Deltafunktion of 4-vectors for energy and momentum coversation

[Ulf]

Homework Statement

for a compton-scattering-problem, i want to show that:

\delta(p_{10}+k_1-p_0-k)=p_0\delta(\underline{k}_1(\underline{p}+\underline{k})-\underline{k}\underline{p})

Homework Equations

the momentum- and energy-conversion-law for two particle scattering.
\underline{p}+\underline{k}=\underline{p}_1+\underline{k}_1

relations of kinematic-invariants
:
\underline{k}\underline{p}=\underline{k}_1\underline{p}_1
\underline{k}_1\underline{p}=\underline{k}\underline{p}_1
\underline{k}_1(\underline{p}+\underline{k})=\underline{k}\underline{p}

here \underline{p} denotes the electron 4-vector \underline{p}=\{p_0,\overline{p}\}, the same for k the 4-vector describing the photon \underline{k}=\{k_0,\overline{k}\}. no subscript and subscribt 1 denote initial and scattered particles respectively.

The Attempt at a Solution

do i have to use: \delta(f(x))=\frac{1}{|f'(x_0)|}\delta(x-x_0)? but how?

 

[grey_earl]

First, it is always a good idea to show explicitely the dimensionality of the \delta distribution, so you want to show
\delta^4(p_{10}+k_1-p_0-k)=p_0\delta^3(\underline{k}_1(\underline{p}+\underline{k})-\underline{k}\underline{p})
Then you already see that both sides are of different dimensionality, so on the right side probably someone integrated without telling you. I would try splitting the \delta distribution, and integrating over \delta(p^0_{10}+k^0_1-p^0_0-k^0) using the on-shell relation \vec{p}² - (p^0)² = m², and similar for the k particle. If your relation is correct, this should work out after some algebra.

 

[Ulf]

first of all: thanks for you answer! the hint to check the dimensions was right. i guess my notation is a bit confusing. so i will choose a new one here in the solution:

\underline{p}=\{p^0,\vec{p}\} and
\underline{k}=\{k^0,\vec{k}\}, where \vec{p} and \vec{k} are the 3-momentum vectors of the electron and the photon respectively. after the scattering we have:
\underline{p}'=\{p^0',\vec{p}'\} and
\underline{k}'=\{k^0',\vec{k}'\}.
so the momentum-energy-law reads:
\underline{p}+\underline{k}=\underline{p}'+\underline{k}'now one can see that both sides of the equation that i want to show are of dim=1.

\delta^{(1)}(p^0'+k^0'-p^0-k^0)=p^0\delta^{(1)}(\underline{k}'(\underline{p}+\underline{k}) -\underline{p}\underline{k}), that is because on the LHS are only the 0-componets, and on the RHS we have scalar-products of 4-vectors, which are also 1-dimensional. now i didnt had to integrate, but to use the formula:

\delta(ax)=\frac{1}{|a|}\delta(x)

in this case p^0=\frac{1}{|a|}. so left to show:

p^0'+k^0'-p^0-k^0=\frac{1}{p^0}(\underline{k}'(\underline{p}+\underline{k}) -\underline{p}\underline{k}) which works out thus the RHS can be written as:

\frac{1}{p^0}(\underline{k}'(\underline{p}+\underline{k}) -\underline{p}'\underline{k}')=\frac{1}{p^0}(k^0'p^0+k^0'k^0-k^0'p^0'-\vec{k}'\vec{p}-\vec{k}'\vec{k}+\vec{k}'\vec{p}')=\frac{1}{p^0}(k^0'p^0+k^0'k^0-k^0'p^0'-\vec{k}'(\vec{p}-\vec{k}+\vec{p}'))

with use of k^0k^0'=p^0p^0' , k^0'p^0'=k^0p^0
\vec{p}+\vec{k}-\vec{p}'=\vec{k}' and ( \vec{k}')^2=(p^0)^2 we get the result.

 

[grey_earl]

Very good! So I also got confused by your notation :)