# Notation confusion for differential volume

Can anyone help me with an explanation of the following notation. I am a bit confused:

Lets say we have some type of integral and in the end we write different differential, such as:

$$d\vec r ,\quad d^3\vec r,\quad dxdydz,\quad dV.$$

How are these with each other related? Can we express the first one in components, or the second one.

Any explanation would help. I am asking because i am looking at this thread:

Speed distribution in 1 dimension

And i don't understand it. First of all PDF (like normal distribution for example) have no differential part like there is for the velocity in the above link. And then what is the difference between maxwell velocity distribution and maxwell speed distribution.

What i know ( please correct me if i am wrong):

$$dP(x)=f(x)dx$$ which physically means that we are searching the probability that $$x$$ is in the interval $$x$$ and $$x + dx$$.

Then for the velocity/speed (i don't know which of the 2 terms to use) distribution (in 3D), in analogy with the above equation we would have:

$$dP(\vec v)=f(\vec v)dv_x dv_y dv_z /$$. Is $$dv_x dv_y dv_z = d\vec v$$ or $$d^3 \vec v$$. I am confused by the notations etc.

• The 3-dimensional (scalar) volume element is $$dV = d^3 \vec{r} = dx dy dz$$. All three symbols are different notation for the same thing, with the main difference that $$dx dy dz$$ commits to using a Cartesian coordinate system. You will use this measure to integrate over a volume. A typical example would be to take a function $$f(\vec{r})$$ and an integral over some region $$\Omega$$ (for instance, $$\Omega$$ the interior of a sphere) $$\begin{equation} \int_\Omega dV f(\vec{r}) = \int_\Omega d^3 \vec{r} f(\vec{r}) = \int dx \int dy \int dz f(x,y,z) \end{equation}$$ Note in the last line I've intentionally expanded the volume integral into three integrals over Cartesian coordinates and expressed the arguments of $$f$$ as 3 coordinate values instead of one vector value.
• A vector-valued line element is given by $$d\vec{r}$$. This is a measure factor for a line integral. A line integral is an integral of a vector function over a one dimensional path $$\Gamma$$ through a larger space. For example, perhaps you are computing the work done when you integrate the force $$\vec{F}$$ along some path $$\Gamma$$ that goes from point $$A$$ to point $$B$$, parameterized by $$\lambda$$. A typical example of a line integral would be $$\begin{equation} \int_\Gamma d\vec{r} \cdot \vec{F}(\vec{r}) = \int d\lambda \left(\frac{d\vec{r}}{d\lambda} \cdot \vec{F}(\vec{r})\right) = \int d \lambda \left(\frac{dx}{d\lambda} F_x + \frac{dy}{d\lambda}F_y + \frac{dz}{d\lambda}F_z\right) \end{equation}$$ where the path $$\Gamma$$ is described by the function $$\vec{r}(\lambda)=x(\lambda) \hat{e}_x + y(\lambda) \hat{e}_y + z(\lambda) \hat{e}_z$$, and $$\hat{e}_i$$ is a unit vector in the $$i$$-th direction.
• Thx for the explanation. The $d^3 \vec r$ part confused me, because i assumed that we have 3 vectors and each of them has 3 components, but i kinda get it now. thx – imbAF Apr 15 at 19:01
• @imbAF Yeah it's basically lazy notation. I would prefer writing $d^3 x$. – Andrew Apr 15 at 19:01
• You can integrate over three vectors like $\sum_{ijk} \int d x^{(1)}_i \int d x^{(2)}_j \int d x^{(3)}_k T^{ijk}$ where $\vec{x}^{(1)}, \vec{x}^{(2)}, \vec{x}^{(3)}$ are three separate vectors and $T^{ijk}$ is a tensor. But... don't worry about that if it doesn't make sense. – Andrew Apr 15 at 19:04