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Inertial frame: In an inertial frame of reference a body with zero nett force acting on it does not accelerate

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🪦 Postulates:

1. The laws of physics (ie results of experiments) are the same in all inertial frames of reference
2. The speed of light (in vacuum) is the same in all frames of reference

Lorentz transformation

💼 Case: Consider two identical frames

• At $t'=t=0$ the origins of $S$ and $S'$ coincide
• We can related these coordinates system using the Lorentz transformations

$$ \begin{aligned} [ct']&=\gamma([ct]-\beta x) \\ x'&= \gamma (-\beta [ct]+x) \\ y'&=y \\ z'&=z \end{aligned} $$

🗒️ Note: $[ct]$ has units of length rending the expressions for Lorentz transformation symmetric

4-vectors

Archetypical example of a $4$-vector

$$ \tilde x = (ct,\vec r)=(ct,x,y,z) = (ct,x^i)=(x^0,x^1,x^2,x^3)=x^\mu $$

where $i=1,2,3$ and the upper script are not power but Einstein convention

⚙️ Properties:

• Has the structure (scalar, vector)

• We arrange for all 4 components to have the same units

• components transform according to the Lorentz Transformation

  $$ x'^0=\gamma (x^0-\beta x') \quad x'^1=\gamma(-\beta x^0+x^1) \quad x'^2=x^2 \quad x'^3=x^3 $$

• Under subtraction $\tilde x_{(1)} - \tilde x_{(2)} = \Delta \tilde x =(c \Delta t, \Delta \vec r)$ which transforms as a 4-vector

• Scalar product of two 4-vectors (mostly negative convention)

  $$ \tilde a \cdot \tilde b \equiv a^0 b^0 - \vec a \cdot \vec b = a^ 0 b^0 - a^1 b^1 -a^2 b^2 - a^3 b^3 $$

• The scalar product has the same value in all inertial frames

  $$ \tilde a' \cdot \tilde b'= \tilde a \cdot \tilde b $$

  🗒️ Note: this is also called “invariant with respect to a Lorentz transformation” “Lorentz invariant” “Lorentz scalar”

  • Proof:

Example

💼 Case: space-time interval between two events in $S$ connected by a signal travelling at the speed of light

• The distance is space between the 2 events divided by the distance in time equals $c$

  $$ \frac{\Delta r}{\Delta t}=c $$

• We can calculate its magnitude of the signal

  $$ (\Delta \tilde x)^2 = (c\Delta t)^2 - (\Delta r)^2 =0 $$

  because it is travelling at the speed of light

• Now since scalar products are invariant that means that in for the other frame we have

  $$ (\Delta \tilde x')^2=(c\Delta t')^2-(\Delta r')^2 =0 $$

💎 Conclusion: thus we get $\frac{\Delta r'}{\Delta t'}=c$ meaning that signal travels at the speed of light in both frames