# Peskin page 23

1. Jan 14, 2013

### silverwhale

Hello everybody!

Looking on solving equation 2.34 I am stuck at the fourth line. What is the calculation to do to go from:

$$\delta^{(3)} (p'- q') \gamma (1 + \beta \frac{dE}{dp_3})$$

to

$$\delta^{(3)} (p'- q') \frac{\gamma}{E} (E + \beta p_3).$$

2. Jan 14, 2013

### Staff: Mentor

Can the function of E(p3) be determined?

I can see how you can mix in a E by multipying by E / E making the last factor become:

β E dE / dp3

hence the step implies E dE / dp3 = p3 which would work if E = p3

3. Jan 14, 2013

### silverwhale

Well I had the same guess! :)

BUT, let's set p^2= E^2 -|p|^2=m^2 equal to 0 for a massless particle, which would be actually ok as we're looking at the KG eq, well then E^2= |p|^2. But then I should set p_1 = p_2 = 0.. that I don't understand..

4. Jan 14, 2013

### Avodyne

$$E=(p_1^2+p_2^2+p_3^2+m^2)^{1/2}$$
$${dE\over dp_3}={1\over2}(p_1^2+p_2^2+p_3^2+m^2)^{-1/2}(2p_3)={p_3\over E}$$
Faster:
$$E^2=p_1^2+p_2^2+p_3^2+m^2$$
$$E\,dE = p_1\,dp_1 + p_2\,dp_2 + p_3\,dp_3$$
Set $dp_1=dp_2=0$, and rearrange to get
$${dE\over dp_3}={p_3\over E}$$

5. Jan 15, 2013

### silverwhale

Got it!! It's that easy...

Many many Thanks! :)