First time asking here, so please forgive me if I'm doing anything wrong. I'm having quite a hard time with this equation: $$\mathbf{\omega} = \frac{1}{2} \left[ \left(\mathbf{i}\wedge\frac{d\mathbf{i}}{dt}\right)+ \left(\mathbf{j}\wedge\frac{d\mathbf{j}}{dt}\right)+ \left(\mathbf{k}\wedge\frac{d\mathbf{k}}{dt}\right) \right].$$ I don't understand where it comes from. I know it represents the angular velocity vector but I have no clue how we got here and neither my textbook nor the internet have proved helpful.

Reference: Elementi di Meccanica Razionale (Maria Letizia Bertotti, Giovanni Modanese).

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    $\begingroup$ So ... do you understand any of it? The meaning of the bolding? The derivatives? The wedge products? Other aspects of the expression? How far back do you expect an answer to start? Do you understand the same concept in some other notation? Remember that readers here don’t have the first clue where you are coming from. You’ve got to tell us those things. $\endgroup$ Commented Oct 31, 2019 at 18:44
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    $\begingroup$ I think more context is needed, not just of the equation, but also the notation. What do the parentheses indicate? Multiplication? Elements of a 3-vector? Does $\bf bold$ indicate a vector (in which case, why isn't $\omega$ in bold)? Are $\bf i$, $\bf j$, and $\bf k$ unit vectors, or something else? As you know, of course, angular velocity is $({\bf r} \times {\bf v})/r^2$, but this notation is unfamiliar to me and I can't quite make sense of it. $\endgroup$
    – Richter65
    Commented Oct 31, 2019 at 18:59
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    $\begingroup$ Which page? Which equations? $\endgroup$
    – Qmechanic
    Commented Nov 1, 2019 at 8:26
  • $\begingroup$ And I do understand the derivative and the wedge products and all of the mathematical expressions. Im having trouble associating this expression with the concept of angular velocity and the textbook provided no introduction to this topic, just this formula with no demonstration or explanation. Thanks again $\endgroup$ Commented Nov 1, 2019 at 12:24
  • $\begingroup$ The only thing that preceded this equation were the three Poisson formulas, which are said to be solved by this vector. $\endgroup$ Commented Nov 1, 2019 at 12:28

2 Answers 2


Looks very much like the formula at this link, except that wedge product is used instead of vector product and unit vectors instead of directional derivatives

EDIT(11/01/2019): So let me derive the formula. It should be noted that it was FUBAR by an editor, who omitted plus signs in the following original formula of the OP: $$\mathbf{\omega}=\frac{1}{2}\left(\mathbf{i}\wedge\frac{d\mathbf{i}}{dt}+\mathbf{j}\wedge\frac{d\mathbf{j}}{dt}+\mathbf{k}\wedge\frac{d\mathbf{k}}{dt}\right).$$

Following the KISS principle, I will use vector products $\times$, rather than wedge products $\wedge$. I assume that vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ are standard unit basis vectors in a rotating frame. The time derivatives of these unit vectors are: $$\frac{d\mathbf{u}}{dt}=\mathbf{\omega}\times\mathbf{u},$$

where $\mathbf{\omega}$ is the vector of angular velocity of the rotating frame and $\mathbf{u}$ is one of the unit basis vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ (I guess the above formula is what the OP calls "Poisson formula").

Let us vector-multiply the above formula by $\mathbf{u}$ from the left and use the formula for vector triple product: $$\mathbf{u}\times\frac{d\mathbf{u}}{dt}=\mathbf{u}\times(\mathbf{\omega}\times\mathbf{u})=\mathbf{\omega}-(\mathbf{u}\cdot\mathbf{\omega})\mathbf{u}.$$

Applying this formula separately to $\mathbf{i},\mathbf{j},\mathbf{k}$ and adding the results, we obtain: $$\mathbf{i}\times\frac{d\mathbf{i}}{dt}+\mathbf{j}\times\frac{d\mathbf{j}}{dt}+\mathbf{k}\times\frac{d\mathbf{k}}{dt}=\\= \mathbf{\omega}-(\mathbf{i}\cdot\mathbf{\omega})\mathbf{i}+\mathbf{\omega}-(\mathbf{j}\cdot\mathbf{\omega})\mathbf{j}+\mathbf{\omega}-(\mathbf{k}\cdot\mathbf{\omega})\mathbf{k}=2\mathbf{\omega}.$$

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    $\begingroup$ It looks nothing like that formula. A curl involves spatial derivatives. $\endgroup$
    – G. Smith
    Commented Nov 1, 2019 at 7:22
  • $\begingroup$ @G.Smith : and unit vectors are indeed spatial derivatives (see the second link in my answer). $\endgroup$
    – akhmeteli
    Commented Nov 1, 2019 at 7:24
  • $\begingroup$ If that is the case here, what are all the spatial and temporal derivatives acting on? $\endgroup$
    – G. Smith
    Commented Nov 1, 2019 at 7:27
  • $\begingroup$ @G.Smith : what is nabla (en.wikipedia.org/wiki/Nabla_symbol) in the formula in the first link of my answer acting on? $\endgroup$
    – akhmeteli
    Commented Nov 1, 2019 at 7:33
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    $\begingroup$ @Richter65 : "the cross product of two parallel vectors (e.g., $\mathbf{i}$ and $\frac{d\mathbf{i}}{dt}$) is zero" Only if $\mathbf{i}$ does not vary with time, as in the case of a frame reference attached to a (rotating) rigid body. $\endgroup$
    – akhmeteli
    Commented Nov 2, 2019 at 2:54

This is not common notation and it is unclear what it means. It looks like something from fringe physics, not mainstream physics. You should ignore it.

The notation is completely at odds with the common usage of $\mathbf i$, $\mathbf j$, and $\mathbf k$ to mean Cartesian unit vectors in the $x$, $y$, and $z$ directions. These unit vectors are constants so their time derivatives are zero, making the formula nonsensical.

The role of the parentheses and how they “combine” the three products is also completely obscure.

The formula appears to be an example of either ignorance or obscurantism, both of which are common attributes of fringe physics.

ADDENDUM: Thank you, Akhmeteli, for revealing the egregious mis-edit that made the formula look fringe-y. It was neither ignorant nor obscure, just mangled beyond recognition by an editor.

  • $\begingroup$ If any downvoter knows what it means, I would be interested in having a lesson about it and learning something new. I will be happy to upvote an answer linking to a mainstream book or paper using this notation. $\endgroup$
    – G. Smith
    Commented Nov 1, 2019 at 6:08
  • $\begingroup$ Thanks for the answer! It actually comes from this textbook amazon.it/Elementi-meccanica-razionale-Letizia-Bertotti/dp/… $\endgroup$ Commented Nov 1, 2019 at 12:25
  • $\begingroup$ I must admit I ignore what "fringe physics" are, but I believe this equation belongs to the field of Analytical Mechanics. $\endgroup$ Commented Nov 1, 2019 at 12:27
  • $\begingroup$ @G.Smith : Please see the EDIT to my answer. $\endgroup$
    – akhmeteli
    Commented Nov 2, 2019 at 5:10
  • $\begingroup$ @akhmeteli I've upvoted your answer and added an addendum to mine. I'm surprised that the OP did not notice the mis-edit. $\endgroup$
    – G. Smith
    Commented Nov 2, 2019 at 5:33

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