Domain of relativistic mechanics

[07PRADEEP]

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Consider the simple formula 4AB=(A+B)²-(A-B)²
We proceed like this If A+B>A-B ,Then B>0
Because mc>mv in all cases
We are going to check whether the numerical values are satisfied
Let us assume above simple formula is analogous to either (m₀ c)²=(mc)²-(mv)²,where L.H.S represents product of rest mass and velocity of light squared,
Or m₀²=(square of m )-(mv/c)², Case 1 (m₀ c)²=(mc)²-(mv)²
Let A=c,B=(c-mv)We get 4(c) (c-mv)=(2c-mv)²-(mv)²
B>0 gives mv<c
For m₀<2kg,mc < 2c units, we obtain m< 2Kg
for (mv)less than(c),Maximum of m=2c,minimum of or atmost v=c, m=2 kg,m₀=2kg,
Case-2
Consider the simple formula 4AB=(A+B)²-(A-B)²
Analogous formula
m₀²=m²-(mv/c) ²,Putting A=c,B=(c-mv/c) and
comparing with 4(c) (c-mv/c)=(2c-mv/c)²-(mv/c)²,B>0 implies
mv< c² , m₀²=m²-(mv/c) ²
m₀<2c kg
mc<2c²,m<2c
mv< c²
therefore maximum of v=c/2 units
Case-3
Consider the simple formula 4AB=(A+B)²-(A-B)²
Analogous formula,m₀²=m²-(mv/c) ²,Putting A=k,B=(k-mv/c) and
comparing with 4(k) (k-mv/c)=(2k-mv/c)²-(mv/c)²,we get
mv<ck,
B>0 givesmv<ck,
m₀<2k where k stands for any fixed value ,
maximum of v=c/2,m<2k


Conclusion

Maximum of v=c/2 units, in RELATIVISTIC MECHANICS DOMAIN namely Case2 and Case3
Case1 mv<c , rest mass m₀ element of ( 0 to 2)kg , m<2 kg
Case 2 mv <c²,rest mass m₀ element of (2 to 2c)kg , m<2c kg
Case3 mv<kc ,rest mass m₀ element of (2c to k)kg m<k kg
Where k stands for any fixed value ,however large