Special Relativity

1 Time Dilation and Length Contraction

Special relativity is based on two postulations: in all inertial frames of reference, regardless of their velocities v,

• all laws of physics are the same, i.e., no experiment can be performed to determine whether one is at rest or moving;

• the speed of light c in vacuum will measure the same.

Consider a beam of light traveling between two mirrors facing each other. Assume, according to an observer A in the same reference frame as the mirrors, the length between the two mirrors is L_{A}, called the rest or proper length, and the time it takes for the light to travel from one mirror to the other and come back is Δt_{A} = 2L_{A}/c, called the rest time or proper time, i.e.,

L_{A} = cΔt_{A}/2  (1)

Consider the time Δt_{B} and length L_{B} of the system according to another observer B at rest, when the system is moving at a constant speed v in either of the following two directions:

• Motion direction is perpendicular to the path of the light

According the observer B, the distance traveled by the light is 2\sqrt{L^{2} + (vΔt_{B}/2)^{2}}, and the speed of light c (the same for both observers A and B) is

c = \frac{2L_{A}}{Δt_{A}} = \frac{2\sqrt{L_{A}^{2} + (v Δt_{B}/2)^{2}}}{Δt_{B}}  (2)

Solving for Δt_{B} we get

Δt_{B} = \frac{2L_{A}}{c} \frac{1}{\sqrt{1 − v^{2}/c^{2}}} = Δt_{A}γ > Δt_{A}  (3)

where γ = 1/\sqrt{1 − v^{2}/c^{2}} > 1 is the gamma factor. This equation indicates that according observer B, the time interval Δt_{A} in the moving frame becomes the dilated time Δt_{B} > Δt_{A}.

• Motion direction is in parallel with the path of the light

According to observer B, the light’s traveling time is Δt_{1} = (L_{B} +vΔt_{1})/c along the direction of motion and Δt_{2} = (L_{B} − vΔt_{2})/c in the opposite direction. Solving these equations we get Δt_{1} = L_{B}/(c − v) and Δt_{2} = L_{B}/(c + v), and the total traveling time is

Δt_{B} = Δt_{1} + Δt_{2} = \frac{2L_{B}}{c} \frac{1}{1 − (v/c)^{2}} = \frac{2L_{B}}{c} γ^{2}  (4)

Solving this for L_{B}, we get the length between the two mirrors according to B:

L_{B} = \frac{cΔt_{B}}{2γ^{2}} = \frac{cΔt_{A}}{2γ} = \frac{L_{A}}{γ} < L_{A}  (5)

where we have used Δt_{B} = γΔt_{A} in Eq. (3) and L_{A} = cΔt_{A}/2 in Eq. (1). This equation indicates that according to B, the proper length L_{A} in the moving frame becomes the contracted length L_{B} < L_{A}.

For example, when v = 0.9 c, γ = 2.2943, according to B at rest, one hour in the moving frome becomes 2.29 hours, and one meter in the moving frame becomes 0.4359 meters.

2 Lorentz Transformation

Consider two reference frames, S at rest and S′ moving along the positive x direction (to the right) at speed v with respect to S. Assuming the two frames coincide at t = 0, then an object at position x in S at moment t is at x′ = x−vt in S according to the Galilean transformation:

\{ \begin{matrix}x′ & = & x − vt \\ t′ & = & t \end{matrix}  (6)

The inverse Galilean transformation is:

\{ \begin{matrix}x & = & x′ + vt \\ t & = & t′ \end{matrix}  (7)

However, based on length contraction the position of the object in the moving frame S’ should be x′/γ according to an observer B in the resting frame S, we therefore have

x′/γ = x − vt, i.e. x′ = γ(x − vt)  (8)

On the other hand, according to an observer A in the moving frame S′, frame S is moving to the left at speed v, and based on length contraction and time dilation, we have

x/γ = vt′ + x′, i.e. t′ = \frac{x/γ − x′}{v}  (9)

Substituting x′ = γ(x − vt), we get

t′ = \frac{x/v − x′}{v} = \frac{x/γ − γ(x − vt)}{v}  (10)

Dividing both sides by γ, we get

\frac{t′}{γ} = \frac{x/γ^{2} − x + vt}{v} = \frac{x(1 − v^{2}/c^{2}) − x + vt}{v}
= \frac{vt − xv^{2}/c^{2}}{v} = t − vx/c^{2}  (11)

Solving for t′ we get

t′ = γ(t − vx/c^{2})  (12)

Now we get the Lorentz transformation for observer B in frame S:

\{ \begin{matrix}x′ & = & γ(x − vt) \\ t′ & = & γ(t − vx/c^{2}) \end{matrix}  (13)

Similarly, the inverse Lorentz transformation for observer A in frame S’:

\{ \begin{matrix}x′ & = & γ(x − vt) \\ t′ & = & γ(t − vx/c^{2}) \end{matrix}  (14)