I know the way of deriving the formula using usual lorentz transformation formulas,,but is there a way out of deriving it using 4-vector notation??please help
Answer
In tensor notation, $\Lambda^\mu{}_\nu$ is the matrix that performs the Lorentz transformations. Now, since $p^\mu$ is a tensor, under an arbitrary Lorentz transformation, it transforms as $$ p^\mu \to p'^\mu = \Lambda^\mu{}_\nu p^\nu $$ Consider now a particle moving entirely in the $x$ direction, with velocity $v_1$. Its 4-momentum is $$ p^\mu = \gamma_1( 1 , \beta_1 , 0 , 0 ) ~~\text{where}~~ \gamma = \frac{1}{ \sqrt{ 1 - v_1^2 / c^2 } } ,~~ \beta_1 = \frac{v_1}{c} $$ Now, we want to reproduce the velocity addition formula, so we are interested in a boost. The Lorentz matrix corresponding to a boost with velocity $-v_2$ is explicitly$\dagger$ $$ \Lambda = \pmatrix{ \gamma_2 & \beta_2 \gamma_2 & 0 & 0 \\ \beta_2 \gamma_2 & \gamma_2 & 0 &0 \\ 0 &0 &1 &0 \\ 0 & 0 & 0 & 1} $$
The new boosted 4-momentum is $$ p'^\mu = \gamma_1 \gamma_2 \left( 1 + \beta_1 \beta_2 , \beta_1 + \beta_2 , 0 , 0 \right) $$ Now, we simply have to extract the transformed velocity from this formula. Firstly note that since this is 4-momentum, it is of the form $$ p'^\mu = \gamma_3 ( 1 , \beta_3 , 0 , 0 ) $$ for some velocity $v_3$. Thus, we have the equation $$ \gamma_1 \gamma_2 ( 1 + \beta_1 \beta_2 ) = \gamma_3 $$ We can use this equation to solve for $v_3$ and we find $$ v_3 = \frac{ v_1 + v_2 }{ 1 + \frac{ v_1 v_2 }{ c^2 } } $$ which we sought out to prove.
$\dagger$Be careful to note that this is the boost matrix corresponding to $-v_2$. This is because I want to reproduce "velocity addition" as opposed to "relative velocity" both of which are obviously related concepts. In others, given a particle moving with speed $v_1$ in one frame, if I boost myself with speed $v_2$ in the opposite direction, then the speed of the particle in the new frame literally adds, i.e. it is proportional to $v_1 + v_2$. This is the reason to choose $-v_2$.
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