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4.5: Force

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    • Explain force and relativity

    Force is a concept that is seldom needed in relativity, and that’s why this section is optional.


    By analogy with Newtonian mechanics, we define a relativistic force vector

    \[F = m\cdot a\]

    where \(a\) is the acceleration four-vector (section 3.5) and \(m\) is the mass of a particle that has that acceleration as a result of the force \(F\). This is equivalent to

    \[F = \frac{\mathrm{d} p}{\mathrm{d} \tau }\]

    where \(p\) is the mass of the particle and \(τ\) its proper time. Since the timelike part of \(p\) is the particle’s mass-energy, the timelike component of the force is related to the power expended by the force. These definitions only work for massive particles, since for a massless particle we can’t define \(a\) or \(τ\). \(F\) has been defined in terms of Lorentz invariants and four-vectors, and therefore it transforms as a god-fearing four-vector itself.

    The force measured by an observer

    The trouble with all this is that \(F\) isn’t what we actually measure when we measure a force, except if we happen to be in a frame of reference that momentarily coincides with the rest frame of the particle. As with velocity and acceleration ( section 3.7), we have a four-vector that has simple, standard transformation properties, but a different \(F_o\), which is what is actually measured by the observer \(o\). It’s defined as

    \[F_o = \frac{\mathrm{d} p}{\mathrm{d} t }\]

    with a \(dt\) in the denominator rather than a \(dτ\). In other words, it measures the rate of transfer of momentum according to the observer, whose time coordinate is \(t\), not \(τ\) — unless the observer happens to be moving along with the particle. Unlike the three-vectors \(v_o\) and \(a_o\), whose timelike components are zero by definition according to observer \(o\), \(F_o\) usually has a nonvanishing timelike component, which is the rate of change of the particle’s mass-energy, i.e., the power. We can refer to the spacelike part of \(F_o\) as the three-force.

    The following two examples show that an object moving at relativistic speeds has less inertia in the transverse direction than in the longitudinal one. A corollary is that the three-acceleration need not be parallel to the three-force.

    Example \(\PageIndex{1}\): Circular motion

    For a particle in uniform circular motion, \(γ\) is constant, and we have

    \[F_o = \frac{\mathrm{d} }{\mathrm{d} t }(m\gamma v) = m\gamma \frac{\mathrm{d} v}{\mathrm{d} t}\]

    The particle’s mass-energy is constant, so the timelike component of \(F_o\) does happen to be zero in this example. In terms of the three-vectors \(v_o\) and \(a_o\) defined in section 3.7, we have

    \[F_o = m\gamma \frac{\mathrm{d} v_o}{\mathrm{d} t} = m\gamma a_o\]

    which is greater than the Newtonian value by the factor \(γ\). As a practical example, in a cathode ray tube(CRT) such as the tube in an old-fashioned oscilloscope or television, a beam of electrons is accelerated upto relativistic speed. To paint a picture on the screen, the beam has to be steered by transverse forces, and since the deflection angles are small, the world-line of the beam is approximately that of uniform circular motion. The force required to deflect the beam is greater by a factor of \(γ\) than would have been expected according to Newton’s laws.

    Example \(\PageIndex{2}\): Linear motion

    For accelerated linear motion in the x direction, ignoring \(y\) and \(z\), we have a velocity vector

    \[v = \frac{\mathrm{d} r}{\mathrm{d} \tau }\]

    whose \(x\) component is \(γv\). Then

    \[\begin{align*} F_{o,x} &= m\frac{\mathrm{d} (\gamma v)}{\mathrm{d} t}\\ &= m\frac{\mathrm{d} (\gamma )}{\mathrm{d} t}v + m\gamma \frac{\mathrm{d} v}{\mathrm{d} t}\\ &= m\frac{\mathrm{d} \gamma }{\mathrm{d} v}\frac{\mathrm{d} v}{\mathrm{d} t} + m\gamma a\\ &= m\left ( v^2 \gamma ^3 a + \gamma a \right )\\ &= ma\gamma ^3 \end{align*}\]

    The particle’s apparent inertia is increased by a factor of \(γ^3\) due to relativity.

    The results of above two examples can be combined as follows:

    \[F_o = m\gamma a_{o,\perp } + m\gamma ^3 a_{o,\parallel }\]

    where the subscripts \(\perp\) and \(\parallel\) refer to the parts of \(a_o\) perpendicular and parallel to \(v_o\).

    Transformation of the force measured by an observer

    Define a frame of reference \(o\) for the inertial frame of reference of an observer who does happen to be moving along with the particle at a particular instant in time. Then \(t\) is the same as \(τ\), and \(F_o\) the same as \(F\). In this frame, the particle is momentarily at rest, so the work being done on it vanishes, and the timelike components of \(F_o\) and \(F\) are both zero.

    Suppose we do a Lorentz transformation from o to a new frame \(o'\), and suppose the boost is parallel to \(F_o\) and \(F\) (which are both purely spatial in frame \(o\)). Call this direction \(x\). Then \(dp = (dp_t,dp_x) = (0,dp_x)\) transforms to \(dp' = (-γv dp_x,γ dp_x)\), so that \(F_{o',x} = dp'_x/dt' = (γ dp_x)/(γ dt) = F_{o,x}\). The two factors of \(γ\) cancel, and we find that \(F_{o',x} = F_{o,x}\).

    Now let’s do the case where the boost is in the \(y\) direction, perpendicular to the force. The Lorentz transformation doesn’t change \(dp_y\), so

    \[\begin{align*} F_{o',y} &= \frac{\mathrm{d} p'_y}{\mathrm{d} t'}\\ &= \frac{\mathrm{d} p_y}{(\gamma \mathrm{d} t)}\\ &= \frac{F_{o',y}}{\gamma } \end{align*}\]

    The summary of our results is as follows. Let \(F_o\) be the force acting on a particle, as measured in a frame instantaneously comoving with the particle. Then in a frame of reference moving relative to this one, we have

    \[F_{o',\parallel } = F_{o,\parallel }\]


    \[F_{o',\perp } = \frac{F_{o,\perp }}{\gamma }\]

    where \(\parallel\) indicates the direction parallel to the relative velocity of the two frames, and \(\perp\) a direction perpendicular to it.


    Consider the one-dimensional version of the three-force, \(F = dp/dt\). An advantantage of this quantity is that it allows us to use the Newtonian form of the (one-dimensional) work-kinetic energy relation \(dE/dx = F\) without correction. Proof:

    \[\frac{\mathrm{d} E}{\mathrm{d} x} = \frac{\mathrm{d} E}{\mathrm{d} p} \frac{\mathrm{d} p}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x} = \frac{\mathrm{d} E}{\mathrm{d} p} \frac{F}{v}\]

    By implicit differentiation of the definition of mass, we find that \(dE/dp = p/E\), and this in turn equals \(v\) by the identity proved in Example 4.3.2. This leads to the claimed result, which is valid for both massless and material particles.