# Do light beams need to be seen to be observed?

I thought I had understood special relativity, at least the idea of the clocks with light pulses and rods from the book The Meaning of Relativity. But going through the basic illustration of the time dilation made me think otherwise.

I don't get one idea in this illustration. What is confusing for me is this sentence: "Keeping the speed of light constant for all inertial observers, requires a lengthening of the period of this clock from the moving observer's perspective".

I have understood that the axiom of special relativity is that the if a pulse of light is reaching the observer's frame of reference, then he would measure the speed of this light as c irrespective of the speed of the source (thinking about the Michelson-Morley experiment). However, in this illustration, the pulse of light they are talking about is the pulse travelling along the line D in the picture. This pulse never reaches the observer.

Then how can one invoke the original special relativity axiom in this context?

• Read What is time dilation really? Messing around with light clocks and pulses of light is a very poor way to understand time dilation. Aug 15, 2018 at 16:30

Forget about observers altogether. The idea of observers is totally unnecessary and often a cause of misunderstandings.

The speed of light is the same in every inertial reference frame, regardless of whether or not anyone observes it.

In the frame of the light clock, the light travels vertically up and down the clock. In the frame in which the clock is moving, the light travels on a longer angled path. Given that the light travels further in the second frame than it does in the first, the time interval between the departure and the return of the light must be greater in the second frame than in the first. Ergo time dilation. No observers are needed.

• Thanks for the late answer. I understood it finally, after I discussed your answer with a friend of mine, together with the first answer above I was always wondering what would happen then if the information could travel instantaneously between a system and an inertial observer travelling with a large constant speed. But then finally I understood that if that is the case, then information (or light) could also instantaneously between those mirrors in the system itself, eliminating the right angled triangle altogether. So, I get that the having a different observer or not has no consequence Dec 12, 2022 at 13:34

Let's give the observers names to make it easier to sort out what is happening. Olivia is the observer in the left-hand figure in your link, and Sarah is the observer in the right-hand figure in your link.

Olivia is using a single clock to measure the time needed for the light to travel along path $D$. Sarah is using two clocks, both stationary, one located where the pulse is generated and the other located where the pulse is received.

Sarah's location in her reference frame isn't entirely relevant, but we can suppose she is located at the position of the first clock, where the light pulse was emitted. She receives a signal transmitted from the second clock, where the light pulse was received. That signal records both the time of reception and the position of the second clock. Sarah then knows the distance between the clocks in her frame, and compares the timestamps to get $\Delta t$ in her frame. (Note that it is not necessary that the speed of the transmitted signal be the speed of light, although we usually imagine that it is.)

However, in this illustration, the pulse of light they are talking about is the pulse travelling along the line D in the picture. This pulse never reaches the observer.

The light doesn't have to 'reach the observer' in order to be observed. There's a (vast) difference between seeing and observing within Special Relativity.

From the Wikipedia article Observer (special relativity):

Physicists use the term "observer" as shorthand for a specific reference frame from which a set of objects or events is being measured. Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from. The effects of special relativity occur whether or not there is a sentient being within the inertial reference frame to witness them.

The measurement of the speed of light along the path $$D$$ is (conceptually) made with rods and clocks at rest in that inertial reference frame moving with respect to the light clock.

• I understand this answer, but I get further confused. I understand the first type of observer because she is having a definite relative velocity w.r.t the moving frame and hence has to resort to the light to get information from the moving frame to see and measure what is going on in the moving frame. Here the invariant speed of light bringing out the adjustments in time and space measurements. But I fail to understand the second type of observer. The axiom only talks about what happens between two frames. It says nothing about this second observer. Aug 16, 2018 at 10:40
• @PermanentGuest, maybe this quote from Schutz will help? "It is important to realize that an "observer" is in fact a huge information-gathering system, not simply one man with binoculars. In fact, we shall remove the human element entirely from our definition, and say that an inertial observer is simply a coordinate system for spacetime, which makes an observation simply by recording the location (x,y,z) and time (t) of any event." Aug 16, 2018 at 13:50