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Let's say we have a double integral in spacetime, \begin{equation} \int d^4 x_1 d^4 x_2 f(x_1, x_2)= \int d^3 \vec{x}_1 d^3 \vec{x}_2 \int d x_1^0 d x_2^0\,\, f(x^0_1, x^0_2,\vec{x}_1, \vec{x}_2) \end{equation} Where $x_1$, $x_2$ are 4-vectors in Minkowski space. On the r.h.s. are we allowed to change the dummy variables $\vec{x}_1\leftrightharpoons \vec{x}_2$ without exchanging the time variables also, $x_1^0, x_2^0$?

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    $\begingroup$ The function $f$ has eight inputs, and you can name the dummy variables whatever you'd like. Did you have anything other than a trivial relabeling in mind? $\endgroup$
    – J. Murray
    Commented Jun 1, 2021 at 3:18
  • $\begingroup$ Thanks for your comment. I was actually considering a second order term of the Dyson series in perturbative QFT, see here: physics.stackexchange.com/questions/640174/…. My initial guess is that you can't exchange the space coordinates without also exchanging the time coordinates, as they are related. x_1 is associated with x_1^0 and so on. If I am relabeling my space coordinates then the associated time coordinates also need to be relabeled. $\endgroup$ Commented Jun 1, 2021 at 15:09
  • $\begingroup$ Well again, you can relabel your coordinates trivially - call $(x_1^0,x_1^1,x_1^2,x_1^3) \equiv (a,b,c,d)$ or $(\star, \circ,\triangle,\square)$ if you'd like. But that won't get you anywhere unless you're also assuming some kind of symmetry of the integrand or something along those lines. $\endgroup$
    – J. Murray
    Commented Jun 1, 2021 at 15:57
  • $\begingroup$ Ok, I realize your point now. I should have been more specific then. How about if I have something like this in the integrand $e^{i(p_1x_1+p_2x_2)}+e^{i(p_1x_2+p_2x_1)}$, where $p_1$, $p_2$ are 4-momentum and $p_1x_1$.... etc are the inner products between energy-momentum 4-vector and space-time 4-vector in Minkowski space? $\endgroup$ Commented Jun 1, 2021 at 16:40

2 Answers 2

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You can certainly relabel your integration variables however you'd like, but that doesn't get you anything all by itself. Dummy variables in an integral are essentially placeholders, and the pen strokes you use to represent them are arbitrary.

How about if I have something like this in the integrand $e^{i(p_1x_1+p_2x_2)}+e^{i(p_1x_2+p_2x_1)}$, where $p_1, p_2$ are 4-momentum and $p_1x_1....$ etc are the inner products between energy-momentum 4-vector and space-time 4-vector in Minkowski space?

You can perform a trivial relabeling of your variables if you'd like, but all that's going to do is muddle the interpretation of the various quantities. For example, looking at the first term, you would have

$$\exp[i(p_1^0 x_1^0 + \vec p_1 \cdot \vec x_2 + p_2^0 x_2^0 + \vec p_2 \cdot \vec x_1)]\equiv \exp[i(p_1 \tilde x + p_2 \tilde y)]$$ where $\tilde x\equiv(x_1^0,\vec x_2)$ and $\tilde y\equiv (x_2^0,\vec x_1)$. In other words, if you relabel $\vec x_1\leftrightarrow \vec x_2$, you're simply choosing to make $\vec x_2$ the position of particle $1$ and $\vec x_1$ the position of particle 2. This is confusing and probably pointless, but not illegal.

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  • $\begingroup$ Okay this makes sense. I think I have been able to clarify what I was asking about. Mathematically we are allowed to relabel anything and play with the definitions of various quantities. But muddling the interpretation of quantities is what I referring to earlier. $p_{1(2)}=(p_{1(2)}^0, \vec{p}_{1(2)})$ are defined quantities, e.g. the energy and momentum of electron 1(2). So probably we are not allowed to change the definition arbitrarily later. Thank you for your time! $\endgroup$ Commented Jun 1, 2021 at 19:37
  • $\begingroup$ @upquark31415 I've edited my answer in light of the fact that the $x$'s, not the $p$'s, are the dummy variables. Relabeling doesn't change the interpretation of the $p$'s, but rather of the $x$'s. $\endgroup$
    – J. Murray
    Commented Jun 1, 2021 at 19:56
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Yes, you are allowed. The product $d^3 \vec{x}_1 d^3 \vec{x}_2$ is just the dot product of two position vectors. This is commutative. So $d^3 \vec{x}_1 d^3 \vec{x}_2=d^3 \vec{x}_2 d^3 \vec{x}_1$. Likewise, the time differentials can be interchanged. These are just scalars.

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    $\begingroup$ $d^3\vec x_1 d^3 \vec x_2$ is certainly not the dot product of two position vectors... $\endgroup$
    – J. Murray
    Commented Jun 1, 2021 at 3:07
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    $\begingroup$ If you want to get particularly technical, a volume element is a pseudoscalar density of weight -1, see e.g. here. $\endgroup$
    – J. Murray
    Commented Jun 1, 2021 at 3:47
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    $\begingroup$ Consider the space-time position Lorentz 4-vector \begin{equation} \boldsymbol{X} \boldsymbol{=}\left(x_0,x_1,x_2,x_3,\right)\boldsymbol{=}\left(x_0,\boldsymbol{x}\right)\qquad x_0\boldsymbol{=}c\,t \tag{01}\label{01} \end{equation} and the space-time infinitesimal displacement Lorentz 4-vector \begin{equation} \mathrm d\boldsymbol{X} \boldsymbol{=}\left(\mathrm dx_0,\mathrm dx_1,\mathrm dx_2,\mathrm dx_3,\right)\boldsymbol{=}\left(\mathrm dx_0,\mathrm d\boldsymbol{x}\right)\qquad \mathrm dx_0\boldsymbol{=}c\,\mathrm dt \tag{02}\label{02} \end{equation} $\endgroup$
    – Frobenius
    Commented Jun 1, 2021 at 4:37
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    $\begingroup$ ...then \begin{equation} \mathrm d\mathcal V \boldsymbol{=} \mathrm d^4\boldsymbol{X}\boldsymbol{=}\mathrm dx_0\mathrm dx_1\mathrm dx_2\mathrm dx_3\boldsymbol{=}\mathrm dx_0\mathrm d^3\boldsymbol{x} \tag{03}\label{03} \end{equation} is the space-time 4-dimensional infinitesimal $''$volume$''$ while \begin{equation} \mathrm d\mathrm v \boldsymbol{=} \mathrm dx_1\mathrm dx_2\mathrm dx_3\boldsymbol{=}\mathrm d^3\boldsymbol{x} \tag{04}\label{04} \end{equation} is the usual 3-dimensional infinitesimal volume. $\endgroup$
    – Frobenius
    Commented Jun 1, 2021 at 4:38
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    $\begingroup$ ...by the way, note that the space-time 4-dimensional infinitesimal $''$volume$''$ $\mathrm d\mathcal V$ is a Lorentz invariant scalar while the usual 3-dimensional infinitesimal volume $\mathrm d\mathrm v$ is not. $\endgroup$
    – Frobenius
    Commented Jun 1, 2021 at 4:39

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