I have been asked several times that “why do we use orthogonal axes in coordinate systems?” and I was always replying that “because of simplicity”. But, today morning, someone asked me that question and after my answer, he said “how?”

I thought and three examples came to mind as how orthogonal coordinate systems make simplicity.

$1$. In a plane (for instance), if we use non-orthogonal coordinate system; we will have:

($\vec e$ is unit vector) enter image description here $$\vec r=r\large{\frac{\sin (\gamma-\theta)}{\sin \gamma}} \vec e_X+r\large{\frac{\sin \theta}{\sin \gamma}}\vec e_Y$$ If $\gamma=90$; then $\sin \gamma =1$ and denominators will be removed as shown below: $$\vec r=r\cos \theta \;\vec e_x+r\sin \theta \;\vec e_y$$ This simplicity is appeared when we use dot product of vectors also. For example when we want to calculate work done by a force (in a plane), we write $W=\int \vec F\cdot\mathrm d\vec x$. If we break force $\vec F$ to two orthogonal vector that one of them is parallel to $\mathrm d\vec x$; then work done by the other force (perpendicular to $\mathrm d\vec x$) will be zero.

$2$. In a plane (for instance), if $\vec a=\vec b+\vec c$, then magnitude of $\vec a$ is equal to $$a=\sqrt {\large{b^2+c^2-2bc\cos \theta}}$$ ($\theta$ is the angle between vectors $\vec b$ and $\vec c$). If $\theta=90$; then $\cos \theta=0$ and we will have $$a=\sqrt {\large{b^2+c^2}}$$

$3$. In cross product of vectors, if we break them to orthogonal vectors, then our calculations will become much easier. Because we know that $$|\vec a\times \vec b|=ab\sin \theta$$ ($\theta$ is the angle between $\vec a$ and $\vec b$) If we use orthogonal coordinates for breaking vectors, the angles that are involved in our problem will be $0$ or $90$ that make our calculations so easy.

So, I have two questions:

$1$. Is there any other reason that why we use orthogonal coordinate systems except simplification? $2$. Is there any other example about simplification?

Thanks a lot.

P.S.$1$. We are talking about classical mechanics only.

P.S.$2$. About first question: I think there is no reason except simplification. Because I believe that physical laws are independent of coordinate systems. I think physical phenomenon occur as they are; regardless of whether we can observe or measure them or not. But, I wanted to ask if I am missing something.

P.S.$3$. About second question: I wanted to ask for more useful examples because I believe that when students ask questions like current question, it will be better for education that we express many examples as much as we can. Because more examples, more clarification and interest for students.

  • $\begingroup$ It's started by Rene Descartes.. $\endgroup$ Jun 14, 2016 at 10:56
  • $\begingroup$ @philip_0008 I don't mean Cartesian coordinate system only. $\endgroup$
    – lucas
    Jun 14, 2016 at 11:09
  • 2
    $\begingroup$ An example: When finding components of forces (and other vector quantities). If coordinate axes are perpendicular, they do not have any components in common. Any force vector can then be split in unique components along each axis. Then in a later step, fx in Newton's laws, we can consider all forces along the y-direction e.g. for itself. For a non-perpendicular coordinatesystem, one component would depend on another component, and we couldn't split the calculations into equations along each axis for themselves. $\endgroup$
    – Steeven
    Jun 14, 2016 at 12:17
  • 1
    $\begingroup$ @Steeven "For a non-perpendicular coordinate system, one component would depend on another component" This is not correct. If $\vec F=\vec F_1+\vec F_2$, then $\vec a=\vec a_1+\vec a_2$ ($\vec a$ is acceleration vector). $\vec a_1$ is independent of $\vec F_2$ and $\vec a_2$ is independent of $\vec F_1$. Because $\vec a_1=\large{\frac 1m}\vec F_1$ and $\vec a_2=\large{\frac 1m}\vec F_2$ $\endgroup$
    – lucas
    Jun 14, 2016 at 12:33
  • $\begingroup$ @lucas Ahh, yes. Sorry for that - What I wanted to describe was the goal of having components that account for unique directions. $\endgroup$
    – Steeven
    Jun 14, 2016 at 13:50

2 Answers 2


I'm not sure if it helps you with your students, but maybe gives you some background: I guess the underlying reason for orthogonal basis vectors is that you are implicitly using a euclidean metric that will just have diagonal values. These would e.g. be

$$g_\mathrm{\mu\nu, ~euclidean}=I=\pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \quad{\text{or}}\quad g_\mathrm{\mu\nu,~spherical}=\pmatrix{1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2 \theta}$$

for euclidean or spherical coordinate systems.

A specific vector $u = (u^1, u^2, u^3)$ with the coordinates $u^1$, $u^2$, $u^3$ belongs to a specific metric (and thus basic vectors). E.g. to describe a specific point, for euclidean, it will be the values $(x,y,z)$, for spherical coordinates it will be other coordinates $(r, \theta, \phi)$ (that you calculate from the transformation, e.g. $r=\sqrt{x^2+y^2+z^2}$, so they describe the same point).

Having only diagonal elements implies an orthogonal basis. If you change your basis (e.g. using axis $Y$ instead of $y$) without changing your metric, your formulas will get more complicated.

You are absolutely correct that the physics and thus the measurements should not depend on the coordinate system you use. A length of 1 meter will be a length of 1 meter (not considering relativity). To measure e.g. the length of a scalar, you have to calculate the inner/scalar product according to your metric, so the specifics of the metric are included there.

E.g. consider the work of a constant force $F$ along $x$. In tensor notation, the scalar product will be

$$W=\left<F,x\right> = g_{\mu\nu}~F^{\mu}x^{\nu}$$

This is, up to this point, independent from your exact metric. If and only if you use your euclidean metric $g_{\mu\nu}=g_\mathrm{\mu\nu,~euclidean}=I$,

$$W=\left<\vec F,\vec x \right> = g_\mathrm{\mu\nu,~euclidean} ~F^{\mu} x^{\nu} = F^1 x^1 + F^2 x^2 + F^3 x^3 = F_x x + F_y y + F_z z$$

this scalar product will be the same as you are used to.

With any other metric, your formula would still look the same up to the point where coordinates come into play (you actually don't even have to know the dimension), and you will get the same scalar result (as long as the specific coordinates belong to the metric/basic vectors), just the metric will look differently. So you wouldn't have to do much if you change from $y$ to $Y$, except for other coordinates and including the metric where it is needed, but your metric would then have values not only on the diagonal. This will give you the same mixed terms you had before (the absolute value can of course not change), but it happens only at the "last step" of the calculation. That is btw. still true for e.g. spherical coordinates that still are orthogonal, they have a "simple" scalar product too.

So the underlying simplification that you use is that you start with an orthogonal base/metric, and that is of course a natural, good and obvious idea (unless you want to torture your students), but it could have been any another way. So the main reason why using an orthogonal basis is simpler is that you (want to) keep your obvious metric. Changing to a non-orthogonal base while keeping the orthogonal metric will make your formulas more complicated. Using tensors (and that basically means changing the base AND the metric) will keep the formulas the same, but will make the metric more complicated.

Formulas in tensor notation are independent of your concrete base. More precise, these formulas are Lorentz-invariant. The very nice thing is that basically all formulas you teach don't look much different as a tensor (because they describe the "concept"), there will just be a difference when you enter specific coordinates. First differences arise when you use vector products, e.g. $\vec\omega = \vec{v} \times \vec{r}$, where another basis might introduce a "-"-sign to the $\vec\omega$, while written as a tensor, the sign will not change (unfortunately, a tensor-$\vec\omega$ it will not be a simple vector with 3, but with 9 components). And maybe not all calculation you do will be "tensor safe" (e.g. writing $\vec F \vec x = F_x x + F_y y + F_z z$ explicitly without adding the metric would not be tensor-safe anymore - not that you should loose any sleep over that).

The unfortunate (or for your students: good) thing is that you probably have no reason to use tensors in your class, so this information will not help you explain it unless you maybe include special relativity (although introducing tensors just for this will be overkill, and there is probably not much to learn from it unless you want to study physics or mathematics).

  • $\begingroup$ Thanks very very much because of your attention and time. Your explanations were so useful and I learnt a lot. Thank you again. $\endgroup$
    – lucas
    Jun 14, 2016 at 17:43

We don't always use orthogonal coordinate frames. For example working with three phase motors it's sometimes convenient to work with a three axis coordinate system in a plane. Convenience, simplicity set aside, the main reason we most often work with orthogonal reference frames is the concept of dimension.

We can express an n-dimensional linear system as a set of n linear independent equations. For a linear system, the number of linearly independent eigenvectors defines the dimension of the system.

  • 1
    $\begingroup$ Thanks for your answer. Are you saying that we use orthogonal coordinate system because of linearity? But, linearity holds in non-orthogonal coordinate system too. $\endgroup$
    – lucas
    Jun 14, 2016 at 14:12
  • $\begingroup$ evident independence, not linearity $\endgroup$
    – hyportnex
    Jun 14, 2016 at 15:54
  • $\begingroup$ @hyportnex right. There are models and there is physical reality. Reality sometimes might appear linear but if you look closer ... My answer, addressing more the world of models. $\endgroup$
    – docscience
    Jun 14, 2016 at 16:15
  • $\begingroup$ @lucas " linearity holds in non-orthogonal coordinate system too" - this can be true, but it makes it more difficult to do eigen-analysis, determine dimensionality using these frames of reference. $\endgroup$
    – docscience
    Jun 14, 2016 at 16:17
  • $\begingroup$ @docscience "it makes it more difficult" So, you agree with me that the only reason is simplicity? $\endgroup$
    – lucas
    Jun 14, 2016 at 17:35

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