This question is based on Box 4.3 in Gravity by James B. Hartle where a cloud appears to move at 10 times the speed of light. I will simplify the example by having the cloud move in one spatial dimension.

Consider an inertial observer $O$ where we use $t$ and $x$ for the time and spatial coordinate in the observer's frame. Let a cloud start at $x=0$ at time $t=0$ and move towards the observer located at $x=L$ with speed $V=\frac{dx}{dt}$. At time $t_0$ light from the cloud is emitted towards the observer, and is received at time $t_{\text{obs}}$. The distance traveled by the light can be calculated in two ways so \begin{equation} c(t_{\text{obs}}-t_0) = L-Vt_0, \end{equation} giving \begin{equation} t_{\text{obs}}= t_0(1-V/c)+L/c, \end{equation} The speed measured by the observer given by $V_O = \frac{dx}{dt_{\text{obs}}}$ is then \begin{equation} V_O = \frac{dx}{dt_0} \frac{dt_0}{dt_{\text{obs}}} = \frac{V}{1-V/c}, \end{equation} explaining why could measure a speed much greater than $c$ even though $V$ is smaller than $c$. My questions are then:

1) Is there a more direct method of calculating the speed $V=\frac{dx}{dt}$ (without using the formula above)?

2) Is this how physicist usually calculate the speed of objects close to the speed of light? For instance at CERN there are particles moving close to $c$.

Note that I'm asking this for special relativity and not general relativity, so we can assume that there are no GR effects.

  • $\begingroup$ I'm having trouble understanding the example. It seems that we're describing everything in the observer's rest frame, and the coordinates are the observer's coordinates. In that case, why are there two different velocities, $V$ and $V_O$? In general, yes there are much simpler ways of measuring velocities, and those velocities are not $>c$. For example, at a linear particle accelerator, one could measure the beam's velocity by picking up an electrical pulse as a beam pulse passes position $x_1$ at time $t_1$, then again at $(t_2,x_2)$, and $v=\Delta x/\Delta t$, and this will be $<c$. $\endgroup$ – Ben Crowell Apr 21 '18 at 15:50
  • $\begingroup$ I think the point being made by Hartle is probably that, according to an observer in a certain frame of reference, the rate at which the distance between two moving objects changes can be $>c$. But that rate is not the velocity of any one thing relative to any other thing. $\endgroup$ – Ben Crowell Apr 21 '18 at 15:52
  • $\begingroup$ @BenCrowell Yes, everything is described with respect to the observer. The velocity $V_O$ is how far the object has moved over the time we receive two photons. $\endgroup$ – N.U. Apr 21 '18 at 16:21
  • $\begingroup$ Isn't there also a point that we can only measure the time $t_{\text{obs}}$ instead of $t_0$? $\endgroup$ – N.U. Apr 21 '18 at 16:35
  • $\begingroup$ The velocity VO is how far the object has moved over the time we receive two photons. I read through the question carefully again, and I don't see any description of two photons, only one photon. $\endgroup$ – Ben Crowell Apr 21 '18 at 18:44

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