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Note : sorry for the poor drawing above.

Note : in what follows, the speed of light is taken as unit : $c =1$ and the trajectory of a light signal sent from $(0,0)$ is $x = ct = t$.

In his Theoretical Minimum's volume devoted to the Special Theory of Relativity, Susskind establishes the equation of the $x'$ axis ( $x$ axis of the moving frame) in the following way:

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(1) 3 observers in the moving frame ( along the $x$ axis with velicity $v$ , one(Art) with trajectory $ x = vt$ , the other one ( Maggy) with trajectory $ x= vt+1$ , the last one( Lenny) with trajectory $ x= vt+2$

(2) at time $t=t'=0$ Art and Lenny send a light signal to Maggy ( who receives the two signals at the sime time, Art and Lenny being equidistant from Maggy )

(3) Art's signal meets Maggy's trajectory at point $a$

(4) Knowing the point at which Lenny's signal meets Maggy ( point $a$, like Art's one) , the fact that the trajectiory of Lenny's signal is of the form $ x+t = K$ with $K$ a constant ( equal to the $x_a+ t_a$) , and finally the fact that point $b$ satisfies Lenny's trajectory $ x = vt+2$ , one can finally determine the coordinates of point $b$, i.e. the point ( of spacetime) at which Lenny sends his signal.

(5) since $b$ must belong to the $x'$ axis ( at which $ t'=0$ ) and since the $x'$ axis passes through $(0,0)$, with these two points one can determine the equation of this axis

$$ t = (t_b/x_b) x = vx$$

My question is simply : is there a more direct ( even if less elegant) way to determine the equation of the $x'$ axis?

Vince Vickler
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1 Answers1

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Essentially, one draws the tangent-line to the hyperbola (the "circle" in Minkowski spacetime)... think "tangent is perpendicular to radius" (which works for Euclidean space, Minkowski spacetime, and Galilean spacetime). You then draw the parallel line through the origin.

From my Is simultaneity in SR merely an artifact of coordinate systems?

  • For SR, here's how Minkowski describes this...

From Minkowski's "Space and Time"...

We decompose any vector, such as that from O to x, y, z, t into four components x, y, z, t. If the directions of two vectors are, respectively, that of a radius vector OR from O to one of the surfaces ∓F = 1, and that of a tangent RS at the point R on the same surface, the vectors are called normal to each other. Accordingly, $$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$ is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to be normal to each other.

See also: Space time diagrams: Length contraction

A related construction involves https://en.wikipedia.org/wiki/Pole_and_polar .

In Minkowski (1+1)-spacetime, two lines are Minkowski-perpendicular when the product of their slopes equals $+1$ (which is the analogue of the Euclidean case, where the product would be $-1$).

robphy
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