March 14, 2014 My pleasure of studying "Theory of relativity"

March 14, 2014 My pleasure of studying "Theory of relativity"

When I read the concept of time, space, and being conceived by Buddha written in the book "Japanese spirituality (日本的霊性) in Japanese)" authored by Mr. Daisetz Suzuki, I thought it resembles very much to that of "Theory of relativity". Since then I have become interested in "Theory of relativity" and have been studying the theory by the translated book by Mr. ERIGUCHI, Ryohji (doctor of science) and Mr. FUTAMASE, Toshifumi (doctor of science) of "A first Course in General Relativity" authored by Mr. Bernard F. Schutz (Professor, Cardiff University).

Hereinafter I write 2 interesting points which, I hope, may allure some of the browsers to study the Einstein's theory.

1. Coordinate transformation between 2 systems of orthogonal coordinates (Each function shown here is written based on the form adopted by EXEL.)

The coordinate transformation between coordinates X-Y-0 and X1-Y1-O (X1 axis is inclined to X axis by θ.) is expressed by
the following equations as we have learned in the senior high school.

X1= X*COSθ+Y*SINθ
Y1=-X*SINθ+Y*COSθ
Therefore POWER（X,2)+POWER（Y,2)=POWER（X1,2)+POWER（Y1,2). This means 2 points which are equally distant from the origin
are on the same circle having its center at the origin.

Coordinate transformation between 2 systems of coordinates (t-X-O (orthogonal coordinates) and t1-X1-O
t-X-O coordinate system stands still and t1-X1-O coordinate system expresses what runs in plus X direction with
velocity v (ratio of the actual velocity "v" to velocity of light, 300,000km/sec).

t1= t*COSH(u)-X*SINH(u)
X1=-t*SINH(u)+X*COSH(u)
But, v=TANH(u)
Therefore -POWER（t,2)+POWER（X,2)=-POWER（t1,2)+POWER（X1,2). This means 2 points which are same in square distance from
the origin are on the curve of the same hyperbola, -POWER(t,2)+POWER(X,2)=Constant.
(In all the other books, the time axis is expressed by "ct" (c is the velocity of light, 300,000km/sec), but the above
equations are based on the concept which defines c= 300,000km/sec=1. Therefore the time axis is expressed by "ct=t".)

2. A rocket having the constant acceleration very near to the Earth gravitation, 10m/POWER(s,2) goes to the center of the
Milky Way which is 21,000 light years away from the Earth. The rocket passes the determination and U-turns and decelerates
with 10m/POWER(s,2) to return to the Earth.
The rocket lands at the Earth 42,001.90254 years(based on the calculation which has 10 significant figures) after it left
the Earth but the persons aboard the rocket are only 20.3490 years older (based on the calculation which has 6 significant
figures) than he was when the rocket left the Earth.

The calculation is as follows;

The coordinate system of the Earth is t-X-O, and that of the rocket is t1-X1-O.
When the rocket (proper) time increases from t1 to t1+dt1, the rocket velocity increases from v to v+a*dt1 (a=10m/POWER
(s,2)).
Therefore v(t1+dt1)=(v+a*dt1)/(1+v*a*dt1)=(v+a*dt1)(1-v*a*dt1)=v+a*dt1(1-POWER(v,2))(because dt1 becomes infinitesimal.)
So, dv=v(t1+dt1)-v(t1)=a*dt1(1-POWER(v,2),
That is, dv/(1-POWER(v,2)=a*dt1, which is 1/2(dv/(1+v)+dv/(1-v))=a*dt1.
We integrate both sides, and get
1/2(LN((1+v)/(1-v)))=a*T1+C=ω　(C=Constant).
Therefore
ATANH(v)=1/2*LN((1+v)/(1-v))=ω
So, v=TANH(ω)
When t1=0,v=0,ω=0, so,C=0.
Finally we get
v=TANH(a*dt1)----------------------------------------------(1)
we have the following equation between t and t1;
dt=dt1/POWER((1-POWER(v,2),1/2)(because X1=0)--------------(2)
Insert the equation (1) into the equation (2),and we get
dt=dt1/POWER((1-POWER(TANH(a*t1),2),1/2)=dt1*COSH(a*t1)----(3)
We integrate both sides and get
t=(1/a)*SINH(a*t1)-----------------------------------------(4)
Therefore a*t=SINH(a*t1)-----------------------------------(5)
We change the equation (1) as follows;
v=SINH(a*t1)/COSH(a*t1)=SINH(a*t1)/POWER((1+POWER(SINH(a*t1),2),1/2)--------(6)
Insert the equation (6) into (4) and we get
v=a*t/POWER((1+POWER(a*t,2),1/2)----------------------------(7)
We set forth "L(=X)" as the distance the rocket reaches during the Earth time "t", so we integrate "vdt" from the equation
(7) and get "L" as follows;
L=(POWER(1+POWER(a*t,2),1/2)-1)/a---------------------------(8)

L=21,000 light years=3.0*POWER(10,8)m/s*(60*60*24*365s/year*21,000years=1.98677*POWER(10,20)m-----(9)
Insert the equation (9) into the equation (8) and we get
t=1.98686*POWER(10,20)m=21,000.95127 years----------------------(10)
This means the rocket passes by the center of the Milky Way 21,000.95127 years (10 significant figures) after it leaves
the Earth.

Insert the equation (10) into the equation (4) and we get
t1=ASINH(a*t)/a=10.1745 years---------------------------------(11)
Therefore the persons aboard the rocket are 10.1745 years older than he was when the rocket left the Earth.

The rocket comes back to the Earth by decelerating the same deceleration, 10m/POWER(s,2), consuming the same times
(t and t1).

If the rocket continues to accelerate by the same acceleration, and the rocket will pass by the Earth consuming the
times (t and t1) as follows;

L= 2 times the equation (9) and we get the times as written below.

t=42,000.95128 years, therefore the time consumed during the return voyage is 21,000.00001 years, which means the rocket
flied virtually with the velocity of light, because 42,000.95128-21,000.95127=21,000.0001.

t1=10.8338 years, therefore the time consumed during the return voyage by the persons aboard the rocket is 0.6593 year,
because 10.8338-10.1745=0.6593.

When I consider that time and light, both very familiar with us, are far beyond our feeling and perception, only
understandable by mathematics and physics, I am deeply impressed how abstruse the universe is.

NAMUAMIDABUTSU NAMUAMIDABUTSU
I thank SHAKA BUDDHA and AMIDA BUDDHA and vow to spread Buddhism.