General derivation Many problems require use of noninertial reference frames, for example, those involving satellites Below are a number of examples applying this result for fictitious forces. More examples can be found in the article on
centrifugal force.
Rotating coordinate systems A common situation in which noninertial reference frames are useful is when the reference frame is rotating. Because such rotational motion is non-inertial, due to the acceleration present in any rotational motion, a fictitious force can always be invoked by using a rotational frame of reference. Despite this complication, the use of fictitious forces often simplifies the calculations involved. To derive expressions for the fictitious forces, derivatives are needed for the apparent time rate of change of vectors that take into account time-variation of the coordinate axes. If the rotation of frame 'B' is represented by a vector
Ω pointed along the axis of rotation with the orientation given by the
right-hand rule, and with magnitude given by |\boldsymbol{\Omega} | = \frac {d \theta }{dt} = \omega (t), then the time derivative of any of the three unit vectors describing frame B is \frac {d \mathbf{u}_j (t)}{dt} = \boldsymbol{\Omega} \times \mathbf{u}_j (t), and \frac {d^2 \mathbf{u}_j (t)}{dt^2}= \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j +\boldsymbol{\Omega} \times \frac{d \mathbf{u}_j (t)}{dt} = \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j+ \boldsymbol{\Omega} \times \left[ \boldsymbol{\Omega} \times \mathbf{u}_j (t) \right], as is verified using the properties of the
vector cross product. These derivative formulas now are applied to the relationship between acceleration in an inertial frame, and that in a coordinate frame rotating with time-varying angular velocity ω(
t). From the previous section, where subscript A refers to the inertial frame and B to the rotating frame, setting
aAB = 0 to remove any translational acceleration, and focusing on only rotational properties (see
Eq. 1): \frac {d^2 \mathbf{x}_\mathrm{A}}{dt^2}=\mathbf{a}_\mathrm{B} + 2\sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \frac{d^2 \mathbf{u}_j}{dt^2}, \begin{align} \mathbf{a}_\mathrm{A} &= \mathbf{a}_\mathrm{B} +\ 2\sum_{j=1}^3 v_j \boldsymbol{\Omega} \times \mathbf{u}_j (t) + \sum_{j=1}^3 x_j \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{u}_j \ + \sum_{j=1}^3 x_j \boldsymbol{\Omega} \times \left[ \boldsymbol{\Omega} \times \mathbf{u}_j (t) \right] \\ &=\mathbf{a}_\mathrm{B} + 2 \boldsymbol{\Omega} \times\sum_{j=1}^3 v_j \mathbf{u}_j (t) + \frac{d\boldsymbol{\Omega}}{dt} \times \sum_{j=1}^3 x_j \mathbf{u}_j + \boldsymbol{\Omega} \times \left[\boldsymbol{\Omega} \times \sum_{j=1}^3 x_j \mathbf{u}_j (t) \right]. \end{align} Collecting terms, the result is the so-called
acceleration transformation formula: \mathbf{a}_\mathrm{A}=\mathbf{a}_\mathrm{B} + 2\boldsymbol{\Omega} \times\mathbf{v}_\mathrm{B} + \frac{d\boldsymbol{\Omega}}{dt} \times \mathbf{x}_\mathrm{B} + \boldsymbol{\Omega} \times \left(\boldsymbol{\Omega} \times \mathbf{x}_\mathrm{B} \right)\, . The
physical acceleration aA due to what observers in the inertial frame A call
real external forces on the object is, therefore, not simply the acceleration
aB seen by observers in the rotational frame B, but has several additional geometric acceleration terms associated with the rotation of B. As seen in the rotational frame, the acceleration
aB of the particle is given by rearrangement of the above equation as: \mathbf{a}_\mathrm{B} = \mathbf{a}_\mathrm{A} - 2\boldsymbol{\Omega} \times \mathbf{v}_\mathrm{B} - \boldsymbol{\Omega} \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B}) - \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}. The net force upon the object according to observers in the rotating frame is
FB =
maB. If their observations are to result in the correct force on the object when using Newton's laws, they must consider that the additional force
Ffict is present, so the result is
FB =
FA +
Ffict. Thus, the fictitious force used by observers in B to get the correct behaviour of the object from Newton's laws equals: \mathbf{F}_{\mathrm{fict}} = - 2 m \boldsymbol\Omega \times \mathbf{v}_\mathrm{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B}) - m \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}. Here, the first term is the
Coriolis force, the second term is the
centrifugal force, and the third term is the
Euler force.
Orbiting coordinate systems As a related example, suppose the moving coordinate system
B rotates with a constant angular speed ω in a circle of radius
R about the fixed origin of inertial frame
A, but maintains its coordinate axes fixed in orientation, as in Figure 3. The acceleration of an observed body is now (see
Eq. 1): \begin{align} \frac {d^2 \mathbf{x}_{A}}{dt^2} &= \mathbf{a}_{AB}+\mathbf{a}_{B} + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2} \\ &=\mathbf{a}_{AB}\ +\mathbf{a}_B\ , \end{align} where the summations are zero inasmuch as the unit vectors have no time dependence. The origin of the system
B is located according to frame
A at: \mathbf{X}_{AB} = R \left( \cos ( \omega t) , \ \sin (\omega t) \right) \ , leading to a velocity of the origin of frame
B as: \mathbf{v}_{AB} = \frac{d}{dt} \mathbf{X}_{AB} = \mathbf{\Omega \times X}_{AB} \ , leading to an acceleration of the origin of
B given by: \mathbf{a}_{AB} = \frac{d^2}{dt^2} \mathbf{X}_{AB} = \mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) = - \omega^2 \mathbf{X}_{AB} \, . Because the first term, which is \mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right)\, , is of the same form as the normal centrifugal force expression: \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\, , it is a natural extension of standard terminology (although there is no standard terminology for this case) to call this term a "centrifugal force". Whatever terminology is adopted, the observers in frame
B must introduce a fictitious force, this time due to the acceleration from the orbital motion of their entire coordinate frame, that is radially outward away from the centre of rotation of the origin of their coordinate system: \mathbf{F}_{\mathrm{fict}} = m \omega^2 \mathbf{X}_{AB} \, , and of magnitude: |\mathbf{F}_{\mathrm{fict}}| = m \omega^2 R \, . This "centrifugal force" has differences from the case of a rotating frame. In the rotating frame the centrifugal force is related to the distance of the object from the origin of frame
B, while in the case of an orbiting frame, the centrifugal force is independent of the distance of the object from the origin of frame
B, but instead depends upon the distance of the origin of frame
B from
its centre of rotation, resulting in the
same centrifugal fictitious force for
all objects observed in frame
B.
Orbiting and rotating As a combination example, Figure 4 shows a coordinate system
B that orbits inertial frame
A as in Figure 3, but the coordinate axes in frame
B turn so unit vector
u1 always points toward the centre of rotation. This example might apply to a test tube in a centrifuge, where vector
u1 points along the axis of the tube toward its opening at its top. It also resembles the Earth–Moon system, where the Moon always presents the same face to the Earth. In this example, unit vector
u3 retains a fixed orientation, while vectors
u1,
u2 rotate at the same rate as the origin of coordinates. That is, \mathbf{u}_1 = (-\cos \omega t ,\ -\sin \omega t )\ ;\ \mathbf{u}_2 = (\sin \omega t ,\ -\cos \omega t ) \, . \frac{d}{dt}\mathbf{u}_1 = \mathbf{\Omega \times u_1}= \omega\mathbf{u}_2\ ; \ \frac{d}{dt}\mathbf{u}_2 = \mathbf{\Omega \times u_2} = -\omega\mathbf{u}_1\ \ . Hence, the acceleration of a moving object is expressed as (see
Eq. 1): \begin{align} \frac {d^2 \mathbf{x}_{A}}{dt^2}&=\mathbf{a}_{AB}+\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j \ \frac{d \mathbf{u}_j}{dt} + \ \sum_{j=1}^3 x_j \ \frac{d^2 \mathbf{u}_j}{dt^2}\\ &=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) +\mathbf{a}_B + 2\ \sum_{j=1}^3 v_j\ \mathbf{\Omega \times u_j} \ +\ \sum_{j=1}^3 x_j\ \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{u}_j \right)\\ &=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times X}_{AB}\right) + \mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ \ +\ \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\\ &=\mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times} (\mathbf{ X}_{AB}+\mathbf{x}_B) \right) + \mathbf{a}_B + 2\ \boldsymbol{\Omega} \times\mathbf{v}_B\ \, , \end{align} where the angular acceleration term is zero for the constant rate of rotation. Because the first term, which is \mathbf{ \Omega \ \times } \left( \mathbf{ \Omega \times} (\mathbf{ X}_{AB}+\mathbf{x}_B) \right)\, , is of the same form as the normal centrifugal force expression: \boldsymbol{\Omega} \times \left( \boldsymbol{\Omega} \times \mathbf{x}_B \right)\, , it is a natural extension of standard terminology (although there is no standard terminology for this case) to call this term the "centrifugal force". Applying this terminology to the example of a tube in a centrifuge, if the tube is far enough from the center of rotation, |
XAB| =
R ≫ |
xB|, all the matter in the test tube sees the same acceleration (the same centrifugal force). Thus, in this case, the fictitious force is primarily a uniform centrifugal force along the axis of the tube, away from the centre of rotation, with a value |
Ffict| = ω2
R, where
R is the distance of the matter in the tube from the centre of the centrifuge. It is the standard specification of a centrifuge to use the "effective" radius of the centrifuge to estimate its ability to provide centrifugal force. Thus, the first estimate of centrifugal force in a centrifuge can be based upon the distance of the tubes from the centre of rotation, and corrections applied if needed. Also, the test tube confines motion to the direction down the length of the tube, so
vB is opposite to
u1 and the Coriolis force is opposite to
u2, that is, against the wall of the tube. If the tube is spun for a long enough time, the velocity
vB drops to zero as the matter comes to an equilibrium distribution. For more details, see the articles on
sedimentation and the
Lamm equation. A related problem is that of centrifugal forces for the Earth–Moon–Sun system, where three rotations appear: the daily rotation of the Earth about its axis, the lunar-month rotation of the Earth–Moon system about its centre of mass, and the annual revolution of the Earth–Moon system about the Sun. These three motions influence the
tides.
Crossing a carousel Figure 5 shows another example comparing the observations of an inertial observer with those of an observer on a rotating
carousel. The carousel rotates at a constant angular velocity represented by the vector
Ω with magnitude
ω, pointing upward according to the
right-hand rule. A rider on the carousel walks radially across it at a constant speed, in what appears to the walker to be the straight line path inclined at 45° in Figure 5. To the stationary observer, however, the walker travels a spiral path. The points identified on both paths in Figure 5 correspond to the same times spaced at equal time intervals. We ask how two observers, one on the carousel and one in an inertial frame, formulate what they see using Newton's laws.
Inertial observer The observer at rest describes the path followed by the walker as a spiral. Adopting the coordinate system shown in Figure 5, the trajectory is described by
r(
t): \mathbf{r}(t) =R(t)\mathbf{u}_R = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \begin{bmatrix} R(t)\cos (\omega t + \pi/4) \\ R(t)\sin (\omega t + \pi/4) \end{bmatrix}, where the added π/4 sets the path angle at 45° to start with (just an arbitrary choice of direction),
uR is a
unit vector in the radial direction pointing from the centre of the carousel to the walker at the time
t. The radial distance
R(
t) increases steadily with time according to: R(t) = s t, with
s the speed of walking. According to simple kinematics, the velocity is then the first derivative of the trajectory: \begin{align} \mathbf{v}(t) &= \frac{dR}{dt} \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} + \omega R(t) \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix} \\ &= \frac{dR}{dt} \mathbf{u}_R + \omega R(t) \mathbf{u}_{\theta}, \end{align} with
uθ a unit vector perpendicular to
uR at time
t (as can be verified by noticing that the vector
dot product with the radial vector is zero) and pointing in the direction of travel. The acceleration is the first derivative of the velocity: \begin{align} \mathbf{a}(t) &= \frac{d^2 R}{dt^2} \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} + 2 \frac {dR}{dt} \omega \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix} - \omega^2 R(t) \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} \\ &=2s\omega \begin{bmatrix} -\sin(\omega t + \pi/4) \\ \cos (\omega t + \pi/4) \end{bmatrix} -\omega^2 R(t) \begin{bmatrix} \cos (\omega t + \pi/4) \\ \sin (\omega t + \pi/4) \end{bmatrix} \\ &=2s\ \omega \ \mathbf{u}_{\theta}-\omega^2 R(t)\ \mathbf{u}_R \, . \end{align} The last term in the acceleration is radially inward of magnitude ω2
R, which is therefore the instantaneous
centripetal acceleration of
circular motion. The first term is perpendicular to the radial direction, and pointing in the direction of travel. Its magnitude is 2
sω, and it represents the acceleration of the walker as the edge of the carousel is neared, and the arc of the circle travelled in a fixed time increases, as can be seen by the increased spacing between points for equal time steps on the spiral in Figure 5 as the outer edge of the carousel is approached. Applying Newton's laws, multiplying the acceleration by the mass of the walker, the inertial observer concludes that the walker is subject to two forces: the inward radially directed centripetal force and another force perpendicular to the radial direction that is proportional to the speed of the walker.
Rotating observer The rotating observer sees the walker travel a straight line from the centre of the carousel to the periphery, as shown in Figure 5. Moreover, the rotating observer sees that the walker moves at a constant speed in the same direction, so applying Newton's law of inertia, there is
zero force upon the walker. These conclusions do not agree with the inertial observer. To obtain agreement, the rotating observer has to introduce fictitious forces that appear to exist in the rotating world, even though there is no apparent reason for them, no apparent gravitational mass, electric charge or what have you, that could account for these fictitious forces. To agree with the inertial observer, the forces applied to the walker must be exactly those found above. They can be related to the general formulas already derived, namely: \mathbf{F}_{\mathrm{fict}} = - 2 m \boldsymbol\Omega \times \mathbf{v}_\mathrm{B} - m \boldsymbol\Omega \times (\boldsymbol\Omega \times \mathbf{x}_\mathrm{B} ) - m \frac{d \boldsymbol\Omega}{dt} \times \mathbf{x}_\mathrm{B}. In this example, the velocity seen in the rotating frame is: \mathbf{v}_\mathrm{B} = s \mathbf{u}_R, with
uR a unit vector in the radial direction. The position of the walker as seen on the carousel is: \mathbf{x}_\mathrm{B} = R(t)\mathbf{u}_R, and the time derivative of
Ω is zero for uniform angular rotation. Noticing that \boldsymbol\Omega \times \mathbf{u}_R =\omega \mathbf{u}_{\theta} and \boldsymbol\Omega \times \mathbf{u}_{\theta} =-\omega \mathbf{u}_R \, , we find: \mathbf{F}_{\mathrm{fict}} = - 2 m \omega s \mathbf{u}_{\theta} + m \omega^2 R(t) \mathbf{u}_R. To obtain a
straight-line motion in the rotating world, a force exactly opposite in sign to the fictitious force must be applied to reduce the net force on the walker to zero, so Newton's law of inertia will predict a straight line motion, in agreement with what the rotating observer sees. The fictitious forces that must be combated are the
Coriolis force (first term) and the
centrifugal force (second term). (These terms are approximate.) By applying forces to counter these two fictitious forces, the rotating observer ends up applying exactly the same forces upon the walker that the inertial observer predicted were needed. Because they differ only by the constant walking velocity, the walker and the rotational observer see the same accelerations. From the walker's perspective, the fictitious force is experienced as real, and combating this force is necessary to stay on a straight line radial path holding a constant speed. It is like battling a crosswind while being thrown to the edge of the carousel.
Observation Notice that this
kinematical discussion does not delve into the mechanism by which the required forces are generated. That is the subject of
kinetics. In the case of the carousel, the kinetic discussion would involve perhaps a study of the walker's shoes and the friction they need to generate against the floor of the carousel, or perhaps the dynamics of skateboarding if the walker switched to travel by skateboard. Whatever the means of travel across the carousel, the forces calculated above must be realized. A very rough analogy is heating your house: you must have a certain temperature to be comfortable, but whether you heat by burning gas or by burning coal is another problem. Kinematics sets the thermostat, kinetics fires the furnace. ==See also==