On page 181 in Peskin & Schroeder they say that we consider the integral (intensity) $$\tag{1}\mathcal{I}(\mathbf{v},\mathbf{v}') = \int\frac{\mathrm{d}\Omega_\hat{k}}{4\pi}\,\frac{2(1-\mathbf{v}\cdot\mathbf{v}')}{(1-\hat{k}\cdot\mathbf{v})(1-\hat{k}\cdot\mathbf{v}')}-\frac{m^2/E^2}{(1-\hat{k}\cdot\mathbf{v}')^2}-\frac{m^2/E^2}{(1-\hat{k}\cdot\mathbf{v})^2}$$ in the extreme relativistic limit (ERL). Then they say that in this limit most of the radiated energy comes from the two peaks in the first term of $(1)$. Is this because in the ERL one can take the mass $m$ to be zero: $m=0 ~(\text{ERL})$ so only the first term in $(1)$ remains?

The next question is what I really want an explanation for: They claim that in (ERL) we break up the integral into a piece for each peak, let $\theta=0$ along the peak in each case. Integrate over a small region around $\theta=0$, as follows: $$\tag{2}\mathcal{I}(\mathbf{v},\mathbf{v}') \approx \int_{\hat{k}\cdot\mathbf{v}= \mathbf{v}'\cdot\mathbf{v}}^{\cos\theta=1}\mathrm{d}\cos\theta\,\frac{(1-\mathbf{v}\cdot\mathbf{v}')}{(1-v\cos\theta)(1-\mathbf{v}\cdot\mathbf{v}')} \\[1cm] +\int_{\hat{k}\cdot\mathbf{v}'= \mathbf{v}'\cdot\mathbf{v}}^{\cos\theta=1}\mathrm{d}\cos\theta\,\frac{(1-\mathbf{v}\cdot\mathbf{v}')}{(1-v'\cos\theta)(1-\mathbf{v}\cdot\mathbf{v}')}. $$

Then they claim that the lower limit are really not that important, but in any case: my question is where the lower limits comes from and how about the replacement in the denominator of the integrand, in other words: How does one go from $(1)$ to $(2)$?

I should add that $\mathbf{v}, \mathbf{v}'$ are the particle velocity before and after interaction. I think one must have access to the book to understand the question unfortunately, other than that, I just want to understand where the lower limits of the integral comes from.

NOTE: PS are working in a frame where $p^0=p^{'0}=E$ which (according to them) implies $$k^\mu=(k,\mathbf{k}),~~p^\mu=E(1,\mathbf{v}),~~p^{'\mu}=E(1,\mathbf{v'}) $$ where (I guess) $k=||\mathbf{k}||. $ Then for instance $(k_\mu p^\mu)^2$ becomes $(Ek)^2\left(1-\frac{\mathbf{k}}{k}\cdot\mathbf{v}\right)^2$ which is (I assume) one of the denominators (up so some factors) in $(1)$. So I guess the correct notation in $(1)$ should be $$\tag{3}\mathcal{I}(\mathbf{v},\mathbf{v}') =\int \dots-\frac{m^2/E^2}{\left(1-\hat{\mathbf{k}}\cdot\mathbf{v}'\right)^2}-\frac{m^2/E^2}{\left(1-\hat{\mathbf{k}}\cdot\mathbf{v}\right)^2}. $$

Overall, bad notation is used IMO on the pages near 181 in PS.


1 Answer 1


I am not sure if my answer is correct, from what I understood:

(i) At the relativistic limit, $m<<E$, so the second and third terms in (6.15) will be negligible, just as you said.

(ii) P&S is aiming at $\hat{k}$ parallel to $\mathbf{v}$ or $\mathbf{v}'$ and integrating around $\theta=0$, since (6.15) peaks there (also ref my comment below this answer). For the lower limit of integration in the first term in your Eq. (2), $\hat{k}$ is parallel with $\mathbf{v}'$. Thus $$\hat{k} \cdot \mathbf{v} = v \cos \theta \approx \cos \theta $$, since at the relativistic limit, $v \approx 1$ (ref bottom of p180). And $$\mathbf{v}' \cdot \mathbf{v}= v' v \cos \theta \approx \cos \theta$$. So the lower limit is valid for any $\theta$.

For the lower limit in the second term in your Eq. (2), $\hat{k}$ is parallel to $\mathbf{v}$. Similar with the first term, the lower limit is valid for any $\theta$ as well.

(iii) Once $$\hat{k} \cdot \mathbf{v} = \mathbf{v}' \cdot \mathbf{v} $$, we replace $ 1- \hat{k} \cdot \mathbf{v} $ as $1- \mathbf{v}' \cdot \mathbf{v} $, we will get $$ \frac{ 1-\mathbf{v}' \cdot \mathbf{v} }{ ( 1- v' \cos \theta)( 1- \mathbf{v}' \cdot \mathbf{v} ) } $$. At the relativistic limit, $v \approx v' \approx 1$. I think we can use $v$ and $v'$ interchangeably. Since P&S is aiming at a small region around $\theta =0$, the integrand will not change. Namely our replacement is valid in this small region.

  • $\begingroup$ OK but can we just choose as we wish to have $k$ parallel to $v'$ and then parallel to $v$? What is the physical interpretation? Could you please elaborate a little on that? $\endgroup$ Feb 24, 2014 at 14:08
  • 1
    $\begingroup$ It's quite confusing here. From what I understood, as P&S said under (6.16), "The integrand of $\mathcal{I}(\mathbf{v},\mathbf{v}')$ peaks when $\hat{k}$ is parallel to either $\mathbf{v}$ or $\mathbf{v'}$". The denominator in the first term of (6.15) is $(1-\hat{k} \cdot \mathbf{v} ) (1-\hat{k} \cdot \mathbf{v}' ) =(1-v \cos \theta_{kv} ) (1-v' \cos \theta_{kv'} ) \approx (1- \cos \theta_{kv} ) (1- \cos \theta_{kv'} )$ . It has to be either $\theta_{kv}=0$ or $\theta_{kv'}=0$ for the peaks. $\endgroup$
    – user26143
    Feb 24, 2014 at 14:24
  • 1
    $\begingroup$ I think the book should explain better, but thanks for your answer. $\endgroup$ Feb 24, 2014 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.