Littlepig said:
yes yes, till there i agree with it, but i wnat to demonstrate it mathematicaly...
ok, my point is:
for have time dilatation we have that:
2 obeservers, A and B, A is moving at speed Va and B is still.
"A" have a light clock(2 perfect mirrors, 1 photosensor, 1 laser making a 90º to mirrors).
as light travels at same speed at diferent referencials, we have that B will see the light moving not at 90º but at a diferent angle.
A will jut see light moving up and down like nothing was hapening...
now, from trignometry, c_1=sqrt(H^2-c_2^2)
as V=d/t,
we have that:
distance that light takes, measured by B is H,
distance travel by light, measured by A is c_1;
distance travel A, measured by B, is c_2;
so as d=V \cdot t;
t_A \cdot c= \sqrt{(t_B \cdot c)^2-(t_b \cdot V_a)^2}<=>t_B= \frac{t_A}{ \sqrt{1 - \frac{v^2}{c^2}}}
now, i demonstrated with maths the time dilatation...how i demonstrate leght contraction and mass dilatation?
ok, after some tries, i think i reach a point it's correct, however, i would like a confirmation of what I've done.
so, to demonstrate a leght contration, I used this logic:
t_B= \frac{t_A}{ \sqrt{1 - \frac{v_a^2}{c^2}}}, V=\frac{d}{t}<=>t=\frac{d}{v}
with d as the leght travel by "a".
so, as v_a is equal to both observers,
\frac{d_b}{v_a}=\frac{d_a/v_a}{ \sqrt{1 - \frac{v_a^2}{c^2}}}
what is equal to:
\frac{d_b}{v_a}=\frac{d_a}{v_a \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}
what is equal to:
d_b=\frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}
what is equal to:
d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}
so, to observer "a", who is moving, leght is everytime lower than to "b", and convertor factor is "Y"
am I right??
to demonstrate mass dilatation, I dout it's correct by my though, however, i putted it here

:
d_a=d_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}
and,
D= \frac{m}{V}
where m is mass, D is density, and V is Volume
so, considering a rectangular object,
m= D \cdot (H \cdot W \cdot d)
where H is High, w is Width, and d is Leght
considering m_a the mass of "a" for the observer "a" and m_b the mass of "a" for the observer "b"
for moving observer "a"
m_a= D \cdot H \cdot W \cdot d_a
for rest observer "b":
m_b= D \cdot H \cdot W \cdot \frac{d_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}
so, taking D off:
\frac{m_a}{H \cdot W \cdot d_a}=\frac{m_b}{H \cdot W \cdot d_a} \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}
as H, W, D and d_a are equal to both observers,
m_a=m_b \cdot \sqrt{1 - \frac{v_a^2}{c^2}}}
so,
m_b=\frac{m_a}{\sqrt{1 - \frac{v_a^2}{c^2}}}}
can it be done with this logic? or we need to use another way??
thanks in advance, and hope wasn't too annoying