# Construction of the Lorentz Transformation

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 Problem The challenge problem posed on the Centre of the Lightcone page was to find the relation — the Lorentz transformation — between the spacetime coordinates $$( t , x , y , z )$$ and $$( t' , x' , y' , z' )$$ of a spacetime interval in Vermilion's versus Cerulean's frames, if Cerulean is moving relative to Vermilion at velocity $$v$$ in the (say) $$x$$ direction. Below follows a geometrical construction of the Lorentz transformation, which achieves the desired goals (1) that both Vermilion and Cerulean consider themselves to be at the centre of the lightcone, and (2) that distances perpendicular ($$y$$ and $$z$$) to the direction of motion remain unchanged.

 Step 1 Draw a spacetime diagram (see the The Paradox of Special Relativity page for a definition) from Vermilion's point of view. For simplicity keep only one spatial dimension, the $$x$$ direction, the direction of Cerulean's velocity relative to Vermilion. The time $$t$$ and distance $$x$$ of events, relative to the origin, are those measured by Vermilion. Draw Vermilion's worldline vertically through the origin, and a horizontal “now” line, also through the origin.

 Step 2 Draw the worldline of Cerulean moving at speed $$v$$ relative to Vermilion (remember that $$c = 1$$ in a spacetime diagram). Use primed coordinates to denote the time $$t'$$ and distance $$x'$$ of events, relative to the origin, measured by Cerulean. Cerulean's worldline lies along $$x = vt$$ in Vermilion's frame, and along $$x´ = 0$$ in his own frame. Put Cerulean somewhere on his worldline. Without loss of generality, take this event to be one tick of Cerulean's clock past the origin. In other words, take Cerulean to be at position $$(t' , x' ) = ( 1 , 0 )$$ from his own point of view. In Vermilion's frame, Cerulean's position relative to the origin is $$( t , x ) = ( \gamma , \gamma v )$$. This is true because, as seen on the previous page on Time Dilation, Vermilion thinks her clock ticks $$t = \gamma$$

 Step 3 Draw a rectangle, sides 45° from vertical, with the origin at the centre and Cerulean at one corner. The rectangle represents the path of light rays which Cerulean uses to define a hypersurface of simultaneity, as described on the simultaneity page, Spacetime diagram illustrating simultaneity from Cerulean's point of view. Draw the extra diagonal across this rectangle. The diagonal is a hypersurface (reduced to a line) of simultaneity, a “now” line, for Cerulean. The now line lies along $$t = vx$$ in Vermilion's frame, and along $$t = 0$$ in Cerulean's frame.

 Step 4 Drop a perpendicular from Cerulean to the horizontal axis, and draw an arc centred on this point, passing through Cerulean. Mark Vermilion where the arc intersects the vertical axis. This puts Vermilion at position $$( t , x ) = ( 1 , 0 )$$ in her own frame, just as Cerulean is at position $$( t' , x' ) = ( 1 , 0 )$$ in his frame. Why does this work? Cerulean is at position $$( t , x ) = ( \gamma , \gamma v )$$. Remember from the discussion of the Lorentz gamma factor on the previous page that a triangle with sides $$1$$, $$\gamma$$, and $$\gamma v$$ forms a right-angle triangle. So the construction puts Vermilion at $$t = 1$$ on the vertical axis, as claimed.

 Step 5 Draw a square, sides 45° from vertical, with the origin at the centre and Vermilion at one corner. The square represents the path of light rays which Vermilion uses to define a hypersurface of simultaneity, as described on the simultaneity page, Spacetime diagram illustrating simultaneity from Vermilion's point of view. As anticipated, Vermilion thinks events on horizontal lines happen simultaneously. Notice that the areas of Vermilion's and Cerulean's light rectangles are equal. This has to be so for fundamental reasons. For suppose that the area were multiplied by some (necessarily positive) factor $$A$$ say; then a Lorentz transformation at velocity $$v$$ followed by a Lorentz transformation at velocity $$v$$ in the opposite direction would multiply the area by $$A^2$$; but such a transformation should get you back where you started, so $$A$$ must equal one. Note that mathematically the area factor $$A$$ is the determinant of the Lorentz transformation matrix.

 Lorentz transform grid Draw grids of lines parallel to the worldlines and now lines of Vermilion and Cerulean. Add a lightcone through the origin. The two grids are Lorentz transforms of each other; that is, they show how Vermilion's and Cerulean's spacetime frames transform into each other. The diagram is to be understood as follows. Put down any event on the spacetime diagram. Read off the coordinates $$( t , x )$$ of the event in Vermilion's frame using Vermilion's axes  . Read off the coordinates $$( t' , x' )$$ of the event in Cerulean's frame using Cerulean's axes  . Watch this Lorentz grid transformation movie (29K GIF), or same movie, double-size on screen (same 29K GIF), which shows how the Lorentz transformation changes with velocity.

 Centre of the lightcone Notice that Vermilion is at the centre of the lightcone from her point of view, while Cerulean is at the centre of the lightcone from his point of view. We have arrived, finally, at the place where we wanted to be.

 Mathematical form of the Lorentz transformation Mathematically, the form of the Lorentz transformations illustrated pictorially above is as follows. The units in these equations are $$c = 1$$, following the convention for spacetime diagrams. Suppose that Cerulean (primed frame) is moving relative to Vermilion (unprimed frame) at velocity $$v$$ in the $$x$$ direction. Then the coordinates $$( t' , x' , y' , z' )$$ of an event in Cerulean's frame are related to the coordinates $$( t , x , y , z )$$ of the same event in Vermilion's frame by $\begin{array}{l} t' = \gamma t - \gamma v x \\ x' = - \, \gamma v t + \gamma x \\ y' = y \\ z' = z \end{array} \ , \quad \begin{array}{l} t = \gamma t' + \gamma v x' \\ x = \gamma v t' + \gamma x' \\ y = y' \\ z = z' \end{array}$ These Lorentz transformation formulae can be written elegantly in matrix notation: \begin{align} \left( \begin{array}{c} t' \\ x' \\ y' \\ z' \end{array} \right) &= \left( \begin{array}{cccc} \gamma & - \gamma v & 0 & 0 \\ - \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} t \\ x \\ y \\ z \end{array} \right) \ , \\ \left( \begin{array}{c} t \\ x \\ y \\ z \end{array} \right) &= \left( \begin{array}{cccc} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{c} t' \\ x' \\ y' \\ z' \end{array} \right) \ . \end{align} Any 4-dimensional quantity $$( t , x , y , z )$$ that transforms according to the rules of Lorentz transformations, as above, is called a 4-vector. A Lorentz transformation at velocity $$v$$ followed by a Lorentz transformation at velocity $$v$$ in the opposite direction, i.e. at velocity $$-v$$ yields the unit transformation, as it should: $\left( \begin{array}{cccc} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \left( \begin{array}{cccc} \gamma & - \gamma v & 0 & 0 \\ - \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right) \ .$ The determinant of the Lorentz transformation is one, as it should be: $\left| \begin{array}{cccc} \gamma & - \gamma v & 0 & 0 \\ - \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right| = \gamma^2 ( 1 - v^2 ) = 1 \ .$ Indeed, requiring that the determinant be one provides another derivation of the formula for the Lorentz gamma factor.

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Updated 29 Apr 1998; converted to mathjax 17 Jan 2013