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Re: The Fatal Achilles Heel of Einstein's Relativity

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  Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-11-30 09:13 -0800
    Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-11-30 19:26 +0100
      Re: The Fatal Achilles Heel of Einstein's Relativity Odd Bodkin <bodkinodd@gmail.com> - 2016-11-30 12:51 -0600
        Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-01 14:35 +0100
          Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-01 14:42 +0100
          Re: The Fatal Achilles Heel of Einstein's Relativity Odd Bodkin <bodkinodd@gmail.com> - 2016-12-01 10:06 -0600
            Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-02 16:40 +0100
              Re: The Fatal Achilles Heel of Einstein's Relativity Odd Bodkin <bodkinodd@gmail.com> - 2016-12-02 09:52 -0600
                Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-02 18:45 +0100
              Re: The Fatal Achilles Heel of Einstein's Relativity Odd Bodkin <bodkinodd@gmail.com> - 2016-12-02 10:37 -0600
      Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-11-30 11:03 -0800
        Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-01 14:21 +0100
          Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-01 15:25 +0100
            Re: The Fatal Achilles Heel of Einstein's Relativity Julian Alewine <lliian@lliianenpo.org> - 2016-12-01 14:31 +0000
          Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-12-02 11:15 -0800
            Re: The Fatal Achilles Heel of Einstein's Relativity Gary Harnagel <hitlong@yahoo.com> - 2016-12-02 12:17 -0800
              Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-12-03 09:47 -0800
                Re: The Fatal Achilles Heel of Einstein's Relativity Gary Harnagel <hitlong@yahoo.com> - 2016-12-03 09:56 -0800
                  Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-12-03 10:49 -0800
                    Re: The Fatal Achilles Heel of Einstein's Relativity Gary Harnagel <hitlong@yahoo.com> - 2016-12-03 10:58 -0800
                      Re: The Fatal Achilles Heel of Einstein's Relativity alsor@interia.pl - 2016-12-03 11:31 -0800
            Re: The Fatal Achilles Heel of Einstein's Relativity Thomas 'PointedEars' Lahn <PointedEars@web.de> - 2016-12-02 22:18 +0100
        Re: The Fatal Achilles Heel of Einstein's Relativity moroney@world.std.spaamtrap.com (Michael Moroney) - 2016-12-01 17:30 +0000

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#400343 — Re: The Fatal Achilles Heel of Einstein's Relativity

W dniu niedziela, 27 listopada 2016 23:42:55 UTC+1 użytkownik Thomas 'PointedEars' Lahn

> It is not the rotation period of the stars (either you have no idea what 
> “rotation period” means or your understanding of English is utterly 
> insufficient or defective).  It is the rotation period of _Earth_ (as you 
> have said yourself), and it is relevant in the sideshow that *you* started 
> (where you hilariously attempted to calculate the orbital speed of a body
> by dividing the orbit’s *radius* by the rotation period of the central 
> body).

In the relativity stupid model it's impossible to distinguise what is moving...
that is just the key of the stupidity: the so-called relativity principle,
and/or constancy of the lightspeed.

In the reality the motion of the sun is just an observed fact.
The case of the motion is quite irrelevant,
esspecialy in the relativity - a caseless model!

Only in the classical science the case is obvious,
and discovered long time ago by Copernicus, Newton and many others.

> > Thus:
> > […]
> > sqrt[(1-v)/(1+v)] = sqrt[(1-vr)/(1-vs) * (1+vs)/(1+vr)]
> > 
> > then: v = (r-s)/(1-s*r);
> > 
> > so, the relativistic formula gives wrong speed, […]

Of course.
The relativity provides a speed: (s-r) / (1-r*s),
thus wrong, because the relative speed is: v = s-r simply.

Relativist introduces an error: r*s/c^2,
so, for small relative speed: r ~= s, what means: v << r;
the error is about: r^2/c^2;

for example:
r = 300 km/s and s = 310 km/s =>
then the relative speed is:
v = 10km/s of course;

but the relativity provides:
v' = v / 1 - rs =~ v (1 + r^2) = v(1 + 1e-6) = 10 km/s + 10mm/s.

so, the error of the relativistic approximation is evident:
dv = 10mm/s; or dv/v = 1e-6;

which is perfectly consistent with the flyby anomalies,
any many other anomalies.

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#400357

alsor@interia.pl wrote:

> […] Thomas 'PointedEars' Lahn
>> It is not the rotation period of the stars (either you have no idea what
>> “rotation period” means or your understanding of English is utterly
>> insufficient or defective).  It is the rotation period of _Earth_ (as you
>> have said yourself), and it is relevant in the sideshow that *you*
>> started (where you hilariously attempted to calculate the orbital speed
>> of a body by dividing the orbit’s *radius* by the rotation period of the
>> central body).

Your quotation has nothing to do with your reply.  Learn to quote.
 
> In the relativity stupid model it's impossible to distinguise what is
> moving...

You probably mean “_distinguish_”, and it is not stupid.  Whether your car 
hits the truck, standing on the road, at e.g. 50 km∕h, or the truck hits
you, standing on the road, at that speed, makes the same damage to personnel 
and material.

And when both of you are moving towards each other, your relative speed is 
higher, and the damage from the collision will be even greater.  Your speeds 
will approximately add, but only *approximately* because your speeds are so 
low compared to the speed of light.  When one or both of you on a collision 
course would travel close to or at the speed of light, your speeds would 
_not_ simply add up, and you could visibly see each other blueshifted 
instead (among other relativistic effects).

<https://en.wikipedia.org/wiki/Velocity-addition_formula>

> that is just the key of the stupidity:

That you do not understand it does not make it stupid.

> [misconceptions]

>> > Thus:
>> > […]
>> > sqrt[(1-v)/(1+v)] = sqrt[(1-vr)/(1-vs) * (1+vs)/(1+vr)]
>> > 
>> > then: v = (r-s)/(1-s*r);
>> > 
>> > so, the relativistic formula gives wrong speed, […]
> 
> Of course.
> The relativity provides a speed: (s-r) / (1-r*s),

Repeating nonsense (and babble it to yourself) does not make it true.

> thus wrong, […]

Ex falso quodlibet.

And you have still not considered or answered the very simple questions 
posed to you.

-- 
PointedEars

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

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#400361

On 11/30/2016 12:26 PM, Thomas 'PointedEars' Lahn wrote:
> alsor@interia.pl wrote:
>
>> In the relativity stupid model it's impossible to distinguise what is
>> moving...
>
>> that is just the key of the stupidity:
>
> That you do not understand it does not make it stupid.
>

I think it's worth asking our Polish friend about taking a flight due 
west out of Oslo, Norway, with the ground speed of the plane being 550 
mph. Is it the plane or the ground that is moving? Certainly the plane 
is moving with respect to the ground, but is it the plane or the ground 
moving that is responsible for that relative motion. If it is the plane 
that's moving, then why when the pilot looks up in the sky, the sun (and 
the other stars) are not progressing across the sky?


-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#400450

Odd Bodkin wrote:

> I think it's worth asking our Polish friend about taking a flight due
> west out of Oslo, Norway, with the ground speed of the plane being 550
> mph. Is it the plane or the ground that is moving? Certainly the plane
> is moving with respect to the ground, but is it the plane or the ground
> moving that is responsible for that relative motion. If it is the plane
> that's moving, then why when the pilot looks up in the sky, the sun (and
> the other stars) are not progressing across the sky?

Why 550 mph?
 
-- 
PointedEars

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

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#400451

Thomas 'PointedEars' Lahn wrote:

> Odd Bodkin wrote:
>> I think it's worth asking our Polish friend about taking a flight due
>> west out of Oslo, Norway, with the ground speed of the plane being 550
>> mph. Is it the plane or the ground that is moving? Certainly the plane
>> is moving with respect to the ground, but is it the plane or the ground
>> moving that is responsible for that relative motion. If it is the plane
>> that's moving, then why when the pilot looks up in the sky, the sun (and
>> the other stars) are not progressing across the sky?
> 
> Why 550 mph?

Ah, Oslo :)

-- 
PointedEars

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

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#400479

On 12/1/2016 7:35 AM, Thomas 'PointedEars' Lahn wrote:
> Odd Bodkin wrote:
>
>> I think it's worth asking our Polish friend about taking a flight due
>> west out of Oslo, Norway, with the ground speed of the plane being 550
>> mph. Is it the plane or the ground that is moving? Certainly the plane
>> is moving with respect to the ground, but is it the plane or the ground
>> moving that is responsible for that relative motion. If it is the plane
>> that's moving, then why when the pilot looks up in the sky, the sun (and
>> the other stars) are not progressing across the sky?
>
> Why 550 mph?
>
>

1. It's a realistic airline travel speed.
2. At the latitude of Oslo, it matches the surface rotational speed of 
the Earth.

-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#400583

Odd Bodkin wrote:

> On 12/1/2016 7:35 AM, Thomas 'PointedEars' Lahn wrote:
>> Odd Bodkin wrote:
>>> I think it's worth asking our Polish friend about taking a flight due
>>> west out of Oslo, Norway, with the ground speed of the plane being 550
>>> mph. Is it the plane or the ground that is moving? Certainly the plane
>>> is moving with respect to the ground, but is it the plane or the ground
>>> moving that is responsible for that relative motion. If it is the plane
>>> that's moving, then why when the pilot looks up in the sky, the sun (and
>>> the other stars) are not progressing across the sky?
>> Why 550 mph?
> 
> 1. It's a realistic airline travel speed.
> 2. At the latitude of Oslo, it matches the surface rotational speed of
> the Earth.

(See my follow-up several hours before yours.)

However, according to my calculations, it exceeds the latter.

The equatorial rotational speed of Terra is ca. v_eq = 1040.3955 mph 
(Cox, 2000).  The latitude L of Oslo, Norway, is ca. 59.95° N (GeoHack).

The rotational speed v_lat of Terra at latitude L is approximately

  v_lat = cos(L) × v_eq

        = cos(59.95°) × 1040.3955 mph

        ≈ 520.98 mph.

So your airplane travels 29.02 mph faster than the Earth rotates at that 
latitude.
 
-- 
PointedEars

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

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#400584

On 12/2/2016 9:40 AM, Thomas 'PointedEars' Lahn wrote:
> Odd Bodkin wrote:
>
>> On 12/1/2016 7:35 AM, Thomas 'PointedEars' Lahn wrote:
>>> Odd Bodkin wrote:
>>>> I think it's worth asking our Polish friend about taking a flight due
>>>> west out of Oslo, Norway, with the ground speed of the plane being 550
>>>> mph. Is it the plane or the ground that is moving? Certainly the plane
>>>> is moving with respect to the ground, but is it the plane or the ground
>>>> moving that is responsible for that relative motion. If it is the plane
>>>> that's moving, then why when the pilot looks up in the sky, the sun (and
>>>> the other stars) are not progressing across the sky?
>>> Why 550 mph?
>>
>> 1. It's a realistic airline travel speed.
>> 2. At the latitude of Oslo, it matches the surface rotational speed of
>> the Earth.
>
> (See my follow-up several hours before yours.)
>
> However, according to my calculations, it exceeds the latter.
>
> The equatorial rotational speed of Terra is ca. v_eq = 1040.3955 mph
> (Cox, 2000).  The latitude L of Oslo, Norway, is ca. 59.95° N (GeoHack).
>
> The rotational speed v_lat of Terra at latitude L is approximately
>
>   v_lat = cos(L) × v_eq
>
>         = cos(59.95°) × 1040.3955 mph
>
>         ≈ 520.98 mph.
>
> So your airplane travels 29.02 mph faster than the Earth rotates at that
> latitude.
>
>

I admittedly made a number of round-offs and only looked for major 
cities in the rough vicinity. A good replacement city on the 50th 
parallel would be a nice addition.

-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#400597

Odd Bodkin wrote:

> I admittedly made a number of round-offs and only looked for major
> cities in the rough vicinity. A good replacement city on the 50th
> parallel would be a nice addition.

Take your pick :)

Vancouver, Canada:      49.193889° N, 123.184444° W (YVR)
Calgary, Canada:        51.113889° N, 114.020278° W (YYC)
Winnipeg, Canada:       49.91°     N,  97.24°     W (YWG)
London, UK:             51.505278° N,   0.055278° E (LCY)
                        51.148056° N,   0.190278  W (LGW)
                        51.4775°   N,   0.461389  W (LHR)
Brussels, Belgium:      50.901389° N,   4.484444° E (BRU)
Cologne, Germany:       50.865833° N,   7.142778° E (CGN)
Frankfurt, Germany:     50.033333° N,   8.570556° E (FRA)
Erfurt, Germany:        50.979722° N,  10.958056° E (ERF)
Leipzig, Germany:       51.423889° N,  12.236389° E (LEJ)
Dresden, Germany:       51.134444° N,  13.768056° E (DRS)
Wrocław, Poland:        51.102683° N,  16.885836° E (WRO)
Lublin, Poland:         51.236111° N,  22.715278° E (LUZ)
Kiev, Ukraine:          50.345°    N,  30.894722° E (KPB)
Saratov, Russia:        51.565°    N,  46.046667° E (RTW)


But I think you mean the 60th parallel [cos(60°) = 0.5].
Airports there are a little harder to come by :)

Anchorage, USA:         61.174361° N, 149.996361° W (ANC)
Reykjavík, Iceland:     63.985°    N,  22.605556° E (KEF)
Stockholm, Sweden:      59.587942° N,  16.627139° E (VST)
Tallinn, Estonia:       59.413317° N,  24.832844° E (TLL)
Helsinki, Finland:      60.317222° N,  24.963333° E (HEL)
St. Petersburg, Russia: 59.800292° N,  30.262503° E (LED)
Yakutsk, Russia:        62.093333° N, 129.770556° E (YKS)

(Apparently no cities or airports at or near 50° or 60° South.)

-- 
PointedEars

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

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#400587

On 12/2/2016 9:40 AM, Thomas 'PointedEars' Lahn wrote:
> Odd Bodkin wrote:
>
>> On 12/1/2016 7:35 AM, Thomas 'PointedEars' Lahn wrote:
>>> Odd Bodkin wrote:
>>>> I think it's worth asking our Polish friend about taking a flight due
>>>> west out of Oslo, Norway, with the ground speed of the plane being 550
>>>> mph. Is it the plane or the ground that is moving? Certainly the plane
>>>> is moving with respect to the ground, but is it the plane or the ground
>>>> moving that is responsible for that relative motion. If it is the plane
>>>> that's moving, then why when the pilot looks up in the sky, the sun (and
>>>> the other stars) are not progressing across the sky?
>>> Why 550 mph?
>>
>> 1. It's a realistic airline travel speed.
>> 2. At the latitude of Oslo, it matches the surface rotational speed of
>> the Earth.
>
> (See my follow-up several hours before yours.)
>
> However, according to my calculations, it exceeds the latter.
>
> The equatorial rotational speed of Terra is ca. v_eq = 1040.3955 mph
> (Cox, 2000).  The latitude L of Oslo, Norway, is ca. 59.95° N (GeoHack).
>
> The rotational speed v_lat of Terra at latitude L is approximately
>
>   v_lat = cos(L) × v_eq
>
>         = cos(59.95°) × 1040.3955 mph
>
>         ≈ 520.98 mph.
>
> So your airplane travels 29.02 mph faster than the Earth rotates at that
> latitude.
>
>

arccos(550 mph/1040 mph) = 58.1° N
Perm, Russia, or Juneau, USA, would suffice.
Both have airports.

-- 
Odd Bodkin --- maker of fine toys, tools, tables

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#400365

W dniu środa, 30 listopada 2016 19:26:02 UTC+1 użytkownik Thomas 'PointedEars' Lahn


> > In the relativity stupid model it's impossible to distinguise what is
> > moving...
> 
> You probably mean “_distinguish_”, and it is not stupid.  Whether your car 
> hits the truck, standing on the road, at e.g. 50 km∕h, or the truck hits
> you, standing on the road, at that speed, makes the same damage to personnel 
> and material.

Not necessarily.
Look at the case of a moving train colision with a car:
the car is totaly destroyed, mixed, crushed and fragmented.

Next: the car moves and hits the same train, which now stays in place.
The car simply contracts somewhat only - it not destroys, nor fragments, unlike in the earlier version.
 
> <https://en.wikipedia.org/wiki/Velocity-addition_formula>

Of course!
That is just the relativistic speed, which is wrong in the general case,
as any approximation, simplification!

> > that is just the key of the stupidity:
> 
> That you do not understand it does not make it stupid.
 
You do not understand the elementary math yet!

Simply:
there is the general classical Doppler formula: D = D(r,s);
and the second one - a relativistic: D-approx = D(v);

where: v = v/(1-rs);

The formula D(r,s) is perfect - universal, good everywere:
not only the case of the light, but in case of any other waves too,
like the sonic waves - for sonars in the water, for a sonic radar (bats), ect!


So, try to use now the relativistic radar formula for the flying bat,
and what do you get?
An error proportional to the (u/c)^2 where: u = speed of the bat, and c - speed of sonic waves in the air.

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#400448

alsor@interia.pl wrote:

> W dniu środa, 30 listopada 2016 19:26:02 UTC+1 użytkownik Thomas
> 'PointedEars' Lahn
>> > In the relativity stupid model it's impossible to distinguise what is
>> > moving...
>> 
>> You probably mean “_distinguish_”, and it is not stupid.  Whether your
>> car hits the truck, standing on the road, at e.g. 50 km∕h, or the truck
>> hits you, standing on the road, at that speed, makes the same damage to
>> personnel and material.
> 
> Not necessarily.

Yes, *necessarily*.

> Look at the case of a moving train colision with a car:
> the car is totaly destroyed, mixed, crushed and fragmented.
> 
> Next: the car moves and hits the same train, which now stays in place.
> The car simply contracts somewhat only - it not destroys, nor fragments,
> unlike in the earlier version.

Incorrect.  If the "moving" car hits the "standing" train at the same angle 
and the same speed, the same thing happens as if the "standing" car would be 
hit by a "moving" train.  Because the magnitude of the energy is the same.
Classical mechanics suffices for the calculation.

I dare you to prove me wrong by finding an *uninhabited* train and crashing 
your car into it at 100 km∕h.  If you do this, you will finally have learned 
something, and if we are very lucky, you will not be able to show your 
stupid face here again anytime soon.
  
>> <https://en.wikipedia.org/wiki/Velocity-addition_formula>
> 
> Of course!
> That is just the relativistic speed, which is wrong in the general case,

No, it is correct in the special *and* the general case.

While your idea is *wrong* in the *general case* (arbitrary speed) to begin 
with, and only *approximately* correct in the special one (speeds much lower 
than the speed of light).

> as any approximation, simplification!

Learn to read!

The *approximation* is that *if* you add/subtract the speeds *instead of* 
*also* considering a tiny Lorentz factor at low speeds – e.g.,

  γ(v = 50 km∕h) = 1∕(1 – ((50∕3.6)²/299'792'458²) ≈ 1.00000000000000107316

–, you get a result that is not *precisely* correct, but suffices for daily 
life (it does not matter in daily life, for example, if the result is off by 
0.000001 km∕h).

But when one or both of the (collision) speeds u and v of the bodies, as 
measured in the rest frame, approach the speed of light (*then* either is 
called “relativistic speed”), simply adding (or subtracting) the speeds gets 
you an *obviously* *wrong* result.  So *wrong* *as* *you* *calculated*.

In the worst case, if u = v = c, and you simply add (according to the 
*classical*, Galilei transformation)

  s = u + v = c + c = 2 c,

the result is *wrong* *by a factor of two*, because the *correct* result is 
(according to the relativistic, _Lorentz_ transformation)

  s = (u + v)∕(1 + u v∕c²) = (c + c)∕(1 + c²∕c²) = 2 c∕2 = c,

where s is the relative speed.  This corresponds to the *experimentally* 
*confirmed* assumption that *nothing* can go faster than c, in *no* frame of 
reference (which also implies that c is the same in all inertial frames of 
reference [SR postulate #2]).
 
>> > that is just the key of the stupidity:
>> That you do not understand it does not make it stupid.
>  
> You do not understand the elementary math yet!

*Your* “elementary math” is *wrong*, refuted by experiment to begin with.

It also does not make any sense, as one can clearly see below.
 
> Simply:

*Too* simple, stupid!

> […] there is the general classical Doppler formula: D = D(r,s);
> and the second one - a relativistic: D-approx = D(v);
> 
> where: v = v/(1-rs);

What a mindbogglingly stupid nonsense.  You are attempting to calculate
the value of a quantity (v) by using the very quantity that you want to 
calculate.

Divide both sides of this equation by v, and you get (assuming v ≠ 0):

  1 = 1∕(1 - rs)

which is only true if r = s = 0, or, considering your *ambiguous*, 
*unscientific* notation, if “rs” = r_s = 0.  So much for generality.

And, of course, what you have posted so far is *neither* the classical *nor* 
the relativistic "Doppler formula".  The classical equation is

  v_{observed} = v_{source} ∕ (1 ± v_{source}∕v_{wave}).

But *it does not apply to _electromagnetic_ waves such as light*!

Go *learn* some elementary math and physics, and do not come back before!

-- 
PointedEars

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

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#400461

Thomas 'PointedEars' Lahn wrote:

> The *approximation* is that *if* you add/subtract the speeds *instead of*
> *also* considering a tiny Lorentz factor at low speeds – e.g.,
> 
>   γ(v = 50 km∕h) = 1∕(1 – ((50∕3.6)²/299'792'458²) ≈

I forgot to write the square root again.  If one does –

  γ(v = 50 km∕h) = 1∕√(1 – ((50∕3.6)²/299'792'458²) ≈

– then one *does* get:

>   1.00000000000000107316

| $ echo '1/sqrt(1-(50/3.6)^2/(299792458)^2)' | bc -l
| 1.00000000000000107316

AFAIK, bc(1) is the only standard program that can calculate this precisely; 
your usual desktop calculator just gives you 1 here.

-- 
PointedEars

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

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#400463

Thomas 'PointedEars' Lahn wrote:

> | $ echo '1/sqrt(1-(50/3.6)^2/(299792458)^2)' | bc -l |
> 1.00000000000000107316
> 
> AFAIK, bc(1) is the only standard program that can calculate this
> precisely; your usual desktop calculator just gives you 1 here.

You are wrong so very much.

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#400615

W dniu czwartek, 1 grudnia 2016 14:21:34 UTC+1 użytkownik Thomas 'PointedEars' Lahn

> > Look at the case of a moving train colision with a car:
> > the car is totaly destroyed, mixed, crushed and fragmented.
> > 
> > Next: the car moves and hits the same train, which now stays in place.
> > The car simply contracts somewhat only - it not destroys, nor fragments,
> > unlike in the earlier version.
> 
> Incorrect.  If the "moving" car hits the "standing" train at the same angle 
> and the same speed, the same thing happens as if the "standing" car would be 
> hit by a "moving" train.  Because the magnitude of the energy is the same.

No, the energies of the collision are tramedously different!

1. Mv^2/2 = 100000kg v^2/2 
2. mv^2/2 = 1000 v^2/2

so, the energy of the collision is 100 times bigger in the first case,
therefore a moving train destroys completely any car.

> Classical mechanics suffices for the calculation.

You are stupid again.
Classical mechanics suffices for any calculation in the reality -
there is no anomalies in the classical science.

> I dare you to prove me wrong by finding an *uninhabited* train and crashing 
> your car into it at 100 km∕h.  If you do this, you will finally have learned 
> something, and if we are very lucky, you will not be able to show your 
> stupid face here again anytime soon.

There are milions of such proofs...
you are just a stupid student - totaly unexperienced!
   
> In the worst case, if u = v = c, and you simply add (according to the 
> *classical*, Galilei transformation)
> 
>   s = u + v = c + c = 2 c,

It's just the correct - mathematical/physical result.
You are fooled by relativity!
 
> the result is *wrong* *by a factor of two*, because the *correct* result is 
> (according to the relativistic, _Lorentz_ transformation)
> 
>   s = (u + v)∕(1 + u v∕c²) = (c + c)∕(1 + c²∕c²) = 2 c∕2 = c,

Of, course.
If you measure a speed proportionaly to the light speed (= max),
then any speed is less than 1, because 1 is just the c,
then you stupid must write unconditionaly:
c+v = c, for any v, including v = -c: c-c = c, too!

And go away idiot.

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#400625

On Friday, December 2, 2016 at 12:15:17 PM UTC-7, al...@interia.pl wrote:
>
> W dniu czwartek, 1 grudnia 2016 14:21:34 UTC+1 użytkownik Thomas 'PointedEars' Lahn
> >
> > > Look at the case of a moving train colision with a car:
> > > the car is totaly destroyed, mixed, crushed and fragmented.
> > > 
> > > Next: the car moves and hits the same train, which now stays in place.
> > > The car simply contracts somewhat only - it not destroys, nor fragments,
> > > unlike in the earlier version.
> > 
> > Incorrect.  If the "moving" car hits the "standing" train at the same angle 
> > and the same speed, the same thing happens as if the "standing" car would
> be hit by a "moving" train.  Because the magnitude of the energy is the same.
> 
> No, the energies of the collision are tramedously different!
> 
> 1. Mv^2/2 = 100000kg v^2/2 
> 2. mv^2/2 = 1000 v^2/2
> 
> so, the energy of the collision is 100 times bigger in the first case,
> therefore a moving train destroys completely any car.

Silly fool!  Kinetic energy is frame dependent.  Your calculations refer to
two different frames.  Everyone knows this ... except you.

You are stupid again.

> Classical mechanics suffices for any calculation in the reality -
> there is no anomalies in the classical science.

Which has nothing whatever to do with any kind of science, except scientology.

>[Remainder of stupid blatherings deleted for mental sanity of readers]

And go away idiot.

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#400697

W dniu piątek, 2 grudnia 2016 21:17:15 UTC+1 użytkownik Gary Harnagel

> > so, the energy of the collision is 100 times bigger in the first case,
> > therefore a moving train destroys completely any car.
> 
> Silly fool!  Kinetic energy is frame dependent.  Your calculations refer to
> two different frames.  Everyone knows this ... except you.

The collision energy is not frame-dependent at all, unfortunately.

Simply: the collision train-car is realised in the frame of the ground,
therefore a moving train destroys easily and completely any car;
why? The train has huge energy wrt the ground, on which the car
is crushed during collision, due to the giant energy of the train.

You are talking about a collision in the free space,
so there is no ground at all, thus the dissipated energy is different.
 
> Which has nothing whatever to do with any kind of science, except scientology.

You are still too stupid to conclude anything - remember that.

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#400699

On Saturday, December 3, 2016 at 10:47:16 AM UTC-7, al...@interia.pl wrote:
>
> W dniu piątek, 2 grudnia 2016 21:17:15 UTC+1 użytkownik Gary Harnagel
> >
> > > so, the energy of the collision is 100 times bigger in the first case,
> > > therefore a moving train destroys completely any car.
> > 
> > Silly fool!  Kinetic energy is frame dependent.  Your calculations refer to
> > two different frames.  Everyone knows this ... except you.
> 
> The collision energy is not frame-dependent at all, unfortunately.

Unfortunately, you are wrong ... and stupider than rocks.

> Simply: the collision train-car is realised in the frame of the ground,
> therefore a moving train destroys easily and completely any car;
> why? The train has huge energy wrt the ground, on which the car
> is crushed during collision, due to the giant energy of the train.

The ground is irrelevant.  The observer on the train sees a much different
energy.  You admitted this yourself, but here you are blathering nonsense.

> You are talking about a collision in the free space,

Not necessarily.  All that is needed is a frame, like the frame of the train.

> so there is no ground at all, thus the dissipated energy is different.

The ground is irrelevant since the car isn't attached to it.  Put some
rocks in your head.  That will improve your intelligence.

> > Which has nothing whatever to do with any kind of science, except
> > scientology.
> 
> You are still too stupid to conclude anything - remember that.

Pot, kettle, black

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#400705

W dniu sobota, 3 grudnia 2016 18:56:02 UTC+1 użytkownik Gary Harnagel napisał:
> On Saturday, December 3, 2016 at 10:47:16 AM UTC-7, al...@interia.pl wrote:
> >
> > W dniu piątek, 2 grudnia 2016 21:17:15 UTC+1 użytkownik Gary Harnagel
> > >
> > > > so, the energy of the collision is 100 times bigger in the first case,
> > > > therefore a moving train destroys completely any car.
> > > 
> > > Silly fool!  Kinetic energy is frame dependent.  Your calculations refer to
> > > two different frames.  Everyone knows this ... except you.
> > 
> > The collision energy is not frame-dependent at all, unfortunately.
> 
> Unfortunately, you are wrong ... and stupider than rocks.
> 
> > Simply: the collision train-car is realised in the frame of the ground,
> > therefore a moving train destroys easily and completely any car;
> > why? The train has huge energy wrt the ground, on which the car
> > is crushed during collision, due to the giant energy of the train.
> 
> The ground is irrelevant.  The observer on the train sees a much different
> energy.  You admitted this yourself, but here you are blathering nonsense.
> 
> > You are talking about a collision in the free space,
> 
> Not necessarily.  All that is needed is a frame, like the frame of the train.
> 
> > so there is no ground at all, thus the dissipated energy is different.
> 
> The ground is irrelevant since the car isn't attached to it.  Put some
> rocks in your head.  That will improve your intelligence.
> 
> > > Which has nothing whatever to do with any kind of science, except
> > > scientology.
> > 
> > You are still too stupid to conclude anything - remember that.
> 
> Pot, kettle, black

Oh! These stupid students again...

The car just colides mainly with the solid ground,
therefore it can be crushed even completely, due the huge energy source - of the train, which is up to 1000 times more than your naive calculations.

You sholuld try to look a litte around firstly, instead of talking
permanently the stupididity - like these relativistics PHds fantastic imbeciles.

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#400707

On Saturday, December 3, 2016 at 11:49:27 AM UTC-7, al...@interia.pl wrote:
>
> The car just colides mainly with the solid ground,

The car isn't moving wrt the ground (before the crash.  Interaction with
the ground is a secondary effect.

> therefore it can be crushed even completely, due the huge energy source
> - of the train, which is up to 1000 times more than your naive calculations.

I didn't do any calculations, rock-brain.

> You sholuld try to look a litte around firstly, instead of talking
> permanently the stupididity - like these relativistics PHds fantastic
> imbeciles.

YOU are the only imbecile running around here, Rocky.

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