meteor would be.-.98c.  Pick whatever value you want between 0 and -c for the RV of the meteor.

You haven’t explained why the RV shouldn’t be calculated that way or any other way to calculate it, so there is no reason to think it is wrong.

You also haven’t explained why you claim the earth and meteor are accelerating the same direction, but the meteor has a slightly less velocity if they both started at the same time, in the same direction, with the same acceleration.

All you’ve done is make vague references to “gravitation and other forces”, but there is no gravitation in SR.  It’s fundamentally inconsistent with it, that’s why there is GR.  You haven’t identified any other forces, so there is no reason to think your claim is true.

In short, nothing you have presented contradicts anything I’ve said.

2

Flat Earth Theory / Re: How do FE meteors work

« on: January 13, 2023, 05:49:31 AM »

[quoteHere's a fun challenge for you: contact your favourite spirit level manufacturer or award-winning physicist and ask the following question: would a spirit level work in a rocket accelerating through space, in a zero-gravity environment? [/quote]

The consensus is no.

https://cr4.globalspec.com/thread/6596/How-would-a-spirit-level-function-in-space

https://www.pistonheads.com/gassing/topic.asp?h=0&f=191&t=670185

https://www.thenakedscientists.com/forum/index.php?topic=42904.0

The issue is the figures you pulled out of thin air - the ones you plugged into your equation. One last chance to justify them! 

The rest of your rambling about how you don't understand the equations you googled is going unread, since you still haven't started fixing your claims.

The equations for hyperbolic motion are well known.I justified the numbers in the explanation that you left unread.  The equation for velocity  in hyperbolic motion uses proper acceleration and coordinate time (ie proper velocity).  You seem to be under the impression that coordinate acceleration should be used to find velocity in hyperbolic motion.  That is wrong.

First of all, we need to be clear what we mean by continuous acceleration at 1g.  The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket 

That is proper acceleration.

Using that equation, the velocity is .77c  for coordinate time 1.19 years and 1 year proper time. The equation for velocity uses proper acceleration and coordinate time (ie proper velocity).. 

It is the exact same result for the formula in the wiki

3

Flat Earth Theory / Re: How do FE meteors work

« on: January 11, 2023, 06:41:41 PM »

Irrelevant to the issue at hand.

How relative velocity is determined is the issue and you keep avoiding it.

Well, you can, it's just pretty redundant

It isn’t just redundant.  Its meaningless.  Its like using a word to define itself.  It’s also pointless, if you already know the relative velocity between two objects, you don’t have to find it.  It’s the very definition of circular logic.

I, and at least one other person, explained why your premise is false (it assumes that UA is the only factor, and it assumes that one body's initial velocity is huge while the other's is 0, without providing any justification for this).

You said:

There is also nothing novel about them colliding with Earth. OP proposes that something has to "cause a meteor to suddenly lose acceleration and 'fall'", but that's blatantly not the case. Both bodies are affected by UA, so their relative velocity will be unaffected by it. The meteoroid will continue to travel with its initial velocity, and will continue to be affected by other factors, like gravitation

The wiki says:

Celestial Gravitation is a part of some Flat Earth models which involve an attraction by all objects of mass on earth to the heavenly bodies. Celestial Gravitation accounts for tides and other gravimetric anomalies across the Earth's plane.

It doesn’t say anything about celestial gravitation effecting objects of mass that are not on the earth.  So what other factors might effect a meteor’s velocity? 

And I did provide justification for the velocities by explaining how relative velocity should be calculated (something you have failed to explain).  And by explaining that in the absence of gravity, objects or observers that are uniformly accelerated  who start from rest at the same time in the same inertial frame will remain stationary to one another (at least in that same inertial frame..in an accelerated frame, the distance between the objects would increase).  There would be no reason for  the earth and a meteor to meet under those circumstances.  Unless, there is some other force at work.  The only option you have offered is Gravitation, but according to the wiki, it doesn’t effect objects not on the earth.  You haven’t offered any other possible alternatives, so why wouldn’t someone assume there is no other force or factor in play?

I didn't say he was wrong. I said I now know where the factor of 2 snuck in

Incorrect - sounds like whatever "do my homework plz" snuck in a factor of 2 somewhere in there. I already told you what the result should be. Please study relativity until you can perform this calculation correctly

Yes, you did.  The result you told me it should be was not .77c. It wasn’t even .71c, which is the velocity your own equation in the wiki gives.

You don’t understand hyperbolic motion or  the difference and relationship between proper acceleration, the physical acceleration experienced by an object and that by definition is measured in a co-moving inertial frame  and coordinate acceleration, which is measured in some other inertial frame that is not co-moving.

The coordinate acceleration will decrease over time, but the proper acceleration remains constant.  If you want to maintain earth’s proper acceleration, at a constant 9.8m/s^2, you can’t use the coordinate acceleration to find the velocity..

Coordinate and proper acceleration are related by the transformation:





If you transform a coordinate acceleration less than g, you get a proper acceleration less than g and the earth wouldn’t be experiencing a constant acceleration of g.  Proper acceleration is Lorentz Invariant. If the acceleration is less than g, in one  frame it is less than g in all of them.

The bottom line is this...

1. Is constant acceleration hyperbolic motion?  Yes
2. Do you have to use the hyperbolic motion equations to find the velocity of an object in hyperbolic motion? Yes
3.  Is the equation for velocity of an object in hyperbolic motion

yes.
4.  Is the acceleration in that  equation proper or coordinate?
Proper

Last but not least
5. Do objects in hyperbolic motion eventually experience an event horizon, beyond which no signal can be sent or received?  Yes.
6. Does that mean a flat earth with constant acceleration of 9.8m/s^2 will experience an event horizon, beyond which no signal can be sent or received??
Yes


https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)#Proper_reference_frame

 However, this does not mean that a spirit level works due to differences in gravitational potential. This has been explained to you time and time again. Perhaps you'll want to read the responses you previously received.

what do you think “This allows the bubble to move to the highest point of the radius as gravity acts on the liquid inside the vial once it is level. “ means?

The curved radius “allows” the bubble to move to the highest point.  It doesn’t “cause” it to move to the highest point. That is like saying the sky causes a balloon to float away.  A curve in a glass vial doesn’t exert a force on the bubble and move it up. Gravity works on the fluid, pulling it down, which allows the bubble to move  to the highest point.  When the potential is constant, the highest point is in the middle, where the curve is. If gravity wasn’t working on the fluid as the potential changes, the bubble wouldn’t move.

A spirit level detects differences in the gravitational potential across a surface. That’s their whole purpose.  If there were never any differences in gravitational potential, they wouldn't work. They only work because those differences exist.

4

Flat Earth Theory / Re: How do FE meteors work

« on: January 10, 2023, 05:56:16 AM »

It absolutely is. You're presenting a hypothetical scenario you've invented. Everything about it is "yours".

Relative velocity between two objects can only be calculated using a third frame as a benchmark.  Once more for the people in the back...you can’t find the relative velocity between objects by using their relative velocity.  That’s common sense.

Incorrect - sounds like whatever "do my homework plz" snuck in a factor of 2 somewhere in there. I already told you what the result should be. Please study relativity until you can perform this calculation correctly.

LOL, you might want to let the internationally known Ph.D , whose written text books and published nearly 100 peer reviewed papers on math and physics that his math is wrong.  I’m sure he’d appreciate it. 

The formula isn’t some super secret.  You can find it in just about any textbook that discusses hyperbolic motion and Rindler coordinates, the coordinate system used for uniform acceleration in flat spacetime, even wikipedia.

Incorrect. I already told you why. Feel free to address it, but just restating your delirious ramblings isn't going to help your case

No you didn’t. You just said it was because the meteor’s velocity would only be marginally different without explaining why.  “You’re wrong, but I’m not going to tell you why” usually indicates one either doesn’t have an explanation or isn’t interested in honest discussion.  If I am wrong, how could explaining why “help my case”?

Well, it's just not how spirit levels work, as multiple FE'ers and RE'ers explained to you. It's also idiotic because your new argument conflates a definition of the word "level" with the mechanism responsible for the operation of a tool that happens to contain this word in its name.

What are you talking about?

Carpenter's level Definition & Meaning - Merriam-Webster
https://www.merriam-webster.com/dictionary/carpenter's level
Webcarpenter's level noun 1 : plumb level 2 : a straight bar (as of aluminum or wood) with a small spirit level embedded in it
 
Geoid=an equipotential surface.  A spirit level (aka carpenter's level  aligns itself to it.

An equipotential surface is level, a spirit level establishes if a surface is level, therefore, a spirit level can establish if a surface is equipotential (has constant gravitational potential).

Which of those statements is wrong?

I guess its a good thing none of those people actually manufacture them, maybe someone should let them know they don’t understand how their own product works

Principles
For all spirit levels, the sensitivity specification is determined by the sensitivity of the vials that are used. This is determined by the radius of curvature within the vials, which the bubble moves across. All spirit level vials and bubble vials contain this curved radius, even though it cannot always be seen. This allows the bubble to move to the highest point of the radius as gravity acts on the liquid inside the vial once it is level. The sensitivity is directly related to the radius of curvature of the vial:

https://www.leveldevelopments.com/2020/10/sensitivity-accuracy-of-spirit-level-vials

5

Flat Earth Theory / Re: How do FE meteors work

« on: January 09, 2023, 12:47:08 AM »

It absolutely is. You're presenting a hypothetical scenario you've invented. Everything about it is "yours"

Can you explain how you go about determining the relative velocity between 2 objects?

Incorrect - sounds like whatever "do my homework plz" snuck in a factor of 2 somewhere in there. I already told you what the result should be. Please study relativity until you can perform this calculation correctly.

John Baez, mathematical physicist at UC Riverside “did the homework”.

https://math.ucr.edu/home/baez/

(click on physics FAQ in the upper right corner)

In no way am I advocating FE, but that premise seems incorrect. 

In the RE case, everything on Earth is affected by the same gravitational acceleration, but things still meet; bats meet balls, cars meet pedestrians, and bullets meet victims. 

You are presuming;
1.  No other forces are involved.
2.  Initial velocities are identical. 

The relative velocity between the earth and a falling object isn’t 0.  Relative to an inertial observer, the velocity of the object would be some value, V.  Relative to an inertial observer, the velocity of the earth would be 0 (it is stationary relative to an observer).  So the relative velocity between the object and the earth would be V+0.

I’m assuming that all celestial bodies in the universe have 0 initial velocity. IOW, all are stationary relative to one another and begin accelerating at the same rate at the same time, in the same direction. That means at any given time all celestial bodies...are still stationary relative to one another.   If that was not the case, we would see distances between the “slower” and “faster” bodies change over time.

If a meteor is initially 10,000 miles above the surface of the earth and each begins accelerating in the same direction at 1g at the same same time, the meteor will always stay 10,000 miles above the surface of the earth.  Unless, as you suggest there is some other force in play.  But according to Pete, there isn’t.

 OP proposes that something has to "cause a meteor to suddenly lose acceleration and 'fall'", but that's blatantly not the case.

Yes, this is obvious to everyone but OP and pearly boy. Let me pre-warn you: this guy's threads usually go like this. He'll exclaim something absolutely idiotic like "spirit levels work because if you don't hold it level, there's a difference in gravitational force affecting two ends of the level, which in turn causes the air bubble to move",

What exactly is idiotic about it?  An equipotential surface has constant gravitational potential, and is level.  Therefore a spirit level can show whether or not the surface has constant gravitational potential.

The geoid is a hypothetical Earth surface that represents the mean sea level in the absence of winds, currents, and most tides. The geoid is a useful reference surface. It defines the horizontal everywhere and gravity acts perpendicular to it. A carpenter’s level aligns itself along the geoid and a carpenter’s plumb bob points down the vertical or perpendicular to the geoid

https://earthobservatory.nasa.gov/features/GRACE/page3.php

level surface—A surface which at every point is perpendicular to the plumb line or the direction in which gravity acts. A level surface is an equipotential surface. *the surface of a body of still water is a level surface. See leveling, water surface of the ocean – if changes caused by tides, currents, winds. atmospheric pressure, etc., are not considered – is a level surface. The surface of the geoid is a level surface

https://learncst.com/level-definitions/.

6

Flat Earth Theory / Re: How do FE meteors work

« on: January 08, 2023, 01:37:32 AM »

Okay. If that's your frame of reference then the Earth would not be accelerating at 1g after an infinitesimal length of time. So that's the correction you'll have to make to your first assumption.

It isn’t “my” frame.  It is how relative velocity is measured.  You have to use a 3rd common  reference frame that is stationary relative to at least one of the moving objects. You can’t determine the relative velocity between two objects by using the relative velocity of the same two objects.

You can use a fixed point on the earth, but you can’ t use the earth itself.

The rate of acceleration doesn’t matter, only the velocity that is produced relative to an inertial observer.

It's an absolutely terrible approximation. Under your assumptions (which are incorrect anyway, but we may as well finish going through this), that figure would be just around 0.46c.

If you want more precise numbers, after one year the velocity would be .77c.  Note that velocity is computed with acceleration relative to an inertial observer. 

First of all, we need to be clear what we mean by continuous acceleration at 1g.
The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket

 

https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

And that number is also close to the FET formula in the wiki, which is .71c So use whichever value you want for a year or either formula, but either way within a short time the velocity will reach .9c.

After all, the meteor is affected by UA just like everything else is, so its velocity relative to your observer will only be marginally different from that of the Earth.

Why would it be different all?   That’s the crux of the OP’s question.

There is also nothing novel about them colliding with Earth. OP proposes that something has to "cause a meteor to suddenly lose acceleration and 'fall'", but that's blatantly not the case. Both bodies are affected by UA, so their relative velocity will be unaffected by it. The meteoroid will continue to travel with its initial velocity, and will continue to be affected by other factors, like gravitation.

If UA effects the meteor, their relative acceleration would be zero and they would never meet.

7

Flat Earth Theory / Re: How do FE meteors work

« on: January 06, 2023, 05:21:39 AM »

Fantastic. Explain, in detail, how you arrived at your argument. Where did you get your initial figures from?

I did. You just aren’t connecting the dots.

Starting with two assumptions.  The earth has been accelerating at 1g for at least a year or so and the meteor and the earth are moving towards one another.

The basic equation for relative velocity can be stated as:

The velocity of A relative to B is the velocity of A relative to something other than B plus the velocity of B relative to something other than  of A.

I shouldn’t have to explain the bolded language, but given RET penchant for circular logic, I’m not so sure.

To find the relative velocity of two moving objects,  you have to introduce a separate frame that is assumed to be inertial, like I did with the car example.  Usually, the earth is assumed to be that frame, but we obviously can’t use it in this situation.  So we have to use a hypothetical inertial observer, in the same way that is done here.

https://www.khanacademy.org/science/physics/special-relativity/einstein-velocity-addition/v/applying-einstein-velocity-addition

He finds the relative velocity of the two moving objects by adding their velocities relative to a third inertial observer. Notice he says that “both of these velocities are in my frame of reference”. The equation could be stated as:

The velocity of A relative to C is the velocity a A relative to B plus the velocity of C relative to B.

What would be the velocity of the earth as measured by the inertial observer?  Since the observer is inertial and the earth has been accelerating for a year or more, its velocity will asymptotically approach c, but never reach it.  So .9c is a good approximation of the velocity relative to the inertial observer.

What would the velocity of the meteor be as measured by the inertial observer? I chose to use 0 because the actual value doesn’t matter.

 As long as the earth is accelerating, the earth’s velocity will continue to increase, but never reach c relative to an inertial observer.  It doesn’t matter if relative to that observer, the earth is only accelerating at .00000000000000001 m/s^2, it’s velocity will continue to increase relative to the inertial observer, just very slowly. Acceleration is the rate of change in velocity, The only way it’s velocity would stop increasing is if it stopped accelerating icompletely.

In Newtonian mechanics,  any velocity greater than .1c added to .9c will exceed c.  That’s why we have to do a Lorentz Transformation to find the relative velocity, but all that does is keep the relative velocity from exceeding c, it doesn’t keep it from increasing.

Here’s a good illustration using the same set up as the Khan Academy Video.  Note that the magnitude will be the same if viewed from either object.

http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/einvel2.html#c1

If the velocity of A is .9c (the velocity of the earth relative to the inertial observer)

and the velocity of B is -.5c (because it is going in the opposite direction), then the relative velocity is -0.96.



 if the velocity of B is -.9c, the relative velocity is -.99c



  if the velocity of B is zero, the relative velocity is -.9c



The relative velocity will always be “ludicrous”, no matter what the velocity of the meteor is relative to the 3rd inertial frame, as long as the velocity of the earth relative to the inertial frame is .9c or greater.

I hope that clears up your confusion.

8

Flat Earth Theory / Re: How do FE meteors work

« on: January 05, 2023, 03:02:24 PM »

Good attempt at deflection, but you don’t address the issue.



Do the math. Do it manually if you want. Show your work if you are confident that the relative velocity wouldn’t be “ludicrous”.

You are right that it’s basic high school math, so you shouldn’t have any problem with it. It’s harder to argue with a program, though, especially one that you don’t have to plug any formulas into and gives the step by step solution. And it only takes minutes. All you have to do is type in “relative velocity”.

*BZZZT* Oh no, your assumptions don't match the scenario you're discussing! You were talking about meteors in FET, which do not travel at 0.9c relative to the Earth. I guess you'll just have to try again

I didn’t assume that.  Relative velocity is calculated using either a rest frame or some other third frame that is assumed to be stationary wrt to the moving objects as a benchmark.  The base equation for relative velocity is

V_r=V_A+V_B

The velocities on the right hand side of the equation have to be relative to something other than the two moving objects.  You can’t use the relative velocity between two objects to determine the relative velocity between them.  That’s a tautology.  Instead, you have relate the velocities of each to a third frame.  If you want to know the relative velocity of two cars approaching head on, you add each of their velocities, relative to the road.  The relative velocity of one car going 75 mph relative to the road and another going 25  mph relative to the road would be 100 mph. Or you could assume rest frames, either way you get the same result.  That’s not even basic high school math.  It’s common sense.

The assumption I made, and the assumption that the calculations are based on is that wrt to some frame that is stationary to both , the velocity of the meteor would be 0 and the velocity of the earth would be .9c.

If you think those assumptions should be different, make your own.

9

Flat Earth Theory / Re: How do FE meteors work

« on: January 05, 2023, 01:06:31 AM »

Right. So when you were saying the number was ludicrously large, you didn't know what the number was?

 I wanted to give you the chance to figure it out yourself.

If the earth’s velocity in the rest frame of the meteor is .9c and the meteor’s velocity is 0 in the earth’s rest frame, the relative velocity would be .9c.

10

Flat Earth Theory / Re: How do FE meteors work

« on: January 04, 2023, 04:09:09 PM »

Define "that velocity".

You’d figure it doing a velocity transformation.



Tl;dr divide the lorentz equation for spatial coordinates by the equation for time coordinates, do the math and end up with

V’=V-U
      1-VU
        c2

V’=velocity measured in the moving frame
V=velocity measured in the stationary frame
U=velocity of moving frame relative to stationary frame

That’s the full derivation, but its easier to just figure the relative velocity using the rest frame of each object and go straight to the velocity addition formula.





You have to change the signs if the objects are moving towards each other, though.

https://en.wikipedia.org/wiki/Relative_velocity

11

Flat Earth Theory / Re: How do FE meteors work

« on: January 04, 2023, 03:58:14 AM »

The bigger issue, is that a meteor colliding with the earth at that velocity would cause a blast of Old Testament proportions.