Graduate Showing Delta^3(p-q) is Not Lorentz Invariant

[TroyElliott]

From page 22 of P&S we want to show that $\delta^{3}(\vec{p}-\vec{q})$ is not Lorentz invariant. Boosting in the 3-direction gives $p_{3}' = \gamma(p_{3}+\beta E)$ and $E' = \gamma(E+\beta p_{3})$. Using the delta function identity $\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),$ we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$

This is where I am confused, how is this step arrived at? From what I can see we have
$$\delta^{3}(\vec{p'}-\vec{q'}) = \delta(p_{1}-q_{1})\delta(p_{2}-q_{2})\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q})).$$

We can further write $$\delta( \gamma((p_{3}-q_{3})+\beta (E_{p}-E_{q}))) = \delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma.$$

From here I don't see how to transform $\delta((p_{3}-q_{3})+\beta (E_{p}-E_{q}))/\gamma$ into the form $\delta(p_{3}-q_{3})\frac{dp_{3}}{dp'_{3}}.$

Any ideas on how to go about this? Thanks!

 

[strangerep]

[...] delta function identity $\delta(f(x)-f(x_{0})) = \frac{1}{|f'(x_{0})|}\delta(x-x_{0}),$ we get $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'}) \frac{dp'_{3}}{dp_{3}}.$$This is where I am confused, how is this step arrived at?

Think of $p'$ as a function of $p$. I.e., substitute $f(x) \to p'(p)$ and $f(x_0) \to p'(q)$. Then the $\frac{dp'_{3}}{dp_{3}}$ is essentially just the term $1/|f'(x_0)|$.

HTH.

 

[TroyElliott]

Think of $p'$ as a function of $p$. I.e., substitute $f(x) \to p'(p)$ and $f(x_0) \to p'(q)$. Then the $\frac{dp'_{3}}{dp_{3}}$ is essentially just the term $1/|f'(x_0)|$.

HTH.

Thanks!

So taking this route we end up with $$\delta^{3}(\vec{p}-\vec{q}) = \delta^{3}(\vec{p'}-\vec{q'})\frac{E'}{E},$$

where the $\frac{dp'_{3}}{dp_{3}}$ term became $\frac{E'}{E}.$ When using the Dirac delta identity above, we should technically evaluate $\frac{dp'_{3}}{dp_{3}}$ at $q_{3}$, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force $p_{3}$ to equal $q_{3}$ when integrated?

Thank you very much.

 

[strangerep]

When using the Dirac delta identity above, we should technically evaluate $\frac{dp'_{3}}{dp_{3}}$ at $q_{3}$, right? Does this not need to be explicitly done since we are dealing with a delta function, which will force $p_{3}$ to equal $q_{3}$ when integrated?

Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

 

[vanhees71]

I think it's most easily to see considering momentum-space volume elements $\mathrm{d}^3 p$. It's important to note that we deal with "on-shell particles" for the asymptotic free states, i.e., $p^0=E_{\vec{p}}=\sqrt{\vec{p}^2+m^2}$. This means you can effectively work with functions dependent on four-momentum with the on-shell constraint understood. Then you can consider the invariant (under proper orthochronous Poincare transformation)
$$\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2)=\frac{\mathrm{d^3} p}{2E_{\vec{p}}},$$
which implies that
$$\delta^{(3)}(\vec{p}-\vec{p}') E_{\vec{p}}$$
is a Lorentz scalar.

 

[realanswers]

Strictly speaking, all that delta fn stuff only makes sense in the context of the theory of distributions (aka generalized functions). But after a relation involving delta fns is established using rigorous integral expressions, most people just use shorthand notation, sans integrals.

I still do not fully understand your answer and do not see how it answers the question. SO do we need to evaluate the derivate at $q_3$?

 

[strangerep]

I still do not fully understand your answer and do not see how it answers the question. So do we need to evaluate the derivate at $q_3$?

No.

If you want to work through this properly, start by proving formula (2.34), i.e.,
$$\delta\Big( f(x) - f(x_0) \Big) ~=~ \frac{1}{|f'(x)|} \; \delta( x - x_0)
~=~ \frac{1}{|f'(x_0)|} \; \delta( x - x_0) ~,$$where I have given you an extra hint by showing the middle step (which is all you really need here to perform the computation at the top of p23).

Btw, the $|\dots|$ should be seen as the modulus of a Jacobian -- this is necessary in general for the multivariate case.