Lorentz Transformations
Lorentz transformations
The Lorentz transformations relate the coordinates of an event in one inertial frame to those in another moving at a constant velocity relative to the first.
They ensure that the speed of light remains constant in all inertial frames, a key postulate of special relativity.
Deriving the Lorentz Transformations
Consider two inertial frames, $S$ and $S'$:
- $S'$ moves with velocity $v$ relative to $c$ along the x-axis.
- The origins of $S$ and $S'$ coincide at $t = t' = 0$.
The Lorentz transformations relate the coordinates $(x, t)$ in $S$ to the coordinates $(x', t')$ in $S'$.
Transformation for Position
- In Galilean relativity, the position transformation is: $$
x' = x - vt
$$ - However, this does not account for the constancy of the speed of light.
- The Lorentz transformation modifies this to: $$
x' = \gamma(x - vt)
$$ where $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$ is the Lorentz factor.
Transformation for Time
- In Galilean relativity, time is absolute ($t' = t$).
- However, in special relativity, time is relative.
- The Lorentz transformation for time is: $$
t' = \gamma\left(t - \frac{vx}{c^2}\right)
$$
The term $\frac{vx}{c^2}$ accounts for the relativity of simultaneity, ensuring that events that are simultaneous in one frame may not be simultaneous in another.
Why the Lorentz Factor?
- The Lorentz factor $\gamma$ ensures that the speed of light is constant in all inertial frames.
- It approaches 1 at low speeds (recovering Galilean transformations) and increases significantly as $v$ approaches $c$.
Let’s apply the Lorentz transformations to an event at $x = 100$ m and $t = 2$ s in frame $S$. If $S'$ moves at $0.8c$ relative to $S$, what are the coordinates in $S'$?
Solution
- Calculate $\gamma$:$$
\gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}}$$ $$ = \frac{1}{\sqrt{1 - 0.64}} = \frac{5}{3}
$$ - Apply the transformations:$$
x' = \gamma(x - vt) = \frac{5}{3}(100 - 0.8 \cdot 3 \cdot 10^8 \cdot 2)$$ $$ = \frac{5}{3}(100 - 4.8 \times 10^8) \approx -8.0 \times 10^8
$$ $$t' = \gamma\left(t - \frac{vx}{c^2}\right) $$ $$= \frac{5}{3}\left(2 - \frac{0.8 \cdot 100}{(3 \cdot 10^8)^2}\right) \approx \frac{10}{3}$$ - The event occurs at $x' \approx -8.0 \times 10^8$ m and $t' \approx \frac{10}{3}$ s in $S'$.
Time Dilation
Time dilation
Time dilation is the phenomenon where time passes more slowly for an observer in motion relative to a stationary observer.
Deriving the Time Dilation Formula
- Consider a clock at rest in frame $S'$.
- It ticks twice, separated by a proper time interval $\Delta t_0$ (measured in the clock’s rest frame).
- An observer in frame $S$ sees the clock moving with velocity $v$.
- Using the Lorentz transformation for time: $$
t' = \gamma\left(t - \frac{vx}{c^2}\right)
$$ - Since the clock is at rest in $S'$, $\Delta x' = 0$.
- Therefore: $$
\Delta t = \gamma \Delta t_0
$$
The time interval measured by the observer in $S$ is $\Delta t$.
- The proper time $\Delta t_0$ is always the shortest time interval.
- Any observer measuring the interval in a frame where the clock is moving will measure a longer interval.
If a muon travels at $0.99c$, what is its lifetime as measured by an observer on Earth?
Solution
- Calculate $\gamma$:$$
\gamma = \frac{1}{\sqrt{1 - \frac{(0.99c)^2}{c^2}}} \approx 7.09
$$ - Apply the time dilation formula:$$
\Delta t = \gamma \Delta t_0 $$ $$= 7.09 \times 2.2 \times 10^{-6} \approx 1.56 \times 10^{-5} \text{ s}$$ - The muon’s lifetime is extended to $1.56 \times 10^{-5}$ s in the Earth’s frame.
Relativity of Simultaneity
In special relativity, the concept of simultaneity is not absolute, events that are simultaneous in one inertial frame may not be simultaneous in another moving frame.
This arises because the measurement of time depends on the relative motion between the observer and the events being measured.
- Consider two events occurring at different positions, $x_1$ and $x_2$, in frame $S$ but at the same time, $t_1 = t_2$.
- An observer in another frame, $S'$, moving with velocity $v$ relative to $S$, will measure the time difference between the events as: $$\Delta t' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \right)$$
- Since $\Delta t = 0$ in the original frame ($S$), the time interval observed in $S'$ becomes: $$\Delta t' = -\gamma \frac{v \Delta x}{c^2}$$
- That equation shows that the time difference between events depends on their spatial separation ($\Delta x$) and the relative velocity ($v$) of the frames.
- Thus, simultaneity is relative and depends on the observer's frame of reference.
- Two fireworks explode 2 km apart and are simultaneous in frame $S$.
- An observer in frame $S'$, moving at $0.6c$ relative to $S$, perceives the explosions as occurring at different times due to the relativity of simultaneity.
- The time difference can be calculated using the Lorentz transformations.
- Don’t assume simultaneity is absolute.
- In special relativity, two observers in relative motion can disagree on whether two events happened at the same time.
- To visualize relativity of simultaneity, consider two lightning strikes occurring simultaneously as seen from a stationary observer.
- A moving observer may perceive one strike happening before the other due to their motion relative to the events.
Spacetime Interval: An Invariant Quantity
- In special relativity, the spacetime interval ($\Delta s$) between two events remains invariant across all inertial reference frames.
- This means that even though different observers may measure different distances and time intervals, the spacetime interval is the same for all observers.
- It is mathematically expressed as: $$(\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2$$ where:
- $\Delta s$ is the spacetime interval,
- $c\Delta t$ represents the separation in time (converted into distance units by multiplying by $c$),
- $\Delta x$ represents the spatial separation between the events.
This equation closely resembles the Pythagorean theorem, but with a minus sign, highlighting the fundamental difference between space and time in relativity.
Why is the Spacetime Interval Invariant?
The invariance of the spacetime interval means that all inertial observers, regardless of their motion, agree on its value, even though they may disagree on $\Delta x$ and $\Delta t$.
This property ensures that the laws of physics remain consistent in all reference frames and underpins the effects of time dilation, length contraction, and the relativity of simultaneity.
Interpretation and Consequences
- If $(\Delta s)^2 > 0$, the interval is timelike:
- $\implies$ The two events are causally connected, and a signal traveling at or below the speed of light could travel between them.
- If $(\Delta s)^2 = 0$, the interval is lightlike:
- $\implies$ The events are connected by a signal traveling exactly at the speed of light.
- If $(\Delta s)^2 < 0$, the interval is spacelike:
- $\implies$ No signal, not even light, could travel between the events, making them causally disconnected.
The spacetime interval remains constant across all inertial reference frames, even when distances and time intervals change for different observers.
- Don’t confuse the spacetime interval with the Euclidean distance formula.
- The negative sign in the equation reflects the fundamental difference between space and time in relativity.
If two events are separated by $\Delta x = 3$ km and occur $\Delta t = 2$ microseconds apart, what is the spacetime interval $\Delta s$?
- Two flashes of light occur 600 meters apart with a time interval of 2 microseconds in one frame.
- Using the spacetime interval formula: $$(\Delta s)^2 = (3.0 \times 10^8 \times 2.0 \times 10^{-6})^2 - (600)^2$$ $$\Delta s^2 = (600)^2 - (600)^2 = 0$$
- Since $\Delta s = 0$, these events are connected by a light signal.
Length Contraction
Length contraction
Length contraction is the phenomenon where an object in motion appears shorter along the direction of motion to a stationary observer.
Deriving the Length Contraction Formula
- Consider a rod at rest in frame $S'$. Its proper length (measured in its rest frame) is $L_0$.
- An observer in frame $S$ measures the length of the rod while it moves with velocity $v$.
To measure the length in $S$, the observer must record the positions of the rod’s ends simultaneously.
- Using the Lorentz transformation for position: $$
x' = \gamma(x - vt)
$$ - The length measured in $S$ is: $$
L = x_2 - x_1
$$ - The proper length is: $$
L_0 = x_2' - x_1' = \gamma(x_2 - vt_2) - \gamma(x_1 - vt_1)
$$ - Since the measurement is simultaneous in $S$ ($t_2 = t_1$), the equation simplifies to: $$
L_0 = \gamma L
$$ - Solving for $L$ gives the length contraction formula: $$
L = \frac{L_0}{\gamma}
$$
- Length contraction only occurs along the direction of motion.
- Perpendicular dimensions remain unchanged.
A spaceship has a proper length of 100 m. What length does an observer measure if the spaceship travels at $0.8c$?
Solution
- Calculate $\gamma$:$$
\gamma = \frac{1}{\sqrt{1 - \frac{(0.8c)^2}{c^2}}} = \frac{5}{3}
$$ - Apply the length contraction formula:$$
L = \frac{L_0}{\gamma} $$ $$= \frac{100}{\frac{5}{3}} = 60 \text{ m}
$$ - The observer measures the spaceship’s length as 60 m.
Relativistic Velocity Addition
- At relativistic speeds, the classical formula for adding velocities ($u = u' + v$) breaks down.
- Instead, the relativistic velocity addition formula is used: $$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}
$$
This formula ensures that the resulting velocity never exceeds the speed of light.
A rocket moves at $0.8c$ relative to the ground. It fires a missile forward at $0.6c$ relative to the rocket. What is the missile’s speed relative to the ground?
Solution
- Use the relativistic velocity addition formula:$$
u = \frac{u' + v}{1 + \frac{u'v}{c^2}}$$ $$= \frac{0.6c + 0.8c}{1 + \frac{0.6c \cdot 0.8c}{c^2}}$$ $$= \frac{1.4c}{1 + 0.48} = \frac{1.4c}{1.48} \approx 0.95c$$ - The missile’s speed relative to the ground is approximately $0.95c$.
- A common mistake is to assume that velocities simply add up.
- Always use the relativistic formula at high speeds.
- How do the Lorentz transformations differ from Galilean transformations?
- What is the proper time, and why is it always the shortest time interval?
- Why does length contraction only occur along the direction of motion?
- How does the relativistic velocity addition formula ensure that no object exceeds the speed of light?
