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4.5: Force

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    Learning Objectives

    • 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.

    This page titled 4.5: Force is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Benjamin Crowell via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

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