Relative Motions

Abstract

We have already seen that the equations of motion depend on the reference frame. The law of motions, and more generally all the laws of Physics, transform, as we say, from one frame to another. This chapter is dedicated to the study of these transformations. We first consider the transformations between frames at rest with one another, namely the translations and rotations of the axes. We then consider frames in relative motion and learn that, when the relative motion is a translation with constant speed, the laws of physics do not change their form. This is the relativity principle established by Galilei. Finally we shall study the motion in accelerated frames and introduce, in particular, the concept of inertia force or pseudo-force.

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In our study of the kinematics of the material point, we have already seen that the equations of motion depend on the reference frame. The law of motions, and more generally all the laws of Physics, transform, as we say, from one frame to another. This chapter is dedicated to the study of these transformations.

Two reference frames may differ in different ways.

The two frames have no relative motion, their co-ordinate homologous axes are parallel, but have different origins; the frames differ for a rigid translation.

The two frames have no relative motion and coincident origins, but the directions of the axes are different; the frames differ for a rigid rotation.

One frame can translate relative to the other in time with uniform or varying velocity, or it can rotate, again with constant or varying angular velocity, or it can translate and rotate contemporarily.

In Sect. 5.1, we shall consider two stationary frames relative to one another, with a relative translation or rotation. We shall see that the laws of Physics have the same form, namely the same mathematical expressions, in both frames. As we say, the laws are covariant under translations and rotations. The meaning of the term will be explained.

We shall then consider frames in relative motion and learn that, when the relative motion is a translation with constant speed, the laws of mechanics are also covariant. This is the relativity principle, a fundamental principle of physics, established by Galilei. For example, experiments done inside a closed room in a ship cannot establish whether the ship is moving in uniform motion or is standing still. One of the consequences is that once we have found an inertial frame, any other frame moving in a uniform translation motion relative to it is also inertial.

In Sect. 5.3, we shall deal with the relative translatory accelerated motion. As already anticipated, in any reference that accelerates relative to an inertial frame, the Newton laws are not valid. For example, a body at rest can start moving without any force acting on it. The motion can be described introducing fictitious forces, which are known by several equivalent names, apparent forces of the relative motions, pseudo-forces and inertial forces. We feel such “force,” for example, when we brake suddenly in a car. In Sect. 5.4, we shall deal with the general case (translation and rotation) and we shall see the relations between velocities and between accelerations in two frames of any relative motion. In Sect. 5.5, we shall discuss several examples of motion in frames rotating relative to an inertial frame.

Any frame at rest in a laboratory on earth does, in fact, move with earth. In initial, and quite good, approximation, these frames can be considered to be inertial. Not completely, however, because earth rotates on its axis and moves along its orbit around the sun, and even the sun moves along its orbit in the galaxy. In Sect. 5.7, we shall study a few effects of the inertial forces in frames at rest relative to earth: the variation with latitude of the magnitude of the weight, the rotation of the oscillation plane of pendulums, the deviation from the vertical of free fall and the circulation of winds.

The inertial forces acting on a body are proportional to its inertial mass, while the gravitational attraction of earth is proportional to its gravitational mass. This observation allows for the realization of very delicate experiments to check whether the two masses are different or equal. We shall describe such an experiment in Sect. 5.8.

5.1 Covariance of the Physical Laws Under Rotations and Translations

Consider two Cartesian reference frames, which are stationary relative to each other, S (coordinates x, y, z, origin O) and S′ (coordinates x′, y′, z′, origin O′).

A physical law is a mathematical equation between physical quantities. The relation between the two frames can be a rigid rotation or a rigid translation . Let us start with rotations.

We choose the origins of the two frames as coincident. For simplicity, we consider their z-axes also to be coincident. The frames differ for a rotation, by an angle θ, around this axis. The rotation is in the common plane xy, as shown in Fig. 5.1.

Fig. 5.1

Two reference frames different for a rigid rotation

Suppose now that an observer in S makes a very simple experiment. He measures, using a balance, the masses of two objects, finding the values m ₁ and m ₂. He finds that the second mass is three times the first. He writes the relation (let us call it the “law”)

$$m_{2} = 3m_{1} .$$

(5.1)

Another observer in S′ performs the same experiment. We indicate with a prime the homologous quantities he finds. In this very simple case, considering that the procedure of measuring the mass with a balance does not depend on the direction of the axes, we can conclude that he will find the same result, namely

$$m_{1}^{\prime} = m_{1} ,\quad m_{2}^{\prime} = m_{2} .$$

(5.2)

The second observer also states that

$$m_{2}^{\prime} = 3m_{1}^{\prime} .$$

(5.3)

Equation (5.3) has the same form as Eq. (5.1). Namely, the two observers describe the same phenomenon with laws of the same form. Indeed, mass is a scalar quantity, which is invariant under rotations of the axes. In general, a relation between scalar quantities, if valid in a frame, is also valid in any frame rotated relative to the first one, because both sides of the equation do not vary going from one frame to the other.

But it is not always so. Suppose that the observer in S measures two components of the velocity of a point, finding the values υ _x and υ _y. He finds, again, the second quantity three times larger than the first and writes the equation

$$\upsilon_{y} = 3\upsilon_{x} .$$

(5.4)

What would the observer in S′ find? We know the answer because we know the relations between velocities in the two frames:

$$\upsilon_{x}^{\prime} = \upsilon_{x} \cos \theta + \upsilon_{y} \sin \theta ;\quad \upsilon_{y}^{\prime} = - \upsilon_{x} \sin \theta + \upsilon_{y} \cos \theta .$$

(5.5)

We calculate the ratio between $\upsilon_{x}^{\prime}$ and $\upsilon_{y}^{\prime}$, also employing Eq. (5.5):

$$\frac{{\upsilon_{y}^{\prime} }}{{\upsilon_{y}^{\prime} }} = \frac{{ - \upsilon_{x} \sin \theta + \upsilon_{y} \cos \theta }}{{\upsilon_{x} \cos \theta + \upsilon_{y} \sin \theta \, }} = \frac{{ - \sin \theta + \frac{{\upsilon_{y} }}{{\upsilon_{x} }}\cos \theta }}{{\cos \theta + \frac{{\upsilon_{y} }}{{\upsilon_{x} }}\sin \theta \, }} = \frac{ - \sin \theta + 3\cos \theta }{\cos \theta + 3\sin \theta \, }.$$

In conclusion, in S′, we have

$$\upsilon_{y}^{\prime} = \frac{ - \sin \theta + 3\cos \theta }{\cos \theta + 3\sin \theta \, }\upsilon_{x}^{\prime} .$$

(5.6)

The form of the “law” is different this time in the two frames, being (5.4) in S and (5.6) in S′. This is an obvious consequence of the fact that the components of a vector transform differently one from another.

But wait a moment, a law may be valid in both frames, even if its sides are not invariant, as in the case of the masses; rather, it is sufficient that, if they vary, in the same way. Let us see what happens for a law linking vector quantities.

The observer in S′, which we assume, for the sake of this example, to be inertial, studies the motion of a material point. He measures the acceleration a (namely its three components), the force acting on the point F (again, the three components) and the mass m. He finds the relation

$${\mathbf{F}} = m{\mathbf{a}}\text{.}$$

(5.7)

More explicitly, this vector relation corresponds to three equations:

$$F_{x} = ma_{x} ,\quad F_{y} = ma_{y} ,\quad F_{z} = ma_{z} .$$

(5.8)

We know how the components of the vectors, such as F and a are, transform from one frame to the other, namely

$$\begin{aligned} F_{x}^{\prime} & = F_{x} \cos \theta + F_{y} \sin \theta ;\quad F_{y}^{\prime} = - F_{x} \sin \theta + F_{y} \cos \theta ;\quad F_{z}^{\prime} = F_{z} \\ a_{x}^{\prime} & = a_{x} \cos \theta + a_{y} \sin \theta ;\quad a_{y}^{\prime} = - a_{x} \sin \theta + a_{y} \cos \theta ;\quad a_{z}^{\prime} = a_{z} \\ \end{aligned}$$

(5.9)

and we can write

$$\begin{aligned} F_{x}^{\prime} & = ma_{x} \cos \theta + ma_{y} \sin \theta = m\left( {a_{x} \cos \theta + a_{y} \sin \theta } \right) \\ F_{y}^{\prime} & = - ma_{x} \sin \theta + ma_{y} \cos \theta = m\left( { - a_{x} \sin \theta + a_{y} \cos \theta } \right) \\ F_{z}^{\prime} & = ma_{z} \\ \end{aligned}$$

and, for Eq. (5.9),

$$F_{x}^{\prime} = ma_{x}^{\prime} ,\quad F_{y}^{\prime} = ma_{y}^{\prime} ,\quad F_{z}^{\prime} = ma_{z}^{\prime} ,$$

(5.10)

which has the same form as Eq. (5.8). Both sides of the equations are different, varying from one frame to the other. However, they vary in the same way, because both sides are vectors. Thus, we say that the equation is covariant .

In conclusion, the laws of Physics keep the same form under rotations of the axes, or, in other words, are covariant under rotations. And yet, from another perspective, it is impossible experimentally to establish any privileged directions of the reference axes. Space should be considered isotropic, without preferential directions.

The case of the translations is very simple. Scalar quantities obviously have the same values in two frames differing for a translation. This is also valid for vectors, which are simply translated; hence, they are the same vector.

5.2 Uniform Relative Translation. Relativity Principle

Consider now two reference frames, S and S’, which are in relative motion. We arbitrarily call one of them S (origin O and coordinates x, y, z) fixed and the other one S’ (origin O’ and coordinates x’, y’, z’) mobile. We consider the case of a uniform translation of S’. All the points of S’ move with the same velocity relative to S, which is constant in magnitude and direction. The frames, for example, might consist of one fixed on the ground, the other on a carriage moving on straight rails, or a frame fixed at the shore and one on a ship moving straight, in both cases with uniform motion.

The axes of the two frames do not change the relative directions and we can take them as being parallel. Fig. 5.2 shows the two frames at a certain instant. At a later time, the mobile frame will be in a different position, more on the right, but its axes will still be parallel to the axes of S. We choose t = 0 as the time at which the axes of the two frames overlap.

Fig. 5.2

Two reference frames in relative uniform translation motion

Figure 5.2 shows the material point P and its trajectory. The position vectors r and r’ of P in the two frames have the well-known relation

$$\mathbf{r} = \mathbf{r^{\prime}} + \mathbf{r}_{O^{\prime}},$$

(5.11)

where rO’ is the position vector of the origin O’ of the mobile frame S’ in the fixed frame O, namely OO’.

A fixed and a mobile observer see the point P moving with different velocities, v and v’. To find their relation, we take the time derivatives of Eq. (5.11), obtaining

$$\mathbf{v} = \mathbf{v^{\prime}} + \mathbf{v}_{O^{\prime}},$$

(5.12)

where v O’ is the velocity of the origin O’ of the mobile frame, and also of all its points (because the motion is a translation) as seen by S. The velocity of an insect flying in the ship in the above example relative to the shore is the vector sum of the velocity of the insect relative to the ship and the velocity of the ship relative to the shore.

A further time derivation gives the relation between accelerations

$$\mathbf{a} = \mathbf{a^{\prime}} + \mathbf{a}_{O^{\prime}},$$

(5.13)

where aO’ is the velocity of the origin O’ of the mobile frame, and also of all its points.

We now consider the important particular case in which the translation of S’ relative to S is uniform, namely the velocity of its origin, and of all its points, seen by S is constant in time

$$\mathbf{v}_{O^{\prime}} = \text{constant}.$$

(5.14)

Then, obviously,

$$\mathbf{a}_{O^{\prime}} = 0$$

(5.15)

and Eq. (5.13) becomes

$$\mathbf{a} = \mathbf{a^{\prime}}.$$

(5.16)

The accelerations in the two frames are equal. The implications of this simple conclusion are extremely important considering inertial frames.

If S is an inertial frame, any material point P not subject to forces moves at constant velocity v (or remains at rest). In other words, its acceleration is zero, a = 0. In the mobile frame, its acceleration a’, which is equal to a, is also zero. Consequently, S’ is inertial too.

We conclude that, given an inertial reference frame, any other frame moving relative to it by a uniform translation is also inertial.

What about the second Newton law? It is valid in the frame S, which is inertial by assumption. Is it also valid in S’? In S, we have

$$\mathbf{F} = {\it m}\mathbf{a}.$$

(5.17)

The observer in S’ measures the same mass (m’=m) and the same force (if, e.g., he uses a dynamometer, the spring stretches by the same amount), F = F’. The acceleration a’ that he measures is also equal to a, but only in the case we are considering of relative translation at constant velocity. Then, in S’, the relation between force, mass and acceleration is

$$\mathbf{F^{\prime}} = {\it m}^{\prime}\mathbf{a^{\prime}}.$$

(5.18)

In other words: the laws of mechanics are covariant under the transformations that link two reference frames in relative uniform translation motion.

As an example, consider a reference S’ fixed on a sailing ship moving on the sea at constant velocity and S a frame fixed to the shore. As above, we choose the axes of the two frames mutually parallel and with coincident origins at t = 0. An experimenter climbs on top of the mast and drops a stone. Fig. 5.3 shows the trajectories of the stone as seen by an observer on the shore, a), and on the ship, b).

Fig. 5.3

Trajectory of a stone dropped from the top of the mast of a ship, as seen from the ship and the shore

For the observer in S, the stone falls under the action of its weight, a constant force (F = m g), directed downwards, opposite to the z-axis (that we have taken to be vertical upwards). The initial velocity of the stone is the velocity of the ship, and we have taken the x-axis in that direction. Hence, the motion of the stone in the z direction is uniformly accelerated, while in the x direction, it is uniform (neglecting the air resistance). The trajectory is a parabola. In the figure, we marked the positions of the stone in time instants separated by the same time interval.

In S’, the forces are the same, but the initial conditions are different; the initial velocity of the stone is zero. Hence, it falls vertically along the z’-axis with a uniformly accelerated motion.

Summarizing, in the two frames, the trajectories are different. The reason for the difference is in the different initial conditions of the motion. On the contrary, both observers describe the motion with the same law, F = m a. The two frames are perfectly equivalent for every dynamic experiment. Each of them can be considered as fixed or movable.

This conclusion is important and is known as the relativity principle. The principle does not deal directly with the phenomena but rather with the laws that describe the phenomena. It states that: the laws of Physics are covariant, namely have the same form, in any reference frame moving of translational uniform relative motion.

In our discussion, we have seen that the relativity principle is valid for the laws of mechanics, which is the physics chapter we are studying. However, its validity is completely general, including, in particular, all fundamental interactions, gravitational, electromagnetic, nuclear strong and weak interactions. In other words, it is impossible experimentally to establish the relative motion, provided it is as uniform translation. Historically, the principle was established by G. Galilei. He did not use that name, which was given to it by Henri Poincaré (1854–1912) in 1904, but Galilei established it in complete generality, describing, in a beautiful page, a series of experiments, some of which were of an electromagnetic nature, below the deck of a large sailing ship. The page of the Dialogue (transalted from Italian into English by the author) is:

Shut yourself with a few friends in the largest room below decks of some large vessel, and have with you flies, butterflies and similar small flying animals. Let a large bowl of water with several small fish in it be the cabin too. Hang also, at a certain height, a bucket pouring out water drop by drop into another vase with a narrow mouth beneath it. When the ship stands still, carefully observe how those flying small animals fly with equal speed towards all sides of the cabin; you will see the fish swim indifferently in all directions; all the drops will fall into the vessel beneath; and you, when throwing something to a friend, will not need throw it more strongly in one direction than another, when the distances are equal; and jumping up feet together, you will pass equal spaces in all directions.

Once you have observed all these things carefully, though there is no doubt that when the vessel is standing they must happen like that, let the vessel move with speed as high as you like. Then (provided the motion is uniform and not unevenly fluctuating) you will not discover the slightest change in any of the named effects, nor you will be able to understand from any of them whether the ship is moving or standing still. In jumping you will pass on the planking the same spaces as before, nor you will make longer jumps toward the stern than toward the prow, as a consequence of the fast motion of the vessel, despite the fact that during the time you are in the air the planking under you is running in a direction opposite to your jump. In throwing something to your companion, no more force will be needed to reach him whether he is on the side of the prow and you of the stern or your positions are inverted. The drops will fall as before in the lower bowl, without a single one dropping towards the stern, although, while the drop is in the air, the vessel runs many palms. The fish in their water will swim toward the forward part of their vase with no more effort than toward the backward part, and will come with equal ease to food placed anywhere on the rim of the vase. And finally the butterflies and the flies will continue their flights indifferently towards every side, nor will ever happen to find them concentrated close to the wall on the side of the stern, as if tired from keeping up with the course of the ship, from which they, remaining in the air, will have been separated for a long time. And if some smoke will be made burning a bit of incense, it will be seen ascending upward and, similar to a little cloud, remaining still and indifferently moving no more toward one side than the other. The cause of all these correspondences of effects is that the motion of the ship is common to all things contained in it, and to the air also.

We notice here that the development of electromagnetism in the last part of the XIX century led to doubts concerning the general validity of the principle. The process of in depth analysis of the physical laws that followed, leading to the relativity theory , showed that the Galilei relativity principle was, as we have stated, valid in general. However, it was found that the transformations of the co-ordinates, and of the time, between reference frames, valid at a small velocity relative to the speed of light, do not hold at high speeds. We shall discuss that in Chap. 5. Here, we simply anticipate the root of the issue. Consider the transformation equations that link the co-ordinates in S′ and in S

$$x^{\prime} = x - \upsilon_{{O^{\prime} }} t;\quad y^{\prime} = y;\quad z^{\prime} = z;\quad t^{\prime} = t,$$

(5.19)

where we have included the relation between times t and t′ measured by the two observers. Indeed, the measurement of a time interval should be, we think, the same on the shore as on the ship (to continue the example). However, this conclusion, coming from our everyday experience and from experiments at the usual velocities, is wrong at velocities not too small compared to the speed of light. Two observers in two frames moving at those speeds measure different time intervals between the same two events; in other words, t and t′ are not equal. A consequence is that the relations between co-ordinates are different from those of Eq. (5.19). The transformation Eq. (5.19), called Galilei transformations , fail at high velocities and must be generalized into the Lorentz transformations , as we shall see in Chap. 5. But the relativity principle remains completely valid.

5.3 Non-uniform Translation. Pseudo Forces

We now consider the case in which the motion of the reference S′ relative to S is still a translation, but with variable velocity. Consider, for example, S′ to be fixed on a trolley moving on straight rails with an accelerated (or decelerated) motion relative to S fixed on the ground. We still consider the motion of the point P in Fig. 5.2 as seen by two observers in the two frames. The relation between the accelerations is

$${\mathbf{a}} = {\mathbf{a}}^{\prime} + {\mathbf{a}}_{{O^{\prime} }} \text{.}$$

(5.20)

As in the previous section, a _O′ is the acceleration in S of the origin of S′ and also of all the points fixed in it (its motion being a translation). Suppose now that S is an inertial frame. If no force acts on P, its acceleration is zero, a = 0. In S′, however, a = – a _O′ ≠ 0. Namely, in S′, a body not subject to forces may accelerate. The law of inertia does not hold. S′ is not inertial. Consider the trolley in the above example initially moving at constant speed. If we put a ball on a horizontal plane, it will not move. If the trolley now suddenly slows down, we shall see the ball accelerating forward, without any force acting. This is the interpretation of the observer in S′. The inertial observer in S thinks that there is no force acting on the ball (suppose friction to be negligible), and that it is just continuing its uniform motion (Fig. 5.3).

If a force F acts on the point P of mass m, the inertial observer in S finds the relation

$${\mathbf{F}} = m{\mathbf{a}}\text{.}$$

(5.21)

The observer in S′ measures the same value of the mass, m′ = m, the same force, F′ = F, but a different acceleration a′, and finds ${\mathbf{F}}^{\prime} = {\mathbf{F}} = m{\mathbf{a}} = m^{\prime} {\mathbf{a}}^{\prime} + m^{\prime} {\mathbf{a}}_{{O^{\prime} }}$, or

$${\mathbf{F}}^{\prime} = m^{\prime} {\mathbf{a}}^{\prime} + m^{\prime} {\mathbf{a}}_{{O^{\prime} }} .$$

(5.22)

We also see that the second Newton law, not only the first one, does not hold for the non-inertial observer S′.

However, the observer in S′ can play a trick. Indeed, he is accustomed to thinking that any acceleration will be due to a force and will imagine that a force has suddenly started to act on the ball on the table. Formally, the trick is by Jean Baptiste d’Alembert (1717–1783); we can re-write Eq. (5.22) moving m′a _O′ to the left-hand side, as

$${\mathbf{F}}^{\prime} - m^{\prime} {\mathbf{a}}_{{O^{\prime} }} = m^{\prime} {\mathbf{a}}^{\prime}$$

(5.23)

and call −m′a _O′ a force, or, more accurately, a fictitious force , or inertial force

$${\mathbf{F}}_{in} = - m^{\prime} {\mathbf{a}}_{{O^{\prime} }}$$

(5.24)

and Eq. (5.23) becomes

$${\mathbf{F}}^{\prime} + {\mathbf{F}}_{in} = m^{\prime} {\mathbf{a}}^{\prime} \text{.}$$

(5.25)

Namely, if we add to the “real” forces the fictitious, or inertial, ones, we re-establish the validity of the 1st and 2nd Newton laws. However, these forces are, as we said, fictitious, not real, because they are not produced by any physical agent. Consequently, there is no corresponding reaction. The 3rd Newton law, the action-reaction law, does not hold for the inertial forces.

Let us go back to the example of a sphere on a table on the trolley. The resultant of the true forces, weight and normal reaction of the plane, is zero. When the velocity of the trolley is constant, the fictitious force F _in is also zero because so is a _O′. But when the trolley slows down, the observer in the S′ sees the effect of a fictitious force as in Eq. (5.24). It is directed forward, opposite to a _O′. He can measure the fictitious force attaching the sphere to a spring and measuring its stretch. In this way, he verifies that Eq. (5.24) is correct.

5.4 Rotation and Translation. Pseudo Forces

Consider now a stationary frame S (origin O and coordinates x, y, z) and a mobile frame S′ (origin O′ and coordinates x′, y′, z′), the motion of which is completely general. It may be a translation, with constant or variable velocity, a rotation, again with constant or varying angular velocity, or both of them together. Figure 5.4 represents the two frames at a certain time. At another time, for example, a bit later, both the position of O′ and the direction of the axes of S′ will be, in general, different.

Fig. 5.4

Reference frame S′ moves in an arbitrary motion relative to S

We begin by finding a formula that will be useful in the following. Consider a vector A, which does not vary with time relative to S′. Examples are the position vector in S′ and the velocity of a point moving in rectilinear uniform motion relative to S′. The vector A is not constant in S. We now find its time derivative. We notice that A varies relative to S only in direction, not in magnitude. More precisely, A rotates with the same angular velocity at which, in that instant, the mobile frame S′ rotates relative to S. We indicate it with ω. Notice that ω can vary in time, which is why we specify “in that instant”. Under these conditions, the time derivative of A is given by the Poisson formula and we have

$$\left( {\frac{{d{\mathbf{A}}}}{dt}} \right)_{S} = {\varvec{\upomega}} \times {\mathbf{A}}\text{,}$$

where the subscript S specifies that it is the rate of change in the reference S.

If the vector A also varies in S′, we have to sum the rate of change in S′, and finally we have

$$\left( {\frac{{d{\mathbf{A}}}}{dt}} \right)_{S} = \left( {\frac{{d{\mathbf{A}}}}{dt}} \right)_{{S^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{A}}\text{,}$$

(5.26)

which is the formula we were looking for. Notice that in the preceding sections, we did not take care to specify in which frame we were taking the derivatives. This was allowed because, being the considered transformations translations, the Cartesian components of the vectors were not modified. This can be immediately verified in Eq. (5.26) in which, if ω = 0, the derivatives in the two frames are equal.

We shall now find the relations between the kinematic quantities in S and in S′. We shall call the former absolute and the latter relative, but we notice that the definition is arbitrary; we could have started calling S′ stationary and S mobile.

See that the relation between the position vectors is always

$${\mathbf{r}} = {\mathbf{r}}^{\prime} + {\mathbf{r}}_{{O^{\prime} }} .$$

(5.27)

To obtain the relation between relative (in S′) and absolute (in S) velocities, we need the time derivatives. To do that, we need to have on each side of the equation only vectors in one frame. Hence, we re-write Eq. (5.27) as

$${\mathbf{r}} - {\mathbf{r}}_{O\prime} = {\mathbf{r}}^{\prime}$$

(5.28)

and derive the vector r – r _O′ using the rule (5.26), obtaining

$$\left( {\frac{{d\left( {{\mathbf{r}} - {\mathbf{r}}_{{O^{\prime} }} } \right)}}{dt}} \right)_{S} = \left( {\frac{{d\left( {{\mathbf{r}} - {\mathbf{r}}_{{O^{\prime} }} } \right)}}{dt}} \right)_{{S^{\prime}}} + {\varvec{\upomega}} \times \left( {{\mathbf{r}} - {\mathbf{r}}_{{O^{\prime} }} } \right).$$

(5.29)

The meaning of the left-hand side of this equation is clear: it is the difference between the absolute velocities of the point P, say v, and of the point O′, say v _O′. We substitute Eq. (5.28) on the right-hand side, obtaining

$${\mathbf{v}} - {\mathbf{v}}_{{O^{\prime} }} = \left( {\frac{{d{\mathbf{r}}^{\prime} }}{dt}} \right)_{{S^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{r}}^{\prime} .$$

(5.30)

Now, we see that the first term on the right-hand side is the rate of change in S′ of the position vector in S′, namely the velocity of P in S′, which we call relative and indicate with v′. We then write

$${\mathbf{v}} = {\mathbf{v}}^{\prime} + {\mathbf{v}}_{{O^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{r}}^{\prime} = {\mathbf{v}}^{\prime} + {\mathbf{v}}_{t} .$$

(5.31)

In other words, the velocity v of the point P in S is the sum of its velocity v′ in S′ and of two more terms that we have grouped in v _t. The meaning of the latter is understood considering the case in which the point does not move in S′, namely if v′ = 0. Then, v _t is the absolute velocity of the point. We can then state that v _t is the velocity of the point fixed in the frame S′, and call it Q, through which the moving point P passes at the considered time. We can think of v _t as the velocity of the moving space. It is called the velocity of transportation . It contains two terms,

$${\mathbf{v}}_{t} \equiv {\mathbf{v}}_{{O^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{r}}^{\prime} ,$$

(5.32)

which we discuss looking at Fig. 5.5. The first one is the velocity of the origin of S′ and corresponds to the translational component of its motion relative to S. The second term is due to the rotation of S′. We can think of this as taking place about an instantaneous rotation axis passing through O′ with angular velocity, in the considered instant, ω. Indeed, the velocity of the point Q stationary in S′ where P is passing is just ${\varvec{\upomega}} \times {\mathbf{r}}^{\prime}$.

Fig. 5.5

The relative velocity in the rotating frame

We shall now find the accelerations, by a further derivative. We shall meet more terms. We start from Eq. (5.31) in the form

$${\mathbf{v - v}}_{{O^{\prime} }} = {\mathbf{v}}^{\prime} + {\varvec{\upomega}} \times {\mathbf{r}}^{\prime}$$

(5.33)

and derive the left-hand side using Eq. (5.26), obtaining

$$\left( {\frac{{d\left( {{\mathbf{v}} - {\mathbf{v}}_{{O^{\prime} }} } \right)}}{dt}} \right)_{S} = \left( {\frac{{d\left( {{\mathbf{v}} - {\mathbf{v}}_{{O^{\prime} }} } \right)}}{dt}} \right)_{S\prime} + {\varvec{\upomega}} \times \left( {{\mathbf{v}} - {\mathbf{v}}_{{O^{\prime} }} } \right).$$

(5.34)

Similarly to above, the left-hand side is the difference between the absolute accelerations of P, say a, and O′, say a _O′. Still analogously, we use Eq. (5.33) to substitute v – v _O′ on the right-hand side, obtaining

$$\begin{aligned} {\mathbf{a}} - {\mathbf{a}}_{{O^{\prime} }} & = \left( {\frac{{d{\mathbf{v}}^{\prime} }}{dt}} \right)_{{S^{\prime} }} + \left( {\frac{{d\left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right)}}{dt}} \right)_{{S^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{v}}^{\prime} + {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right) \\ & = \left( {\frac{{d{\mathbf{v}}^{\prime} }}{dt}} \right)_{{S^{\prime} }} + \left( {\frac{{d{\varvec{\upomega}}}}{dt}} \right)_{{S^{\prime} }} \times {\mathbf{r}}^{\prime} + {\varvec{\upomega}} \times \left( {\frac{{d{\mathbf{r}}^{\prime} }}{dt}} \right)_{{S^{\prime} }} + {\varvec{\upomega}} \times {\mathbf{v}}^{\prime} + {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right). \\ \end{aligned}$$

(5.35)

The last term looks a bit complicated, but its terms have well-defined physical meanings. Let us examine them. The first term is the acceleration of P in S, namely the relative acceleration, say a′. In the second term, the angular acceleration of the motion of S′ relative to S appears. We shall name it

$${\varvec{\upalpha}} = \left( {\frac{{d{\varvec{\upomega}}}}{dt}} \right)_{{S^{\prime} }} .$$

(5.36)

The next two terms are equal. We put them together and also group some other terms, writing

$${\mathbf{a}} = {\mathbf{a}}^{\prime} + \left[ {{\mathbf{a}}_{{O^{\prime} }} + {\varvec{\upalpha}} \times {\mathbf{r}}^{\prime} + {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right)} \right] + 2{\varvec{\upomega}} \times {\mathbf{v}}^{\prime} ,$$

(5.37)

which expresses the Coriolis theorem, after Gustave de Coriolis (1792–1843). We now define

$${\mathbf{a}}_{t} \equiv {\mathbf{a}}_{O\prime} + {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right) + {\varvec{\upalpha}} \times {\mathbf{r}}^{\prime} ,$$

(5.38)

which is called the acceleration of transportation and

$${\mathbf{a}}_{Co} \equiv 2{\varvec{\upomega}} \times {\mathbf{v}}^{\prime} ,$$

(5.39)

which is called the Coriolis . Finally, we write Eq. (5.37) as

$${\mathbf{a}} = {\mathbf{a}}^{\prime} + {\mathbf{a}}_{t} + {\mathbf{a}}_{Co} .$$

(5.40)

The meaning of the acceleration of transportation a _t is analogous to that of the velocity of transportation v _t. Indeed, if both velocity and acceleration of P in S′ are zero, then its absolute acceleration is a _t, as the other two terms on the right-hand side of Eq. (5.40) are then zero. The term a _t is the absolute acceleration of the point stationary in S′ through which the point P (call it Q again) is passing at the considered instant. It is the sum of three terms. The first is the acceleration relative to S of the origin of the mobile frame S′. The second term is the absolute acceleration of Q due to the rotation of S′ relative to S. The situation is shown in Fig. 5.6. Indeed, the velocity of Q (of position vector r′) due to the rotation is ${\varvec{\upomega}} \times {\mathbf{r}}^{\prime}$. In turn, this velocity varies in time, and its rate of change is, by the same formula ${\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right)$. This is simply the centripetal acceleration of the point Q. Indeed, as we understand looking at Fig. 5.6, we have

Fig. 5.6

Geometry in a rotating frame

$$\left| {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right| = \omega r^{\prime} \sin \theta = \omega d,$$

where d is the curvature radius (the radius of the osculating circle) of the curve Q is describing. And further

$$\left| {{\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right)} \right| = \omega^{2} d,$$

which is the centripetal acceleration of Q.

Consider now the third term in Eq. (5.36). If the angular velocity ω is constant, the absolute velocity of Q varies only in direction, and this term is zero. If ω is not constant, the magnitude of the absolute velocity of Q also varies. This acceleration is given by the third term, ${\varvec{\upalpha}} \times {\mathbf{r}}^{\prime}$.

As for the Coriolis acceleration a _Co, we see in Eq. (5.39) that it is zero in three cases: when, in the considered instant, the point P does not move in S′ (v′ = 0), when the mobile frame does not rotate (ω = 0) and when the velocity of the point P is parallel to the angular velocity. The Coriolis acceleration does not depend on the position of P but does depend on its relative velocity and becomes larger for larger relative velocities. It is always directed perpendicularly to the motion and consequently is a cause of change in its direction, rather than of its magnitude. We shall see examples in the next section.

We shall now assume that S is an inertial frame. As we have seen in the previous section, if S′ accelerates relative to S, it is consequently not inertial. In other words, the Newton laws in S do not hold. Let us look at the details.

In the inertial frame S, the law of motion of the mass m under the action of the force F is F = m a. This can be written, using Eq. (5.40), as

$${\mathbf{F}} = m{\mathbf{a}}^{\prime} + m{\mathbf{a}}_{t} + m{\mathbf{a}}_{Co} .$$

The observer in S′ measures the acceleration a′ and wants to have that on the right-hand side. We move the other terms to the left-hand side, obtaining

$${\mathbf{F}} - m{\mathbf{a}}_{t} - m{\mathbf{a}}_{Co} = m{\mathbf{a}}^{\prime} .$$

(5.41)

We get the Newton law back formally by defining two fictitious forces

$${\mathbf{F}}_{t} \equiv - m{\mathbf{a}}_{t}$$

(5.42)

and

$${\mathbf{F}}_{Co} = - m{\mathbf{a}}_{Co} ,$$

(5.43)

which is called the Coriolis force , and we subsequently get

$${\mathbf{F}} + {\mathbf{F}}_{t} + {\mathbf{F}}_{Co} = m{\mathbf{a}}^{\prime} .$$

(5.44)

We can then state that, in a frame mobile with an arbitrary motion relative to an inertial frame, the product of the mass times the acceleration is equal to the resultant of both true and fictitious forces. However, as already stated, the fictitious forces are not real and are not due to any physical agent. Consequently, the action-reaction law is not satisfied.

5.5 Motion in a Rotating Frame

Consider now the simple case in which the reference frame S′ rotates relative to the inertial frame S with angular velocity ω constant in magnitude and direction. For example, S′ may be fixed on a rotating platform, for example, a merry-go-round, and S stationary on earth. As we shall see in the next section, such a frame is not exactly inertial due to the rotation of earth on its axis and its revolution around the sun, but the effects of the difference are quite small and we shall disregard them here.

We choose the origin of both frames in the center of the platform, their z and z′ axes vertical upwards and, consequently, x, y and x′, y′ in the horizontal plane of the platform, as shown in Fig. 5.7. The axes x and y are stationary relative to the ground, while x′ and y′ rotate. With our choice of co-ordinates, the position vectors in the two frames coincide, r = r′.

Fig. 5.7

The S reference frame is stationary to the ground, S′ rotates with constant angular velocity

In the particular case we are considering, the relevant expressions for velocities and accelerations simplify in

$${\mathbf{v}} = {\mathbf{v}}^{\prime} + {\mathbf{v}}_{t} ,$$

(5.45)

$${\mathbf{v}}_{t} = {\varvec{\upomega}} \times {\mathbf{r}}^{\prime} = {\varvec{\upomega}} \times {\mathbf{r}}\text{,}$$

(5.46)

$${\mathbf{a}} = {\mathbf{a}}^{\prime} + {\mathbf{a}}_{t} + {\mathbf{a}}_{Co} ,$$

(5.47)

$${\mathbf{a}}_{t} = {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}^{\prime} } \right) = {\varvec{\upomega}} \times \left( {{\varvec{\upomega}} \times {\mathbf{r}}} \right),$$

(5.48)

$${\mathbf{a}}_{Co} = 2{\varvec{\upomega}} \times {\mathbf{v}}^{\prime} .$$

(5.49)

Let us consider the velocities. In general, the point P is not necessarily on the platform. In Fig. 5.8, we have drawn it somewhat higher up. In general, the vectors ω and r are not parallel. Recalling that v _t is the velocity of the point Q stationary in S′ in the instant position of P, we see that it is tangent to the circle thorough P normal to the rotation axis and with its center on the axis. This circle is the trajectory of Q. The magnitude of the velocity of transportation is then

Fig. 5.8

The velocity v _t of the point Q

$$\upsilon_{t} = \omega r\sin \theta = \omega d,$$

(5.50)

where d is the radius of the circle, namely the distance from the rotation axis. v _t is then simply the velocity of Q in its circular motion.

We now consider the accelerations. We immediately see that the a _t term is simply the centripetal acceleration of the point Q as seen in the inertial frame S.

Let us now consider a point P of mass m to be standing still, relative to S′, on the platform at the distance r from the axis. Suppose that the friction is negligible and that P is kept in position by a rubber band attached to a small ring around the axis.

The inertial observer in S sees P moving in uniform circular motion with velocity ωr. He knows that the motion has an acceleration towards the center, the centripetal acceleration, of magnitude ω ² r (this is the absolute acceleration in this case). The (centripetal) force causing the acceleration is due to the rubber band. The observer can check that measuring the stretch of the rubber band.

The non-inertial observer in S′, on the platform, also sees that the rubber band is stretched, determining that a centripetal force is acting on P. He measures it and finds the same result as the inertial observer. The mobile observer now insists on having the first Newton law be valid and concludes that a second force, equal and opposite to that of the rubber band, must exist. This is the inertial force, due to the acceleration of transportation, –m a _t, the direction of which is opposite to the centripetal force. In this case, the force is centrifugal . In this case, and always, the centrifugal forces are not real forces, but pseudo forces of the relative motion. They appear only when we pretend to describe the motion in a non-inertial, rotating frame as if it were inertial. However, the centrifugal force is felt as a real force, such as, for example, in a fast rotating merry-go-round.

We now discuss the Coriolis acceleration (Eq. 5.49) and the effects of the corresponding fictitious Coriolis force

$${\mathbf{F}}_{Co} = - m{\mathbf{a}}_{Co} = - 2m{\varvec{\upomega}} \times {\mathbf{v}}^{\prime} .$$

(5.51)

Consider again the point P lying on the rotating platform. If P does not move relative to the platform, the Coriolis acceleration is null, as in the case just discussed. Let v′ be this velocity, which we assume, for simplicity, to be parallel to the platform. As we have already noticed, the Coriolis acceleration, and consequently the Coriolis force , does not depend on the position of P on the platform and is in any case perpendicular to the relative velocity. Consider Fig. 5.9. If the angular velocity ω is directed out of the plane of the figure, as in Fig. 5.9a, we see the platform turning counter-clockwise. In this case, the Coriolis acceleration is directed towards the left of the motion, and the Coriolis force to the right. Suppose you are the point P waking or running on the platform. You will feel a push to the right of your speed, in whatever direction you move. Contrastingly, if ω is directed inside the drawing, as in Fig. 5.9b, and the rotation is clockwise, the Coriolis force pushes to the left of the speed.

Fig. 5.9

Coriolis acceleration and (pseudo)force on a platform rotating. a Counter-clockwise, b Clockwise

If we were to look at the earth from some distance from its surface on the axis, we would see the northern hemisphere rotating counter-clockwise if we were above the North pole, and the southern one clockwise if we were above the South pole. The Coriolis forces are the dominant causes of the circulation of winds in the atmosphere and cyclonic and anticyclonic phenomena. We shall discuss that in the next section.

Consider now another example, namely a material point P, standing in equilibrium above the platform in a fixed position relative to S, i.e., to the ground. We might think about a fly located just above the platform. The observer in S sees P at rest. Knowing that it is subject to its weight, he understands that another force, equal and opposite to the weight, should exist. The force is exerted by the beating of the fly’s wings.

For the observer in S′, the description is more complicated. He sees P moving in a circular uniform motion on a circle of radius r with velocity ωr. The motion is accelerated with a centripetal acceleration ω ² r. He deduces that a force mω ² r should act on the fly. However, he also knows, as the result of experiments he has done in the past, such as the one we just discussed, that a centrifugal force exists on the platform, namely a force of magnitude mω ² r directed outwards. Considering that the point moves on a circle, he concludes that the centripetal force on the fly must be twice as large, namely 2 mω ² r. From where is this force is coming? It is the Coriolis force . In this case, ω and v′ are mutually perpendicular; Eq. (5.49) says that the magnitude of this force is just 2 mω ² r and that its direction is radial, towards the center. Physics is difficult in non-inertial frames, but the factor two is needed!

As a final example, let us go back to the first one, in which the point P is kept still on the platform by a rubber band attached to the axis. The motion seen by S is circular uniform. At a certain instant when we cut the band, S will see P sliding on the platform of a straight uniform motion at the velocity it had at the moment of the cut, directed as the tangent to the circle in that moment. Indeed, there is no net force acting on P.

How does the observer in S′ describe the motion? To be concrete, assume the rotation to be counter-clockwise. When the rubber band is cut, the force that is needed in the rotating system to keep the objects standing disappears, and we might expect to see the point P moving outside along the radius of the platform. But this is not what we observe; rather, the point moves outside describing a curve. The reason is the Coriolis force. Before the rubber band was cut, P did not move on the platform, and the Coriolis force was null, but it is not so any longer since P has started moving. The Coriolis force acts, pushing P to the right all along its trajectory. Observing from outside, we can better understand what is going on. When the rubber band is cut, P moves with the same velocity as the point of the platform on which it is seated. While moving outwards, P reaches points of the platform having higher speeds, because they are farther from the axis, and consequently is left behind by them.

5.6 The Inertial Frame

As we have already stated, a reference frame is defined as inertial, if in that frame the first Newton law is valid. We have also seen that if a reference frame is inertial, any other one moving in uniform translation motion relative to it is also inertial. Indeed, the relativity principle we saw in Sect. 5.2 states that no experiment can distinguish between them. In other words, there is no absolute reference frame. Finally, we have seen that the Newton laws are covariant under the Galilei transformations.

However, nature does not necessarily behave according to our definition, and inertial reference frames might just not exist. The answer must come, as always, from the experiment. Basically, we need to check if we can find one reference frame in which material points not subject to forces, or, better yet, subject to forces of null resultant, always move in a rectilinear uniform motion. As a matter of fact, as is often the case in physics, we proceed through successive approximations. We can find reference frames that can be considered inertial, within a certain approximation, namely for experiments of a certain sensitivity or precision. For more precise experiments, we must search for frames that are closer to the inertial one, and we can find them.

Indeed, the largest fraction of the experiments takes place on earth, and is described in a stationary frame relative to the walls of our laboratory. These frames can be considered inertial within a quite good approximation, although not perfect. Indeed, earth rotates on its axis, making a turn (2π angle) in a day (84 600 s). The corresponding angular velocity, directed from the South to the North pole, is ω _rot = 7.3 × 10^–5 s^–1. Figure 5.10a, for example, shows a stationary reference frame on earth at a certain latitude λ. In this frame, the transportation and Coriolis acceleration are present.

Fig. 5.10

Three reference frames with, acceleration towards a the rotation axis of earth a ₁ = 2.4 × 10^–2 ms^–2, b the sun a ₂ = 5.9 × 10^–3 ms^–2, c the center of the Galaxy a ₃ = 10^–10 ms^–2

Let us analyze the first one, to which the centrifugal (pseudo)force corresponds. The magnitude of this force on a point P of mass m is the product of the mass, the square of the angular velocity (equal everywhere on earth) and the radius of the circle on which P moves. The latter is the distance from the axis, Rcosλ, where R is the earth radius and λ is the latitude of P. Calling a ₁ the acceleration, the magnitude of the force is

$$F_{1} = ma_{1} = m\omega_{\text{rot}}^{2} R\cos \lambda .$$

(5.52)

Let us look at the numbers. Recalling that R = 6.4 × 10⁶ m and taking, for example, λ = 45˚, the acceleration, which is also the force per unit mass, is

$$a_{1} = 2.4 \times 10^{ - 2} \;{\text{ms}}^{{{-}2}} ,$$

(5.53)

which is quite small, less than a per mille of the gravitational acceleration. However, for precise measurements, it can be relevant. The Coriolis force is usually smaller. However, it is important for large-scale phenomena, as we shall see in the next section.

However, a stationary reference frame on earth differs from an inertial frame for a second reason, to even smaller effect. Indeed, earth moves along its orbit, turning around in a year, with an angular velocity of ω _riv = 2×10^–7 s^–1 on an orbit of radius R _orb = 1.49 × 10¹¹ m (Fig. 5.10b). The centripetal acceleration is

$$a_{2} = \omega_{\text{riv}}^{2} R_{\text{orb}} = 5.9 \times 10^{ - 3} \;{\text{ms}}^{{{-}2}} ,$$

(5.54)

which is an order of magnitude smaller than a ₁. The effects of the corresponding pseudo force are negligible, if not for the most precise measurements. Usually, the Coriolis force is even smaller.

Even these small effects, however, can be eliminated by choosing a reference frame with its origin in the sun and directions of the axes stationary to the fixed stars. This frame is inertial to an extremely good approximation, although not perfect. Indeed, the sun is located at the periphery of our spiral galaxy (10¹¹ stars in order of magnitude). The sun turns around the center of the galaxy in an orbit of radius R _S ≈ 2.4 × 10²⁰ m over a period of about 150 million years, corresponding to the angular velocity of ω _S = 7.9 × 10^–16 s^–1. The corresponding centripetal acceleration is

$$a_{3} = \omega_{S}^{2} R_{\text{S}} \sim 10^{ - 10} \;{\text{ms}}^{{{-}2}} .$$

(5.55)

This is very small indeed. However, experiments exist that are so sensitive, they are able to detect deviations from the state of inertia even at these extremely small levels. As a matter of fact, our galaxy moves too, in a non-uniform motion. However, when needed, we know how to eliminate the effects.

In conclusion, inertial reference frames exist in nature at every level of approximation we need.

5.7 Earth, as a Non-inertial Frame

As we just saw, the rotation of earth on its axis, with the angular velocity, ω _rot = 7.3 × 10^–5 s^–1. This implies that in reference frame stationary on earth dynamical effects of the transportation and Coriolis fictitious forces exist. We shall discuss the principal ones in this section.

We take a reference system S with the origin in the center of earth and stationary with it. We shall use the symbols v and a for velocities and accelerations in this frame, omitting the prime we used in the previous sections.

The acceleration of transportation is

$${\mathbf{a}}_{t} = {\mathbf{a}}_{O} - \omega_{{_{\text{rot}} }}^{2} {\mathbf{r}}_{E} ,$$

where a _O is the acceleration of the earth’s center. This is the centripetal acceleration of its motion around the sun in a very good approximation, and r _E is the radius of the orbit, as shown in Fig. 5.11a. The Coriolis acceleration on a point moving with velocity v in S is

Fig. 5.11

a Forces and pseudo forces on matter point P; b Displacement to east in the free fall (exaggerated)

$${\mathbf{a}}_{Co} = 2{\varvec{\upomega}}_{\text{rot}} \times {\mathbf{v}}\text{.}$$

In S, the equation of motion of a point with mass m subject to the real force F _true is then

$$m{\mathbf{a}} = {\mathbf{F}}_{t} + {\mathbf{F}}_{Co} + {\mathbf{F}}_{\text{true}} = - m{\mathbf{a}}_{O} + m\omega_{\text{rot}}^{2} {\mathbf{r}}_{E} - 2m{\varvec{\upomega}}_{\text{rot}} \times {\mathbf{v}} + {\mathbf{F}}_{\text{true}} .$$

(5.56)

We can distinguish the following contributions to the true force F _true: the gravitational attraction of earth F _E, the gravitational attraction of all the other heavenly bodies F _O, and of any other force that might be present (air resistance, tension of a wire, etc.), with resultant F. We re-write Eq. (5.56), grouping the terms according to their causes,

$$m{\mathbf{a}} = \left( {{\mathbf{F}}_{E} + m\omega_{\text{rot}}^{2} {\mathbf{r}}_{E} } \right) - 2m{\varvec{\upomega}}_{\text{rot}} \times {\mathbf{v}} + \left( {{\mathbf{F}}_{O} - m{\mathbf{a}}_{O} } \right) + {\mathbf{F}}\text{.}$$

(5.57)

The gravitational force F _O is due to all the heavenly bodies different from earth, but is largely dominated by the sun. As the diameter of earth is much smaller than the distance from the sun, in a first approximation, we can consider F _O equal in all the points of the earth. However, the small differences that are present are one of the causes of the tides, as we shall see in Sect. 6.4. The acceleration produced by F _O on every body is proportional to the mass of the body. Consequently, it is the same on the surface of the earth and in its center. In other words, it is the acceleration a _O, of the earth herself. Hence, F _O–m a _O = 0.

We have reached an important conclusion, which is true as long as F _O can be considered not to vary on the points of the earth, that the gravitational forces of the sun, the moon end of the other heavenly bodies do not appear in the equations of motion in reference frames stationary on earth. These forces are exactly balanced by the inertial forces resulting from the acceleration that those agents impart to the earth.

We can simplify Eq. (5.57) as

$$m{\mathbf{a}} = \left( {{\mathbf{F}}_{E} + m\omega_{\text{rot}}^{2} {\mathbf{r}}_{E} } \right) - 2m{\varvec{\upomega}}_{\text{rot}} \times {\mathbf{v}} + {\mathbf{F}}\text{.}$$

(5.58)

Now, we are ready to consider several important examples.

The first case is of a body at rest, and F is simply its weight. This is the force we measure with a balance and that we have written as

$${\mathbf{F}}_{w} = m{\mathbf{g}}\text{,}$$

(5.59)

where g is a vector quantity, which is equal for all the bodies in a given position. Up to now, we have talked of it as gravitational acceleration, but we are now ready to see that it is only approximately so. Equation (5.58) indicates that the force pushing a body downwards that does not move (v = 0, a = 0) is ${\mathbf{F}}_{E} + m\omega_{\text{rot}}^{2} {\mathbf{r}}_{E}$. We can say that the gravitational force of the earth on the body is

$${\mathbf{F}}_{G} = m{\mathbf{G}}$$

(5.60)

and write

$${\mathbf{F}}_{w} = m{\mathbf{g}} = m\left( {{\mathbf{G}} + \omega_{\text{rot}}^{2} {\mathbf{r}}_{E} } \right),$$

(5.61)

where G is the gravitational field of earth, and

$${\mathbf{g}} = {\mathbf{G}} + \omega_{\text{rot}}^{2} {\mathbf{r}}_{E} .$$

(5.62)

The acceleration is the same for all the bodies in the same location. Equation (5.58) shows that a body dropped in absence of any force other than its weight, from a position of rest, v = 0, moves with an acceleration a = g. We can say that g is the acceleration of the free-fall of any body, provided its velocity is null in the considered instant. If v ≠ 0, the Coriolis acceleration is, in general, present too.

In any case, Eq. (5.61) tells us that the weight is the sum of two contributions: the gravitational attraction m G of the earth, which largely dominates, and the centrifugal force due to the rotation of earth, which is much smaller and varies with the position. We will now discuss the observable consequences of that.

The local value of g. Suppose we take a plumb and fix it at a support. In the equilibrium position, its weight F _w, given by Eq. (5.59), and the tension of the wire are equal and opposite. The direction is given by the wire. The distance from the rotation axis of a point P on the surface at the latitude λ is r _E = Rcosλ, where R is the earth radius (Fig. 5.11a). The weight F _w can be decomposed in a component, let us call it F _w,r, directed to the center of earth, and a component, F _w,θ, in the direction of the meridian, to the North in the northern hemisphere and to the South in the southern one. The two components are

$$\begin{aligned} F_{wr} & = mG - F_{t} \cos \lambda = m\left( {G - \omega_{\text{rot}}^{2} R\cos^{2} \lambda } \right) \\ F_{w\theta } & = F_{t} \sin \lambda = m\omega_{\text{rot}}^{2} R\sin \lambda \cos \lambda . \\ \end{aligned}$$

(5.63)

The centrifugal term, the first one, is zero at the poles and maximum at the Equator. The tangential component is zero both at the poles and at the Equator. In these locations, but not elsewhere, the weight is precisely directed to the center of earth. As for the magnitude, the measured values are g = 9.832 ms^–2 at the poles and g = 9.780 ms^–2 at the Equator. If we approximate the shape of the earth surface with a sphere, all its points are at the same distance from the center, and if the mass distribution inside the earth is spherically symmetric, the gravitational term G is equal everywhere. It should be equal to g at the poles, G = 9.780 ms^–2. Let us check by giving an estimate, starting from g at the Equator.

$$G = g + \omega_{\text{rot}}^{2} R = 9.780 + \left( {7.3 \times 10^{ - 5} } \right)^{2} \times 6.378 \times 10^{6} = 9.814\;{\text{ms}}^{{{-}2}} .$$

This value is close, but still a bit smaller than what we found from g at the poles. The main reason for that is that earth is not really spherical but somewhat squeezed at the pole, an effect of the centrifugal forces. Consequently, the poles are a bit closer to the center than the Equator.

Notice however, that small differences on the value of g in the different points of the surface are present, due to the local geology.

Absence of weight. If we measure the weight of an object with a balance on the space station, we find it to be zero. Such is also the weight of all the objects in the station, and in every artificial satellite. The arguments we just made are still valid, if we put the station in the place of earth, and consider the earth as an external body, as the sun, the moon and the other planets are. The spaceship is small enough for the gravitational force of those bodies to be considered equal at all the points of the ship. This force is exactly balanced by the inertial force to the acceleration of the spaceship. If its engines are shot, the ship freely falls under the action of gravitation. In this case, the equivalent of the weight on earth, namely the gravitational attraction of the ship on the body inside it, is completely negligible, F _w = 0. The centrifugal term to the weight in the space ship is also negligible because the ship does not rotate appreciably. The weight in the ship is zero.

Eastwards shift in the free-fall . If a material point P of mass m is dropped with null initial velocity at a height h from the ground, it initially falls under the action of the weight, F _w. However, as soon as its velocity, v, is appreciably different from zero, a second inertial force, the Coriolis force , enters into action. It is

$${\mathbf{F}}_{Co} = - 2m\omega_{\text{rot}}^{{}} \times {\mathbf{v}}\text{.}$$

(5.64)

The velocity v relative to earth is in the plane containing the earth’s axis and point P, namely the plane PON in Fig. 5.11a. Consequently, the Coriolis force is perpendicular to this plane. Considering that the direction of the angular velocity is from South to North, and that v is downwards, we see that the Coriolis force is toward East in both hemispheres. The situation is shown in Fig. 5.11b, where AB is the direction of the plumb, i.e., the direction of F _w (no Coriolis force on the plumb that does not move) and C is the point in which the body reaches the ground, falling from the height h. The shift from the vertical BC is very small, and exaggerated in the figure. Let us calculate it.

We take a reference with the z-axis vertical, i.e., in the local direction of the plumb, and the x-axis horizontal towards the East. Within a good approximation, we can take the magnitude of the velocity to be υ = gt, as in the vertical fall. Its direction is opposite to the z-axis. The equation of the component of the motion on the x-axis is

$$m\frac{{d^{2} x}}{{dt^{2} }} = 2m\omega_{\text{rot}} gt\cos \lambda .$$

We solve the equation by integrating twice on time and imposing the initial conditions x(t) = 0, (dx/dt(0)) = 0, obtaining

$$x = \frac{1}{3}g\omega_{\text{rot}} t^{3} \cos \lambda .$$

(5.65)

The time of the fall is, with good approximation, $t = \sqrt {2h/g}$, and we have

$$x = \frac{2\sqrt 2 }{3}g^{ - 1/2} \omega_{\text{rot}} h^{3/2} \cos \lambda .$$

(5.66)

For example, at the latitude of 45˚ and a fall from h = 50 m, the eastward shift is x ~ 5 mm, which is quite small, but has been measured, carefully eliminating perturbing effects.

Horizontal wind circulation . As is well known, the earth’s atmosphere in a certain instant contains zones of high pressure and zones of low pressure. Naively, one would expect winds to blow from the former to the latter in the direction of the pressure gradient. However, the direction of the winds is substantially perpendicular to that, moving along the isobars, as you can see watching weather forecasts on TV. The effect is due to the Coriolis force .

Figure 5.12 summarizes the situation. H is the pressure maximum, L a pressure minimum, in the Northern hemisphere. Hence, the earth’s angular velocity direction is out of the paper and the Coriolis force is directed, perpendicular to the velocity, to the right. Consider, for simplicity, a horizontal wind at constant velocity (in magnitude). Suppose we insulate a small mass of air within an ideal film and follow its motion. Two vertical and two horizontal forces act on our mass. The vertical ones are the weight and the Archimedes force. As the motion is horizontal, they are equal and opposite. The horizontal forces are the pressure (true) force and the Coriolis (pseudo) force. The pressure force acts on the surfaces of our gas mass. The pressure on its left-hand face pushes to the right, while the pressure on the right face pushes to the left. If the pressure were equal on the two sides, the neat force would be null. However, if there is a pressure maximum on the right of the gas mass we are following, as in Fig. 5.12a, there is a neat pressure force F ^(P) pushing to the left. The Coriolis force has an equal and opposite direction. Consequently, the two forces may balance each other, or result in the right value being the centripetal force for the curvature of the wind trajectory. This can happen only if the wind circulates in a counter-clockwise direction around a pressure maximum (anticyclone). Contrastingly, it must circulate clockwise around a minimum (cyclone), as in Fig. 5.12b. The two situations are inverted in the southern hemisphere.

Fig. 5.12

Isobars around pressure maximum (left) and minimum (right-hand), in the Northern hemisphere and the forces on a mass of air

Let us look at the orders of magnitudes. The magnitude of the Coriolis force on an air mass m moving with horizontal speed υ at the latitude λ is

$$F_{Co} = 2m\omega_{\text{rot}} \upsilon \sin \lambda .$$

(5.67)

For example, the force on a kilogram of air, which is about 1 m³, moving at 10 m/s at 45˚ is about 10^–3 N. This should be compared to the pressure forces on the same volume. To be of the same order of magnitude, the pressure forces on two opposite sides of our cubic meter volume should be different by 10^–3 N. This corresponds to a pressure difference of 10^–3 Pa, being the surface unitary. Hence, the pressure gradient should be of 10^–3 Pa/m, corresponding, say, to a distance of 100 km between two isobars of 100 Pa difference. This is reasonable (have a look at the weather maps).

The Foucault pendulum . A simple pendulum abandoned in a non-equilibrium position with null velocity oscillates in a vertical plane. However, if we watch carefully for a long enough time, along the order of one hour, we can see that the oscillation plane rotates relative to the laboratory, i.e., relative to a reference fixed on earth. The reason for the rotation is, once more, that the frame is not exactly inertial. As a matter of fact, the oscillation plane is fixed in an inertial frame, relative to which the earth rotates, as in Fig. 5.13.

Fig. 5.13

The Foucault pendulum

While the effect has been known since its first observation by Vincenzo Viviani (1622–1703) in 1661, the main experiment and its correct interpretation were done by Léon Foucault (1819–1868) in 1851 in the Pantheon of Paris. His pendulum was 67 m long and had a 28 kg mass.

A similar situation, shown in Fig. 5.14, helps in our understanding. There, we have a pendulum, supported on a turning platform. If we put the pendulum in oscillation and the platform in rotation, we observe the oscillation plane remaining fixed, as expected, and the platform rotating under the pendulum. We can easily imagine what an observer on the platform would see, namely the plane of oscillation rotating in the opposite direction.

Fig. 5.14

Pendulum on a rotating platform

In this way, we easily understand what happens on earth if we are on a pole. Here, the angular velocity vector ω _rot is normal to the “platform”, the earth surface, exactly as in that experiment. From the point of view of an inertial observer, the oscillation plane is constant, and he sees the earth turning relative to it. He understands why an observer at the pole sees the oscillation plane rotating and making a complete turn in 24 h. In the reference fixed to earth, the equation of motion is, once more, (5.58), with F the tension of the wire (neglecting air resistance). The Coriolis force ${\mathbf{F}}_{Co} = - 2m{\varvec{\upomega}}_{\text{rot}} \times {\mathbf{v}}$ is normal to the oscillation plane, and causes its rotation.

If the experiment is done not at the pole but at a latitude λ, we must pay attention to the vector characteristic of ω _rot. We decompose it in a horizontal component, namely parallel to the ground in our position, ω _h, and a vertical one ω _v: ω _rot = ω _h + ω _v. We further decompose the horizontal component, which is still a vector, in its components parallel, ω _p, and normal, ω _n, to the oscillation plane and write Eq. (5.56) as

$$m{\mathbf{a}} = m{\mathbf{g}} - 2m{\varvec{\upomega}}_{\text{v}} \times {\mathbf{v}} - 2m{\varvec{\upomega}}_{\text{p}} \times {\mathbf{v}} - 2m{\varvec{\upomega}}_{\text{n}} \times {\mathbf{v}} + {\mathbf{F}}\text{.}$$

(5.68)

The term $- 2m{\varvec{\upomega}}_{\text{n}} \times {\mathbf{v}}$ has the direction of the wire. Its effect is to change the tension a bit, but it has no effect on the oscillation plane. The term $- 2m{\varvec{\upomega}}_{\text{v}} \times {\mathbf{v}}$ is perpendicular to the oscillation plane and causes its rotation. The third term $- 2m{\varvec{\upomega}}_{\text{n}} \times {\mathbf{v}}$ is also perpendicular to the rotation plane, but is very small. Indeed, as we can see in Fig. 5.13b, it is proportional to sinθ, where θ is the angle between the wire and the vertical and is small, for small oscillations. We can then simplify Eq. (5.68) by writing

$$m{\mathbf{a}} = m{\mathbf{g}} - 2m{\varvec{\upomega}}_{\text{v}} \times {\mathbf{v}} + {\mathbf{F}}\text{.}$$

(5.69)

In concussion, the motion is similar to that at the pole with the sole difference being that in place of the angular velocity ω, we must consider its component along the local vertical, of magnitude

$$\omega_{\text{v}} = \omega_{\text{rot}} \sin \lambda .$$

(5.70)

The oscillation plane makes a complete turn in the period

$$T_{\text{rot}} = \frac{2\pi }{{\omega_{\text{rot}} \sin \lambda }} = \frac{{24\;{\text{hr}}}}{\sin \lambda }.$$

(5.71)

At 45˚ latitude, in one hour, the plane rotates by 10.6˚.

Figure 5.13c shows the projection on the horizontal plane of the trajectory of the Foucault pendulum. The vector ω _v is normal to the drawing towards the observer. The Foucault force is always directed normally to the velocity to the right of the direction of motion. The force bends the trajectory, as shown with exaggeration in Fig. 5.13c. Suppose that the pendulum is initially in A and abandoned with null velocity. Initially, when the Coriolis force is very small, the pendulum heads to A′. But as soon as the velocity becomes appreciable, the Coriolis force pushes to the right, bending the trajectory. The pendulum reaches point B, where it stops. When the velocity has again sufficiently increased, but in the opposite direction, the Coriolis force pushes in the opposite direction too, although still to the right of the motion. The pendulum reaches C, etc.

In the Foucault experiment, the length of the pendulum was large, l = 67 m, corresponding to a period T = 16.4 s. With such a long period, the lateral shift can already be observed in a single oscillation. The oscillation amplitude was A = 3 m. At the Paris latitude, sinλ = 0.753 and the rotation period is T _rot = 3.8 h = 14 480 s = 31,8 h = 14480 s. In an oscillation period T, the plane rotates at the angle 2πT/T _rot. Hence, the shift of the oscillation extreme in one period is s = 2πAT/T _rot = 2.7 mm.

Moreover, the length is important for another reason to which we can only hint. In practice, it happens that the stress forces always present in the wire and in the hook supporting the pendulum result in a spurious rotation of the oscillation plane. The effect is slow, but important for observations of several hours. It can be shown, however, that it is smaller for longer lengths.

5.8 The Eötvös Experiment

In Sect. 2.5, we have seen how Galilei and then Newton experimentally established the identity between inertial and gravitational mass. This is a very fundamental issue, and experiments have been done, and are still being done, to increase the precision within which the equality is verified. We discuss here the beautiful experiments conducted by the Hungarian physicist Loránd Eötvös (1848–1919) in the last years of the XIX century.

In this chapter we gave a number of examples of the effects of the inertial forces, the pseudo forces that appear in non-inertial frames. The inertial forces acting on a mass are proportional to the mass, just like the gravitational force. There is an important difference, however, as inertial forces are proportional to the inertial mass m _i, and the gravitational force is proportional to the gravitational mass m _g. Suppose the ratio between the two types of mass to be different for different substances. We could then hang spheres made of the two substances to two wires and look for any small difference in the directions of the wires. In this section, we shall use different symbols, m _i and m _g, for the two types of mass.

Consider a body hanging from a wire fixed in Ω, as in Fig. 5.15, at a point at the latitude λ. The distance from the axis is r _E Rcosλ, where R is the earth’s radius. The centrifugal force has a direction perpendicular to the axis outwards and magnitude

Fig. 5.15

The basis of the Eötvös experiment

$$F_{c} = m_{i} \omega_{\text{rot}}^{2} R\cos \lambda .$$

(5.72)

and the gravitational force

$$F_{G} = m_{g} g = G_{N} \frac{{m_{G} M_{E} }}{{R^{2} }}.$$

(5.73)

If for two substances, m _i and m _g are different, the angle between the two forces is also different, and so is the direction of the wire. As we saw in Sect. 5.6, the centrifugal acceleration on the earth’s surface is of the order of the per mille of the gravity acceleration. Correspondingly, the sought-after effects can be very small.

The Eötvös experiment directly compares the angles of wires to which spheres of different substances are attached. The two wires are attached to the extremes of a rigid bar. The bar is suspended by a metal wire that acts as a torsion balance, as shown in Fig. 5.16, similar to what we described in Sect. 4.7.

Fig. 5.16

The scheme of the Eötvös experiment

Figure 5.16a shows the system in perspective, with Fig. 5.16b looking at it parallel to the bar. If the ratio m _i/m_g is different for the two spheres, the directions α and β of the two tensions are a bit different. This produces a moment on the bar, due to the horizontal components of the two tensions, that rotates it about the wire from which it hangs. Under rotation, the wire develops an elastic moment, which increases with the angle. At the equilibrium angle, the two moments are equal and opposite. Measuring the angle, the torsion balance gives the moment.

The result of the very sensitive Eötvös experiment was null, allowing him to give the upper limit $m_{g} /m_{i} - 1 < 5 \times 10^{ - 9}$, namely that the difference, if any, is less than 5 parts per billion. An experiment of the same type by Robert Henry Dicke (1916–1997) in the 1960s established the even smaller limit of $m_{g} /m_{i} - 1 < 3 \times 10^{ - 11}$.

5.9 Problems

• 5.1. A kid sits in a carriage moving on straight rails. (a) If the speed of the carriage is constant, in which direction should he launch a ball to take it back in his hand without moving? In which direction if the carriage accelerates forwards?

• 5.2. A train travels on straight horizontal rails at the velocity υ ₀ = 30 m/s. Reaching a station, its stops, with constant acceleration, in s = 150 m. A suitcase of mass m = 10 kg lies on the floor, with dynamical friction coefficient µ _d = 0.20. During the braking, it slides along the corridor. (a) How much is its acceleration relative to the ground during this time? (b) Which is the velocity of the suitcase when the train stops? (c) After the train stops, the suitcase continues to slide for a while and then itself stops. Which was the total displacement of the suitcase on the floor?

• 5.3. A man measures his weight in a lift, which is at rest, using a spring and balance, and finds it to be 700 N. With the lift moving, he repeats the measurement and finds it to be 500 N. What can he determine about the lift acceleration? And about its velocity?

• 5.4. A person sits in a chair standing on the platform of a merry-go-round, which is turning. The person holds a plumb. Draw separate force diagrams for the plumb, the wire, the person, the chair and the platform. Describe each of the forces in words. Identify the action reaction pairs, both for a frame stationary on earth and for one stationary on the platform. In the latter case, specify which of the forces are inertial.

• 5.5. A kid sits on a merry-go-round that turns at angular velocity ω, while his friend is on the ground. The resultant of the forces on the latter is zero. (a) What is the motion of the second kid seen by the first? (b) What is his acceleration? (c) What are the forces causing it?

• 5.6. An old vinyl disk rotates at 33 turns per minute. Its radius is r = 15 cm. An insect walks from the center towards the border. Will it be able to reach it if the static friction coefficient is µ _s = 0.1?

• 5.7. A tennis player at 45˚ latitude is imparting to the ball a speed of 100 km/s, which we assume to be initially horizontal. Willing to hit ground at a distance of 50 m, should he take into account the Coriolis force?