[Physics] Aether theory discussion

Wed Dec 21 07:12:10 CET 2016

Hello Tom.

Thank you for your reply. I visited also again your full explanation on
http://myweb.tiscali.co.uk/carmam/Hollings.html#lorentz  It's good to see
that you are struggling with the Lorentz Transformation too. I am happy to
see, that you also think it's admissible to relate the Lorentz
Transformation to mass increase. About the calculation of your rocket: I
agree on "The mass increase is therefore 0.0000015 Kg or 0.0015 gram". The
calculation is correct. Now the question is: If we apply this to the fall
acceleration on the surface of the earth: would this lead to an*
accelerated* increase? I think yes: because the outcome is an accelerated
velocity in m/s^2. Long time ago, when I started to study these phenomena I
learned that the orbital velocity of the planets in fact are accelerated
speeds, because they are composed from two directions: one centrifugal
(falling) to the earth and one straight forward. Neverteles they are
expressed in m/s and v^2 is expressed in m^2/s^2.
[image: Inline image 5]
http://www.shmoop.com/forces-motion/gravity-orbital.html
Velocity, acceleration, and force are vector quantities. In centripetal
motion, the velocity is tangential to the orbit, and perpendicular to the
force and acceleration which are in the same direction, as usual, as
related through .

The Newtonian Gravitational Constant is expressed in m^3/kg/s^2. You can
read this as a change in cubic meter over the mass in kg and in an
accelerated way. So the result is: *accelerated linear change*.
Transforming the Newtonian Gravitational Constant in to the Lorentz
Transformation of mass-energy I keep the same units m^3/kg.s^2 resulting
also in an acceleration on the surface 9,8 m/s^2. Although this is
"mind-blowing" and has many implications, this is still an option (for me
and also for Stavros Tassos and Tufail Abbas).

I think you are very well informed about the Lorentz Transformation and I
appreciate it very much if you would read Mind-blowing Gravitation. You can
see that a very small factor gamma can have big results if applied to big
masses as our planets. But also applied to our atoms they result in 9,8
m/s^2 acceleration on the surface. Here are three different derivations
related to the radius of the earth and the quantity of atoms in one line on
the radius (as a sort of educated guess):
[image: Inline image 4]

I know the chosen magnitudes are a little bit fictitious. It's just to show
how the growing of the earth could be a part of the complete "growing"
image with also particles involved. That's where we need geologist,
chemists and quantum physics.

Note: I am not good in thought experiments about the age of twins and the
influence of black holes etc. I strive to keep my reasoning as close as
possible to my environment and daily experience. I don't say that it is not
useful to extrapolate thoughts to the nearly unimaginable world, but it's
not my preferred restframe😉 I just want to know: why does something fall
to the ground if I drop it.

Best regards.

Ruud Loeffen.

On Wed, Dec 21, 2016 at 1:59 AM, carmam at tiscali.co.uk <carmam at tiscali.co.uk>
wrote:

> Ruud, From your latest post (that I have seen) :-
> "I read several opinions about the way Lorentz Transformation of
> mass-energy is useful or not. That’s why I put a question on Quora (and
> gave also an answer): Was Julian Schwinger wrong when he stated that
> increasing speed leads to increasing Mass in accordance with Lorentz
> transformation of mass-energy?
> <https://www.quora.com/Was-Julian-Schwinger-wrong-when-he-stated-that-increasing-speed-leads-to-increasing-Mass-in-accordance-with-Lorentz-transformation-of-mass-energy/answer/Ruud-Loeffen>
>  "
>
> Increased speed leads to an increase in mass as seen by the essential
> observer. That is a very important distinction, and one that is
> overlooked. It is assumed that the essential observer is at the starting
> point of the rocket (I am using a space rocket here, it could be a particle
> in a particle accelerator). In a particle accelerator, it is true that the
> increase in mass is relative to the particle accelerator and therefore
> the starting point of the particle. With a rocket, the increase in mass is
> relative to the exhaust, not to the starting point. For a full explanation
> please go to my web page at http://myweb.tiscali.co.uk/
> carmam/Hollings.html#lorentz . There is an abreviated version here :-
>
> Imagine now a space rocket, which is propelled by ejecting a small amount
> of matter (the rocket exhaust) at high speed from the rear, so imparting a
> thrust in the opposite direction. We will assume that the exhaust velocity
> is 3,000 m/s and the mass of the rocket is 30,000 Kg (very similar to
> NASA's Mercury-Redstone rockets). Now we can use the Lorentz transformation
> to find the new mass. The velocity between exhaust and rocket is 3000 m/s,
> so :-
>
> m = m0 / sqrt( 1 - ( v / c )^2)
> m = mass of rocket at velocity v as measured by the essential observer
> (Remember that Einstein's observer, properly called the essential
> observer, is always at rest relative to the motive force. In this example
> therefore, the essential observer is in the same frame as the rocket
> exhaust).
> m0 = 30,000 Kg (proper mass of rocket or rest mass when v = 0)
> v = 3,000 m/s - rocket's velocity relative to the exhaust
> c = 300,000,000 m/s
> m = 30000 / sqrt( 1 - (3000 / 3e8)^2) = 30000.0000015000000001125 Kg
>
> The mass increase is therefore 0.0000015 Kg or 0.0015 gram which is simply
> not measurable compared to 30,000 kilograms. For all intents and purposes
> the mass increase is zero. A further point to note here is that the mass
> increase is measured against the exhaust which is providing the motive
> force, and no matter what the velocity of the rocket when measured against
> its starting point (or anything else for that matter), the velocity between
> rocket and exhaust never changes, so the rocket mass is always
> 10,000.0000015 Kg (disregarding the loss of mass due to fuel used). In
> other words, the mass is fixed at 10,000.0000015 Kg for the values used
> above between rocket and exhaust, and the extra 0.0000015 Kg is an
> insignificant amount. As there is no significant mass increase with
> velocity, and certainly no accumulative mass increase, there is no
> theoretical upper limit to the velocity of the rocket.
> It therefore follows that as the mass increase is zero, m aproximates very
> closely to m0. If the acceleration is regulated to 1g for the comfort of
> the crew, the space ship can reach an enormous velocity, and time on this
> space ship will pass at exactly the same rate as back at home on earth.
> "The effects of gravity are indistinguishable from the effects of
> acceleration" [AE] (with the qualification in section 5).
>
> Tom Hollings.
>
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>

-- 
*Ruud Loeffen*
Paardestraat32
6131HC Sittard
http://www.human-DNA.org
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