2

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 $V'$?)

Qmechanic
  • 220,844
Rescy_
  • 942

4 Answers4

10

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.

Selene Routley
  • 90,184
  • 7
  • 198
  • 428
1

This came up in one of the first homework problems in my daughter’s intro to relativity class. She wanted help because she thought she was messing up the algebra.

Setup: Rocketship flies by “stationary” planet at a constant speed. You essentially have six variables: distance, elapsed time, and velocity, in both frames (3x2). Two of those variables were given. The relationship between distance in one frame vs the other is described by the Lorenz equation. And in both frames v = delta x / delta t. That’s five equations — what gives I thought? The answer is the reciprocity principle, which says that the two velocities are just inverses of each other. Your 6th equation.

I am a macroeconomist and last time I did anything with relativity was decades ago in a (good, but not perfect) high school physics class. But I can count variables and equations and once I catalogued them I can do the algebra.

So basically, thank you for the answer! I hope I am contributing something by giving an explicit example (albeit a simple homework problem, possibly the simplest?) where this immediately figures in.

Curious Macroeconomist
  • 11
0

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 .

Goutham
  • 69
  • 7
0

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.

Asher
  • 4,936