Questions about Lorentz Invariance of Integrals

Question

I have this really basic (probably silly) question about Lorentz transformation on integrals that I couldn't understand:

So I read on Peskin that the Lorentz invariant measure is $$\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p} = \int \frac{d^4p}{(2\pi)^4}(2\pi)\delta(p^2-m^2).$$ And I have a few things that I don't understand about this integral/measure:

1. How should those integral measures (or if they even should) transform under Lorentz transformation?

For example. An integration $\int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p} f(\vec{p})= \int \frac{d^4p}{(2\pi)^4}(2\pi)\delta(p^2-m^2)f(p)$. Under Lorentz transformation, should it transform to

$\to \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{p}} f(\vec{\Lambda p})= \int \frac{d^4p}{(2\pi)^4}(2\pi)\delta(p^2-m^2)f(\Lambda p)$ or

$\to \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\Lambda p}} f(\vec{\Lambda p})= \int \frac{d^4p}{(2\pi)^4}(2\pi)\delta((\Lambda p)^2-m^2)f(\Lambda p)$ or

$\to \int \frac{d^3\Lambda p}{(2\pi)^3}\frac{1}{2E_{\Lambda p}} f(\vec{\Lambda p})= \int \frac{d^4(\Lambda p)}{(2\pi)^4}(2\pi)\delta((\Lambda p)^2-m^2)f(\Lambda p)$

I am thinking maybe it should be the second because if we Lorentz transform the system (thus the function $f(p)$), there is no reason to also transform the coordinate($d^4p$ and $d^3p$ I mean)as well right? I know realistically it might not matter since $d^4p$ transforms to itself. And we don't have to worry about the integration limits since its over the whole space(?) However, conceptually, which would be correct, to transform $d^4p$ or not?

2. The above sort of lead to my next question. Given that under Lorentz transformation, $d^4p \to d^4(\Lambda p) = d^4p$, isn't it that every integration $\int d^4p f(p)$ should be Lorentz invariant, whether $f(p)$ is invariant under Lorentz transformation or not? Because under Lorentz transformation:

$\int d^4p f(p) \to \int d^4p f(\Lambda p) = \int d^4(\Lambda p) f(\Lambda p) = \int d^4p f(p)$

Because that's basically a relabeling of the original expression?

I know this can't be right because when we are constructing Lorentz invariant Lagrangian we made sure we use the Lorentz invariant measure and made sure the lagrangian is manifestly Lorentz invariant. But why is that necessary? What was wrong about my above statements? Is Lorentz invariance not defined as getting the same value for the integration after a Lorentz transformation?

Answer 1

Part 1

Let $p^\mu\rightarrow(p')^\mu = \Lambda^{\mu}_{\ \ \nu} p^\nu$ for some Lorentz transformation $\Lambda$.

Under this transformation, the measure transforms as usual with a Jacobian factor \begin{equation} {\rm d}^4 p \rightarrow \left|\frac{\partial p'^\mu}{\partial p^\nu}\right| {\rm d}^4 p' = |\det \Lambda| {\rm d}^4 p' = {\rm d}^4 p' \end{equation} since $|\det \Lambda|=1$ for a Lorentz transformation.

Meanwhile the delta function transforms in the usual and opposite way with 1 over a Jacobian factor \begin{equation} \delta(p^2 - m^2) \rightarrow \frac{1}{|\det \Lambda|} \delta((p')^2 - m^2) = \delta((p')^2 - m^2) \end{equation} again using $|\det \Lambda|=1$. (Also note $p^2=(p')^2$).

Part 2

The premise of your question is that $f(p)$ is not invariant under a Lorentz transformation. However, the first step you write down $\int {\rm d}^4 p f(p) \rightarrow \int {\rm d}^4 p f(\Lambda p)$ implicitly assumes that $f$ is transforming as a scalar under Lorentz transformations.

Think bigger: suppose $f(p) \rightarrow f(\Lambda p) + g(\Lambda,p)$ under a Lorentz transformation, for some function $g(\Lambda,p)$. Then the integral will clearly be different in the new frame in a way you can't get rid of by relabeling dummy variables. This is the kind of thing that we don't want to happen in relativistic field theories. (Well, sort of. You actually can't avoid a thing like $g$ when you start building theories of spin 1 particles and you need to include gauge invariance, but I'll let you discover that one later).

Answer 2

Thanks @Andrew for the great answer. I think the answer was very clear itself. I just want to add this answer to the second part of my own question which is sort of how I thought the problem through after I read Andrew's comment. In case anyone or maybe myself in the future get stuck on the same kind of problem.

So I think I am a bit lost on how things like an integral should transform under Lorentz transformation. An example helped me think it out:

Consider a field $\phi$ on a 2D $x-y$ plane. And think only about the simplest case of a rotation of the frame. Suppose I want to do an integral like $\int_{\Omega} d^2\vec{x} \phi^2(\vec{x })$ where I set $\Omega$ to be the first quadrant of the system.

When we rotate the frame(the $x-y$ plane), then we should look at things in the new frame. The field $\phi$ is transformed in this new frame to $\phi'$. And the coordinate of the new frame let's call it $x'$.

So in this new frame, if we want to calculate the same quantity(integral), we should still integrate over the first quadrant of my new frame. And the variables to be integrated should be the coordinates of my new frame $x'$. So in the new frame, the integral should be like: $\int_\Omega d^2\vec{x'} \phi'(\vec{x'})$. Here $\Omega$ is the first quadrant of my new frame.

And if we are integrating over a scalar field, the field transforms trivially. This post address this problem very clearly:https://physics.stackexchange.com/a/315649/278544. Basically, because the field stay at the same space time point no matter which frame you are in, the field strength at coordinate $\vec{x}$ should equal the field strength at coordinate $\vec{x'}$. Because $\vec{x}$ and $\vec{x'}$ are representing the same space time point but in two different coordinate systems. That is to say $\phi'(\vec{x'}) = \phi(\vec{x})$. And because the transformation of coordinate satisfies $\vec{x}=R^{-1}\vec{x'}$. So we have the relationship between $\phi$ and $\phi'$: $\phi'(\vec{x'})=\phi(R^{-1}\vec{x'})$. Put this back to our integration, the integration becomes:

$\int_\Omega d^2\vec{x'} \phi(R^{-1}\vec{x'})$ where $\Omega$ represents the first quadrant.

To see the relationship(or difference) between the integral in the rotated frame and our original integral, we can just relabel $\vec{x'}$ to $\vec{x}$. And the integral becomes:

$\int_\Omega d^2\vec{x} \phi(R^{-1}\vec{x})$

compare at the original:

$\int_\Omega d^2\vec{x} \phi(\vec{x})$

They are different because the integral is not over the whole space.

Alternatively, we can also express all the transformed $\vec{x'}$ with our old coordinate $\vec{x}$ using $R\vec{x}=\vec{x'}$. So the integral can also be written as:

$\int_\Omega d^2(R\vec{x}) \phi(R^{-1}R\vec{x}) = \int_{R^{-1}\Omega} d^2\vec{x} \phi(\vec{x})$ Where $\det{R}=1$ is used.

This is to say that the integral in the new frame is equivalent to an integral in the old frame but over a rotated integration area.

I think the last point of view is what got me confused. So in my original question, I think I should first transform all the quantities(in the case of my question, $f$ and any measures like $E_p$ and $\delta (p^2-m^2)$) to the new frame. Then I should integrate over the new coordinate $p'$. Then I can substitute $p'$ with $p$ by $p'^\mu=\Lambda^\mu_{\ \nu} p^\nu$. And that should be what was described by @Andrew.