How to derive the components $\Lambda^i_j$ of a Lorentz-transformation given a boost $\vec{v}$?

Question

I am stuck in deriving a specific formula concerning Lorentz-boosts. In my Classical Mechanics skript there is a chapter dealing with special relativity. In this chapter the Lorentz transformations are defined as being those linear transformations $\Lambda_\nu^\mu:{\mathbb R}^4 \rightarrow \mathbb R^4$, $x^\mu \rightarrow x'^\mu=\Lambda_\nu^\mu x^\nu+\rho^\mu$ that satisfy the equality $g_{\nu\mu}\Lambda_\sigma^\mu\Lambda_\rho^\nu=g_{\sigma\rho}$, where $g_{\nu\mu}$ is the metric tensor for flat space time.

Furthermore a specific transformation between two inertial systems O and O' is considered. This transformation relates the system O' and O if O' moves with velocity $\vec{v}$ relatively to O.
Then, for this specific situation, the formulas $\Lambda_0^0=\gamma$ and $\Lambda_0^i=\gamma \frac{v^i}{c}$ for $i=1,2,3$ are derived. (I appended the derivation in a picture below.)

Now to my question: My script states, that if the axis of both systems are paralell to each other it is also possible to derive formulas for the remaining components of $\Lambda_\nu^\mu$. These formulas are stated to be: $\Lambda_j^0=\gamma \frac{v^j}{c}$ and $\Lambda_j^i=\delta^i_j+\frac{v_iv_j}{\vec{v}^2}\gamma$. The proof for this equations is not carried out in the script but left as an exercis. I have tried really hard but was not able to do the derivation. I consider my problems to be rooted in my inability to figure out how to use the paralellity of the axes in the derivation. I would be really glad if someone could tell me how it is done correctly.

In the following you can find the derivation for $\Lambda_0^i$ and $\Lambda_0^0$ from my skript.

Answer 1

There is an error in your expression for the $\Lambda^i_{\;j}$ components. They actually read $$ \Lambda^i_{\;j} = \delta^i_j + (\gamma -1)\frac{v^i}{v}\frac{v_j}{v} = \delta^i_j + \frac{\gamma -1}{\beta^2}\frac{v^i}{c}\frac{v_j}{c} $$ with $\beta = v/c$, $v^2 = v^iv_i$.

Keeping this in mind, separate once again timelike and spacelike coordinates in the original transformation and use the already known components of $ \Lambda$: $$ dx'^0 = \Lambda^0_{\;0} \;dx^0 + \Lambda^0_{\;j} \;dx^j = dx^0 \left[ \gamma + \Lambda^0_{\;j} \frac{u^j}{c} \right]\\ dx'^i = \Lambda^i_{\;0} \;dx^0 + \Lambda^i_{\;j} \;dx^j = dx^0 \left[ \gamma\frac{v_i}{c} + \Lambda^i_{\;j} \frac{u^j}{c} \right] $$ where $ u^j/c = dx^j/dx^0$. Then take appropriate ratios to obtain the corresponding velocity transformations: $$ \frac{u'^i}{c} = \frac{\gamma\frac{v^i}{c} + \Lambda^i_{\;j} \frac{u^j}{c}}{\gamma + \Lambda^0_{\;j} \frac{u^j}{c}} $$ Consider now those displacements in the unprimed system that are along the direction of the relative velocity, such that $$ \frac{u^i}{c} = a\frac{v^i}{c},\;\;\;u^iu_i = ac^2 $$ for some $a\in \mathbb R$, and observe that the corresponding displacements in the primed system must also read $$ \frac{u'^i}{c} = a'\frac{v^i}{c},\;\;\;u'^iu'_i = a'c^2 $$ for some other $a'\in \mathbb R$. Substituting in the velocity transformations above yields a general relationship of the form $$ a'\frac{v^i}{c} = \frac{\gamma\frac{v^i}{c} + a \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma + a \Lambda^0_{\;j}\frac{v^j}{c}}\;\;\;\;\;\;\;\;\;(1) $$ The information necessary to derive the $\Lambda$ components comes from the following particular cases:

• From the isotropy of space it follows that $a=-1$ requires $a' = 0$, which can only happen provided $$ \Lambda^i_{\;j} \frac{v^j}{c} = \gamma\frac{v^i}{c}\;\;\;\;\;\;\;\;\;(2) $$

for any $v^i$.

• The invariance of the speed of light further requires that for $a = \pm c/v\Leftrightarrow u^iu_i = c^2$, we must also have $a' = \pm c/v\Leftrightarrow u'^iu'_i = c^2$, and therefore $$ \pm \frac{c}{v}\frac{v^i}{c} = \frac{\gamma\frac{v_i}{c} \pm \frac{c}{v} \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma \pm \frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}} = \frac{\gamma\frac{v_i}{c} \pm \frac{c}{v} \gamma\frac{v^i}{c}}{\gamma \pm\frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}} = \frac{\gamma(1 \pm \frac{c}{v} )}{\gamma \pm \frac{c}{v} \Lambda^0_{\;j}\frac{v^j}{c}}\frac{v^i}{c} $$ Solving for $ \Lambda^0_{\;j}\frac{v^j}{c}$ gives $$ \Lambda^0_{\;j}\frac{v^j}{c} = \gamma \frac{v_j}{c}\frac{v^j}{c}\\ \left[ \Lambda^0_{\;j} - \gamma \frac{v_j}{c}\right] \frac{v^j}{c} = 0 $$ and since the boost transformation is symmetric, we can conclude that $$ \Lambda^0_{\;j} = \gamma \frac{v_j}{c} $$

• The above expressions for $ \Lambda^0_{\;j}$ simplify eq.(1) to $$ a'\frac{v^i}{c} = \frac{\gamma\frac{v^i}{c} + a \Lambda^i_{\;j} \frac{v^j}{c}}{\gamma\left(1 + a \frac{v^2}{c}\right)}\;\;\;\;\;\;\;\;\;(3) $$ and condition (2) further implies $$ a' = \frac{a+1}{1+a\frac{v^2}{c^2}} $$ However, eq.(3) can be rearranged eventually as $$ \left[ \Lambda^i_{\;j} - \left( \frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)}\delta^i_{\;j} + \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}}\frac{v^i}{c}\frac{v_j}{c}\right)\right]\frac{v^j}{c} = 0 $$ which has to hold for any $a \in \mathbb R$. Since $\Lambda^i_{\;j}$ may depend only on the relative velocity, this means that it must be of the form $$ \Lambda^i_{\;j} = Q \delta^i_{\;j} + P \frac{v^i}{c}\frac{v_j}{c} $$ with $Q$ and $P$ some functions of the relative velocity such that $$ \left[ \left[ Q - \frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)} \right] + \left[ P - \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}} \right] \frac{v^2}{c^2} \right] \frac{v^i}{c} = 0 $$ From this we have successively $$ Q + P \frac{v^2}{c^2} = \left[\frac{1}{\gamma\left(1+a\frac{v^2}{c^2}\right)} + \frac{\gamma(a+1)}{1+a\frac{v^2}{c^2}} \frac{v^2}{c^2}\right] = \gamma\\ P = \frac{\gamma - Q}{\beta^2}\\ \Lambda^i_{\;j} = Q \delta^i_{\;j} + \frac{\gamma - Q}{\beta^2} \frac{v^i}{c}\frac{v_j}{c} $$ The exact form of Q may now be determined from $g_{\mu\nu}\Lambda^\mu_\rho\Lambda^\nu_\sigma = g_{\rho\sigma}$.