What would be the apparent velocity of an object approaching us at an actual velocity of $v$? I know $cv/c-v$ is a possible answer but here's another argument - shouldn't the velocity be the Doppler shift

$$v\left(\frac{c+v}{c-v} \right)^{1/2}$$

times the actual velocity?


2 Answers 2


Imagine a stealthed1 alien vessel is approaching the earth with relative velocity $\beta = \frac{v}{c}$. At a distance of two light-years as measured in the Earth's frame2 the craft switches on its radio and send the Earth a ultimatum. We call this moment time zero in the Earth's frame.

Then it turn the radio off until it reaches a distance of one light-year (again, as measured in the Earth frame), when it turns the radio on again to repeat the challenge.

Now consider the timing as measured in the Earth frame.

At $t_1 = 0$ the first message is sent.
At $t_{1'} = t_1 + (2\,\mathrm{year})$ the first message is received.
At $t_2 = t_1 + \frac{(1\,\mathrm{year})}{\beta}$ the second message is sent.
At $t_{2'} = t_2 + (1\,\mathrm{year}) = t_1 + \frac{(1\,\mathrm{year})}{\beta} + (1\,\mathrm{year})$ the second message is received.

Earth bound observers see two messages from a single vessel at separated by a distance or one light-year that arrive with a time separation of $$\Delta t = t_{2'} - t_{1'} = \frac{1\,\mathrm{year}}{\beta} - (1 \,\mathrm{year}) \,.$$ From that they compute an apparent velocity for the approaching craft of $$ \begin{align} v_{app} &= \frac{\Delta x}{\Delta t} \\ &= \frac{1 \,\mathrm{lightyear}}{\frac{1\,\mathrm{year}}{\beta} - (1 \,\mathrm{year})} \\ & = \frac{c}{\frac{1}{\beta} - 1} \\ & = \frac{\beta c}{1 - \beta} \\ & = \frac{v}{1 - v/c}\,. \end{align} $$

For $v \ll c$ we get $v_{app} \approx v$, but as $v$ becomes a significant fraction of $c$ things change. In fact for $v = \frac{c}{2}$ we get $$v_{app}(0.5c) = \frac{0.5c}{1 - 0.5} = c \,,$$ and for approach velocities in excess of one-half c the apparent velocity exceed the speed of light.

This has been observed in jets directly strongly along our line of sight.

The effect is symmetric, the aliens see Earth approaching just as fast.

This effect is also separate of the Doppler shift.

There is an intimate relationship between this analysis and the "fast forward" (or "slow motion") effect on time as observed ahead or behind you when moving fast relative the thing you are observing.

(1) Just so we can concentrate on exactly two signals, this does not affect the results.

(2) Radio observatories on Earth can measure the distance to the origin of the signal by direct parallax or by interferometric methods.

  • $\begingroup$ Given that a lightyear is a distance, the dimensions of some of your formulas don't work. $\endgroup$
    – Dr Chuck
    Commented Apr 6, 2014 at 20:35
  • 2
    $\begingroup$ Well, I've been a little sloppy in a way that makes sense to particle physicists (we live in a $c=1$ world). I'll be explicit. $\endgroup$ Commented Apr 6, 2014 at 20:43
  • 1
    $\begingroup$ I love my $c=\hbar=1$ world! Units don't work out? Throw in some constants until they do. Looking for a force but somehow came out with power? Well good news, they're the same thing! $\endgroup$
    – Jim
    Commented Apr 6, 2014 at 21:01
  • $\begingroup$ dmckee: "(2) Radio observatories on Earth can measure the distance to the origin of the signal by direct parallax or by interferometric methods." -- If it is not explicitly required that "the origin of the signal" and "the receiver of the signal" be at rest wrt. each other, allowing both to be characterized jointly and in mutual agreement by óne distance value, then some quantity pertaining to them together, as evaluated "by direct parallax or by interferometric methods", ought not to be called "distance", but rather "apparent separation" or "unilateral separation". $\endgroup$
    – user12262
    Commented Dec 10, 2016 at 14:18
  • $\begingroup$ @user12262 From the origin of the footnote: "as measured in the Earth's frame". $\endgroup$ Commented Dec 10, 2016 at 17:26

You probably know already that there is no derivation of the velocity transform in Wikipedia, or anywhere else, along with other transforms' derivations for special relativity (at least I did not find any). The sub-section on the principle of relativity states, however, that $v' = -v$.

So, according to Wikipedia, you should assume that the apparent velocity you are asking about is:$$v = - v'$$ (which is quite puzzling though, since the variables x and t change under the transforms in inverse proportion - hence time dilatation and length contraction - and therefore the values for v and v' should differ).


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