Relativistic Dynamics: Force and Acceleration
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You should be comfortable with the definitions of four-momentum \( P^\mu = (\gamma mc, \gamma m\mathbf{u}) \) and the Lorentz factor \( \gamma = (1 - u^2/c^2)^{-1/2} \). Familiarity with vector calculus, specifically the product rule for differentiation, is required.
Why this matters
Newton's second law \( \mathbf{f} = m\mathbf{a} \) fails at relativistic speeds because momentum, \( \mathbf{p} = \gamma m\mathbf{u} \), is no longer proportional to velocity.
In classical mechanics, force is simply the rate of change of momentum, and since mass is invariant, this reduces to mass times acceleration. In special relativity, momentum depends on velocity in a non-linear way. Consequently, the relationship between force and acceleration becomes direction-dependent and velocity-dependent. Understanding this distinction is crucial for predicting the motion of charged particles in accelerators or astrophysical jets, where simple Newtonian intuition leads to significant errors in trajectory and energy calculations.
The Breakdown of Newton's Intuition
In everyday life, if you push a box with a constant force, it accelerates at a constant rate. This works because the box's resistance to motion (mass) is constant. However, as an object approaches the speed of light, its energy increases without bound, even though its speed cannot exceed \( c \). This means that adding more energy results in less and less increase in speed. Therefore, the "effective inertia" of the object increases as it moves faster. In relativity, we do not say the mass increases; rather, we recognize that the relationship between force (the push) and acceleration (the change in speed) is no longer a simple proportionality. The force required to produce a given acceleration depends on how fast the object is already moving and, critically, the direction of the force relative to the velocity.
Four-Force and Three-Force
To maintain covariance, we define the four-force \( F^\mu \) as the derivative of four-momentum \( P^\mu \) with respect to proper time \( \tau \). For a particle of constant rest mass \( m \), this is equivalent to mass times four-acceleration \( A^\mu \). The time component of the four-force relates to the rate of change of energy, while the spatial components relate to the three-force \( \mathbf{f} \). The connection is given by \( F^\mu = \gamma ( \frac{\mathbf{f} \cdot \mathbf{u}}{c}, \mathbf{f} ) \). This shows that the three-force \( \mathbf{f} \) is not simply \( m\mathbf{a} \), but is embedded in a larger geometric structure where energy and momentum change together.
Expanding this derivative using the product rule reveals the complex dependence on velocity. The result is a vector equation that separates into components parallel and perpendicular to the velocity vector.
Parallel vs Perpendicular Acceleration
The most significant consequence of relativistic dynamics is that the acceleration produced by a force depends on the angle between the force and the velocity. If the force is applied parallel to the velocity (\( \mathbf{f} \parallel \mathbf{u} \)), the acceleration is suppressed by a factor of \( \gamma^3 \). If the force is applied perpendicular to the velocity (\( \mathbf{f} \perp \mathbf{u} \)), the acceleration is suppressed by a factor of \( \gamma \). This anisotropy in inertia is a direct result of the geometry of spacetime.
Notice that \( \gamma^3 \) is always greater than or equal to \( \gamma \) for \( u > 0 \). This means it is harder to accelerate a particle in the direction of its motion than in a direction perpendicular to it. This effect becomes pronounced as \( u \) approaches \( c \), where \( \gamma \) becomes very large.
Power and Energy Transfer
The rate at which work is done on a particle, known as power \( P \), is the dot product of force and velocity: \( P = \mathbf{f} \cdot \mathbf{u} \). In relativity, this power is directly related to the rate of change of the particle's total energy \( E \). Since \( E = \gamma mc^2 \), differentiating with respect to time shows that \( \frac{dE}{dt} = \mathbf{f} \cdot \mathbf{u} \). This confirms that force does not just change momentum; it changes the energy of the system. When the force is perpendicular to the velocity, \( \mathbf{f} \cdot \mathbf{u} = 0 \), meaning no work is done, and the particle's speed remains constant, even though its direction changes. This is the principle behind magnetic confinement in particle accelerators.
The derivation
Relativistic Force and Four-Acceleration
We define the four-force as the proper-time derivative of four-momentum, Fμ = dPμ/dτ, show that for a particle of constant rest mass it equals mass times four-acceleration, Fμ = mAμ, and reduce it to the three-force law f = mγa + mγ3(u·a)u/c2, whose parallel and perpendicular projections give f∥ = γ3ma∥ and f⊥ = γma⊥.
A common misconception is that "relativistic mass" \( m_{\text{rel}} = \gamma m \) is a scalar quantity that simply replaces \( m \) in \( \mathbf{f} = m_{\text{rel}} \mathbf{a} \). This is incorrect because the inertia is tensorial, not scalar. The resistance to acceleration is different in different directions, so a single scalar "mass" cannot describe the dynamics. Always use the component-wise relations \( f_{\parallel} \) and \( f_{\perp} \) or the full vector form to avoid errors in trajectory calculations. C
Worked example
Problem. An electron with rest mass \( m = 9.11 \times 10^{-31}\ \mathrm{kg} \) is moving at \( u = 0.8c \). A force \( \mathbf{f} \) is applied strictly perpendicular to its velocity. Calculate the magnitude of the acceleration \( a \) produced by a force of \( 1.0 \times 10^{-10}\ \mathrm{N} \).
- Calculate the Lorentz factor \( \gamma \) for \( u = 0.8c \): \[ \gamma = \frac{1}{\sqrt{1 - 0.8^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3} \approx 1.67 \]
- Identify the relevant force-acceleration relation. Since the force is perpendicular to velocity, we use \( f_{\perp} = \gamma m a_{\perp} \).
- Rearrange to solve for acceleration \( a \): \[ a = \frac{f_{\perp}}{\gamma m} \]
- Substitute the values: \[ a = \frac{1.0 \times 10^{-10}\ \mathrm{N}}{(5/3) \times (9.11 \times 10^{-31}\ \mathrm{kg})} \]
- Compute the denominator: \[ \gamma m = 1.667 \times 9.11 \times 10^{-31} \approx 1.518 \times 10^{-30}\ \mathrm{kg} \]
- Compute the acceleration: \[ a = \frac{1.0 \times 10^{-10}}{1.518 \times 10^{-30}} \approx 6.59 \times 10^{19}\ \mathrm{m/s^2} \]
Units check. Force is in Newtons (\( \mathrm{kg \cdot m/s^2} \)). Mass is in kg. The ratio \( \mathrm{N/kg} \) yields \( \mathrm{m/s^2} \), which is the correct unit for acceleration.
Three quick tests
- If a particle is moving at \( 0.9c \) and a force is applied parallel to its motion, by what factor is the acceleration reduced compared to the Newtonian prediction \( f/m \)?
Answer
The factor is \( \gamma^3 \). For \( v=0.9c \), \( \gamma = 1/\sqrt{1-0.81} = 1/\sqrt{0.19} \approx 2.29 \). Thus \( \gamma^3 \approx 12.0 \). The acceleration is roughly 12 times smaller than the Newtonian prediction. - Why does a magnetic field change the direction of a charged particle's velocity but not its speed?
Answer
The magnetic force \( \mathbf{f} = q\mathbf{u} \times \mathbf{B} \) is always perpendicular to the velocity \( \mathbf{u} \). Therefore, the power \( P = \mathbf{f} \cdot \mathbf{u} \) is zero. No work is done, so the kinetic energy and speed remain constant, while the direction changes due to the perpendicular acceleration. - Is the four-force \( F^\mu \) always orthogonal to the four-velocity \( U^\mu \)?
Answer
Yes. The norm of the four-velocity is constant (\( U^\mu U_\mu = c^2 \)). Differentiating this scalar with respect to proper time gives \( 2 U_\mu \frac{dU^\mu}{d\tau} = 0 \), which implies \( U_\mu A^\mu = 0 \). Since \( F^\mu = m A^\mu \) for constant mass, \( U_\mu F^\mu = 0 \).
Problem set for PU-105 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.