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In physics and engineering it is common to convert between different coordinate systems - spherical, polar, Cartesian, e.t.c. - depending on the problem. Physically, they are all clearly equivalent and it shouldn't matter which one we use.

Nevertheless, when solving problems involving volume or surface integrals, we always have to add the Jacobian matrix to the expression if we are any other coordinate system but Cartesian, just as if we were converting from the Cartesian system.

Take the expression: $$ \iiint_V \nabla \cdot \textbf{u}\ dV $$

It seems intuitive that we should be able to go from this expression and express $V$, $\textbf{u}$, $\nabla$, and $dV$ in whatever coordinate system we want at this point, but plugging the spherical coordinates straight into the expression yields the wrong answer - one has to add the Jacobian just as if the first step was expressing everything in Cartesians and converting to polars. Why do Cartesians get such special treatment? My expectation would be because the eigenvectors are constant throughout space, but I would appreciate a more thorough explanation.

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  • $\begingroup$ I admit you have kinda lost me, (an easy thing to do), but when you go back to the derivation of Gauss, Stokes laws, etc, what coordinate system do they use in explaining it, in the early stages of introduction to these equations? $\endgroup$ – user175021 Nov 12 '17 at 14:06
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    $\begingroup$ For Cartesian coordinates you have to include the Jacobian too -- it just happens to be trivial. $\endgroup$ – AccidentalFourierTransform Nov 12 '17 at 15:50
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    $\begingroup$ Why do Cartesians get such special treatment? The volume element is defined as $dV=\prod_k dx_k$. It's the definition. $\endgroup$ – StephenG Nov 14 '17 at 1:09
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Because our world is (locally) Euclidean.

In an Euclidean world straight lines are favored and we subconsciously defined things like distance, area and volume using straight lines. If the real object we are studding is not made of straight lines, we cut it into small pieces, until we get approximately straight lines and then we sum up these small pieces to get the “real” value.

In an Euclidean world there is a direct correlation between Cartesian coordinates and lengths in the real world. For example, a circle in Cartesian coordinates is a circle in the real world; but a circle in the real world is a rectangle in polar coordinates.

Consider a surface bounded by the following curves in our world: $y=a_0 /x$, $y=a_1 /x$, $y=b_0 x$, $y=b_1 x$

enter image description here

In a world where their vertical lines correspond to a curve defined as $u=xy$ in ours and horizontal lines correspond to the curves $v=y/x$, this complex figure becomes simply a square with sides $a_1-a_0$ and $b_1-b_0$ and area $A=(a_1-a_0)(b_1-b_0) \, u.a$.

enter image description here

In our world it is not that simple and the area bounded by those lines is definitely not that. Now the Jacobian is the tool we use to convert the value of a measurement from one coordinate system to the value that would be obtained if the measurement were performed in Cartesian coordinates. It represents the infinitesimal relation between lengths of an object when drawing in one system to the other. Infinitesimal lengths can always be considered straight lines.

It is important to clarify that Cartesian coordinates is not the only “straight” coordinate system. Any system formed by non orthogonal straight lines can be considered. The thing is that we would consider those systems as sheered and then not necessarily natural.

PS: The Jacobian is always there. We just define $|J|=1$ for a Cartesian coordinate system because the results from this system have direct relation with our world. Some alien species living in a world with a very strong gravitational field (next to a black hole for example ;-), would have defined their $|J|=1$ to a different system.


EDIT:

Using $$I_c= \int \int dx \, dy$$ As the double integral in Cartesian coordinates, and $$I_p= \int \int dϱ \, dθ$$ As the double integral in polar coordinates.

Because of the correspondence of Cartesian coordinates and the “real world”, one could notice, either experimentally or by logic (using methods like squaring the circle and cutting the circle) that $I_c$ gives the same face value as the area bounded by the limiting curves. Therefore, one can define the area as $$A= \int \int \, dA=\int \int dx \, dy$$ And the element of area $$dA=dx \, dy$$ However, $I_p$ gives a different value (in this case the perimeter of the circle). From this point onwards, 2 things can be done;

1. Define $│J│=1$ in the starting system

With this definition, applying the Jacobian when passing from the original system to the next will allow you to get the same numeric value, but double integrals will have all sort of meaning, depending on the meaning it has on the starting system, making it a little ambiguous at minimum.

2. Define $│J│=1$ in the Cartesian system

With this definition, all systems will have not only the same numeric value, but all double integrals will result in area of the surface bounded by the curves.

The same for volume integrals.

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  • $\begingroup$ Are you suggesting we can't use spherical coordinates in a Cartesian world? $\endgroup$ – ZeroTheHero Nov 12 '17 at 15:48
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    $\begingroup$ @ZeroTheHero No. What I’m saying is that, if you use polar coordinates, you must be aware to use the “length converter” (Jacobian) otherwise the end result will not match to the “real world”. $\endgroup$ – J. Manuel Nov 12 '17 at 15:57
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The Jacobian is the multiple integral analogue of the u-substitution method. For example, if you want to make the substitution $x = 2u$ in an integral you are effectively introducing a change of coordinates from $x$ to $u$ and you have to put $dx = 2 \space du$ in place of of $dx$. Similarly for the multidimensional case you make the replacement.

$$dxdy= |J(x(u,v),y(u,v))| dudv$$

Without a change of variables theres no need. You could use the Jacobian if you wanted by considering yourself to be making the substitution $$x(x,y) = x,\space y(x,y)=y$$ but the Jacobian would just be the identity matrix which

has a determinant of $1$ giving you $dxdy=dxdy$

Similarly, in the one variable case the substitution $x=x$ gives $dx=dx$

If your integral was initially in, say, polar coordinates you would have to use the Jacobian to convert back to Euclidean. Euclidean coordinates don't get a pass, its just that generally you switch from Euclidean coordinates to spherical/polar etc instead of the other way around.

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enter image description here

Because the infinitesimal surface element and the infinitesimal volume element, that in Cartesian coordinates are $\:\mathrm{dS}=\mathrm dx\mathrm dy\:$ and $\:\mathrm{dV}=\mathrm dx\mathrm dy\mathrm dz\:$, in general curvilinear coordinates are not $\:\mathrm{dS}=\mathrm du\mathrm dv\:$ and $\:\mathrm{dV}=\mathrm du\mathrm dv\mathrm dw$. The Jacobians are the ''correction'' coefficients. See Figure 01 for example.

Also, an everywhere non-zero Jacobian is a necessary and sufficient condition for the inversibility of coordinates transformations.

So, if you want to find the double integral in Figure 02 : \begin{equation} A=\iint\limits_{\mathrm S} f\mathrm {dS} \tag{01} \end{equation} then by Cartesian coordinates \begin{equation} A=\iint\limits_{\mathrm S} f \!\left(x,y\right)\mathrm dx\mathrm dy \tag{02} \end{equation} since in these coordinates \begin{equation} \mathrm {dS}=\mathrm dx\mathrm dy \tag{03} \end{equation} There is no need of a Jacobian here.

But if it's convenient to use other curvilinear in general coordinates $\left(u,v\right)$, then under the coordinate transformation : \begin{equation} x=x\left(u,v\right)\,, \quad y=y\left(u,v\right) \tag{04} \end{equation} we have \begin{equation} \mathrm {d S}=\:\underbrace{\left\vert \! \left(\dfrac{\partial \mathbf r}{\partial u}\mathrm d u\right)\!\! \boldsymbol{\times}\!\!\left( \dfrac{\partial \mathbf r}{\partial v}\mathrm d v \right)\!\right\vert}_{\textbf{algebraic magnitude}} \!=\underbrace{\left\vert \dfrac{\partial \mathbf r}{\partial u}\boldsymbol{\times}\dfrac{\partial \mathbf r}{\partial v}\right\vert}_{\textbf{algeb. magn.}}\,\mathrm d u\,\mathrm dv = \underbrace{ \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u}\\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} \end{vmatrix}}_{\textbf{Jacobian}} \mathrm du\,\mathrm dv=\dfrac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\mathrm du\,\mathrm dv \tag{05} \end{equation} Also we have \begin{equation} g \left(u,v \right)=f\left[x\left(u,v\right),y\left(u,v\right)\right] \tag{06} \end{equation} Inserting expressions (05) and (06) in (01) we have \begin{equation} A=\iint\limits_{\mathrm S}g\left(u,v\right)\dfrac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\mathrm du\mathrm dv=\iint\limits_{\mathrm S}g\left(u,v\right)\left(\dfrac{\partial x}{\partial u}\dfrac{\partial y}{\partial v}-\dfrac{\partial y}{\partial u}\dfrac{\partial x}{\partial v}\right)\mathrm du\mathrm dv \tag{07} \end{equation} We need a Jacobian here.

enter image description here

Figure 02 (3D version)

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  • $\begingroup$ By saying: “Because the infinitesimal surface element and the infinitesimal volume element, that in Cartesian coordinates are $dS=dxdy$ and $dV=dxdydz$", I think that you are just rephrasing the OP’s question, since you are admitting that for Cartesian coordinates $│J│=1$. What the OP is asking is why Cartesian coordinates have such a privilege in first place, why it is so special. $\endgroup$ – J. Manuel Nov 14 '17 at 7:52
  • $\begingroup$ @J. Manuel : I apologize, maybe you must see my answer as a failure. $\endgroup$ – Frobenius Nov 14 '17 at 9:31
  • $\begingroup$ Sorry for any misunderstanding. I see your answer as a valid one. Perhaps you should elaborate on why $│J│=1$ for Cartesian coordinates, and the rest of your answer will hold perfectly. $\endgroup$ – J. Manuel Nov 14 '17 at 12:11
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The original question: "Why do you have to include the Jacobian for every coordinate system, but the Cartesian?"

Perhaps you should elaborate on why │J│=1 for Cartesian coordinates, and the rest of your answer will hold perfectly.

Why does some one not write $1 x = 2$ and instead of $x = 2$? This is the same question.

Most people define extent of a rectangular object in $\Bbb{R}^n$ in a Cartesian sense as the product of the $\Bbb{R}^1$ extents. In that case of the transformation of an Cartesian original n-dimensional space to Cartesian $\Bbb{R}^n$ is the identity matrix, which is the matrix equivalent of $1$ for coordinate transformations. But, why bother with putting a 1 in your formulas?

$$ \int _{\varphi _1}^{\varphi _2}\int _{\theta _1}^{\theta _2}\int _{\tau _1}^{\tau _2}f((\left( \begin{array}{ccc} \tau & \theta & \varphi \\ \end{array} \right)\cdot\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right))^\mathsf{T})d\tau d\theta d\varphi $$ where $f$ is a real valued function of a row 3-dimensional vector. But, like not writing the $1$ in the algebra equation, why not just use $f(\left( \begin{array}{ccc} \tau & \theta & \varphi \\ \end{array} \right))$? All the matrix multiplication is doing is relabeling $\left(\begin{array}{ccc} \tau & \theta & \varphi \\ \end{array} \right)$ as $\left(\begin{array}{ccc} x & y & z \\ \end{array}\right)$.

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