# $\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$ Does it have any physical meaning?

I was reading a Physics book. Then I saw an equation which was looking like this :

$$\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$$

I was thinking it from just a Calculus book. I can see lots of variable (momentum) changing (adding all the "pieces") respect to their position. When I saw the equation a question came to my mind which is : Does it have any physical meaning if I just think it from Mathematics? Is there possible way to evaluate it?

Usually the book wrote that

$$\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z=V4\pi (\frac{hv}{c})^2 h\frac{dv}{c}$$

I am not sure if I wrote it correctly cause I had taken the image of that book illegally that's why I couldn't take whole equation pic but I didn't notice it that time.

I wonder which I was reading that wrote single integral. But it I was directly searching through online I found that another had wrote 6 integral

$$\int\int\int\int\int\int dx\, dy\, dz\, d p_x\, dp_y\, dp_z$$

which proves that they are taking integral for each function.

Note : I am honestly saying I don't understand anything of it now. And I may not understand answer properly also. But I am just leaving the question to read in future.. :)

• Welcome to stackexchange. I voted to close your question because it is not about mathematics. If you saw this in a physics book then I am sure it has physical meaning. If you have a real question about how the integral is calculated, ask that, with enough detail about what you have tried. Commented Sep 21, 2021 at 16:11
• Without bounds or an integrand, there's no way to evaluate it. But yes, integrating over position and momentum together is physically meaningful, another way of saying it is integration over phase space (in this case the phase space of a single particle). The use of just one $\int$ sign is a common thing in physics when dealing with integrals over more than three scalar variables.
– Ian
Commented Sep 21, 2021 at 16:11
• @lan Can't we deal without bounds? If not than why? While evaluating indefinite-integrals I just get a constant with the main one.. +1 for your edit of your last comment... Commented Sep 21, 2021 at 16:15
• @EthanBolker I am leaving the comment for moderator attention since they take a look at every comments. If it's really off-topic here than I will say to migrate it to Physics SE and I am not going to cross post it there (I don't care of reputation)... Commented Sep 21, 2021 at 16:17
• Just without bounds would be fine, since then it's presumed to be integrated over everything, but then you need an integrand to provide decay (just the integral of 1 over $\mathbb{R}^6$ is infinite which doesn't mean anything).
– Ian
Commented Sep 21, 2021 at 16:44

## 1 Answer

It is a common notation to indicate the integration on the 6-dimensional phase space $$(PS)$$ of a single particle with finite mass (otherwise the momentum $$\mathbf{p}$$ is not the canonical conjugate of the position $$\mathbf{x}$$).

In the end, $$d^3x \, d^3p$$ is just a way to indicate the phase-space volume element (or, better, the volume 6-form that provides the natural integration measure on the PS).

Edit: the integration bounds are omitted because they are physically "obvious": the whole volume $$V \subset \mathbb{R}^3$$ in which the position $$\mathbf{x}$$ of the particle is allowed to vary and the physically possible values of the conjugate momentum $$\mathbf{p}$$. Typically $$\mathbf{p}\in \mathbb{R}^3$$ and $$\mathbf{x}\in V$$. We could formally write that $$PS = V\times \mathbb{R}^3$$, where $$\times$$ is the usual cartesian product.

Therefore: $$\int d^3x = V$$, where $$V$$ indicates also the measure of the position domain $$V \subset \mathbb{R}^3$$. Some examples of how you should interpret this kind of integrations (that are just standard Riemann multidimensional integrals):

$$\int_{PS} f(\mathbf{x}, \mathbf{p}) d^3x d^3p = \text{Integrate f over the whole PS.}$$

$$\int_{PS} f( \mathbf{p}) d^3x d^3p = V \int_{\mathbb{R}^3} f( \mathbf{p}) \, d^3p$$

$$\int_{PS} f(p) d^3x d^3p = 4 \pi V \int_{\mathbb{R}^+} f( p) \, p^2dp \quad (\text{where } p=|\mathbf{p}|)$$

Note that the last equality should help you to understand the example reported in the question (even though I do not understand the factors $$c$$ and $$h$$ there and the meaning of $$v$$, but I guess that $$p = \frac{h}{c}v$$):

$$\int_{PS} f(p) d^3x d^3p = \int_{\mathbb{R}^+} f( p) \, V4\pi \left(\frac{hv}{c}\right)^2 \frac{h\, dv}{c} \quad (\text{where } p=|\mathbf{p}|=\frac{h\, v}{c})$$

Final note: here are some Terence Tao's notes on the phase space (that is a useful concept in classical mechanics, kinetic theory, statistical physics, quantum mechanics, chaos theory, dynamical systems).

• not accepting it as correct answer. Cause it doesn't helpful a lot. But it is understandable for me... that's why I have upvoted.. Commented Sep 21, 2021 at 16:38
• @IstiakShovon What are you expecting? You got a specific answer to a vague question of unspecified origin. Commented Sep 21, 2021 at 16:43
• Thank you Istiak. You do not have to accept it, it's too early... Maybe some better answer will appear. I am a bit confused about your doubt. What exactly is not clear? The notation? or the physical meaning? the reason behind why sometimes we are interested in doing phase space integration? Commented Sep 21, 2021 at 16:44
• @IstiakShovon I updated the question with some explicit examples, I hope it' more useful now. Cheers. Commented Sep 21, 2021 at 23:42
• Ow! It was really helpful. I didn't have any specific question. But it was very understandable someone said that,"The equation might in masters Physics" but I understood your derivation without getting to University either.. (I just self-study.. I don't go to anywhere to learn.. take a random book and read theories..) Thanks anyway. Enjoy your reputation :D Commented Sep 22, 2021 at 8:10