# Why do we say that light travels at a speed? [duplicate]

According to Einstein's theory of relativity, the more speed something has the slower that time passes for it; and presumably when traveling at the speed of light, time stops entirely. So this means that when a photon is created, the rest of existence virtually pauses until the photon ceases to exist, and then the rest of existence begins again. If time is stopped while the photon exists, then what are we measuring when we measure light? In order to measure the actual photon, we would need to take the measurement while time was standing still. Speed requires time, so why do we say that light travels at a speed?

## marked as duplicate by Brandon Enright, John Rennie, alemi, Kyle Kanos, AliJul 16 '14 at 14:39

• Time dilation is a relative phenomenon, in your question you seem to ignore the notion of a frame of reference. Light travels at a speed, which is fixed in any frame of reference. The frame of reference'' of the photon doesn't exist. – fqq Jul 15 '14 at 14:16
• What do you mean that the frame of reference of the photon doesn't exist? Because the photon is not an 'observer'? – Derek Roberts Jul 15 '14 at 14:20
• Just that there is no rest frame for photonos. See physics.stackexchange.com/questions/16018/… – fqq Jul 15 '14 at 14:25
• Claim: There are no frames of reference where a photon is at rest (i.e. there are no frames travelling at the speed of light.) Proof: The speed of light is constant in all frames. In the hypothetical photon frame, the speed of the photon is 0, yet its speed is the speed of light. Contradiction Thus, there are no frames in which a photon is at rest. – ACuriousMind Jul 15 '14 at 14:25
• The speed of light can only ever be approached. It doesn't really make sense to try to understand how a photon experiences the passage of time, because photons are integral to the way that time is defined. – KidElephant Jul 15 '14 at 14:25

You're quite correct that time dilation tends to infinity as the relative velocity tends to $c$, but we don't care what's happening in the rest frame of the fast moving object. When we measure a velocity we measure the time in our frame and the distance in our frame.
• @JerrySchirmer well $#!+... – Derek Roberts Jul 15 '14 at 15:03 If I (on the Earth) and you (on a very fast rocket) measure the time that takes you to travel a certain distance we would not agree. Still, nothing forbids me to compute your velocity using my measures of time and distance. The interesting point is that if your rocket is made of photons, even a third person trying to chasing you will agree with my measured speed. This happens because of the combination of time dilation and length contraction (the person moving with respect to me sees a shorter distance, but his time also goes slower, both of the same factor$\gamma$), he will always agree with me unless he reaches your photonic speed somehow braking relativity. That is just how relativity works: time is not absolute, speed of light is! • If my rocket were made of photons, when it was caused to exist, you and me and the person chasing me would all become frozen in time, and the photon rocket would run the course of it's purpose, and then it would cease to exist, and then time would continue after the photon was gone, and all you would see is visible light rippling through time behind the photon rocket, so how would you measure the photon rocket? You're only measuring visible light, not photons. – Derek Roberts Jul 15 '14 at 15:50 • The time would just cease to exist only for you into your rocket. But who cares of you? You are too fast, both I and the third person measure the same speed and that's enough for us. (You can also "measure" single photons, especially at high energy) – DarioP Jul 15 '14 at 15:59 • But once time ceased to exist for me, in my rocket, you are no longer able to measure, me or my rocket, because we no longer exist in your time. You can only measure my exhaust, because my exhaust is left behind, in your time. But me and my rocket, we don't exist in your reality anymore. – Derek Roberts Jul 15 '14 at 16:06 • For example, you might make an observation of where I started, and then make an observation of where I ended, and then calculate the distance between the points and the time that you observed me taking to travel that distance, and you come up with a velocity. But, since I existed outside of time, I could have traveled an infinite distance to infinite points in existence between the points where you observed me beginning and ending, because I would have made the entire trip and been at my destination while you were still observing me at my start point. How could you possibly know what happened? – Derek Roberts Jul 15 '14 at 16:15 • If you were able to measure me, you would measure that I was all three, I would be at the start point, I would be covering the entire length of the distance, and I would be at the end point, all in your single measurement. So you wouldn't be measuring my speed over a distance, you would be measuring my length over an infinitely small unit of time. – Derek Roberts Jul 15 '14 at 16:48 what are we measuring when we measure light? When considering "(average) speed of light (in vacuum)", or rather primarily: "(average) speed of a signal front" this means the quotient of • the distance of a signal source and a receiver between each other (i.e. under the condition that these two participants had been at rest with respect to each other), divided by • the duration of the signal source from its signal indication, until its indication simultaneous to the reception indication of the receiver (or equally the duration of the receiver from its indication simultaneous to the signal indication of the source, until it's (the receiver') reception indication). By the definition of (how to measure) distance between two participants who were and remained at rest with respect to each other as $$c/2 \text{ ping duration},$$ where$c$is foremost some symbolic parameter which is distinct from Zero, therefore the value of the "(average) speed of a signal front" can be directly calculated and obtained as$1 ~ c$; or for short:$c$. (That is to say there is no reasonable measurement involved at all, but the result value is obtained as a theorem.) According to Einstein's theory of relativity, the more speed something has the slower that time passes for it; That's an unfortunate, imprecise, improper statement. Let's better say that geometric and kinematic relations between participants (such as speed, or curvature) must be accounted for, in order to compare their individual durations to each other. The details of how to carry out such comparisons can of course be derived; beginning with comparisons between participants who move "uniformly" (straight, and with constant average speeds) with respect to each other. As a consequence, some non-zero value of duration can not be attributed to a signal front "itself". Addendum: To address the title question: Why do we say that light travels at a speed? Strictly, we don't say that "light travels", which might suggest some particular "path, or curve" being "travelled along"; but rather that signals had been "exchanged" (between some particular source and receiver), without any further presumption of geometry, but in order to establish geometric relations (involving that source and receiver, and perhaps additional identifiable participants) in the first place. • Neil deGrasse Tyson: "As you travel faster, or if you find yourself in the vicinity of a higher source of gravity, time ticks more slowly for you then for those who are not". – Derek Roberts Jul 16 '14 at 5:02 • The speed of the signal front is not the same as the speed of the photon. The speed of the wake is not the same as the speed of the boat. So, you set up your experiment to measure photons. You have all the equipment ready to gather data. You press "start". In that same single moment when you create the photon that you are attempting to measure, the photon that you have created also ceases to exist. Then the experiment begins, and all the data is gathered. But the photon is gone already. So what you're measuring is not light, it's the aftermath of light. – Derek Roberts Jul 16 '14 at 5:09 • Derek Roberts: "How do you account [...]" -- As far as I understand your comment (which, since I first read it, has been removed) the only "accounting" is pretty much summed up in the derivation of the factor$\sqrt{1 - \beta^2}$, which requires (or presumes) mutually equal determination of the value$\beta^2$. The determination of$\beta^2 = 1\$ is only unilateral, since signal fronts don't exchange signals between each other, therefore signal fronts cannot be said "being at rest to each other" (or constituting an "inertial frame"), etc. – user12262 Jul 16 '14 at 5:14