# Special relativity cheatsheet

by Michael Brunnbauer, 2021-11-04

Check out Spacetime Physics, Second Edition by Edwin F. Taylor and John Archibald Wheeler for a deep and simple overview of the principles of relativity.

### Conversion of units

Measure time in meters (of distance travelled by light in that time):

Quantity Old unit New unit Conversion
Time $$s$$ $$m$$ $$t = t_{old} \cdot c$$
Velocity $$\frac{m}{s}$$ unitless $$v = \frac{v_{old}}{c}$$
Momentum $$\frac{kg \cdot m}{s}$$ $$kg$$ $$p = \frac{p_{old}}{c}$$
Energy $$\frac{kg \cdot m^2}{s^2}$$ $$kg$$ $$E = \frac{E_{old}}{c^2}$$

### Metric of flat spacetime

Inertial observers agree on spacetime distance between events, which is the metric of flat spacetime: $$\tau^2 = \Delta t^2 - \Delta x^2 \quad (\Delta t > \Delta x \text{, timelike separation})$$ $$0 = \Delta t^2 - \Delta x^2 \quad (\Delta t = \Delta x \text{, lightlike separation})$$ $$\sigma^2 = \Delta x^2 - \Delta t^2 \quad (\Delta x > \Delta t \text{, spacelike separation})$$ For coordinate systems with the same origin (reference event), the $$\Delta$$ can be omitted.

$$\tau$$ is the proper time (also called wristwatch time) between the two events - as measured by a clock that is present at both events and travels uniformly between them ($$\Delta x = 0$$).

$$\sigma$$ is the ruler distance (also called proper distance) between the two events - as measured by an observer in whose frame the two events occur at the same time ($$\Delta t = 0$$).

All inertial observers agree on whether two events have timelike, lightlike or spacelike separation and on the amount of their separation in spacetime (proper time, 0 or ruler distance).

### Time dilation

When inertial frame 2 moves with relative velocity $$v$$ to inertial frame 1, time dilation follows from two events separated only by time in frame 2 ($$x_2 = 0$$, $$x_1 = v \cdot t_1$$ - picture a Muon at rest in frame 2 decaying after $$t_2$$ in frame 2 and $$t_1$$ in frame 1): $$t_2^2 - x_2^2 = t_1^2 - x_1^2$$ $$t_2^2 - 0 = t_1^2 - v^2 \cdot t_1^2$$ $$t_2 = t_1 \cdot \sqrt{ 1 - v^2 }$$

### Length contraction

As frame 1 moves with velocity $$-v$$ relative to frame 2, length contraction must occur in frame 2 for the distance travelled: $$d_2 = v \cdot t_2 = v \cdot t_1 \cdot \sqrt{ 1 - v^2 } = x_1 \cdot \sqrt{ 1 - v^2 }$$

### Relativity of simultaneity

By symmetry, frame 2 will observe time dilation in frame 1 and frame 1 will observe length contraction in frame 2. How can frame 2 read $$t_1 > t_2$$ when passing a clock in frame 1 at $$x_1$$ while at the same time that clock runs slower as seen from frame 2? The clock at $$x_1$$ must show a certain future time t in frame 2 at time 0 for frame 2 (relativity of simultaneity): $$t = t_1 - t_2 \cdot \sqrt{ 1 - v^2 }$$ $$= \frac{x_1}{v} - \frac{d_2 \cdot \sqrt{ 1 - v^2 }}{v}$$ $$= \frac{d_2}{v \cdot \sqrt{ 1 - v^2 }} - \frac{d_2 \cdot \sqrt{ 1 - v^2 }}{v}$$ $$= \frac{d_2}{v \cdot \sqrt{ 1 - v^2 }} - \frac{d_2 \cdot ( 1 - v^2 )}{v \cdot \sqrt{ 1 - v^2}}$$ $$= \frac{v^2 \cdot d_2}{v \cdot \sqrt{ 1 - v^2 }}$$ $$= \frac{v \cdot d_2}{\sqrt{ 1 - v^2 }}$$

### Lorentz transformation

Combining all effects (time dilation, length contraction, relativity of simultaneity and relative movement) enables coordinate transformation of an arbitrary event $$t, x$$ in frame 2 to coordinates $$t', x'$$ in frame 1: $$\gamma \equiv \frac{1}{\sqrt{ 1 - v^2 }}$$ $$t' = \gamma \cdot t + \gamma \cdot v \cdot x = \gamma \cdot ( t + v \cdot x )$$ $$x' = \gamma \cdot x + \gamma \cdot v \cdot t = \gamma \cdot ( x + v \cdot t )$$ We have to account for the fact that $$v$$ is the velocity of frame 2 relative to frame 1. To express the transformation from frame 2 ($$t, x$$) to frame 1 ($$t', x'$$) with the velocity of frame 1 relative to frame 2, the sign of $$v$$ has to be reversed: $$t' = \gamma \cdot ( t - v \cdot x )$$ $$x' = \gamma \cdot ( x - v \cdot t )$$ This is the Lorentz transformation. By symmetry, the transformation from frame 1 to frame 2 (still using the velocity of frame 1 relative to frame 2!) is: $$t = \gamma \cdot ( t' + v \cdot x' )$$ $$x = \gamma \cdot ( x' + v \cdot t' )$$ Let's see if our results can be reproduced. The velocity of frame 1 relative to frame 2 in our example is $$-v$$. Let $$x_1 = v \cdot t_1$$. Then $$t_2 = t = \gamma \cdot ( t_1 - v \cdot x_1 ) = \gamma \cdot t_1 \cdot ( 1 - v \cdot \frac{x_1}{t_1} ) = \gamma \cdot t_1 \cdot ( 1 - v^2 ) = \gamma \cdot t_1 \cdot \gamma^{-2} = t_1 \cdot \gamma^{-1}$$ $$x_2 = x = \gamma \cdot ( x_1 - v \cdot t_1 ) = \gamma \cdot ( x_1 - x_1 ) = 0$$ Now for the distance travelled in frame 2 at $$t_2$$, one cannot simply set $$x' = 0$$ and $$t' = t_1$$ because the events $$t_1, 0$$ and $$t_1, x_1$$ are not simultaneous in frame 2. We need to find an event with $$x' = 0$$ and $$t = t_2 = t_1 \cdot \gamma^{-1}$$: $$t = \gamma \cdot ( t' - v \cdot x' )$$ $$t_1 \cdot \gamma^{-1} = \gamma \cdot t'$$ $$t' = t_1 \cdot \gamma^{-2}$$ Insert $$t'$$ into the second equation: $$x = \gamma \cdot ( x' - v \cdot t' )$$ $$x = - \gamma \cdot v \cdot t_1 \cdot \gamma^{-2} = - x_1 \cdot \gamma^{-1} = - d_2$$ What does the clock at rest in frame 1 at $$x' = x_1$$ show at $$t = 0$$ in frame 2? $$t = \gamma \cdot ( t' - v \cdot x' )$$ $$0 = \gamma \cdot ( t' - v \cdot x_1 )$$ $$\gamma \cdot v \cdot x_1 = \gamma \cdot t'$$ $$t' = v \cdot x_1 = v \cdot d_2 \cdot \gamma$$ What does a clock at rest in frame 1 at $$x' = 0$$ show at $$t = 0$$ in frame 2? $$t = \gamma \cdot ( t' - v \cdot x' )$$ $$0 = \gamma \cdot t'$$ $$t' = 0$$ Are those clocks not properly synchronized? Not at all! Slow down to $$v = 0$$ and check for yourself.

This chart illustrates how events in spacetime are affected by the Lorentz transformation. The blue lines are the trajectories of light:

Drag the slider for v to see how
• events move along the hyperbolas defined by the spacetime metric
• every event stays timelike, lightlike or spacelike regardless of speed
• horizontal lines connect different simultaneous events at different speeds (relativity of simultaneity)
• these events had a greater spacial separation at $$v=0$$ (length contraction)
• events connected by vertical lines at $$v=0$$ have greater time separation at other speeds (time dilation)

Suppose a projectile is fired in frame 2 at $$t = t' = 0, x = x' = 0$$ with velocity $$v_p$$. At $$t$$, it will be found at $$v_p \cdot t$$ in frame 2. The velocity of the projectile in frame 1 is: $$\frac{x'}{t'} = \frac{\gamma \cdot (v_p \cdot t + v \cdot t)}{\gamma \cdot ( t + v \cdot v_p \cdot t )} = \frac{v_p + v}{1 + v \cdot v_p}$$

### The stationary-action principle in classical mechanics

The stationary-action principle in classical mechanics states that trajectories are stationary points of the system's action functional (the function's derivative is zero).

Let's consider a mass of 1kg moving from event $$(t_1,x_1)$$ to event $$(t_3,x_3)$$ passing intermediary event $$(t_2,x_2)$$. The velocity between each pair of events is constant:

The bar on the right is the action - calculated as momentum times distance added for the two constant velocity stages of the trip. You can drag around the intermediate event to see how the total action for the trip changes. It is easy to see that the action is at a minimum when the mass travels on a straight line between the two events. Uninfluenced objects travel at constant velocity.

The action is calculated as: $$S = m \cdot \frac{x_2-x_1}{t_2-t_1} \cdot (x_2-x_1) + m \cdot \frac{x_3-x_2}{t_3-t_2} \cdot (x_3-x_2)$$ The trajectory is a stationary point with regard to $$t_2$$ and $$x_2$$: $$\frac{dS}{dt_2} = 0 \quad \text{(Time translation symmetry)}$$ $$\frac{dS}{dx_2} = 0 \quad \text{(Translational symmetry)}$$ According to Noether's theorem, two conserved quantities should arise. The quantity associated with time translation symmetry is energy and the quantity associated with translational symmetry is momentum: $$\frac{dS}{dt_2} = 0 = - m \cdot \frac{(x_2-x_1)^2}{(t_2-t_1)^2} + m \cdot \frac{(x_3-x_2)^2}{(t_3-t_2)^2}$$ $$m \cdot v_1^2 = m \cdot v_2^2$$ $$\frac{dS}{dx_2} = 0 = 2 \cdot m \cdot \frac{x_2-x_1}{t_2-t_1} - 2 \cdot m \cdot \frac{x_3-x_2}{t_3-t_2}$$ $$m \cdot v_1 = m \cdot v_2$$

### The principle of maximal aging in relativity

The principle of maximal aging in relativity states that trajectories of an undisturbed object are maxima of the objects proper time functional. $$\tau_1 = \sqrt{(t_2-t_1)^2 - (x_2-x_1)^2}$$ $$\tau_2 = \sqrt{(t_3-t_2)^2 - (x_3-x_2)^2}$$ $$\frac{d\tau_1}{dt_2} = \frac{t_2-t_1}{\sqrt{(t_2-t_1)^2 - (x_2-x_1)^2}} = \frac{t_2-t_1}{\tau_1}$$ $$\frac{d\tau_2}{dt_2} = - \frac{t_3-t_2}{\sqrt{(t_3-t_2)^2 - (x_3-x_2)^2}} = - \frac{t_3-t_2}{\tau_2}$$ $$\frac{d\tau_1}{dt_2} + \frac{d\tau_2}{dt_2} = 0 = \frac{t_2-t_1}{\tau_1} - \frac{t_3-t_2}{\tau_2}$$ $$\frac{t_2-t_1}{\tau_1} = \frac{t_3-t_2}{\tau_2}$$ $$\gamma_1 = \gamma_2$$ It can be shown that the second derivation of $$\tau_1$$ and $$\tau_2$$ is always negative - so $$\gamma_1 = \gamma_2$$ is a maximum for the total proper time and describes the trajectory. Uninfluenced objects travel at constant velocity in flat spacetime.

### Relativistic energy

Relativistic energy $$E$$ must be proportional to $$\gamma$$ and it should be possible to recover the classical formula for kinetic energy for very low s peeds ( $$v \ll 1$$ ). Let's try $$E = m \cdot \gamma$$. Use the approximation $$(1+\epsilon)^n \approx 1 + n \cdot \epsilon$$ for $$| \epsilon | \ll 1$$ and $$| n \cdot \epsilon | \ll 1$$: $$E \equiv m \cdot \gamma = m \cdot (1-v^2)^{-\frac{1}{2}} \approx m + m \cdot \frac{v^2}{2} \quad (v \ll 1)$$ At low speeds, the relativistic energy is approximately the classical kinetic energy plus the mass. At rest, it is exactly the mass. This makes sense insofar as we use the same unit kg for energy and mass.

### Relativistic momentum

Apply again the principle of maximal aging: $$\frac{d\tau_1}{dx_2} = - \frac{x_2-x_1}{\sqrt{(t_2-t_1)^2 - (x_2-x_1)^2}} = - \frac{x_2-x_1}{\tau_1} = - \frac{(x_2-x_1) \cdot (t_2-t_1)}{(t_2-t_1) \cdot \tau_1} = - v_1 \cdot \gamma_1$$ $$\frac{d\tau_2}{dx_2} = \frac{x_3-x_2}{\sqrt{(t_3-t_2)^2 - (x_3-x_2)^2}} = \frac{x_3-x_2}{\tau_2} = \frac{(x_3-x_2) \cdot (t_3-t_2)}{(t_3-t_2) \cdot \tau_2} = v_2 \cdot \gamma_2$$ $$\frac{d\tau_1}{dt_2} + \frac{d\tau_2}{dt_2} = 0 = v_2 \cdot \gamma_2 - v_1 \cdot \gamma_1$$ $$v_1 \cdot \gamma_1 = v_2 \cdot \gamma_2$$ Again, the mass should be included in the product to get the classical formula for very low speeds: $$p \equiv m \cdot v \cdot \gamma \approx m \cdot v \quad (v \ll 1)$$

### Relativistic mass

The terms relativistic mass and rest mass have become unpopular in modern literature because they tend to create confusion. Objects in motion have more inertia than objects at rest, though. Consider an object moving with constant velocity and a force applied perpendicular to the direction of movement. Force is the time-derivative of momentum. As long as the force stays perpendicular, $$v$$ does not change in magnitude so $$\gamma$$ does not change with time: $$\vec F = \frac{d \vec p}{dt} = \frac{d(m \cdot \vec v \cdot \gamma)}{dt} = m \cdot \vec a \cdot \gamma$$ The object behaves as if it had a mass of $$m \cdot \gamma$$. However - if a force is applied in the direction of movement, the object behaves as if it had even more mass, because $$\gamma$$ is now time-dependent: $$\vec F = \frac{d(m \cdot \vec v \cdot (1-v^2)^{-\frac{1}{2}})}{dt} = m \cdot \vec a \cdot \gamma + m \cdot \vec v \cdot v \cdot \gamma^3 \cdot a$$ $$F = m \cdot a \cdot \gamma + m \cdot v \cdot v \cdot \gamma^3 \cdot a = m \cdot a \cdot \gamma \cdot ( 1 + \frac{v^2}{1-v^2} ) = m \cdot a \cdot \gamma^3$$ In this document, mass or $$m$$ always means rest mass - a quantity all observers will agree on.

### Unifiying mass, momentum and energy: momentum-energy

Mass, momentum and energy all have the same unit. In different inertial systems, momentum and energy will vary but mass is invariant. Furthermore, energy is related to time and momentum is related to space. The situation is similar to time and space in flat spacetime and in fact, the same metric applies: $$E^2 - p^2 = m^2 \cdot \gamma^2 - m^2 \cdot \gamma^2 \cdot v^2 = m^2 \cdot \gamma^2 \cdot ( 1 - v^2 ) = m^2 \cdot \frac{\gamma^2}{\gamma^2} = m^2$$ Also: $$v = \frac{p}{E}$$ Relativistic energy and momentum form a vector. Its "time" component is the relativistic energy and its "space" components (1-3) are the relativistic momentum. The magnitude of the vector is the invariant mass. The direction of the vector is the spacetime direction of movement. Within an inertial system, both components of the vector are conserved quantities - so the vector as a whole is conserved. It is called momentum-energy (short: momenergy) or four-momentum. Like events, the momentum-energy vector moves along a hyperbola when the relative speed is changed:

Due to the euclidean nature of the drawing, the length of the vector does not correctly represent its magnitude $$m$$ for $$v \neq 0$$. The blue lines represent possible values for the momentum-energy of light ( $$E = p, m = 0$$ ).

### Momentum-energy of a system (the mass of heat)

The momentum-energy vector of a system is the sum of the momentum-energy vectors of its constituents. The mass of a system is the sum of the mass of its constituents only if they all move at the same speed. Consider two objects of equal mass moving in opposite directions at the same speed: The total momentum is zero so the mass of the system is equal to the energy of the system, which is the sum of the energy of the two objects. This sum is higher than the combined mass of the objects. The difference is the "mass of heat". Even a system of light particles can have mass although its constituents are massless.

### Center of momentum frame

It is always possible to find a "center of momentum" inertial frame where the momentum of a system is zero:

• $$m_1, m_2$$ mass of the objects
• $$u_1, u_2$$ velocity of the objects
• $$E = E_1 + E_2$$ energy of the objects
• $$p = p_1 + p_2$$ momentum of the objects
• $$v_c$$ velocity of the "center of momentum" inertial frame
• $$u_1', u_2'$$ velocity of the objects in the "center of momentum" frame
$$u_1' = \frac{u_1-v_c}{1-u_1 \cdot v_c}$$ $$u_2' = \frac{u_2-v_c}{1-u_2 \cdot v_c}$$ $$\gamma_1 = \frac{1}{\sqrt{1-u_1^2}} = \frac{1}{\sqrt{1-\frac{(u_1-v_c)^2}{(1-u_1 \cdot v_c)^2}}} = \frac{1}{\sqrt{\frac{(1-u_1 \cdot v_c)^2 - (u_1-v_c)^2}{(1-u_1 \cdot v_c)^2}}} = \frac{1-u_1 \cdot v_c}{\sqrt{(1-u_1 \cdot v_c)^2 - (u_1-v_c)^2}}$$ $$= \frac{1-u_1 \cdot v_c}{\sqrt{1+u_1^2 \cdot v_c^2 - u_1^2 - v_c^2}} = \frac{1-u_1 \cdot v_c}{\sqrt{(u_1^2-1) \cdot (v_c^2-1)}} = \frac{1-u_1 \cdot v_c}{\sqrt{1-u_1^2} \cdot \sqrt{1-v_c^2}}$$ $$\gamma_2 = \frac{1-u_2 \cdot v_c}{\sqrt{1-u_2^2} \cdot \sqrt{1-v_c^2}}$$ $$m_1 \cdot u_1' \cdot \gamma_1 + m_2 \cdot u_2' \cdot \gamma_2 = 0$$ $$\frac{m_1 \cdot (u_1-v_c) \cdot (1-u_1 \cdot v_c )} {(1-u_1 \cdot v_c ) \cdot \sqrt{1-u_1^2} \cdot \sqrt{1-v_c^2}} + \frac{m_2 \cdot (u_2-v_c) \cdot (1-u_2 \cdot v_c )} {(1-u_2 \cdot v_c ) \cdot \sqrt{1-u_2^2} \cdot \sqrt{1-v_c^2}} = 0$$ $$\frac{m_1 \cdot (u_1-v_c)}{\sqrt{1-u_1^2}} + \frac{m_2 \cdot (u_2-v_c)}{\sqrt{1-u_2^2}} = 0$$ $$E_1 \cdot (u_1-v_c) + E_2 \cdot (u_2-v_c) = 0$$ $$E_1 \cdot u_1 - E_1 \cdot v_c + E_2 \cdot u_2 - E_2 \cdot v_c = 0$$ $$p_1 - E_1 \cdot v_c + p_2 - E_2 \cdot v_c = 0$$ $$p_1 + p_2 = v_c \cdot ( E_1 + E_2 )$$ $$v_c = \frac{p_1 + p_2}{E_1 + E_2} = \frac{p}{E}$$ This proof generalizes to any number of objects and three spacial dimensions.

### Relativistic collision

In the case of elastic collision of two objects, their momenta must be equal and opposite in the center of momentum frame. The collision will just reverse the sign of the opposite momenta (system momentum is conserved). As the masses stay unchanged, the velocities simply change sign in the center of momentum frame:

• $$v_1' = -u_1', v_2' = -u_2'$$ velocity of the objects after collision in the "center of momentum" frame
• $$v_1, v_2$$ velocity of the objects after collision
$$v_1 = \frac{v_1' + v_c}{1+v_1' \cdot v_c} = \frac{v_c - u_1'}{1-u_1' \cdot v_c}$$ $$v_2 = \frac{v_2' + v_c}{1+v_2' \cdot v_c} = \frac{v_c - u_2'}{1-u_2' \cdot v_c}$$ In the case of inelastic collision of objects, the resulting object will have a velocity of 0 in the center of momentum frame and of $$v_c$$ in the initial frame.

### Particle physics

Ingoing and outgoing particles need not be the same - not even their number. But momentum, energy and mass of the system will be conserved. This fact is used heavily in particle physics, where particles are smashed into another at high speeds to create new particles.