Newton's First Law of Motion

Newton's first law of motion states that an object subject to zero net external force moves in a straight-line with a constant speed (i.e., it does not accelerate). However, this is only true in special frames of reference called inertial frames. Indeed, we can think of Newton's first law as the definition of an inertial frame: i.e., an inertial frame of reference is one in which a point object subject to zero net external force moves in a straight-line with constant speed.

Suppose that we have found an inertial frame of reference. Let us set up a Cartesian coordinate system in this frame. The motion of a point object can now be specified by giving its position vector, ${\bf r}=(x,y,z)$, with respect to the origin of our coordinate system, as a function of time, $t$. Consider a second frame of reference moving with some constant velocity ${\bf u}$ with respect to our first frame. Without loss of generality, we can suppose that the Cartesian axes in the second frame are parallel to the corresponding axes in the first frame, and that ${\bf u} = (u,0,0)$, and, finally, that the origins of the two frames instantaneously coincide at $t=0$ (see Fig. 16). Suppose that the position vector of our point object is ${\bf r}'=(x',y',z')$ in the second frame of reference. It is fairly obvious, from Fig. 16, that at any given time, $t$, the coordinates of the object in the two reference frames satisfy

$$x' = x - u\,t,$$ (99)

$$y' = y,$$ (100)

$$z' = z.$$ (101)

This transformation law is generally known as the Galilean transformation, after Galileo.

Figure 16: A Galilean transformation.

The instantaneous velocity of the object in our first reference frame is given by ${\bf v} = d{\bf r}/dt = (dx/dt,dy/dt,dz/dt)$, with an analogous expression for the velocity, ${\bf v}'$, in the second frame. It follows from Eqs. (99)-(101) that the velocity components in the two frames satisfy

$$v_x' = v_x - u,$$ (102)

$$v_y' = v_y ,$$ (103)

$$v_z' = v_z.$$ (104)

These equations can be written more succinctly as

$${\bf v}' = {\bf v} - {\bf u}.$$ (105)

Finally, the instantaneous acceleration of the object in our first reference frame is given by ${\bf a} = d{\bf v}/dt = (dv_x/dt,dv_y/dt,dv_z/dt)$, with an analogous expression for the acceleration, ${\bf a}'$, in the second frame. It follows from Eqs. (102)-(104) that the acceleration components in the two frames satisfy

$$a_x' = a_x,$$ (106)

$$a_y' = a_y ,$$ (107)

$$a_z' = a_z.$$ (108)

These equations can be written more succinctly as

$${\bf a}' = {\bf a}.$$ (109)

According to Eqs. (105) and (109), if an object is moving in a straight-line with constant speed in our original inertial frame (i.e., if ${\bf a}={\bf0}$) then it also moves in a (different) straight-line with (a different) constant speed in the second frame of reference (i.e., ${\bf a}'= {\bf0}$). Hence, we conclude that the second frame of reference is also an inertial frame.

A simple extension of the above argument allows us to conclude that there are an infinite number of different inertial frames moving with constant velocities with respect to one another.

But, what happens if the second frame of reference accelerates with respect to the first? In this case, it is not hard to see that Eq. (109) generalizes to

$${\bf a}' = {\bf a} - \frac{d{\bf u}}{dt},$$ (110)

where ${\bf u}(t)$ is the instantaneous velocity of the second frame with respect to the first. According to the above formula, if an object is moving in a straight-line with constant speed in the first frame (i.e., if ${\bf a}={\bf0}$) then it does not move in a straight-line with constant speed in the second frame (i.e., ${\bf a}'\neq{\bf0}$). Hence, if the first frame is an inertial frame then the second is not.

A simple extension of the above argument allows us to conclude that any frame of reference which accelerates with respect to any inertial frame is not an inertial frame.