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)
Addition of velocities
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
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.