# Will observer measure a faster-than-light object in this example?

Have two ships $S$ and $T$. For simplification, lets make everything 1-dimension. In addition, instead of working with variables, I will work with numbers directly (easier). Shall we start?

Assumptions:

• (A) Both ships are not moving relative to themselves.
• (B) Distance between the ships are $10ls$, that is, $10$ light-seconds.
• (C) The ships has cute names, that is, $S$ and $T$. $S$ will fire relativistic projectile at $T$.
• (D) $T$ can measure in real time, the position of the projectile (because projectile emits light pulses (maybe simply because reflects light from sun)).

Lets calculate. I'll be numbering my reasoning for easier identification. =).

• (1) At $t=0$, ship $S$ fires a projectile with speed $v = 0.9c$ aimed at $T$.
• (2) The light of this event, will take then $10$ seconds to go from $S$ to $T$.
• (3) $T$ only knows $S$ has fired, $10$ seconds after $S$ has actually fired.
• (4) However, in $t=10$ seconds, the projectile is $9$ light-seconds away from $S$. That is, only $1$ light seconds close to $T$.
• (5) In this point, the projectile will hit $T$ after $1/0.9 = 1.111$ seconds.
• (6) From point of view of $T$: Nothing happens in the first $10$ seconds, then $S$ fires. $1.111$ seconds later, projectile hits $T$.

The average velocity measured by $T$ from the light-pulses: $$v = \frac{\Delta x}{\Delta t} = \frac{10ls}{1.111s} = \frac{10ls\cdot 0.9}{s} = 9c$$

As you can see, $T$ will measure a projectile with speed of $9c$ (because crossed $10$ light-seconds of space, in $1/0.9$ seconds. Its up to notice that in reality, nothing has really gone faster than the speed of light. But that's not what $T$ is gonna see.

Is this true? Will $T$ indeed measure this faster-than-light speed? Or I have some reasoning wrong?

• The search term you need is "apparent velocity". Apparent velocities can exceed $c$, but this is because they don't represent the actual velocity as measured by $T$ which is found by collecting range information as well and computing where the object was where the light was emitted. – dmckee Dec 8 '16 at 16:35
• @dmckee Ohh!! Well, in all books of relativity I've read, I never ever saw the term "apparent velocity". Seems to be very important for truly understanding relativity. Any resource recommendation? =). – Physicist137 Dec 8 '16 at 20:55
• All the introductory texts on relativity or "modern physics" that I'm aware of cover it. – dmckee Dec 8 '16 at 20:57
• @dmckee All introductory texts on relativity or "modern physics" that I'm aware of does not cover it. This one for instance, does not. This one does not too. And so on. I could list a lot. Can you thus provide me some? – Physicist137 Dec 8 '16 at 21:06
• Don't know what to tell you. I think of it as a standard part of the introductory curriculum (it makes it crystal clear that the light-speed retardation is separate from time-dilation for one thing), but I don't have those books. I've seen it relegated to a problem occasionally. – dmckee Dec 8 '16 at 21:15

[Consider] two ships [...] The ships [have] cute names, that is, $S$ and $T$.
Assumptions:

• (A) Both ships are not moving relative to [each other].
• (B) Distance between the ships are $10~\text{ls}$, that is, 10 light-seconds.

Consequently $S$ finds constant ping duration of $20$ seconds wrt. $T$; and $T$ finds constant ping duration of $20$ seconds wrt. $S$.

• (C) [...] $S$ will fire relativistic projectile at $T$.

The projectile ought to have a name, too; I'd suggest: $P$.

• (D) $T$ can measure in real time, the position of the projectile (because projectile emits light pulses (maybe simply because reflects light from sun)).

Surely $T$ may observe $P$ throughout the trial, and even specificly recognize $P$ being released from $X$. Surely $P$ may observe $T$ in turn, too (e.g. $T$ emitting radar pulses); consequently $T$ may determine its ping durations wrt. $P$ (and likewise $P$ might determine its ping durations wrt. $T$).

[Let's] calculate.
- (1) [...] ship $S$ fires a projectile [$P$] with speed $v = 0.9~c$ aimed at $T$.

Whether the speed of $P$ wrt. the system constituted of $S$ and $T$ remained at $v = 0.9~\text c$ on average and throughout the trip could only be confirmed after $P$ actually had arrived at $T$.

• (2) The light of this event, will take then $10$ seconds to go from $S$ to $T$.

Saying that "light goes" from sender to receiver seems too suggestive of "light being being on the way" and "light being observable inbetween", too.

But certainly, by the setup prescription from above, it will take $S$ $10$ seconds from its having indicated the release of $P$ until its indication simultaneous to $T$'s indication of being reached (or passed) by $P$; and likewise it will take $T$ $10$ seconds from its indication simultaneous to $S$'s indication of sending off $P$ until its ($T$'s) indication of being reached (or passed) by $P$.

• (3) $T$ only knows $S$ has fired, 10 seconds after $S$ has actually fired.

Right; based on the simultaneity relation which $S$ and $T$ can determine between each other, since (or as far as) they had been and had remained at rest wrt. each other.

• (4) However, in [...] $10$ seconds, the projectile is $9$ light-seconds away from $S$. That is, only $1$ light seconds close to $T$.

Saying that the projectile $P$ had "been a certain distance away" from $S$, or from $T$, seems too suggestive of $P$ having been and remained at rest wrt. $S$ and $T$ (contrary to the prescription above).

Rather, projectile $P$ may have been passing some suitable additional participant, let's say ship $H$, who was and remained at rest wrt. $S$ as well as $T$, $9$ light-seconds away from $S$, $1$ light-second away from $T$, and whose indication of being passed by $P$ was simultaneous to $T$'s indication of seeing and first knowing that $S$ had fired.

• (5) In this point, the projectile will hit $T$ after $1/0.9=1.111$ seconds.

More explicitly, and from the perspective of hindsight: $T$'s duration from its indication simultaneous to $H$'s indication of being passed by $P$ until its ($T$'s) indication of being reached/passed by $P$ is $\tau T[~\circledS H \!\circ\! P, \circ P~] \equiv \tau T[~\circledR S \!\circ\! P, \circ P~] = 1~\text{ls}/0.9~{\text c} = 1.111$ seconds.

Or, perhaps even more practical, with the goal of measuring the projectile speed in the first place:

From the distance between $H$ and $T$ and from $T$'s relevant duration the (average) speed of $P$'s motion from $H$ to $T$ is obtained as

$$\overline v_{HT}[~P~] := \frac{d[~H, T~]}{\tau T[~\circledR S \!\circ\! P, \circ P~]} = \frac{1~\text{ls}}{1.111~\text s} = 0.9~\text{c},$$

where of course the distance between $H$ and $T$ is explicitly determined from their ping durations wrt. each other:

$$d[~H, T~] := \frac{1}{2}~\text {c}~2~\text{s} = 1~\text{ls}.$$

(Of course, the overall average speed of $P$'s motion from $S$ to $T$, $\overline v_{ST}[~P~]$, would be obtained similarly from the distance between $S$ and $T$ and from the duration of the "track" being occupied by $P$.)

• (6) From point of view of $T$: Nothing happens in the first 10 seconds, then $S$ fires. 1.111 seconds later, projectile hits $T$.

Correct.

The average velocity measured by $T$ from the light-pulses: [...] $= \frac{10~\text{ls}}{1.111~\text s} = 9~c$.

No, this is not a calculation of $P$'s average velocity (or average speed) wrt. the system of $S$ and $T$ (and $H$); but this is the calculation only of some other, rather unimportant quantity, which happens have the same dimension as a speed value.

Instead, the average speed of $P$'s motion from $S$ to $T$ is obtained in terms of durations of $T$ as

$$\overline v_{ST}[~P~] := \frac{d[~S, T~]}{\tau T[~\circledS S\!\circ\! P, \circ P~]} = \left(\frac{1}{2}~\text{c}~20~\text{s}\right)~\left(\frac{1}{\frac{1}{2}~20~\text{s} + 1.111~\text{s}}\right) = 0.9~\text{c}.$$