Symmetry and Relative Velocity

Time Dilation & Length Contraction (Symmetry)

Symmetry of Observers:
- A moving and stationary observer both perceive the other’s clock and rod differently.
- For clocks:
  - A stationary observer will measure a moving clock to have a longer time interval between consecutive ticks (this is time dilation).
  - A moving observer will measure a stationary clock to have longer time intervals between ticks.
- For rods:
  - A moving observer will measure a stationary rod to be shorter than its proper (rest) length (length contraction).
  - A stationary observer will measure a moving rod to be shorter than its rest length.

Time Dilation Derivation

Consider a clock at rest in the $K$ system that produces signals at regular intervals $Δ t = t (2) - t (1)$ .

An observer in a moving system $K^{'}$ measures a time interval $Δ t^{'}$ on the same clock using the Lorentz transformation:

t^{'} = γ (t - \frac{v x_{1}}{c^{2}})

For two events happening at the same position $x_{1}$ in the $K$ system, $x_{1} (2) = x_{1} (1)$ . Therefore, the time interval measured in $K^{'}$ is given by:

Δ t^{'} = γ Δ t = \frac{Δ t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

Thus, the time interval appears longer in the moving frame $K^{'}$ , which explains why moving clocks run more slowly compared to clocks at rest in the observer's frame.

Length Contraction Derivation

Consider a rod of length $L_{0}$ lying along the $x$ -axis of an inertial frame $K$ . The observer in system $K^{'}$ moving with uniform speed $v$ along the $x$ -axis measures the length of the rod as $L^{'}$ in the moving system $K^{'}$ .

Using the Lorentz transformation:

x^{'} (2) - x^{'} (1) = \frac{x_{2} (2) - x_{1} (1)}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

where $x_{2} - x_{1} = L_{0}$ . The length measured in the moving frame is given by:

L^{'} = L_{0} \sqrt{1 - \frac{v^{2}}{c^{2}}}

Thus, the moving rod appears shorter in the frame $K^{'}$ compared to its proper length $L_{0}$ in the rest frame $K$ .

Relative Velocity in Lorentz Transformation

Concept: When a particle moves with velocity $u_{x}$ in the $K$ frame, we want to find its velocity $u_{x}^{'}$ in the $K^{'}$ frame, which moves with velocity $v$ relative to $K$ .

Steps:

Consider a particle moving in the $x$ -direction with velocity $u_{x}$ in frame $K$ .
Using the Lorentz transformation for position and time:

x^{'} = γ (x - v t)

t^{'} = γ (t - \frac{x v}{c^{2}})

To find the velocity in the $K^{'}$ frame, differentiate $x^{'}$ and $t^{'}$ with respect to $t$ :

\frac{d x^{'}}{d t} = γ (\frac{d x}{d t} - v) = γ (u_{x} - v)

\frac{d t^{'}}{d t} = γ (1 - \frac{u_{x} v}{c^{2}})

Therefore, the velocity $u_{x}^{'}$ in the $K^{'}$ frame is:

u_{x}^{'} = \frac{d x^{'}}{d t^{'}} = \frac{u_{x} - v}{1 - \frac{u_{x} v}{c^{2}}}

For $y$ - and $z$ -components:

The transverse components of velocity transform differently:

u_{y}^{'} = \frac{u_{y}}{γ (1 - \frac{u_{x} v}{c^{2}})}

u_{z}^{'} = \frac{u_{z}}{γ (1 - \frac{u_{x} v}{c^{2}})}

Conclusion:

The velocity of the particle in the moving frame $K^{'}$ is related to its velocity in $K$ by:

u_{x}^{'} = \frac{u_{x} - v}{1 - \frac{u_{x} v}{c^{2}}}