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Hello everybody the things I understand are the transformation properties of a four vector given as $\tag{1} p^\mu = \Lambda^\mu_{\,\,\nu} p^\nu.$ and an arbitrarily boosted transformation is of the form. enter image description here

what I am not getting is that how to convert this matrix into terms containing $E$ and $p$. As we read that a particle at rest has $p^\mu = (E,0,0,0)$ observed in boosted frame with negative velocity has $p^\mu = (E,p_x,p_y,p_z)$. So we should have $$\left(\begin{matrix} E \\ p_x \\ p_y \\ p_z \end{matrix}\right) = L \,\left(\begin{matrix} E \\ 0 \\ 0 \\ 0 \end{matrix}\right).$$ So What is $L$? I know that some thing is wrong with my concepts. What is that?

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    $\begingroup$ Change $(ct,x,y,z)$ to $(E/c,p_x,p_y,p_z)$. $\endgroup$
    – jinawee
    Commented Feb 2, 2015 at 12:14
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    $\begingroup$ The matrix is the one you wrote, i.e. $L= \Lambda$. What is it that you don't understand exactly? The transformation rule $$ V^\mu \longrightarrow \Lambda^\mu_\nu V^\nu$$ holds for any 4-vector $V^\mu$, with the $\Lambda$ you wrote. $\endgroup$
    – glS
    Commented Feb 2, 2015 at 12:14
  • $\begingroup$ No dear i don't understand how to write the matrix in terms of $E$ and $p$. $\endgroup$
    – Zohaib Aarfi
    Commented Feb 2, 2015 at 12:54

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Be careful: in general the $\beta_i=v_i/c$ and $\gamma=(1-\beta^2)^{-\frac{1}{2}}$ coefficients that appear in the transformation matrix don't depend on the particle's velocity or energy ($v$ is the relative velocity beetween the frames).

Seems like you are trying to switch to the particle's rest frame; in that case $v$ is indeed the particle's velocity. Since the 4-momentum is defined by: $$p^\mu=mu^\mu=(\gamma mc,\gamma m\mathbf v),$$ what you get is $\gamma=p^0/mc=E/mc^2$ and $\beta _i = v_i/c = p^i/p^0$ (where $i=x,y,z$, note that these are not equations beetween tetravectors).

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  • $\begingroup$ how did you define $\gamma$ and $\beta^i$. Kindly tell me little basics. $\endgroup$
    – Zohaib Aarfi
    Commented Feb 2, 2015 at 16:01
  • $\begingroup$ Hi, $\gamma$ and $\beta _i$ are the same quantities that appear in your matrix and are defined here: en.wikipedia.org/wiki/Lorentz_transformation#Matrix_forms. I had wrote $\beta ^i$ with an upper index before, but that's only confusing since $\beta$ is not a four-vector, so please ignore it. $\endgroup$
    – pppqqq
    Commented Feb 2, 2015 at 16:33
  • $\begingroup$ I got it now. How would i relate this matrix to $e^{-i\omega_{\mu\nu}J^{\mu\nu}/2}$ $\endgroup$
    – Zohaib Aarfi
    Commented Feb 2, 2015 at 17:10
  • $\begingroup$ @ZohaibAarfi sorry but I have no idea of what does $e^{-i \text{stuff}} $ mean. I can try to add that in the question, explaining what does the notation mean. $\endgroup$
    – pppqqq
    Commented Feb 2, 2015 at 17:23
  • $\begingroup$ Thanx a lot. u ve been very helpful. @pppqqq $\endgroup$
    – Zohaib Aarfi
    Commented Feb 2, 2015 at 17:27

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