Imagine if you will, a strand of fiber optic cable 186,000 miles long. A pulse of light is sent through the stationary cable: it takes 1 second for light to travel the entire length of the cable.

Now again, imagine this same fiber optic cable travelling at 98,000 miles per second and a pulse of light is sent through the cable. In this instance both the cable and the pulse of light are travelling in the same direction. How long does it take the light to travel the length of the cable? Would it take the light 2 seconds to travel the 1 light second of distance through the cable?

Carl Sagan, "Thou shalt not add thy speed to the speed of light"

Yet again, picture this same fiber optic cable is travelling at 98,000 miles per second.. but this time the pulse of light is sent in the opposite direction that the cable is moving. Does the light take only half a second to travel the 1 light second of distance?

if the cable is moving at 99.9% the speed of light and a pulse of light is sent perpendicular to the course of the cable... is the information lost because the light cannot move in the direction the cable is moving AND its path within the fiber optic cable?

  • 1
    $\begingroup$ Possible duplicate of If I walk along the aisle of a bus traveling at the speed of light, can I travel faster than it? $\endgroup$
    – Muze
    Jan 19, 2016 at 8:43
  • 1
    $\begingroup$ That other question asks whether the resulting speed is faster than light, whereas this one asks about timing. I'm not sure if it's quite a duplicate. It's definitely very close though. (And I do think this is a duplicate of something, if not that one.) $\endgroup$
    – David Z
    Jan 19, 2016 at 9:07
  • $\begingroup$ I am not trying to break the cosmic speed limit here. this has everything to do with the amount of time light takes to move within mass itself moving at a high velocity. Jen sites the example of a person moving inside a moving bus, but wants to know if the person goes faster than light. my question in that context would be how long does it take light to travel from the back of the bus to the front of the bus if the bus is moving at half the speed of light. $\endgroup$ Jan 19, 2016 at 10:10
  • $\begingroup$ For one thing the speed of light in the cable is not the speed of light in vacuum, for another, the cable breaks Lorentz symmetry, so neither Sagan's law nor comparisons with similar sounding relativistic problems (speed of light in a rocket etc.) can be applied naively. $\endgroup$
    – CuriousOne
    Jan 19, 2016 at 10:12
  • $\begingroup$ It was not my intent to ask a naïve question, maybe it comes naturally? I am simply looking for an answer to the math. does it take twice the amount of time for light to travel a given distance if the mass it is in is travelling the same direction. would it take half as long if the light were travelling the opposite direction... and finally, by applying the pythagorean theorem what would happen to the information if the light could not move fast enough to both keep up with the cable moving forward and through the cable (perpendicular to the direction being travelled) $\endgroup$ Jan 19, 2016 at 10:23

6 Answers 6


Let's replace the fiber optic cable with one laser source and one photodetector a distance $L=186,000$ mi apart in vacuum and at rest relative to each other. The laser source is pointed straight at the photodetector. Alice observes the laser source and the detector moving at constant velocity $v = 93,000$ mi/s $= c/2$ with respect to her inertial frame, in the positive $x$ direction. The source fires a pulse at the exact moment $t=0$ when it passes by Alice. So according to Alice, how long does it takes the light pulse to reach the detector? What if the source and the detector switch places? Or if the source-detector direction is perpendicular to the direction of motion?

Short answer: For $v=c/2$, Alice sees the light pulse hit the detector after $c\Delta t = L \sqrt{3}$ or $\Delta t = \sqrt{3} s$. When the source and the detector switch places, she observes the detection after $c\Delta t = L /\sqrt{3}$ or $\Delta t = 1/\sqrt{3} s$, while for a setup perpendicular to the direction of motion she sees $c\Delta t = 2L /\sqrt{3}$ or $\Delta t = 2/\sqrt{3} s$.

Longish details:

1. Forward propagating light pulse

Due to length contraction, at $t = 0$ Alice sees the detector at position $x = L/\gamma$, with $\gamma = (1-\beta^2)^{-1/2}$ the time dilation factor and $\beta = v/c$. So she finds that it travels according to $$ x_{\text{detector}}(t) = \frac{L}{\gamma} + \beta ct $$ She also sees the light pulse traveling at light speed, according to $$ x_{\text{pulse}}(t) = ct $$ The pulse reaches the detector when $x_{\text{pulse}}(\Delta t) = x_{\text{detector}}(\Delta t)$, meaning $$ c\Delta t = \frac{L}{\gamma} + \beta c\Delta t \;\; \Rightarrow \;\;c\Delta t = \frac{L}{(1-\beta)\gamma} = L\sqrt{\frac{1+\beta}{1-\beta}} $$ For $\beta = 1/2$ this gives $$ c\Delta t = L \sqrt{3} $$

Here's the confusing issue:

In the source-detector frame the corresponding duration is clearly $c\Delta t' = L$. But if time in the source-detector frame appears time-dilated to Alice, such that $c\Delta t' = c\Delta t/\gamma$, then how is it that for $c\Delta t'=L$ we find $c\Delta t = \frac{L}{(1-\beta)\gamma}$ instead of $c\Delta t = L\gamma$?

If we look carefully, Alice determines the pulse propagation time as the time during which she observes the detector moving from $$ x_{\text{detector}} = \frac{L}{\gamma}\;\;\text{at}\;\;ct = 0 $$ to $$ x_{\text{detector}} = \frac{L}{\gamma} + \frac{\beta L}{\gamma(1-\beta)} = \frac{L}{\gamma(1-\beta)}\;\;\text{at}\;\;ct = \frac{L}{\gamma(1-\beta)} $$ In the source-detector frame, the end point indeed corresponds to coordinates $$ x'_{\text{detector}}\Big|_{ct = \frac{L}{\gamma(1-\beta)}} = \gamma\left(\frac{L}{\gamma(1-\beta)} - \beta \frac{L}{\gamma(1-\beta)} \right) = L\\ ct'_{\text{detector}}\Big|_{ct = \frac{L}{\gamma(1-\beta)}} = \gamma\left(\frac{L}{\gamma(1-\beta)} - \beta \frac{L}{\gamma(1-\beta)} \right) = L $$ as we'd expect for a light pulse propagating across distance $L$ starting at $ct=0$, but the start point has coordinates $$ \begin{eqnarray} x'_{\text{detector}}\Big|_{ct=0} &=& \gamma\left(\frac{L}{\gamma} - \beta \cdot 0\right) = L \\ ct'_{\text{detector}}\Big|_{ct=0} &=& \gamma(0 - \beta \frac{L}{\gamma}) = -\beta L\;<\;0 \;\;(!!) \end{eqnarray} $$ According to source-detector time, Alice starts monitoring the detector at a time $ct'= -\beta L$, before the pulse was ever emitted! The detector's proper time duration between Alice's start and end of observation events is then $$ c\bar{\Delta t'} = L - (-\beta L) = (1+\beta) L \;>\;L $$ which is indeed just the time dilated propagation time observed by Alice, as it has to be: $$ c\bar{\Delta t'} = \frac{c\Delta t}{\gamma} = \frac{L}{\gamma^2(1-\beta)} = L(1+\beta) $$ On the other hand, the start of observation event in the source-detector frame is at $x'=L$ and $ct'= 0$, which to Alice means $$ x = \gamma(L + \beta \cdot 0) = \gamma L\\ ct = \gamma(0 +\beta L) = \beta\gamma L $$ So for Alice the time interval from this event until the light pulse reaches the detector is $$ c\bar{\Delta t} = \frac{L}{\gamma(1-\beta)} - \beta\gamma L = \frac{\gamma L}{(1-\beta)}\left(\frac{1}{\gamma^2} - \beta(1-\beta) \right) = \gamma L $$ as expected from time dilation for $c\Delta t' = L$.

Note that Alice sees the light pulse closing in on the detector at an apparent relative velocity $$ \Delta v = \frac{d}{dt}\left[x_{\text{detector}}(t) - x_{\text{pulse}}(t) \right] = -(1-\beta)c = v-c $$

2. Backward propagating light-pulse

Suppose now the source and detector switch places. The source fires at $t=0$ in Alice's time, when it passes the $L/\gamma$ mark on her $x$-axis, and in the negative $x$ direction. At the same moment Alice sees the detector at $x=0$. She observes the light pulse propagating according to $$ x_{\text{pulse}}(t) = \frac{L}{\gamma} - ct $$ and the detector according to $$ x_{\text{detector}}(t) = \beta ct $$ This means, as before, that the light pulse hits the detector when $x_{\text{pulse}}(\Delta t) = x_{\text{detector}}(\Delta t)$ or $$ \frac{L}{\gamma} - c\Delta t = \beta c\Delta t \;\;\Rightarrow \;\; c\Delta t = \frac{L}{\gamma (1 + \beta)} $$ For $\beta = 1/2$, $\gamma = 2/\sqrt{3}$ we now have $$ c\Delta t = \frac{L}{\sqrt{3}} $$ Again, this is different from $c\Delta t = L\gamma$ as would be expected from simple time dilation, and for a similar reason.

In the source-detector frame the source fires from location $x'=L$, but at time $ct' = -\beta L$, and the pulse propagates until it meets the detector at $$ \begin{eqnarray} x' &=& \gamma\left(\frac{\beta L}{\gamma(1+\beta)} - \beta \frac{L}{\gamma (1 + \beta)}\right) = 0\\ ct' &=& \gamma\left(\frac{L}{\gamma (1 + \beta)} - \beta\frac{\beta L}{\gamma(1+\beta)}\right) = (1-\beta) L \end{eqnarray} $$ so the proper propagation time is indeed $$ c\Delta t' = (1-\beta)L - (-\beta L) = L $$ But Alice's start event corresponds to the detector being at $x=0$ for $t=0$, which is just $x'=0$ for $t'=0$ in the other frame. In the latter her observation time $c\Delta t = \frac{L}{\gamma (1 + \beta)}$ corresponds to $$ c\bar{\Delta t'} = (1-\beta)L - 0 = (1-\beta)L $$ which is, of course, the expected time dilated value, $c\bar{\Delta t'} = c\Delta t /\gamma$.

Note that now Alice observes the light pulse closing in on the detector at apparent relative velocity $$ \Delta v = \frac{d}{dt}\left[x_{\text{detector}}(t) - x_{\text{pulse}}(t) \right] = \beta c - (- c) = v+c \;>\; c $$ However, this is not superluminal propagation! It is just the rate at which the distance between two simultaneous events in Alice's frame (observation of detector and observation of light pulse) changes in time. There are no objects moving at velocity $v+c$.

3. Light pulse propagating perpendicular to the direction of motion

Finally, when the source and the detector are arranged perpendicular to Alice's direction of motion, we have a variant of the light clock setup. In this case Alice observes the light pulse going at light speed in a direction tilted relative to the direction of motion of the source at a slope $\frac{1}{\beta\gamma}$ and hitting the detector after a time $c\Delta t = \gamma L$. This answer explains in more detail why this happens.

If you prefer to consider a fiber optic cable where light propagates at $v \sim 2c/3 < c$, as suggested in one of the comments, simply use the velocity addition formula to obtain the velocity of a pulse as seen by Alice, then apply the same reasoning.


There are two things: The light and the cable. Light moves at speed 18600 miles/second to the north, cable moves at speed 98000 miles/second to the south. The two things have speeds they are allowed to have.

The distance between the light and the cable changes 284000 miles/second.

Now let me remind that there were two things with those speeds that the two things have. The last speed is not a speed of any thing, as both things have another speed than 284000 miles/second, and there are no more things than two.

284000 miles/second is a speed. Use it to calculate how fast a distance changes, just as you did.

I repeat once more: 284000 miles/second was not the speed of either of the two things. And nothing at all is wrong with a speed like billion miles/second. For example a fiber optic cable factory is allowed to produces optic cable at speed billion miles/second.


There are a few related things here. Let the light move right and the cable move left at speed $v$.

First, suppose the light is not in the cable. If the light is moving right and the cable is moving left, can the light clear the length of the (light-second long) cable in less than one second?

Yes! The relative velocity of two objects can be greater than $c$, and it's computed just by adding their velocities together, getting $v+c$. The relativistic velocity addition formula doesn't apply here since everything is in one reference frame.

Now, suppose the light is in the cable, and propagates through the cable with speed $c/n$. What do you see?

Propagation speed means 'speed in the frame of the cable itself'. So the light moves with velocity $c/n$ in the cable's frame, and the cable moves with speed $v$ in your frame. You can find how fast the light is going, according to you, by adding these two quantities with the relativistic addition formula. What you'll find is that the light will move with velocity $$v_{light} = \frac{c/n-v}{1-v/nc}.$$ For small $v$, the motion of the cable affects $v_{light}$ less than you would expect. For high $v$, the light can appear to be standing still in your frame, or even moving backwards.

Despite this, no matter what the cable is doing, the light will always pop out the other end. To see this, note that in the cable's frame of reference, the cable is still and the light is moving right at speed $c/n$. If it makes it out in the cable's frame, it also will in yours.


The answers to this problem can be more easily obtained by having the light source "attached" to the optic cable. This way the frame of reference for all cases, is the optic cable, and its motion will have no effect on the light pulse propagation speed in the cable. Under this condition, it is easy to see that the pulse will take 1 second to travel the length of the cable, regardless of the direction and speed the optic cable has (cases 1 & 2). For the case were the pulse is sent perpendicular to the cable motion, the light pulse never "enters" the cable, the information is lost, because there is no component of the pulse in the direction of the cable.


There are two things: The light and the cable. Light moves at speed 18600 miles/second to the north, cable moves at speed 98000 miles/second to the south. The light has nothing to do with the speed of cable. If the observer stands at the end of north, he will read the speed equals to speed of light only. It is to be understood that the light is a cosmological phenomenon and the cable is physical phenomenon.

  • $\begingroup$ Your answer was fine and clear until that last sentence. What do you mean by it? Can you add more details, such as explainig why a cosmological phenomenon is non-physical? $\endgroup$ Jan 22, 2016 at 12:07
  • $\begingroup$ The substance having mass is physical phenomenon, and the substance having no mass is cosmological phenomenon. The cosmological world may not follow the physical world. Constancy of speed of light irrespective any reference frame, time dilation, dark matter etc. are many subjects, which can be answered with cosmological science only. We should separate physics with the cosmology. Photons are just a flow of heat, which is provided by the space and is a part of cosmology. You cannot add or substrate the physical speed in it. If you need more detail, please write. $\endgroup$ Jan 22, 2016 at 19:02

I'm going to assume that you somehow built a vaccum-fiber optic cable. Although you CONVENTIONALLY don't add your speed to the speed of light, when your speed approaches the speed of light, your speed does come into play. If your cable is flying at 99% of the speed of light, and the light beam is right next to it,relative to the cable, the light is traveling at 1% of the speed of light. Because the cable is traveling so quickly, it would take the light longer to travel that length. Although light travels fast, it still follows laws of motion: T=D/V (time equals distance devided by the velocity). The effective distance is doubled if the cable travels at 1/2 the speed of light.


If a flashlight is traveling at the speed of light and you turn the flashlight on, the light from the flashlight will not travel faster than the speed of light.

  • $\begingroup$ An example of the above: An electrical pulse for a clock in a GPS satellite. Let's say NASA were to break a decaying radioactive isotope into 2 pieces. 1 piece is placed in solar orbit at 25 miles per second, and another kept here as a control. $\endgroup$ Jan 8, 2017 at 0:12
  • $\begingroup$ When the piece in orbit returns, by current scientific beliefs it should have decayed less than the control piece. if so, how does this affect the calculations used to determine the age of the earth, or indeed our solar system? but if the decay is equal, would that not contradict everything we are taught? $\endgroup$ Jan 8, 2017 at 0:22

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