1. Author was seeing a foot ball match. Ball & players are moving all over the ground. Author struck something different. In matter electrons & other particles are also moving randomly & vibrating all over the matter similar to players in ground. (Similar to gas molecules in box) This random velocity increases individual mass of particle; cumulative effect will be increase in total mass of matter. When this matter moves with relative to other observer, mass of matter again increases due to relativity. Author consider that at the place of total matter if we consider each individual moving particles in matter & if we consider relative increase in mass of each particle individually & sum the all relative masses with relative to observer, we will get relative mass of matter A with relative to observer. Means, relative cumulative mass of each constituent of matter A is equal to the relative mass of matter A with relative to the observer. To calculate such a individual mass of particles is impossible in real world. So, Author created similar situation.
Consider cabin A of train moving with relative to the man on platform with velocity Vx. Consider at the centre of mass of cabin horizontal metal plank is fixed in such a way that this plank is perpendicular to velocity of train. On this plank two balls of same mass are moving in opposite directions with velocity Vy from same distance in such a way that centre of zero moment of cabin is same as centre of mass of cabin.
Part 1 :-
Let, consider mass of each ball (at rest) = mb
Consider rest mass of cabin A excluding two ball = mc
Then, Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy2/C2)0.5}----------(1)
This mass will act at the centre of mass of train cabin which is also a point where summation of total moment of substances in cabin is zero.
When this cabin A moves with velocity Vx with relative to man on platform.
Mass of cabin with relative to man on platform = Mr. {1/(1-Vx2/C2) 0.5}
By equation (1) put value of Mr
Mass of cabin with relative to man on platform with two balls= [mc + 2 x mb x {1/(1-Vy2/C2) 0.5}]. [1/(1-Vx2/C2) 0.5]
= mc x [1/(1-Vx2/C2) 0.5] + 2 x mb x [1/(1-Vy2/C2) 0.5]. [1/(1-Vx2/C2) 0.5]
----------------------------------------------------------equa (2)
Part II :- Each individual moving matter has taken separately
Now, consider relative mass of balls & cabin separately with relative to man on platform.
Relative mass of only cabin= mc x [1/(1-Vx2/C2) 0.5] ---------------------(a)
Velocity of each ball with relative to man on platform is sum of velocity Vx & Vy in perpendicular direction.
So, Relative velocity of ball with relative to observer on platform = (Vx2+Vy2) 0.5
Relative mass of each ball = mb x [1/{1-(Vx2 +Vy2)/C2} 0.5] ---------------------(a)
Total relative mass of cabin with balls =mass of cabin+2 x mass of individual ball
=eq(1) + 2 eq(2)
= mc x [1/(1-Vx2/C2) 0.5]+2 x mb x [1/{1-(Vx2 +Vy2)/C2} 0.5]
--------------eq(3)
As eq(2) & eq(3) gives same mass of cabin with balls, so R.H.S. results can be equated
i.e. mc x [1/(1-Vx2/C2) 0.5] + 2 x mb x [1/(1-Vy2/C2) 0.5]. [1/(1-Vx2/C2) 0.5] = mc x [1/(1-Vx2/C2) 0.5]+2 x mb x
[1/{1-(Vx2 +Vy2)/C2} 0.5]
[1/(1-Vy2/C2) 0.5]. [1/(1-Vx2/C2) 0.5] = [1/{1-(Vx2 +Vy2)/C2} 0.5]
(1-Vy2/C2) 0.5. (1-Vx2/C2) 0.5 = {1-(Vx2 +Vy2)/C2} 0.5
(1-Vy2/C2). (1-Vx2/C2) = {1-(Vx2 +Vy2)/C2}
1- Vy2/C2-Vx2/C2+(Vy2/C2 ).(Vx2/C2) ) =1-Vx2 /C2-Vy2/C2
(Vy2/C2 ).(Vx2/C2) ) =0
Vy2.Vx2 =0
Vy.Vx =0
But as Vy & Vx is not zero. Vx. Vy is not equal to zero.
So, L.H.S. is not equal to R.H.S.
This will create inconsistency.
Is anything wrong in above calculation?

2.

3. Oh boy, not again

What will it take for you to finally realize that there are NO contradictions in special relativity ? It matters not how many ridiculous scenarios you dream up, you won't find any contradictions, you are only displaying your own ignorance of how relativity works. So once again, here is the GENERAL PROOF that special relativity is a self-consistent system, and that it is impossible to arrive at any contradiction within its axioms :

General Proof that Special Relativity is Self-Consistent

With this your calculation - which, by the way is utterly incomprehensible the way you present it - has been shown to be wrong by default.
This is now the third time that you are being told this ! It makes me wonder why you are even here, because it sure is not to learn anything.

For all readers less versed in relativity : there exists a quantity called the invariant mass. It is defined as the magnitude of the momenergy 4-vector ( in natural units with c=1 ) :

The invariant mass is thus

or written in a slightly different, easier to understand way

For particles at rest ( p=0 ), this reduces to the well known E=mc^2. Since the momenergy 4-vector is invariant any Lorentz transformations, its magnitude ( i.e. the invariant mass ) is the same for all inertial observers. The point is that total energy has a rest mass and a momentum component, so no matter what state of relative motion is considered, the invariant mass stays just that - invariant.

As for your own calculation, I have only the following to say :
1. It is presented so badly that it is basically incomprehensible to me
2. I have shown you in the general case that there are no contradictions is SR, hence your maths are wrong by default
3. I cannot function as your substitute maths & physics teacher. Last time I tried to do that it only led to much frustration for me, and you obviously hadn't learned a thing. I have shown you the general physical facts, you can choose to take them on board or not. It is up to you.

4. Originally Posted by Mahesh khati
Is anything wrong in above calculation?
Yes.

5. Formulae given in reply i.e. E =(mo2c4+p2c2)0.5 & m = mo/(1-V2/C2)0.5 canbe derive from one another & is the same thing.
My problem is very simple;
Part 1:- first I consider railway cabin is closed system as consider in relativity.
In closed system, the relativistic mass is the sum of the relativistic masses (or energies) of the parts, because energies are additive in closed systems
So, in closed system or in railway cabin mass of cabin is the sum of relativistic masses of its parts i.e. cabin mass, relative ball masses A & B.
Then, Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy2/C2)0.5}----------(1)
Now, this cabin is moving with velocity Vx for man on platform, then
Mass of cabin (with balls) with relative to man on platform = Mr. {1/(1-Vx2/C2) 0.5}
Or = mc x [1/(1-Vx2/C2) 0.5] + 2 x mb x [1/(1-Vy2/C2) 0.5]. [1/(1-Vx2/C2) 0.5]
----------------------------------------------------------equa (2)
Part 2 :- Just consider for calculation that railway cabin is not closed system & find out increase in the mass of each parts of system directly with relative to man on platform.
Velocity of only rail cabin with relative to man on platform is Vx
Relative mass of cabin = mc x [1/(1-Vx2/C2) 0.5]
Similarly relative mass of one ball = mb x [1/{1-(Vx2 +Vy2)/C2} 0.5]
As Vx is perpendicular to Vy & relative velocity V =(Vx2 +Vy2) 0.5
So, Total relative mass of cabin with balls = mc x [1/(1-Vx2/C2) 0.5]+2 x mb x [1/{1-(Vx2 +Vy2)/C2} 0.5]
--------------eq(3)

Equation (2) & (3) both gives relative mass of railway cabin with both balls with relative to same observer then both equation must be equal but I found both are different.
What is wrong in it?
Once system is considered closed system has to be considered closed forever.

6. What is wrong in it?
Let's have a look here. The invariant mass, as mentioned before, is the magnitude of the momenergy 4-vector; thus we write

The total rest mass of the cabin/ball/system of particles is just the sum of the rest masses of the individual components, hence

Within the train's frame of reference no movement is being detected, thus as seen by an observer in that frame we have

For the observer on the platform the train is moving, and the rest mass turns into relativistic mass, i.e. takes on the Gamma factor :

which is the exact same as for the observer riding on the train.
The conclusion is simply that while the rest mass of the cabin changes depending on its state of inertial motion, the invariant mass of the system stays constant for all observers, as already shown in post 2.

Originally Posted by Mahesh khati
My problem is very simple;
Indeed it is - you just don't understand relativity.

7. Just because I was in the mood I made a little exercise of it for myself - so here is the proof for the general case.

Consider a system of masses which moves uniformly at some arbitrary speed; evidently that is an inertial frame, and the system can be described by its momenergy 4-vector

The magnitude of this vector is the system's invariant mass :

Now we introduce a Lorentz transformation ( i.e. we go into some other inertial frame ), like so :

and insert this into the expression for the magnitude above :

which is the exact same, unless my tensor algebra deserts me completely.
Physically, this simply means that the magnitude of the momenergy 4-vector does not change if going from one inertial frame to another - the invariant mass of a given system is the same for all inertial observers. QED.

8. I was busy for publication of my marathi poem book. So, I can’t react to your reply early. I want to share one nice my English poem with all.

CHALLENGE
Challenge in life
makes us charming.
Solution of it,
makes world happy.
Without challenge,
world will be finished.
When it disappears
Joy also disappears
& makes us
dull, dull and dull
All systems of world are closed systems at micro level. Mass of closed system like gas container is not only summation of rest masses of molecules of gas but more than that. It is the summation of relativistic masses of molecules of gas in the container & this mass act at the centre of mass or point where total moment of molecules is zero.
In closed system, the mass of system is the sum of the relativistic masses (or energies) of the parts, because energies are additive in closed systems.
Means, For mass of train cabin = summation of relative masses of cabin, ball A & ball B
We can get this relation from your well known equation also,
Refer, The mass of composite systems in http :// en.wikipedia.org/wiki/Mass_in_special_relativity

M2C4 =(summation of E)2 – (summation ofP ) 2x C2
Summation of total moment is zero at mass centre ,,,,
So, (P ) 2x C2=0 & E = [mo/(1-V2/C2)0.5] . C2
M2C4 = [summation of mo/(1-V2/C2)0.5 C2]2 . - 0
M2C4 = [summation of mo/(1-V2/C2)0.5 ]2 . C4

M2 = [summation of mo/(1-V2/C2)0.5]2
M = summation of [mo/(1-V2/C2)0.5]

So, Rest mass of cabin act at point where summation of moment is zero is given by
Mass = relative mass of cabin + relative mass of ball A+ relative mass of ball B

TOP VIEW OF RAIL CABIN
-----------------------------------
l...........................................l
l...........................................l
l................Centre................l
l...... A o--- > + < --o B.........l
l............. Vy....... -Vy........... l
l...........................................l
l.......................................... l
-----------------------------------
....................l.......................
....................l Vx..................
...................\l/......................
At mass centre, summation of moments of two balls is zero.
Then, Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy2/C2)0.5}----------(1)
When this cabin moves with velocity Vx with relative to man on platform.
Mass of cabin with relative to man on platform = Mr. {1/(1-Vx2/C2) 0.5}
From equation (1), put value of Mr in above equation.
Mass of cabin with relative to man on platform with two balls= [mc + 2 x mb x {1/(1-Vy2/C2) 0.5}]. [1/(1-Vx2/C2) 0.5]
= mc x [1/(1-Vx2/C2) 0.5] + 2 x mb x [1/(1-Vy2/C2) 0.5]. [1/(1-Vx2/C2) 0.5]
----------------------------------------------------------equa (2)
Part II :- Each moving matter has taken separately, individual & directly then
Now, consider relative mass of balls & cabin separately with relative to man on platform.
Relative mass of only cabin= mc x [1/(1-Vx2/C2) 0.5] ---------------------(a)

Vy
---------
.\ ...... l
..\...... l
.V \ ... l Vx
.....\ .. l
......\ . l
........\ l
.........\l

So, Relative velocity of ball with relative to observer on platform V2= (Vx2+Vy2)
Relative mass of each ball = mb x [1/{1-(Vx2 +Vy2)/C2} 0.5] ---------------------(b)
Total relative mass of cabin with balls =relative mass of cabin+2 x relative mass of individual ball
=eq(a) + 2 eq(b)
= mc x [1/(1-Vx2/C2) 0.5]+2 x mb x [1/{1-(Vx2 +Vy2)/C2} 0.5]
--------------eq(3)
Eq (2) & (3) both gives relative mass of cabin with two balls then both must be equal but both are not equal.
If we make both equal, we get wrong result
i.e. Vx. Vy = 0

Where is it wrong? Each moving matter has taken separately in part II is wrong.

9. Where is it wrong?
Already answered in posts 5 & 6.

10. TOP VIEW OF RAIL CABIN
-----------------------------------
l...........................................l
l...........................................l
l................Centre................l
l...... A o--- > + < --o B.........l
l............. Vy....... -Vy........... l
l...........................................l
l.......................................... l
-----------------------------------
....................l.......................
....................l Vx..................
...................\l/......................

In train cabin frame, as shown in top view, balls are not at rest but moving with velocity Vy & -Vy on horizontal plank in train cabin, in such a way that mass center of cabin is not disturbed then rest mass of composite system is not summation of rest masses of cabin, ball A & ball B as given in your post 6 but more than that i.e. rest mass of cabin + relative mass of ball A +relative mass of ball B. As proved by me in post 7 &

The mass of composite systems in http :// en.wikipedia.org/wiki/Mass_in_special_relativity[/h] is
M2C4 =(summation of E)2 – ll (summation ofP ) 2 llx C2 ,this equation will act in this situation.

By using above equation, I have proved that

Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy2/C2)0.5}----------(1)

where, Mc = rest mass of only cabin , mb = rest mass of ball A = rest mass of ball B
Other thing is explain in post 7
Is anything wrong in my above calculation?

11. Originally Posted by Mahesh khati
By using above equation, I have proved that

Total rest mass of cabin including two balls, Mr = mc + 2 x mb x {1/(1-Vy2/C2)0.5}----------(1)

where, Mc = rest mass of only cabin , mb = rest mass of ball A = rest mass of ball B
Other thing is explain in post 7
Is anything wrong in my above calculation?
See posts 5 and 6 for a general demonstration that your attempts are all flawed. There is no point in dissecting each and every one of your flawed calculations to discover the errors. It is sufficient to show that your enterprise is doomed from the outset. There are more valuable ways to spend time, and there are certainly more fun ways to waste it.

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