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 $\Delta t$.
Tip- 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.Note
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.
