Question

Show that the energy-momentum relationship given by the equation below follows from the expressions below

Answer 1

\[E^2=p^2c^2+(mc^2)\]
\[E=\gamma mc^2\]
\[p=\gamma mu\]

Answer 2

My attempt at a solution was to use \[E=k+mc^2\] where \[k=\gamma mc^2-mc^2\] and \[\gamma = (1-(u/c)^2)^{-1/2}\] but the relationship I got at the end wasn't quite correct. It did not involve the mass for momentum

Answer 3

First step is to find gamma in terms of momentum,p but independent of u. This can be achieved by squaring the momentum relation,putting the value of gamma and finding the value of u2/c2 in terms of p. Then put this value back into the expression for gamma and you will get gamma in terms of p. Finally insert this newly found value of gamma in energy relation and you will see the relation taking shape.

Answer 4

I'm not sure I am following

Answer 5

\[p=\gamma m u\]
\[p=m u/\sqrt{1-u ^{2}/c ^{2}}\]
\[u ^{2}/c ^{2} = p ^{2}\div (m ^{2}c ^{2}+p ^{2})\]
\[E ^{2}=\gamma ^{2} m ^{2}c ^{4}\]
\[E ^{2}=m ^{2}c ^{4}\div (1-u ^{2}/c ^{2})\]
\[E ^{2}=m ^{2}c ^{4}\div (1-p ^{2}/(m ^{2}c ^{2}+p ^{2}))\]

Now simplify.
Also i noticed a mistake in your expression.The correct expression is \[E ^{2}=p ^{2}c ^{2}+(mc ^{2})^{2} \]