A particle is moving at velocity v. A stationary observer tries to measure its velocity. From the observer reference frame, he will measure a shorter distance travel as compared to what the particle will measure due to length contraction. The observer will also measure a greater time compared to what the particle will measure. Won't this mean that the observer measured velocity of the particle is different from what the particle measure?

Edit: I think the problem here is that the particle can't measure its own velocity as it must be wrt something. But I thought about what if it is moving from point x to point y. Surely it will be able to measure its velocity through that.

  • 1
    $\begingroup$ What is a stationary observer? Stationary relative to what? $\endgroup$
    – CuriousOne
    Jul 3, 2016 at 5:59
  • $\begingroup$ Particle is moving away from the observer who is stationary $\endgroup$
    – newbie125
    Jul 3, 2016 at 6:22
  • $\begingroup$ What is the observer stationary to? $\endgroup$
    – CuriousOne
    Jul 3, 2016 at 6:24
  • $\begingroup$ Stationary wrt the starting point. $\endgroup$
    – newbie125
    Jul 3, 2016 at 6:42
  • 1
    $\begingroup$ You mean the starting point where you nailed a zero-marker into the fabric of spacetime? Where can I buy those nails? $\endgroup$
    – CuriousOne
    Jul 3, 2016 at 6:43

2 Answers 2


It's a good question and it isn't obvious that both observers will agree on the magnitude of their relative velocity. It's fairly straightforward to show this using the Lorentz transformations, but let me attempt a more intuitive approach (if you want to see the maths comment to this answer and I can append it).

You have chosen an inertial frame $S_1$ where observer $A$ is stationary and observer $B$ is moving at some velocity $-v_1$ (we'll take velocities to the right to be positive and to the left to be negative):

Rest frame A

In $B$'s rest frame $S_2$ things look like this:

Rest frame B

And the point you're making is that it isn't intuitively obvious that $v_1 = v_2$ i.e. that both $A$ and $B$ measure the same relative speed.

To see why $v_1 = v_2$ consider that there must be some third observer $C$ who sees both $A$ and $B$ to have equal and opposite velocities i.e. in $C$'s rest frame $S_3$ the situation looks like this:

Rest frame C

In $C$'s frame the motion is symmetrical so the velocities of $A$ and $B$ are equal and opposite. The symmetry is the key point here. There's nothing special about being to the left of observer $C$ or to the right of observer $C$ so we can exchange $A$ and $B$ without changing anything - the velocity of the observer to the left remains $v_3$ and the velocity of the observer to the right remains $-v_3$. That means both $A$ and $B$ must observe their opposite number to have the same velocity (though in opposite directions). If $A$ or $B$ measured their opposite numbers to have different speeds it would break the symmetry.

As requested, here's the maths

We'll start in frame $S_1$ i.e. the frame in which observer $A$ is at rest. In this frame $A$ observes $B$ to be travelling at a velocity $v_A$. Our task is to work out what speed $B$ observes $A$ to be moving, and show that this is also $v_A$ (in the opposite direction) thereby proving that both $A$ and $B$ agree on each others relative speed.

To find out where spacetime points in frame $S_1$ map to in $B$'s frame $S_2$ we need to use the Lorentz transformations:

$$\begin{align} t' &= \gamma \left( t - \frac{vx}{c^2} \right ) \\ x' &= \gamma \left( x - vt \right) \end{align}$$

We can choose our coordinates so that $A$ and $B$ meet at the origin at time zero i.e. $(t=0, x=0)$ it the same point in each frame.

We do the calculation by choosing the two points in $A$'s frame $(t=0, x=0)$ and $(t=\tau, x=0)$. That is $A$ starts at time zero at $x=0$ and a time $\tau$ seconds later $A$ is still at $x=0$, because after all we are working in $A$'s rest frame. If we can find out where these two points are in $B$'s frame then we can work out how far $B$ observes $A$ to move and in what time, and the speed observed by $B$ is then just the distance divided by the time.

This sounds hard but it's actually a very easy calculation. We already know the starting point $(0,0)$ is the same in both frames, so we just need to work out where the point $(\tau,0)$ is in $B$'s frame. If we take the Lorentz transformations and plug in $t=\tau$ and $x=0$ we get:

$$\begin{align} t' &= \gamma \left( \tau - \frac{v_A\,0}{c^2} \right ) = \gamma \tau \\ x' &= \gamma \left( 0 - v_A\tau \right) = -\gamma\tau v_A \end{align}$$

So in $B$'s rest frame $A$ starts at $(0,0)$ and moves to $(\gamma\tau, \gamma\tau v_A)$ i.e. $A$ has moved a distance $\gamma\tau\ v_A$ in a time $\gamma\tau$. The velocity observed by $B$ is just distance moved divided by time, so:

$$ v_B = \frac{-\gamma\tau v_A}{\gamma\tau} = -v_A $$

And there's your answer. If $A$ observes $B$ to be moving at $v_A$ then $B$ observes $A$ to be moving at $-v_A$.

  • $\begingroup$ The symmetry is an experimental fact, it's not given by some deductive reasoning. If we try to make the same construction with acoustic waves, it will fail. $\endgroup$
    – CuriousOne
    Jul 3, 2016 at 6:54
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    $\begingroup$ @CuriousOne it's because you have a preferred reference frame - the one of the air - and the situation isn't symmetric in that frame. It's an axiom of general relativity that there are no preferred inertial frames (even though the cosmic microwave background kinda undermines that), and the Lorentz transformation is derived from that. $\endgroup$ Jul 3, 2016 at 9:50
  • $\begingroup$ I am sorry. I dont quite get it. Do you mind showing the math? That will be really helpful $\endgroup$
    – newbie125
    Jul 3, 2016 at 13:54
  • $\begingroup$ @newbie125: OK, I've added the calculation, but I'm not sure how illuminating it will be ... $\endgroup$ Jul 3, 2016 at 15:44
  • $\begingroup$ Thanks but what if B measures its velocity as it moves from point x to point y and compare it with what A measures B's velocity wrt its stationary self? Won't the result differ? $\endgroup$
    – newbie125
    Jul 3, 2016 at 17:47

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In Figure 1 the system $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$ is moving with velocity $\mathbf{b}=(b,0,0)$ relatively to $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$.

In Figure 2 we build two systems $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ and $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ as shown therein. The system $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ is built from $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$ by reversing the axes $\:O'x',O'y'\:$. The system $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ is built from $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$ by reversing the axes $\:Ox,Oy\:$.

In Figure 3 we see these two built systems alone.

The situation in Figure 3 is physically identical to that of Figure 1, so there is no good reason not to accept that $\bbox[#99FFFF,2px]{\Sigma'\equiv K'\rm{u'v'w'}}$ is moving with velocity $\:\mathbf{b}=(b,0,0)\:$ relatively to $\bbox[#E0E0E0,2px]{\Sigma\equiv K\rm{uvw}}$ and, returning back to Figures 2 and 1 with this series, to accept that $\bbox[#E0E0E0,2px]{S\equiv Oxyz}$ is moving with velocity $\:-\mathbf{b}=(-b,0,0)\:$ relatively to $\bbox[#99FFFF,2px]{S'\equiv O'x'y'z'}$.


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