Re: Was Einstein the most ignorant, immoral, stupid, dishonest plagerist,

"Stamenin" <tasko.s@xxxxxxxxxxx> wrote in message
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On Jan 3, 5:05 pm, "Jeckyl" <no...@xxxxxxxxxxx> wrote:
"Stamenin" <task...@xxxxxxxxxxx> wrote in message

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If my calculation is incorrect can you show how would be it correct
according to your opinion? I did the calculation for two systems of
coordinates and that is my result. To say that the calculation is not
corect it means nothing if you do not show your calculation.

It means your equations ar eincorrect because they are applied
incorrectly.

K--------------------K'-----------a-----b------->x,x'
Now K is a coordinate system attached to the earth.
K' is coordinate system attached to the train
ab is the rod of 1m and is moving with v' relative to K'.

Ok .. so ab is a 1m rod .. so L" = 1 (using K" to mean the iFoR of the
rod
itself)

A HoR can,t be used because for it you can't express the coordinates x
and x'.

WTF are you talking about?

In K' it would have a length of

L' = L" * sqrt(1 - v'^2/c^2) < L"

And here you use the inverse relation of the LT. So the rod should
become infinite when it is in a motion with v=c,

No .. zero .. canb;'t you understand that L' = L" * sqrt(1 - v'^2/c^2) means
L' approaches zero as v approaches c .. its pretty simply math.

and it is in accord
with yours theory
because the mass of the rod becomes infinite.

The mass doesn't change. There is a term called relativistic massm but this
is little used .. you may be thinking of that.

Do you
agree that Einstein has done a mistake here?

No

If not you have to explain this paradox now.

There is no paradox .. I don't see where you would see one

So although the rod is 1m long, in K' it is measured as < 1m
K' is moving relative to K with the speed v.
So, the length of the rod in K is
L = L' * sqrt(1 - v^2/c^2) < L' < L
So the rod is even shorter when measured from K
So what is the problem you see there?

Obviously you don't see one, as you did not answer.


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