4.5: Force

Learning Objectives

• 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.