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 :)