I was reading and extract from "Fundamentals Laws of Mechanics", 1980, by I.E Irodov, § 1.2. 'Kinematics of a Solid' and came across this insane text where the author asks me to imagine a 'solid performing two elementary rotations'.

Moreover, the vector introduced ($d\boldsymbol{\phi}$) can be shown to satisfy the basic property of vectors, that is, vector addition. Indeed, imagine a solid performing two elementary rotations, $d\boldsymbol{\phi}_1$ and $d\boldsymbol{\phi}_2$, about different axes crossing at a stationary point O. The resultant displacement $d\mathbf{r}$ of an arbitrary point A of the body, whose radius vector with respect to the point O is equal to $\mathbf{r}$, can be represented as follows: $$d\mathbf{r} = d\mathbf{r}_1 + d\mathbf{r}_2 = [d\boldsymbol{\phi}_1, \mathbf{r}] + [d\boldsymbol{\phi}_2, \mathbf{r}] = [d\boldsymbol{\phi}, \mathbf{r}]$$ $where$ $$d\boldsymbol{\phi}= d\boldsymbol{\phi}_1 + d\boldsymbol{\phi}_2 \tag{1.12} $$ i.e: the two given rotations, $d\boldsymbol{\phi}_1$ and $d\boldsymbol{\phi}_2$, are equivalent to one rotation through the angle $d\boldsymbol{\phi}= d\boldsymbol{\phi}_1 + d\boldsymbol{\phi}_2$ about the exis coinciding with the vector $d\boldsymbol{\phi}$ and passing through the point O.

enter image description here

I need help trying to imagine/visualize two 'infinitesimal rotations'. I have managed to prove that finite rotations do not obey the law of vector addition by doing two finite rotations on an imaginary object but I am unable to do the same with infinitesimal rotations and verify that infinitesimal rotations, indeed, obey the law of vector addition.

Please explain what the author is trying to convey. I do understand that infinitesimal rotations are vectors but I find it really hard to comprehend the passage given above.

  • $\begingroup$ Hi Yashas. Can you clarify what you are asking? Your question seems to be about how to visualise two infinitesimal rotations. $\endgroup$ Jun 13, 2016 at 6:01
  • $\begingroup$ Yes, I always try to understand everything in the textbook. I have problems visualising infinitesimal rotations. The author says imagine which I am not able to do. I can imagine and prove that finite rotations do not obey the law of vector addition but I am not able to do the same with infinitesimal rotations and verify that they obey the law of vector addition. $\endgroup$
    – Yashas
    Jun 13, 2016 at 6:11
  • 1
    $\begingroup$ In non-standard analysis, which is the right formal system to take, the product of two infinitesimals is infinitely smaller than any of them. Thus, dϕdr=0 or dϕdθ=0 and you only keep the infinitesimals to first order $\endgroup$
    – user65081
    Jun 13, 2016 at 7:03
  • $\begingroup$ Yes but the author says "imagine a solid performing two elementary rotations". How can someone imagine something which isn't even real? I don't what was running in the author's mind when he wrote it and I am trying so hard to understand that. $\endgroup$
    – Yashas
    Jun 13, 2016 at 7:10
  • $\begingroup$ Do you even think if there is an imaginative way to verify that infinitesimal rotations are first order tensors? $\endgroup$
    – Yashas
    Jun 13, 2016 at 7:11

1 Answer 1


enter image description here

All vectors, except $\:\mathbf{r}\:$, are infinitesimals.
I wonder if the author (Irodov) makes use of this result anywhere in his textbook.


The infinitesimal rotation of a vector $\:\mathbf{r}\:$ around the direction of a unit vector $\:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right)\:$ by an infinitesimal angle $\:\mathrm{d}\theta\:$ may be represented as follows :

\begin{equation} \mathbf{r}' =\mathbf{r}+\mathrm{d} \mathbf{r} =\mathbf{r}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}\mathbf{r}=\left(\mathrm{I}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}\right)\mathbf{r} \tag{01} \end{equation}

The infinitesimal rotation matrix is explicitly expressed as

\begin{equation} \mathrm{I}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} +\mathrm{d}\theta \begin{bmatrix} 0 & -n_{3} & n_{2} \\ n_{3} & 0 & -n_{1} \\ -n_{2}& n_{1} & 0 \end{bmatrix} = \begin{bmatrix} 1 & -n_{3}\mathrm{d}\theta & n_{2}\mathrm{d}\theta \\ n_{3}\mathrm{d}\theta & 1 & -n_{1}\mathrm{d}\theta \\ -n_{2}\mathrm{d}\theta& n_{1}\mathrm{d}\theta & 1 \end{bmatrix} \tag{02} \end{equation}

Now, let the two infinitesimal rotations of the question :

\begin{align} \mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times} & =\mathrm{I}+d\phi_{1} \mathbf{n}_{1}\boldsymbol{\times} \tag{03a} \\ \mathrm{I}+d\boldsymbol{\phi}_{2}\boldsymbol{\times} & =\mathrm{I}+d\phi_{2} \mathbf{n}_{2}\boldsymbol{\times} \tag{03b} \end{align}

The composition of these rotations is : \begin{align} & \left(\mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}\right)\left(\mathrm{I}+d\boldsymbol{\phi}_{2}\boldsymbol{\times}\right) \\ & =\mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}+d\boldsymbol{\phi}_{2}\boldsymbol{\times}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}(d\boldsymbol{\phi}_{2}\boldsymbol{\times}\\ &=\mathrm{I}+\underbrace{\left(d\boldsymbol{\phi}_{1}+d\boldsymbol{\phi}_{2}\right)}_{d\boldsymbol{\phi}}\boldsymbol{\times}+d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \\ &=\mathrm{I}+d\boldsymbol{\phi}\boldsymbol{\times}+d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \tag{04} \end{align}

where $\:F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right)\:$ the following finite linear transformation \begin{equation} F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right)\mathbf{r} \equiv \left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{r} \right)\mathbf{n}_{2}-\left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\mathbf{r} \tag{05} \end{equation}

In equation (04) we have the 2nd order term \begin{equation} d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \ne d\phi_{2} d\phi_{1}F\left(\mathbf{n}_{2},\mathbf{n}_{1}\right) \tag{06} \end{equation}

So, to 2nd order the two infinitesimal rotations don't commute. But to 1rst order ($d\phi_{1} d\phi_{2} \approx 0$) since \begin{equation} d\boldsymbol{\phi} = d\boldsymbol{\phi}_{1}+d\boldsymbol{\phi}_{2} = d\boldsymbol{\phi}_{2}+d\boldsymbol{\phi}_{1} \tag{07} \end{equation}

the two rotations commute and we have a form of vector addition of infinitesimal rotations.

  • $\begingroup$ "All vectors, except r, are infinitesimals." I did not understand that. Infinitesimal means something very small (infinitely small). I understand infinitesimal rotation as a very small rotation that doesn't really rotate because its too small but gives a finite rotation when integrated. $\endgroup$
    – Yashas
    Jun 13, 2016 at 13:59
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    $\begingroup$ @Yashas Samaga : Infinitesimals are the magnitudes, not the directions. So, for example, $\:d\boldsymbol{\phi}=d\phi \mathbf{n}, d\boldsymbol{\phi}_{1}=d\phi_{1} \mathbf{n}_{1}, d\boldsymbol{\phi}_{2}=d\phi_{2} \mathbf{n}_{2}\:$ where $\:d\phi ,d\phi_{1} ,d\phi_{2} \:$ are infinitesimal scalars and $\:\mathbf{n}, \mathbf{n}_{1}, \mathbf{n}_{2}\:$ unit vectors. But you don't ask what an infinitesimal vector is, you ask how to imagine equation (1.12) and this is what is included in the Figure. $\endgroup$
    – Frobenius
    Jun 13, 2016 at 16:29
  • $\begingroup$ @ Frobenius: (Sorry to intervene in your post but I had no other way to communicate with you.) Please can you help me in my questions on "Ampere's Force Law" from Maxwell's treatise?So please (if possible) can you give me your rough gmail id so that I can give you my questions there through my gmail ([email protected]). By rough gmail id, I mean that gmail id which you don't use for any personal or official purposes. If you have no rough gmail, you can make a new one(if possible). $\endgroup$
    – N.G.Tyson
    Jun 13, 2016 at 18:25
  • $\begingroup$ @ Frobenius: If you don't like the idea, then let me know. I will find out another way to ask questions.(You already helped me a lot). $\endgroup$
    – N.G.Tyson
    Jun 13, 2016 at 18:26

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