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Re: The Song of Lt. Cdr. Data

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  Re: The Song of Lt. Cdr. Data Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2017-09-20 13:37 +0200
    Re: The Song of Lt. Cdr. Data Arindam Banerjee <banerjeeadda1234@gmail.com> - 2017-09-20 06:26 -0700
      Failures of the Galilean transformation (was: The Song of Lt. Cdr. Data) Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2017-09-20 17:23 +0200
        Re: Failures of the Galilean transformation (was: The Song of Lt. Cdr. Data) Arindam Banerjee <banerjeeadda1234@gmail.com> - 2017-09-20 16:19 -0700
          Re: Failures of the Galilean transformation Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2017-09-22 14:00 +0200
            Re: Failures of the Galilean transformation Arindam Banerjee <banerjeeadda1234@gmail.com> - 2017-09-22 05:10 -0700

#650859 — Re: The Song of Lt. Cdr. Data

Arindam Banerjee wrote:

> On Wednesday, September 13, 2017 at 2:06:56 PM UTC+10, sci.physics wrote:
>> How much energy will it take to get these spaceships to travel faster
>> than light?
>> 
>> Where does this energy come from?
> 
> If the acceleration of the ship is g or say 10m./sec^2 then from rest the
> ship will reach the speed of light

*You cannot reach the speed of light c using conventional propulsion* (i.e. 
not based on dynamic spacetime), because the kinetic energy you would need 
to generate approaches infinity.

Because in special relativity, the square of the total energy as derived 
from the norm of the 4-momentum P⃗ = (E∕c, p⃗), is:

  ||P⃗||² = −E²∕c² + p² = −m²c²

      E² = m² c⁴ + p² c²

where with 3-momentum for massive objects (m ≠ 0)

       p = γ m u

and Lorentz factor

       γ = 1∕√(1 − u²∕c²),

where u is the magnitude of 3-velocity, we find:

      E² = m²c⁴ + (γ m u)²c²
         = m²c⁴ + γ²m²u²c²
         = m²c⁴ + m²u²c²∕(1 − u²∕c²)
         = (m²c⁴ (1 − u²∕c²) + m²u²c²)∕(1 − u²∕c²)
         = (m²c⁴ − m²c²u² + m²u²c²)∕(1 − u²∕c²)
         = (m²c⁴)∕(1 − u²∕c²)
      E² = γ²m²c⁴
      E  = γ m c².

The rest energy is defined as the total energy when u = 0:

      E₀ = m c².

Since the total energy is the sum of the rest energy and kinetic energy –

       E = E₀ + Eₖ

– we find for the kinetic energy

      Eₖ = E − E₀
         = γ m c² − m c
         = m c² (γ − 1)
      Eₖ = m c² (1∕√(1 − u²∕c²) − 1)

    [the kinetic energy is zero at u = 0, proving that this is correct],

and therefore

  lim Eₖ = +∞.
  u→c

Likewise, the magnitude of the momentum that you would need to impart to a 
vehicle to for it to reach the speed of light c (see above) is

   lim p = +∞.
   u→c

Which is why *it cannot be done the conventional way*.

As you have referred to /Star Trek/:

Its creator(s), wanting to portray *science*-fiction, something that would 
be believable and entertaining to scientifically interested or even 
knowledgable people as to have some relation to current science, portraying 
a possible extension of current science, knew that (or learned it from the 
scientific advisors that they employed), and "invented" the “warp drive”: 

Inside a region of “warped” spacetime as created by the warp drive, relative 
to that “warp bubble” the starship never travels at speeds close to the 
speed of light (“thrusters” are at “station-keeping”), but the warp bubble 
can travel at any speed, as allowed by general relativity.  In effect, the 
ship would have an *apparent* speed that is faster than the speed of light.  
(This would also avoid time-dilation effects and enable interstellar 
alliances, but would not avoid dramatically useful violations of causality, 
such as with the “Picard maneuver”.)

Later, Miguel Alcubierre showed that the idea of the warp drive has a 
scientifically sound basis.  Later work by other scientists show that the 
practical difficulties, although still enormous, can be reduced:

<https://en.wikipedia.org/wiki/Alcubierre_drive>

> in c=gt or t=3*10^8/10 = 3*10^7 seconds
> That is 347 days.

Wrong.

> If the ship weighs 200,000 Kg then the force required will be 2 million
> Newtons.

No, as the magnitude of the force required is

  F = ∂p∕dt

it will be infinite, too:

  lim F = +∞
  u→c

Which is why *it cannot be done the conventional way*.

> F=LI^2 as per my derivation in IFE-7.

Then it is not even wrong.  You would probably realize that if you tried to 
publish a paper about it instead of only making YouTube videos.

> […]
> So a 100MW reactor operating for 347 days will raise a craft to light
> speed.

Not even remotely.

-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#650863

On Wednesday, September 20, 2017 at 9:37:46 PM UTC+10, Thomas 'PointedEars' Lahn wrote:
> Arindam Banerjee wrote:
> 
> > On Wednesday, September 13, 2017 at 2:06:56 PM UTC+10, sci.physics wrote:
> >> How much energy will it take to get these spaceships to travel faster
> >> than light?
> >> 
> >> Where does this energy come from?
> > 
> > If the acceleration of the ship is g or say 10m./sec^2 then from rest the
> > ship will reach the speed of light

In 347 days
At 100m/s/s 34.7 days which is quite feasible with modern technology
and with 1000m/s/s 3.47 days which should happen within 100 years
> 
> *You cannot reach the speed of light c using conventional propulsion* (i.e. 
> not based on dynamic spacetime), because the kinetic energy you would need 
> to generate approaches infinity.

Idiotic e=mcc babbling. As c depends upon the velocity of the emitter, the 
first postulate for deriving e=mcc is knocked out so the whole theory is 
absolute bunkum. As proven by the correct analysis of the MMI experiment which
I did back in 2005, and evidenced by the Doppler effect.
 
> Because in special relativity,

The greatest nonsense in science which has been the most popularly accepted
since the flat earth theory, and that preceded by the Aristotle model of the
universe...

When one junks all that as crap, then the issues that matter are:
how much resistance can be expected from the atoms in space?
what are the chances of getting hit by some solid matter like meteors?

A single hydrogen atom will give it the reverse momentum of 
1.67*10^-27 * 9*10^20 = 1.5*10^-6 MKS units which is not very significant
and that is good for there is opinion that free hydrogen atoms may be there
in deep space. 
As for getting hit by a solid object, of say mass 1 Kg the momentum will be
great, of the order of 10^21 MKS units which will be destructive.
On the other hand when travelling at 100c the chances of getting hit are small.
Simulation should help here, going by what we can know about the existence of
solid matter density in outer space.

Cheers,
Arindam Banerjee

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#650876 — Failures of the Galilean transformation (was: The Song of Lt. Cdr. Data)

[X-Post & F’up2 <news:sci.physics.relativity>]

Arindam Banerjee wrote in <news:sci.physics>:

> On Wednesday, September 20, 2017 at 9:37:46 PM UTC+10, Thomas
> 'PointedEars' Lahn wrote:
>> Arindam Banerjee wrote:
>> > On Wednesday, September 13, 2017 at 2:06:56 PM UTC+10, sci.physics
>> > wrote:
>> >> How much energy will it take to get these spaceships to travel faster
>> >> than light?
>> >> 
>> >> Where does this energy come from?
>> > 
>> > If the acceleration of the ship is g or say 10m./sec^2 then from rest
>> > the ship will reach the speed of light
> 
> In 347 days

Argumentum ad nauseam.

No, it will not *ever* reach the speed of light using conventional 
propulsion.

> At 100m/s/s 34.7 days which is quite feasible with modern technology
> and with 1000m/s/s 3.47 days which should happen within 100 years

Newtonian mechanics does not work anymore when the relative speeds approach 
the speed of light (or masses are large, see GR).  For two theoretical 
objects moving at the speed of light, the error is *100 %*:

  u(v, u') = (v + u')∕(1 + v u'∕c²)    Relativistic speed composition

  u(c, c)  = (c + c)/(1 + c²/c²) = c

vs.

  v(v₁, v₂) = v₁ + v₂

  v(c,  c)  = c + c = 2 c              Galilean/Newtonian speed composition,

where u is the speed of K as observed by A, and u' is the speed of K as 
observed by B which is moving relative to A at speed v.

When the speeds are much less than the speed of light, then the 
Galilean/Newtonian speed composition also only works *approximately* as only

  lim      u(v, u') = v + u'.
  v u' → 0

<http://www.wolframalpha.com/input/?dataset=&i=plot+%7B2+v%7D,
+%7B2v%2F(1+%2B+v%5E2)%7D+for+v+%3D+0+to+1>
 
>> *You cannot reach the speed of light c using conventional propulsion*
>> (i.e. not based on dynamic spacetime), because the kinetic energy you
>> would need to generate approaches infinity.
> 
> Idiotic e=mcc babbling.

I did not write “e=mcc”.

> As c depends upon the velocity of the emitter,

It does not.  Experimentally confirmed a thousandfold, if that is even 
enough.

> the first postulate for deriving e=mcc is knocked out

Einstein defines the first postulate of special relativity as follows:

,-<http://einsteinpapers.press.princeton.edu/vol2-trans/154>
| 
| […] lead to the conjecture that not only in mechanics, but in
| electrodynamics as well, the phenomena do not have any properties 
| corresponding to the concept of absolute rest, but that in all coordinate 
| systems in which the mechanical equations are valid, also the same 
| electrodynamic and optical laws are valid, as has already been [3] shown 
| for quantities of the first order. We shall raise this conjecture [4] 
| (whose content will be called "the principle of relativity" hereafter) to 
| the status of a postulate […]

So you are claiming that not “in all coordinate systems in which the 
mechanical equations are valid, also the same electrodynamic and optical 
laws are valid”?

> so the whole theory is absolute bunkum.

If that is true, then, for example, why has the protons’ relative speed in 
the LHC not already surpassed the speed of light *by orders of magnitude*?

If the Galilean transformation would hold in all regimes, then the mass–
energy equivalence E₀ = m c² would be wrong, the protons would have no 
energy when at rest (E₀_{Galilean} = 0), so with the collision energy
(= Galilean kinetic energy) per beam of ca. 6.5 TeV [1], and a proton mass 
of ca. 938 MeV∕c² [2], a proton’s speed relative to the accelerator ring 
should be

  E_{Galilean} = m v²/2

             v = √(2 E_{Galilean}/m)

             v ≈ √((2 × 6.5 TeV)/(938 MeV∕c²) ≈ 118 c.

But it is _not_.

Instead, it is *measured* to be just

             E = γ m c²
             E = m c²/√(1 − u²/c²)
  √(1 − u²/c²) = m c²/E
    1 − u²/c²  = m²c⁴/E²
        u²/c²  = 1 − m²c⁴/E²
        u²     = c² − m²c⁶/E²
        u²     = (E²c² − m²c⁶)/E²
        u²     = c² (E² − m²c⁴)/E²
        u      = c √(E² − m²c⁴)/E
        u      = c √((6.5 TeV)² − (938 MeV∕c²)²c⁴)∕(6.5 TeV)

             u ≈ 0.9999999896 c,

*precisely* as special relativity predicts.

How do you explain that?

> As proven by the correct analysis of the MMI experiment which I did back
> in 2005, 

“MMI”?

> and evidenced by the Doppler effect.

On the contrary, the Doppler effect is striking evidence that the phase 
speed of *any* wave does _not_ change (i.e., is invariant) if the wave 
source is moving relative to the observer.  Because c = λ f = const., where 
c is phase speed, λ is wavelength, and f is frequency.  This applies even if 
you insist on pre-relativistic, even pre-Maxwellian, mechanics.

<https://en.wikipedia.org/wiki/Doppler_effect>

Conclusion: You have not the slightest clue what you are talking about.

_______
[1] <https://home.cern/about/updates/2015/05/protons-set-collide-13-tev-prepare-physics>
[2] <http://pdg.lbl.gov/2017/tables/rpp2017-sum-baryons.pdf>
-- 
PointedEars

Twitter: @PointedEars2
Please do not cc me. / Bitte keine Kopien per E-Mail.

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#650976 — Re: Failures of the Galilean transformation (was: The Song of Lt. Cdr. Data)

On Thursday, September 21, 2017 at 1:23:34 AM UTC+10, Thomas 'PointedEars' Lahn wrote:
> [X-Post & F’up2 <news:sci.physics.relativity>]
> 
> Arindam Banerjee wrote in <news:sci.physics>:
> 
> > On Wednesday, September 20, 2017 at 9:37:46 PM UTC+10, Thomas
> > 'PointedEars' Lahn wrote:
> >> Arindam Banerjee wrote:
> >> > On Wednesday, September 13, 2017 at 2:06:56 PM UTC+10, sci.physics
> >> > wrote:
> >> >> How much energy will it take to get these spaceships to travel faster
> >> >> than light?
> >> >> 
> >> >> Where does this energy come from?
> >> > 
> >> > If the acceleration of the ship is g or say 10m./sec^2 then from rest
> >> > the ship will reach the speed of light
> > 
> > In 347 days
> 
> Argumentum ad nauseam.
> 
> No, it will not *ever* reach the speed of light using conventional 
> propulsion.

True, it will need new physics based upon the fact that the Lorenz force
does not have a reaction.

Check out the following video films.
IFE-4,5,6,7,8 deal with the theory and design of the new spaceships which 
we may call vimans following ancient Indian tradition.

https://www.youtube.com/watch?v=hqBfwAClVlg
IFE - 1 Ground Experiments

https://www.youtube.com/watch?v=w9eGq4Oiv9s
IFE - 2 Experimental setups

https://www.youtube.com/watch?v=V3hC48BMrno
IFE - 3 Pendulum experiments

https://www.youtube.com/watch?v=3sSPxGsLkws
IFE - 4 Evolution of spaceship

https://www.youtube.com/watch?v=pJdM6UDPauU
IFE - 5 Hydrogen Transmission Network

https://www.youtube.com/watch?v=TUAcx7rAplc
IFE - 6 Spaceship Design

https://www.youtube.com/watch?v=H5Zbpvc3fdA
IFE - 7 Anti-Gravity

https://www.youtube.com/watch?v=VA9LUwqMhxY
IFE - 8 New Physics

- snip -

Cheers,
Arindam Banerjee

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#651221 — Re: Failures of the Galilean transformation

Arindam Banerjee wrote:

> […] Thomas 'PointedEars' Lahn wrote:
>> Arindam Banerjee wrote in <news:sci.physics>:
>> > On Wednesday, September 20, 2017 at 9:37:46 PM UTC+10, Thomas
>> > 'PointedEars' Lahn wrote:
>> >> Arindam Banerjee wrote:
>> >> > If the acceleration of the ship is g or say 10m./sec^2 then from
>> >> > rest the ship will reach the speed of light
>> > In 347 days
>> Argumentum ad nauseam.
>> 
>> No, it will not *ever* reach the speed of light using conventional
>> propulsion.
> 
> True, it will need new physics […]

I said conventional *propulsion*, _not_ conventional physics.

> based upon the fact that the Lorenz force does not have a reaction.

Not even wrong.


PointedEars
-- 
“Science is empirical: knowing the answer means nothing;
 testing your knowledge means everything.”
   —Dr. Lawrence M. Krauss, theoretical physicist,
    in “A Universe from Nothing” (2009)

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#651222 — Re: Failures of the Galilean transformation

On Friday, September 22, 2017 at 10:00:17 PM UTC+10, Thomas 'PointedEars' Lahn wrote:
> Arindam Banerjee wrote:
> 
> > […] Thomas 'PointedEars' Lahn wrote:
> >> Arindam Banerjee wrote in <news:sci.physics>:
> >> > On Wednesday, September 20, 2017 at 9:37:46 PM UTC+10, Thomas
> >> > 'PointedEars' Lahn wrote:
> >> >> Arindam Banerjee wrote:
> >> >> > If the acceleration of the ship is g or say 10m./sec^2 then from
> >> >> > rest the ship will reach the speed of light
> >> > In 347 days
> >> Argumentum ad nauseam.
> >> 
> >> No, it will not *ever* reach the speed of light using conventional
> >> propulsion.
> > 
> > True, it will need new physics […]
> 
> I said conventional *propulsion*, _not_ conventional physics.
> 
> > based upon the fact that the Lorenz force does not have a reaction.
> 
> Not even wrong.

heh
> 
> 
> PointedEars
> -- 
> “Science is empirical: knowing the answer means nothing;
>  testing your knowledge means everything.”
>    —Dr. Lawrence M. Krauss, theoretical physicist,
>     in “A Universe from Nothing” (2009)

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