I read the mathematical derivation of fictitious forces on Wikipedia and I'm having trouble understanding it. I went on a few other sites looking for a better derivation, but they're all basically the same. I've also seen this answer already.

What I'm having trouble understanding is that, in every derivation, at some point, the expression $\sum_{i=1}^3 \hat{u}_i x_i$ (where $\hat{u}_i$ represents a unit vector of the non-inertial frame in the inertial frame, and $x_i$ is a coordinate of the particle in the non-inertial frame) is equated with "$\vec{x}_A$" or something similar (the position of the particle in the non-inertial frame). Likewise $\sum_{i=1}^3 \hat{u}_i \frac{dx_i}{dt}$ is equated with $\vec{v}_A$ and $\sum_{i=1}^3 \hat{u}_i \frac{d^2x_i}{dt^2}$ with $\vec{a}_A$.

This doesn't make any sense to me. The position vector in the accelerating frame is just $(x_1, x_2, x_3)$ since $x_1, x_2, x_3$ were defined to be the coordinates of the particle in the accelerating frame. It is not $\sum_{i=1}^3 \hat{u}_i x_i$; that gives you the position vector in the inertial frame.

For example here's what Wikipedia says:

The interpretation of this equation [$\mathbf{x}_{\mathrm{B}} = \sum_{i=1}^3 x_j \mathbf{u}_j$] is that $\mathbf{x}_{\mathrm{B}}$ is the vector displacement of the particle as expressed in terms of the coordinates in frame B at time t.

Even though the derivation clearly gives the correct answers, I simply can't understand this step.

  • $\begingroup$ I deleted an inappropriate comment and its response. $\endgroup$
    – David Z
    Jun 14, 2013 at 5:15

2 Answers 2


I personally think the descriptions on Wikipedia are rather confusing, so I'm going to write a self-contained derivation in my own words; hopefully this helps. Note: I'll use Einstein summation notation throughout.

In order to understand what's really going on in the derivation, I'm going to attempt to separate pure mathematics from physics. In particular, I'm going to derive a purely mathematically result, and then interpret that result at the end.

Some Pure Math

Let $\{\mathbf e_i\}$ denote the standard ordered basis on $\mathbb R^3$. Namely $\mathbf e_1 = (1,0,0)$ etc. For each real number $t$, let $\{\mathbf u_i(t)\}$ denote an orthonormal basis on $\mathbb R^3$, possibly different from the standard one at any given $t$, generated by a one-parameter family of rotations (elements of $\mathrm{SO}(3)$) $B(t)$; $$ \mathbf u_i(t) = B(t)\mathbf e_i. $$ Notice that any vector function $\mathbf w(t)$ can be written in either basis, namely there exists real functions $w^i(t)$ and $w^i_B(t)$ for which $$ \mathbf w(t) = w^i(t)\mathbf e_i = w^i_B(t)\mathbf u_i(t) $$ For notational convenience, for any such vector function we define $$ \mathbf w_B(t) = w^i_B(t)\mathbf e_i $$ and with this notation we have in particular that $$ \mathbf w(t) = B(t)\mathbf w_B(t) $$ Now, let any vector functions $\mathbf r(t), \mathbf X(t)$ be given, and define a vector function $\mathbf x(t)$ by $$ \mathbf x(t) = \mathbf X(t) +\mathbf r(t) $$ We let an overdot denote the derivative of any function with respect to its argument, and we note that we can write any vector function that appears here in either of the two bases defined above. Let's write $\mathbf r(t)$ in the $\mathbf u_i$ basis; $$ \mathbf x(t) = \mathbf X(t) + r_B^i(t)\mathbf u_i(t) $$ then we have $$ \dot{\mathbf x}(t) = \dot{\mathbf X}(t) + \dot r^i_B(t)\mathbf u_i(t) + r^i_B(t)\dot{\mathbf u}_i(t) $$ and $$ \ddot{\mathbf x}(t) = \ddot{\mathbf X}(t) + \ddot r^i_B(t)\mathbf u_i(t) + 2\dot r^i_B(t)\dot{\mathbf u}_i(t) + r^i_B(t)\ddot{\mathbf u}_i(t) $$ Now, before we go into physical intepretations, I find it useful to cast this equation in terms of the one-parameter family of rotations $B(t)$.
$$ \ddot{\mathbf x}(t) = \ddot{\mathbf X}(t) + B(t)(\ddot r^i_B(t)\mathbf e_i) + 2\dot B(t)(\dot r^i_B(t)\mathbf e_i) + \ddot B(t)(r^i_B(t)\mathbf e_i) $$ As with the notation above for a general vector $\mathbf w$, we define $$ \mathbf r_B(t) = r^i_B(t)\mathbf e_i, \qquad \dot{\mathbf r}_B(t) = \dot r^i_B(t)\mathbf e_i, \qquad \ddot{\mathbf r}_B(t) = \ddot r^i_B(t)\mathbf e_i $$ Then we can write the equation we derived as $$ \ddot{\mathbf x}(t) = \ddot{\mathbf X}(t) + B(t)\ddot{\mathbf r}_B(t) + 2\dot B(t)\dot{\mathbf r}_B(t) + \ddot B(t)\mathbf r_B(t) $$ Notice that we can rearrange this equation using the orthogonality of $B$ as follows (I omit the time argument from here on for the sake of notational brevity). \begin{align} \ddot {\mathbf r}_B = B^t\ddot{\mathbf x} - B^t\ddot{\mathbf X} - 2B^t\dot B\dot{\mathbf r}_B - B^t\ddot B\mathbf r_B \end{align} Now define \begin{align} \Omega_B = B^t\dot B \end{align} then it is not hard to verify that \begin{align} B^t\ddot B = \Omega_B^2 + \dot \Omega_B \end{align} so that we get \begin{align} \ddot {\mathbf r}_B = B^t\ddot{\mathbf x} - B^t\ddot{\mathbf X} - 2\Omega_B\dot{\mathbf r}_B - \Omega_B^2\mathbf r_B - \dot \Omega_B \mathbf r_B \end{align} Since $B$ is orthogonal, $\Omega_B$ is a skew symmetric matrix, so there is some vector $\mathbf\Omega_B$ such that for any vector $\mathbf w$, we have \begin{align} \Omega_B \mathbf w = \mathbf\Omega_B\times\mathbf w \end{align} and consequently \begin{align} \Omega_B^2\mathbf w = \mathbf\Omega_B\times(\mathbf\Omega_B\times\mathbf w), \qquad \dot\Omega_B \mathbf w = \dot{\mathbf\Omega}_B\times\mathbf w \end{align} so we finally get \begin{align} \boxed{\ddot {\mathbf r}_B = B^t\ddot{\mathbf x} - B^t\ddot{\mathbf X} - 2\mathbf\Omega_B\times\dot{\mathbf r}_B - \mathbf\Omega_B\times(\mathbf\Omega_B\times\mathbf r_B) - \dot{\mathbf \Omega}_B\times \mathbf r_B} \end{align}

What it all means physically.

Let a physicist named Alice set up a set of cartesian axes in an inertial frame, and let Bob set up cartesian axes in a non-inertial frame. The triples \begin{align} \mathbf x(t) &= (x^1(t) ,x^2(t), x^3(t)) \\ \mathbf X(t) &= (X^1(t), X^2(t), X^3(t)) \\ \mathbf r(t) &= (r^1(t), r^2(t), r^3(t)) \end{align} represent the real numbers that Alice would measure for the position of a particle, the position of the origin of Bob's coordinates, and the difference between those two respectively. The triple $$ \mathbf r_B(t) = (r^1_B(t), r^2_B(t), r^3_B(t)) $$ represents the real numbers that Bob would measure for the position of the same particle. In particular, the triples \begin{align} \dot{\mathbf r}_B(t) &= (\dot r^1_B(t), \dot r^2_B(t),\dot r^3_B(t)) \\ \ddot{\mathbf r}_B(t) &= (\ddot r^1_B(t), \ddot r^2_B(t), \ddot r^3_B(t)) \end{align} represent the real numbers that Bob would measure that give the components of the velocity and acceleration of the particle.

With all of this in mind, let's interpret the boxed equation. On the left is the acceleration of the particle as measured by Bob. On the right, the first term is just the acceleration $\ddot{\mathbf x}$ of the particle as measured by Alice with an extra rotation $B^t$ to account for the difference in the orientations of the axes of the two frames. The second term is the acceleration $\ddot{\mathbf X}$ of the origin of Bob's frame as measured by Alice with an extra rotation $B^t$ to account for the difference in orientations of the axes of the two frames. The third term is the familiar expression for the Coriolis acceleration, the fourth term is the centrifugal acceleration, and the last term is the Euler acceleration. In particular, multiplying through by the mass $m$ of the particle, each of the expressions on the right gives the standard expression for each of the various corresponding fictitious forces. \begin{align} m\ddot {\mathbf r}_B = B^t(m\ddot{\mathbf x}) - B^t(m\ddot{\mathbf X}) \underbrace{- 2m\mathbf\Omega_B\times\dot{\mathbf r}_B}_{\text{Coriolis}} \,\, \underbrace{- m\mathbf\Omega_B\times(\mathbf\Omega_B\times\mathbf r_B)}_{\text{centrifugal}} \,\, \underbrace{- m\dot{\mathbf \Omega}_B\times \mathbf r_B}_{\text{Euler}} \end{align}

The Vector $\mathbf \Omega_B$ - Important Subtlety.

Note that I defined the vector $\mathbf\Omega_B$ via the skew-symmetric matrix $\Omega_B = B^t\dot B$. In particular, $\mathbf\Omega_B$ is the unique vector for which $$ \Omega_B \mathbf w = \mathbf\Omega_B\times\mathbf w $$ However, notice that on Wikipedia, the analogous vector which I'll call $\mathbf \Omega_\mathrm{wiki}$ is defined as the vector for which $$ \dot{\mathbf u}_i = \mathbf\Omega_\mathrm{wiki}\times\mathbf u. $$ But notice that since $\mathbf u_i = B\mathbf e_i$, we have $$ \dot{\mathbf u}_i = \dot B \mathbf e_i = \dot B B^tB\mathbf e_i = \dot B B^t \mathbf u $$ so if we define $\Omega_\mathrm{wiki} = \dot B B^t$, then we see that Wikipedia's $\mathbf\Omega_\mathrm{wiki}$ is precisely that for which $$ \Omega_\mathrm{wiki}\mathbf w = \mathbf\Omega_\mathrm{wiki}\times \mathbf w $$ In particular, my $\Omega_B$ and Wikipedia's $\Omega_\mathrm{wiki}$ are related by similarity transformation; $$ \Omega_B = B^t\Omega_\mathrm{wiki} B $$ from which it follows, as you can show, that $$ \mathbf\Omega_\mathrm{wiki} = B\mathbf\Omega_B $$ In fact, in Wikipedia's convention, $\mathbf\Omega_\mathrm{wiki}$ represents the angular velocity components in the non-rotating basis while in my convention $\mathbf\Omega_B$ represents the angular velocity components in the rotating basis which is why I initially included a subscript $B$ when I defined it.

I hope this was better than Wikipedia. I think this is all pretty clear in my own head, let me know if my wording and notation was clear. If not, I'll attempt to edit for clarity.

  • $\begingroup$ Awesome, detailed answer. Just one physical interpretation question - if $B$ transforms between an inertial and purely rotating frame, could the terms on the right in the boxed equation naturally be identified with, say centrifugal force, some "baseline" force, Coriolis force, and Euler force, respectively? Or am I trying to read too much into it? $\endgroup$
    – user10851
    Jun 15, 2013 at 0:18
  • $\begingroup$ @ChrisWhite Thanks! That means a lot coming from you. I actually edited the end of the pure math and interpretations sections a bit to address your question. The new boxed equation explicitly displays how the various fictitious accelerations are obtained from the mathematical expression I had boxed previously. Let me know what you think. Also, feel free to point out errors if you find any; I didn't proofread all of this too closely. $\endgroup$ Jun 15, 2013 at 2:30
  • $\begingroup$ Thank you for taking the time to write out the derivation in detail and explain each step. I think on Wikipedia (and the other sites) they are possibly sloppy about which quantities are coordinate triples and which are "true" vectors (as in arrows somewhere out there in space). Your explanation is much easier to understand. $\endgroup$
    – Brian Bi
    Jun 15, 2013 at 4:25
  • $\begingroup$ Additionally, it's interesting that this derivation relies on $B$ being an orthogonal matrix (otherwise you don't get the nice property that you can solve for $\ddot{\mathbf{r}}_B$ by multiplying by $B^t$). Wikipedia fails to mention at all the requirement that the non-inertial frame have an orthogonal coordinate system. $\endgroup$
    – Brian Bi
    Jun 15, 2013 at 4:28
  • $\begingroup$ @BrianBi I'm glad it helped. By the way I just added a section at the end which addresses a subtlety regarding conventions for the vector $\mathbf\Omega$ which becomes significant if you want to do actual computation. Also, even if $B$ were not orthogonal, you could still do the inversion by multiplying by $B^{-1}$ which exists provided $B$ is a change of basis matrix, but the computations involving $\Omega$, particularly that fact that it is skew-symmetric, rely on orthogonality of the change of basis matrix. $\endgroup$ Jun 15, 2013 at 5:25

Note that you changed the notation, inverting $A$ and $B$. For me, as for Wikipedia, $B$ represent the non-inertial frame quantities.

I think your confusion is that $\hat{u}^{i}$ represents some unit vector of the non-inertial frame, they are time-dependent in frame $A$ since they move for observer in $A$, and move with the non-inertial frame $B$. An observer in $B$ will not see the $\hat{u}^{i}$'s moving, though. This is clear from the first equation on the Wikipedia page you cited, as well as on the figure there.


$$\mathbf{x}_{B} = x_{i}\hat{u}^{i}$$

is the position of the particle in the non-inertial frame $B$. As usual, the velocity is defined by

$$\mathbf{v}_{B} = \dot{x}_{i}\hat{u}^{i}$$

and the same for the acceleration. These definitions are true $in$ the frame $B$.

But for the observer in $A$, both the $x_{i}$ $and$ the $\hat{u}^{i}$ will move in time, so the more complete relation that Wikipedia gives.

  • 1
    $\begingroup$ Right, the actual unit vectors of the coordinate system in B appear to move to an observer in A, but of course they do not appear to move to an observer in B since the observer in B is using those unit vectors themselves for measurement. But what do the symbols $\hat{u}^i$ represent? Do they represent the unit vectors in the A frame, or do they represent the unit vectors in the B frame? $\endgroup$
    – Brian Bi
    Jun 14, 2013 at 7:03
  • $\begingroup$ @BrianBi Well, if you check the wikipedia page en.wikipedia.org/wiki/… you cited, you will clearly see they are defined in the $B$ frame. Note that my $\hat{u}^{i}$ is the $\hat{u}_{i}$ on Wikipedia. Also, $x_{i}\hat{u}^{i}=\sum_{i=1}^{3} x_{i} \hat{u}_{i}$. Sorry for the possible troubles regarding my notation. $\endgroup$
    – FraSchelle
    Jun 14, 2013 at 7:32

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