Tuesday, 27 August 2019

homework and exercises - Getting from E2p2c2=m2c4 to E=gammamc2




What is each mathematical step (in detail) that one would take to get from:


E2p2c2=m2c4


to


E=γmc2,


where γ is the relativistic dilation factor.


This is for an object in motion.


NOTE: in the answer, I would like full explanation. E.g. when explaining how to derive x from x+22=4, rather than giving an answer of "x+22=4, x+2=8, x=6" give one where you describe each step, like "times 2 both sides, -2 both sides" but of course still with the numbers on display. (You'd be surprised at how people would assume not to describe in this detail).



Answer



Starting with your given equation, we add p2c2 to both sides to get E2=m2c4+p2c2

now using the definition of relativistic momentum p=γmv we substitute that in above to get E2=m2c4+(γmv)2c2=m2c4+γ2m2v2c2
Now, factoring out a common m2c4 from both terms on the RHS in anticipation of the answer we get E2=m2c4(1+v2c2γ2)
Now using the definition of γ as γ=11v2c2
and substituting this in for γ we get E2=m2c4(1+v2c21v2c2)
and making a common denominator for the item in parenthesis we get E2=m2c4(11v2c2)=m2c4γ2
Taking the square root of both sides gives E=±γmc2
Hope this helps.



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