1.
Domain of relativistic mechanics
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Consider the simple formula 4AB=(A+B)-(A-B)˛ or AB=(A+B)˛/4 -(A-B)˛/4
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)˛=(mc)˛-(mv)˛,where L.H.S represents product of rest mass and velocity of light squared,
Or m₀˛=(square of m )-(mv/c)˛, Case 1 (m₀ c)˛=(mc)˛-(mv)˛
Let A=c,B=(c-mv)We get 4(c) (c-mv)=(2c-mv)˛-(mv)˛
B>0 gives mv<c
For m₀<2kg,mc < 2c units, we obtain m< 2Kg
for (mv)less than(c),Maximum of m=2,minimum of or atmost v=c, m=2 kg,m₀=2kg,
Case-2
Consider the simple formula 4AB=(A+B)˛-(A-B)˛ orAB=(A+B)˛/4 -(A-B)˛/4
Analogous formula
m₀˛=m˛-(mv/c) ˛,Putting A=c,B=(c-mv/c) and
comparing with (c) (c-mv/c)=(c-mv/2c)˛-(mv/2c)˛,B>0 implies
mv< c˛ and mv<2c˛hence conclusion is mv< c˛, in m₀˛=m˛-(mv/c) ˛
m₀<c kg
mc<c˛,m<c
Now mv< c˛/2
therefore maximum of v=c/2 units
Case-3
Consider the simple formula 4AB=(A+B)˛-(A-B)˛orAB=(A+B)˛/4 -(A-B)˛/4
Analogous formula,m₀˛=m˛-(mv/c) ˛,Putting A=k,B=(k-mv/c) and
comparing with (k) (k-mv/c)=(k-mv/2c)˛-(mv/2c)˛,we get
mv<ck,
B>0 givesmv<2ck,
m₀<k where k stands for any fixed value,it can assume any value ,
maximum of v=c/2,m<k Since mv<ck,mv<2ck conclusion is mv<ck Also since
mv/2c leads to k/2,Comparing it with the analogous equation,
k/2=kv/c,leads to v=c/2
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 <c˛,rest mass m₀ element of (2 to c)kg , m<c kg
Case3 mv<kc ,rest mass m₀ element of (c to k)kg m<k kg
Where k stands for any fixed value ,however large


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