Question #b3589

2 Answers
Apr 20, 2016

Start with the relativistic momentum equation:
#p = (m_0 v)/sqrt(1-v^2/c^2 # square and multiple top and bottom by #c^2#
#p^2c^2 = (m_0^2v^2c^2)/(1-v^2/c^2) = (m_0^2v^2c^4/c^2)/(1-v^2/c^2# re-arranger add and subtracting a term and write:
#= m_0^2c^4[v^2/c^2-1]/(1-v^2/c^2) + (m_0^2c^4)/(1-v^2/c^2) #
# = -m_0^2c^4[cancel(1-v^2/c^2]/cancel(1-v^2/c^2)] + cancel(m_0^2/(1-v^2/c^2))^(m^2) c^4 #
# = -m_0^2c^4 + color(red)((mc^2)^2) = -m_0^2c^4 + color(red)(E^2)#
bring the negative term to the left rearrange and you have:
#color(red)(E^2) = (pc)^2 + (m_0c^2)^2#
#m_0 ne m# OK?!
You should note that #=>m^2=m_0^2/(1-v^2/c^2)#
Also I want to point out that this effectively a pythagorean identity with hypotenuse of #color(red)(E)# and the cateti #pc and m_0c^2#

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Cheers!

Apr 20, 2016

Answer:

Follow the explanation.

Explanation:

#E = (mc^2)/sqrt(1-(v/c)^2)#

#so, E^2 = (m^2c^4)/(1-(v/c)^2)=(m^2c^6)/(c^2-v^2)#

In the same way

#p = (mv)/sqrt(1-(v/c)^2)#

#so, p^2c^2 = (m^2v^2c^2)/(1-(v/c)^2)=(m^2v^2c^4)/(c^2-v^2)#

So,

#E^2-p^2c^2 = (m^2c^6)/(c^2-v^2)- (m^2v^2c^4)/(c^2-v^2) = m^2c^4 *((c^2-v^2)/(c^2-v^2)) =m^2c^4 = (mc^2)^2 #

#=>E^2-p^2c^2= (mc^2)^2 #

#=>E^2=p^2c^2 + (mc^2)^2 #