# Boosting massless particles

How does one calculate the boost matrix to go from a photon of (standard) four-momentum $k^\mu = (k,0,0,k)$ to $p^\mu = (p,0,0,p)$? (in terms of $|p|/|k|$)

Weinberg in his Quantum Field Theory Vol.1 has written an expression, which I suspect involves some Taylor expansion involving the $m \to 0$.

• Don't you just apply the standard Lorentz transformation? Start from $k^{\mu}$, apply a boost, and set the result equal to $p^{\mu}$. Then you solve for the boost matrix? – Zhengyan Shi Oct 13 '15 at 16:47
• Take an arbitrary Lorentz transformation matrix $\Lambda$ (check Wikipedia for the general form). Then simply solve $p = \Lambda \cdot k$. – Prahar Oct 13 '15 at 16:59

Just boost with the Lorentz matrix: $$\begin{bmatrix} p\\0\\0\\p \end{bmatrix}=\begin{bmatrix} cosh(\lambda) & 0 & 0 & sinh(\lambda) \\ 0&1&0&0 \\ 0&0&1&0 \\ sinh(\lambda) & 0 & 0 & cosh(\lambda) \end{bmatrix} \begin{bmatrix} k\\0\\0\\k \end{bmatrix}=\begin{bmatrix} (cosh(\lambda)+sinh(\lambda))k\\0\\0\\(cosh(\lambda)+sinh(\lambda))k \end{bmatrix}=e^{\lambda}\begin{bmatrix} k\\0\\0\\k \end{bmatrix}$$ where ${v\over c}=tanh(\lambda)$. The $\lambda$ is the Lorentz Boost Parameter (aka rapidity).
• What about in terms of $|p|/|k|$ – EEEB Oct 13 '15 at 18:28
• Of course you are correct, however the answer in Weinberg (vol.1 page 73) with regards to yours is $\cosh(\lambda) = \dfrac{(u^2 + 1)}{2u}$, where $u \equiv |p|/|k|$. How do we show that? – EEEB Oct 13 '15 at 18:35
• Notice from the first and last vector above that ${p\over k}=e^{\lambda}$ which you call $u$. Then $cosh(\lambda)={{e^{2\lambda}+1}\over {2e^{\lambda}}}={{e^{\lambda}+e^{-\lambda}}\over 2}$...which is the definition of cosh. – Gary Godfrey Oct 13 '15 at 20:17