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# 4.5: Force

• • Contributed by Benjamin Crowell
• Professor (Physics) at Fullerton College

Skills to Develop

• Explain force and relativity

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

# Four-force

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

$F_{o',\parallel } = F_{o,\parallel }$

and

$F_{o',\perp } = \frac{F_{o,\perp }}{\gamma }$

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

# Work

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.