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Consider two frames $S$ and $S'$, moving relative to each other. If I stand still in frame $S$ and watch frame $S'$, I can measure its speed to be $V$. However, why is this speed the same as the observer in $S'$ measured for $S$? (denote as $V'$?)

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  • $\begingroup$ Because the Lorentz boost from $S$ into $S'$ is exactly of the form that the "zero velocity" in $S$ becomes $V$ in $S'$? Just plug it in. $\endgroup$ – ACuriousMind Nov 19 '15 at 14:30
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You are asking about the so-called reciprocity principle of special relativity, and it is actually not as trivial as many people make out. Indeed, a careful examination of the postulates we make about the relativities of inertial frames and the transformations that result from these postulates show that at least three postulates must work together to ensure that the velocity $v_{AB}$ of observer $A$ relative to observer $B$ is related to that $v_{BA}$ between $B$ and $A$ by $v_{AB} = - v_{BA}$. These postulates are:

  1. Isotropy of space;
  2. The Galileo relativity postulate;
  3. The continuity of the transformation law, as a function of the relative velocities $v_{AB}$, $v_{BA}$.

See the paper:

Berzi & Gorini, "Reciprocity Principle and the Lorentz Transformations", J. Math. Phys. 10, 1968

for more details. The explanation is really only forthcoming by a careful examination of the equations in section 2 of this paper.

Many first courses in relativity simply take this reciprocity as a postulate, and it would seem thoroughly reasonable to do so.

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This is so because the distance between those two frame of references measured from both perspectives should be the same at any given point of time. Speed is just the 'change of distance' measured b/w those two frames per unit time, which should obviously also be the same from both .

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If you are using scalar speeds, the issue becomes clear: $V$ and $V'$ always equal each other because they both represent the rate of change of separation between two objects, which is a property of the two-object system, not of either object individually. In fact, every observer watching $S$ and $S'$ will calculate $V$ and $V'$ to be equal, regardless of the observer's frame, because $V$ and $V'$ both represent the same value.

Now, if you use vector velocities instead, it gets only a touch more complicated: the relative velocities are opposite in any given coordinate system, because if $S$ is traveling relatively left, $S'$ is traveling relatively right, but the magnitude of relative velocity will still be equal.

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  • $\begingroup$ But why the separation between two frames are measured to be the same? $\endgroup$ – Rescy_ Nov 19 '15 at 21:47
  • $\begingroup$ Why wouldn't it be? Is there some reason to think that one frame would measure a different separation than the other? If A is 3m from B, then B is 3m from A. You'll never have a case where the distance from one to the other is measured to be longer than the reverse by the same observer. $\endgroup$ – Asher Nov 20 '15 at 1:13

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