# Derivation of simple Lorentz boost identity, $dp'_z/dp_z$

Given a Lorentz boost in the $$z$$-direction, in natural units we can write

$$p'_z = \gamma(p_z+\beta E)\,,\;\;\;\;\;E' = \gamma(E+\beta p_z)\,.$$

In Peskins & Schroeder's Introduction to QFT, p. 23, the following claim is made:

$$\frac{dp'_z}{dp_z} \equiv \frac{E'}{E}$$

I cannot make this happen. My attempt is as follows.

$$\begin{split} \frac{dp'_z}{dp_z} &= \gamma\left(1 + \beta\frac{dE}{dp_z}\right)\,;\\ \frac{dE}{dp_z} &= \frac{d}{dp_z}\left(\frac{p}{\beta}\right) \\ &= \frac{d}{dp_z}\left(\frac{[p_x^2+p_y^2+p_z^2]^{1/2}}{\beta}\right) \\ &= \frac{p_z}{\beta [p_x^2+p_y^2+p_z^2]^{1/2}} = \frac{p_z}{\beta p}\\ \implies \frac{dp'_z}{dp_z} &= \gamma\left(1 + \frac{p_z}{p}\right) \end{split}$$

Since $$E = \gamma m$$ and $$p = \gamma m\beta$$, we have $$p = \beta E$$, therefore

$$\frac{dp'_z}{dp_z} = \gamma\left(1 + \frac{1}{\beta}\frac{p_z}{E}\right)\,.$$

By the definition of $$E'$$, we have

$$\frac{E'}{E} = \gamma\left(1 + \beta\frac{p_z}{E}\right) \neq \frac{dp'_z}{dp_z}\,.$$

Have I done something stupid?

For reference, the book derivation in question is the proof that the quantity $$\delta^{(3)}(\mathbf{p}-\mathbf{q})$$ is not Lorentz-invariant, where $$\mathbf{p}$$ and $$\mathbf{q}$$ are momenta. Omitting a couple of lines, they say that

$$\delta^{(3)}(\mathbf{p}-\mathbf{q}) = \delta^{(3)}(\mathbf{p'}-\mathbf{q'})\cdot\frac{dp'_z}{dp_z} = \delta^{(3)}(\mathbf{p'}-\mathbf{q'})\cdot\frac{E'}{E}\,.$$

• Yes. Try seeing $dE/dp_z= p_z/E$. Dec 29 '19 at 21:16
• Thank you for taking the time to comment, Professor (I should say...!) I see it using $E = \sqrt{p^2 + m^2}$. Dec 29 '19 at 21:32

The parameter $$\beta$$ is related to the boost between the two frames, so in my solution, relating the unprimed $$E$$ and $$p$$ using $$\beta$$ was (wildly) incorrect. The correct method uses the following.
$$\begin{split} E &= \sqrt{p^2 + m^2} \\ \implies \frac{dE}{dp_z} &= \frac{p_z}{\sqrt{p^2 + m^2}} = \frac{p_z}{E} \end{split}$$