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Soft Bremsstrahlung: why $\hat{k}\cdot\mathbf{v}= \mathbf{v}'\cdot\mathbf{v}$?

+ 4 like - 0 dislike
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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.

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user Love Learning
asked Feb 23, 2014 in Theoretical Physics by Love Learning (160 points) [ no revision ]

1 Answer

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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.

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user user26143
answered Feb 24, 2014 by user26143 (340 points) [ no revision ]
Thanks, I'll go through your answer tomorrow.

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user Love Learning
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?

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user Love Learning
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.

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user user26143
I think the book should explain better, but thanks for your answer.

This post imported from StackExchange Physics at 2014-03-30 03:09 (UCT), posted by SE-user Love Learning

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