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 - ( x1-Bx0)^2

=> x0^2 - 2Bx1 x0 + B^2 x1^2 - ( x1^2-2Bx1x0 + B^2x0^2 ) all divided by / 1-B^2

I've replaced ^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 =

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/ , x1/ ....... )

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/ remember because x0 = ct

The second element is called P one

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

we alreadly known that t/=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

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 - ( x1-Bx0)^2

=> x0^2 - 2Bx1 x0 + B^2 x1^2 - ( x1^2-2Bx1x0 + B^2x0^2 ) all divided by / 1-B^2

I've replaced ^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 =

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/ , x1/ ....... )

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/ remember because x0 = ct

The second element is called P one

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

we alreadly known that t/=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)^2 - (mv)^2 => m^2 ^2 ( c^2 - v^2)=> m^2 ^2 c^2 (1-v^2/c^2 )

Since 1-v^2/c^2 = 1/^2

=> m^2^2 c^2 1/^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

x0^2 - x1^2 is invariant lets consider in four vector where

P0^2- P1^2 = (mc)^2 - (mv)^2 => m^2 ^2 ( c^2 - v^2)

Since 1-v^2/c^2 = 1/^2

=> m^2^2 c^2 1/^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

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