## Thursday, October 3, 2013

### Special Relativity Advance Explanation

I would like to to start this post considering that you have already aware of  length contraction , simultaneity and casualty , However Time Dilation and Proper time (Tau) and four vector momentum will be going to explain in below . Now special relativity consider when object or the observer moving at a constant velocity or  in stationary  and never accelerating. Also the observer or the object is moving in stragth line not curve or nor in circle .

Time dilation :
Lets take an event where there will be two S's  , S and S'(prime) . here Speed will be as v=x/t in each of their frame of references .

Now on S frame of reference  , let suppose it's duration is in T seconds and T is a duration of event measured by S where S is stationery there fore  X=0.

However  S' says that t'= T-(u/c^2 *x) / Sqrt 1-v^2/c^2

Since x is zero  so (u/c^2*x) will cancel out  therefore we will be left as

t'=T/Sqrt 1-v^2/c^2       <----   This is Time Dilation equation

The Square root is between 0 and 1 therefore t' will be  greater than T, so S' says that the S clock is running slow .

Four Vector :

We all known that three dimensions are explained as  X , Y ,  Z   and  now including Time we'll consider it now as four dimension

here X is known for four vector and it's describe as
X= (t,x,y,z)
also known as
X= (x0 , x1 , x2 , x3 )                       here x0 ( x not ) = time and since it is in time we have to change it in to
same distances vector . Now   x0=ct  and now all are in distance now.
Also  v/c = B(beta )

now  X1' says   that x1'=x1-B*x0/Sqrt 1 - B^2
x0'= x0 - B*x1 /Sqrt 1-B^2

In this case we will ignore X2 and X3 and consider x0 as time and x1 as space .

There is an other term that I'm going to create  S^2 = x0^2 - x1^2                   remember x0^2=( ct)^2

Now lets look at
x0'^2 - x1'^2 = ( x0 - Bx1 )^2 $\gamma$ - ( x1-Bx0)^2 $\gamma$
=> x0^2 - 2Bx1 x0 + B^2 x1^2  - ( x1^2-2Bx1x0 + B^2x0^2 )  all divided by / 1-B^2

I've replaced $\gamma$ ^2  with 1/Sqrt 1- B^2

=> x0^2 + B^2 x1^2 - x1^2  - B^2 x0^2 / 1- B^2 =  x0^2 (1-B^2 ) - x1 ^2(1-B^2) / 1-B^2

what this shows that     x0'^2 - x1'^2 = x0^2 - x1^2

This term is invariant in any frame of reference .

SO lets consider our  S^2 term    S^2 = x0^2 - x1^2                    we know that x0= ct

S^2 = (ct)^2 - x1^2   dividing through by  c^2 t^2

S^2 = c^2t^2 ( 1 - x1^2 / c^2 t^2  )                                        since v =x/t
S = ct ( Sqrt 1- v^2/ c^2)                         Now dividing both sides by c
S/c =t(Sqrt 1-v^2/c^2 )  and now this is called Tau = $\tau$ $\tau$ is called proper time  , It is a time which is invariant  since S^2 is invariant then S/c must be invariant  because  c  is constant    . Tau is turn out to be the proper time  in which observer is traveling.

THE FOUR MOMENTUM VECTOR

Lets call Capital   P=m( x0/ $\tau$  , x1/ $\tau$  ....... )

Here we are only considering  time and Space dimension  ignoring others  like x2 or x3 .

So first vector of element  shall be called P zero

P0= m ct/ $\tau$                         remember because x0 = ct

The second element is called P one

P1= m x1/ $\tau$ which is same or equal as t/ $\tau$

we  alreadly known that       t/ $\tau$=1/Sqrt 1 -V^2 / C^2

so  P0 = mc / Sqrt 1 - V^2/ C^2

Also we know that x1 /t = v

So P1 = mv/Sqrt 1-v^2/c^2

Now using famous maths expansion  (1+x)^n  and is we expand we will get  1 + nx + .....
If we use this Expansion for  (1-v^2/c^2) ^-1/2  = 1 +v^2 / 2c^2 + .......

and this means that our second term   P1=mv+mv^3/2c^2  where if we recognize mv= ordinary momentum
where mv^3 / 2c^2 it's some kind of relativistic  correcting when speed approach  the speed of light .

For our first vector  if we apply expansion  in it we get
P0 = mc +1/2c^2 (mv2) +......

and if we multiply with c we will  get

cP0 = mc^2 +1/2 mv^2

This tells us that  mc^2 means some rest mass energy while  1/2 mv^2  is K.e

so cP0 must be some energy term  and this leads us to  E=mc^2
Lets us now look into an other vector  quantity to find an invariant
x0^2 - x1^2  is invariant  lets consider in four vector where
P0^2- P1^2  = (mc $\gamma$)^2 - (mv $\gamma$)^2  => m^2 $\gamma$^2 ( c^2 - v^2)  => m^2 $\gamma$^2 c^2 (1-v^2/c^2 )

Since 1-v^2/c^2 = 1/ $\gamma$^2

=> m^2 $\gamma$^2 c^2  1/ $\gamma$^2    the gammas cancel out each other and we get
= m^2c^2   so we find that rest mass and c are constant and p0^2 - P1^2  both are invariant  terms

However

P0^2 - P1^2         where  P0^2  = ( E/c)^2   and P1^2= m^2 c ^2  so P1 is general momentum  so P1 = p

so this becomes
(E/c)^2 - p^2 = m^2 c^2   multiplying it by c^2

E^2= c^2 p^2 + m^2 c ^4     this is the general term to find energy  with momentum

If p=0 it comes back to  E^2=m^2c^4
and reduces by square root  we get    E=mc^2..

Please let me know on the comments below if you don't understand anything  I would love to help and as always

THANKS FOR READING :D

PS : sorry about the Tau and gamma  its just Tau and gamma  under neath its nothing :D :D