Four-Vectors and the Geometry of Spacetime
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You should be comfortable with the Lorentz transformation and the concept of rapidity as a hyperbolic rotation. Familiarity with basic vector algebra in three dimensions is assumed.
Why this matters
Four-vectors provide the geometric language that unifies space and time into a single coherent structure.
In previous lectures, we treated time dilation and length contraction as separate kinematic effects. While correct, this approach obscures the underlying symmetry of spacetime. By adopting the four-vector formalism, we recognize that physical laws are not merely invariant under Lorentz transformations; they are geometric statements about the structure of Minkowski space. This perspective eliminates the "paradox" of relativity, revealing it as simple geometry where the invariant interval plays the role of distance in Euclidean space.
The Geometric View
Imagine spacetime not as a stage where events happen, but as a four-dimensional entity where events are points. In Euclidean geometry, the distance between two points is invariant regardless of how you rotate your coordinate axes. Similarly, in special relativity, the "distance" between two events is not the spatial distance or the time difference, but the spacetime interval. This interval is the fundamental geometric quantity. A four-vector is simply an object with four components that transform in a specific way to preserve this interval. Just as a physical arrow in 3D space has a length that doesn't change when you rotate your axes, a four-vector has a "length" (its Minkowski norm) that doesn't change when you change inertial frames. This geometric invariance is the core of relativistic physics.
The Spacetime Interval
The foundation of four-vector calculus is the invariance of the spacetime interval. For two events separated by coordinate differences \(\Delta t, \Delta x, \Delta y, \Delta z\), the interval \(s^2\) is defined as:
This quantity is identical in all inertial frames. The sign of \(s^2\) determines the causal relationship between events. If \(s^2 < 0\), the separation is timelike, meaning one event could have caused the other. If \(s^2 > 0\), the separation is spacelike, meaning no signal traveling at or below the speed of light can connect them. If \(s^2 = 0\), the separation is null, corresponding to lightlike paths. This classification is absolute; no observer can disagree on whether two events are causally connected.
Proper Time and Four-Velocity
To describe motion in this geometry, we need a parameter that is invariant. The proper time \(\tau\) is the time measured by a clock moving with the particle. It is related to the coordinate time \(t\) by the Lorentz factor \(\gamma\). The four-velocity \(U^\mu\) is defined as the derivative of the four-position \(x^\mu = (ct, x, y, z)\) with respect to proper time \(\tau\):
A crucial property of the four-velocity is that its Minkowski norm is constant. Using the metric signature \((-+++)\), we calculate:
With the opposite signature \((+---)\) the same norm is written \(+c^2\); in either convention its magnitude is fixed. This constancy is the relativistic analogue of the fact that the speed of a particle in circular motion is constant in Euclidean space. It reflects the fact that all massive particles move through spacetime at the "speed" \(c\), trading spatial velocity for temporal velocity.
Four-Momentum
Extending this logic to mechanics, we define the four-momentum \(P^\mu\) as the product of rest mass \(m\) and four-velocity \(U^\mu\):
Here, the time component is the total energy divided by \(c\), and the spatial components are the relativistic three-momentum. The norm of the four-momentum is invariant:
Rearranging this yields the famous energy-momentum relation \(E^2 = p^2 c^2 + m^2 c^4\). This equation is not a dynamical law derived from forces; it is a geometric identity, just as \(x^2 + y^2 = r^2\) is a geometric identity in Euclidean space. It holds for all particles, massive or massless, and in all inertial frames.
The derivations
Invariance of the Spacetime Interval
For any two events in flat spacetime, the quantity \( s^2 = -c^2\,\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \) takes the same numerical value in every inertial frame related by a Lorentz transformation (boost, rotation, or any composition thereof). This frame-independent \( s^2 \) is the spacetime interval, and its sign partitions event pairs into timelike (\(s^2<0\)), null (\(s^2=0\)), and spacelike (\(s^2>0\)) classes that fix the causal structure of spacetime.
Proper Time and the Four-Velocity
Along any timelike worldline the invariant quantity c dτ defined by c² dτ² = ds² measures elapsed proper time — the time read by a clock carried on that worldline. Differentiating the coordinates with respect to this invariant scalar yields the four-velocity Uμ = dxμ/dτ = γ(c, v), a timelike four-vector whose Minkowski norm is fixed for every observer at UμUμ = c², independent of the state of motion.
Students often confuse the four-velocity \(U^\mu\) with the ordinary velocity \(\mathbf{v}\). Remember that \(U^\mu\) is a four-vector with units of velocity, but its "length" is fixed at \(c\). The ordinary velocity \(\mathbf{v}\) is just the spatial part of \(U^\mu\) divided by \(\gamma\). Mixing these up leads to incorrect transformations of acceleration and force. Always check the units and the index position (covariant vs contravariant) when manipulating these quantities. C
Worked example
Problem. A particle of rest mass \(m = 2.0 \times 10^{-27} \ \mathrm{kg}\) moves with a speed \(v = 0.6c\). Calculate its total energy \(E\) and momentum magnitude \(p\). Verify the invariant mass relation.
- Calculate the Lorentz factor \(\gamma\): \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - 0.36}} = \frac{1}{\sqrt{0.64}} = \frac{1}{0.8} = 1.25\).
- Calculate the total energy \(E = \gamma mc^2\): \(E = 1.25 \times (2.0 \times 10^{-27} \ \mathrm{kg}) \times (3.0 \times 10^8 \ \mathrm{m/s})^2 = 1.25 \times 2.0 \times 10^{-27} \times 9.0 \times 10^{16} = 22.5 \times 10^{-11} \ \mathrm{J} = 2.25 \times 10^{-10} \ \mathrm{J}\).
- Calculate the momentum \(p = \gamma mv\): \(p = 1.25 \times (2.0 \times 10^{-27} \ \mathrm{kg}) \times (0.6 \times 3.0 \times 10^8 \ \mathrm{m/s}) = 1.25 \times 2.0 \times 10^{-27} \times 1.8 \times 10^8 = 4.5 \times 10^{-19} \ \mathrm{kg \cdot m/s}\).
- Verify \(E^2 - p^2 c^2 = m^2 c^4\): \(E^2 = (2.25 \times 10^{-10})^2 = 5.0625 \times 10^{-20} \ \mathrm{J^2}\). \(p^2 c^2 = (4.5 \times 10^{-19})^2 \times (3.0 \times 10^8)^2 = 20.25 \times 10^{-38} \times 9.0 \times 10^{16} = 182.25 \times 10^{-22} = 1.8225 \times 10^{-20} \ \mathrm{J^2}\). \(E^2 - p^2 c^2 = (5.0625 - 1.8225) \times 10^{-20} = 3.24 \times 10^{-20} \ \mathrm{J^2}\). \(m^2 c^4 = (2.0 \times 10^{-27})^2 \times (3.0 \times 10^8)^4 = 4.0 \times 10^{-54} \times 81 \times 10^{32} = 324 \times 10^{-22} = 3.24 \times 10^{-20} \ \mathrm{J^2}\). The values match.
Units check. Energy is in Joules (\(\mathrm{kg \cdot m^2/s^2}\)). Momentum is in \(\mathrm{kg \cdot m/s}\). The term \(pc\) has units \(\mathrm{kg \cdot m^2/s^2}\), which matches Energy. The invariant mass term \(mc^2\) also has units of Energy. The calculation is dimensionally consistent.
Three quick tests
- Is the spacetime interval invariant under a Lorentz boost?
Answer
Yes, it is the defining property of the interval. - What is the norm of the four-velocity for a massive particle?
Answer
\(-c^2\) (or \(c^2\) depending on metric signature convention). - Can a massless particle have a rest frame?
Answer
No, because its four-momentum is null, and no Lorentz transformation can make a null vector timelike.
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.
Derivations cited: D-091 · D-097 · Threads: force, energy, light