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I am trying to understand the idea that the speed of light is constant for each observer regardless of their motion. But I don't understand how come. I also don't understand how this was measured.

Let's say there is a distance from point A to point B and the way to measure it would be to use a stopwatch (at point A) that snapshots the time when the laser beam was shot and another stop watch (at point B) where it was received. That I understand.

Now, let's say point B is dynamic and it is moving away from point A. Of course it would take a little more time to reach that point B; because during that period of time when the laser was shot and the time when it reached point B, point B moved a little bit already, which means it would take extra time for the laser beam to cover that gap. This experiment still would show that the speed of light is the same.

Let's say that the moving point B is a train wagon and the point B is actually it's front side. Let's say there is another point, point C which is the wagon's back side. The laser beam that would hit point B, would also go through the point C first. Also, let's say there is another stopwatch that tracks the moment when the laser beam passes through that point C. It makes me think that since the distance increases (the wagon is moving) then it would take more time for the laser beam to reach the point B. Which means that for the observer (who sits in that wagon) it would seem that light travels slower!

When it is said that the speed of light is constant for each observer, is it meant the objective light speed is the constant or the perceived by the observer one? If it is the latter then how come it is happening? If it is happening, then there is a mistake in my thought chain.

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    $\begingroup$ Related dummies.com/article/academics-the-arts/science/general-science/… . $\endgroup$
    – anna v
    Dec 8, 2021 at 12:45
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    $\begingroup$ The problem is that the clock carried by the observer in the wagon doesn't tick at the same rate (nor does the internal biological clock of the observer, etc.), because they are in motion - that's the key insight. See also my answer to a related question here, shedding some light on the relativity of simultaneity. $\endgroup$ Dec 8, 2021 at 21:40
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    $\begingroup$ Isn't this a duplicate many, many times over? There is also the Veritasium kerfuffle. $\endgroup$ Dec 9, 2021 at 2:39
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    $\begingroup$ @PeterMortensen This isn't another "one-way speed of light question" à la Veritasium, even though the light is traveling in only one direction. $\endgroup$ Dec 9, 2021 at 2:46
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    $\begingroup$ In relativity, as I mention here, it's important to distinguish between what is observed and what is measured after we compensate for time delays caused by the finite speed of light. $\endgroup$
    – PM 2Ring
    Dec 9, 2021 at 3:18

12 Answers 12

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In order to understand special relativity you must remove from your mind the idea of objective velocity, or time of an event or duration or distance. By doing this, Einstein was able to formulate a consistent theory in which light would always be observed (calculated) to be travelling at the same velocity ($c$) by any observer. There is no point in space that can be considered still in objective terms but only still relative to an observer in the same "inertial frame", i.e. who is traveling at the same velocity.

Light (in a vacuum) is always observed to be traveling at one velocity ($c$) in any experiment, whether it is in your own inertial frame or in another which is moving at high speed relative to yours. This is of course counter-intuitive. In Galilean (common sense) relativity a bullet shot from the front of a fast moving train would have a velocity relative to the track that is the sum of those of the train relative to the track and the bullet relative to the train. That is not the case when a beam of light is fired from the front of a fast-moving spaceship. An observer who is still (not moving with the spaceship) would calculate the light velocity to be $c$, the same as the observer on the spaceship.

To resolve this apparent paradox it is necessary to make use of time dilation, distance contraction, and lastly the non-simultaneity of events separated by distance in the direction of travel in a moving inertial frame. That would include that, clocks at the front and back of a long, fast-moving spaceship, which have been synchronised by a traveller on the spaceship will not be considereded to be synchronised by an observer at rest.

It should be mentioned that the train and the bullet in the above example are not exempt from the special theory of relativity, but their velocities would be such that the relativistic effects on them would be extremely small, and probably too small to measure by any conceivable experiment.

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    $\begingroup$ I would like it explicitly mentioned that in special relativity, the velocity of the bullet fired from the train, as measured from the ground, does not in general equal the sum of the two intermediate velocities. It's implied here, but not outright stated. $\endgroup$
    – Arthur
    Dec 10, 2021 at 7:33
  • $\begingroup$ Well the speed of light is a property of space while the speed of bullet or train is not. So these are actually different things you are trying to relate. What you should expIain is that light travels the same speed in space while observed speed of light is same for all obsrevers irrespective of their frame due to_______. $\endgroup$
    – user316791
    Dec 10, 2021 at 8:59
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    $\begingroup$ For me the key point was realizing that the whole concept of simultanity is relative - i.e. depends on the magnitude and direction of velocity. Then the idea that measuring that speeds are different in different reference systems becomes totally logical and necessary. But losing the concept of "absolute time" is hard - Hollywood will clearly never get it. $\endgroup$
    – Mike Wise
    Dec 10, 2021 at 15:13
  • $\begingroup$ Paragraph added in response to Arthur. Yes the speed of light is a property of space in a sense but that doesn't effect the logic of the answer. $\endgroup$ Dec 11, 2021 at 20:49
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In the frame of the train, the light arrives at Point C, then travels the length of the carriage before arriving at Point B. In the frame of the track, the light has travelled a greater distance, since the far end of the carriage (ie Point B) moves along the track before the light reaches it. Since the speed of light is the same in both frames, that means the time taken for light to traverse the carriage must be greater in the frame of the track than it is in the frame of the train.

The effect is a real one, not some kind of illusion. It arises from the relativity of simultaneity, which means that the plane of constant time on the train is tilted relative to the plane of constant time on the track.

The effect is entirely reciprocal. To see this, imagine standing together with a friend who flashes a light. The light moves off to the left and right at speed c so that at any instant it is equidistant from you in both directions. Now imagine you walk after the light to the right, while your friend remains in position. Since both you and your friend experience the same speed of light, that means that at any instant in your frame the light is equidistant from you, while at any given instant in your friends frame the light is equidistant from them- you each think you are at the centre of an expanding sphere of light. The only way that can work is if you and your friend disagree about what constitutes 'a given instant'.

To illustrate the effect, let's assume light travels a foot per nano second. From your perspective, after a nanosecond in your frame, the light is a foot away from you in either direction. From your friend's perspective, however, light that is a foot away from you to the right has travelled slightly more than a foot, since you have been walking to the right, while the light that is a foot away from you to the left has travelled slightly less than a foot from your friend. Therefore what for you are two simultaneous instants, both a nanosecond after the light has flashed, are two separate instants to your friend, one happening just before a nanosecond in your friend's frame and one happening just after.

Likewise, when your friend sees the light a foot away in each direction after a nanosecond in their frame, you will see the light slightly further away to the left and slightly closer to the right, so that will happen at two separate instants in your frame.

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  • $\begingroup$ Yes, that is right. The relativistic Doppler effect depends upon whether you are travelling away from, or towards, a source of light. In the first case, time signals you receive appear to be slowed down- in the second they seem to be speeded up. $\endgroup$ Dec 10, 2021 at 10:42
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When it is said that the speed of light is constant for each observer, is it meant the objective light speed is the constant or the perceived by the observer one?

It means that the speed of light measured by each observer has the same value, about $3\times 10^{8}$ m/s.

In your example where $A$ is on land and $B$ and $C$ are on the moving train. If $A$ sent out a pulse of light from a laser beam, and could measure the speed of light by, for example, measuring the time for part of the beam to cover the first meter, then $A$ would measure $3\times 10^{8}$ m/s.

The beam would then pass $C$ on its way to $B$. If $C$ could measure the speed of the pulse as it passed, by measuring the time for part of the beam to cover the first meter, then $C$ would also measure $3\times 10^{8}$ m/s.

Similarly when the pulse arrived at $B$, he/she would also measure the same speed.

If it is the latter then how come it is happening? If it is happening, then there is a mistake in my thought chain.

It's a postulate of special relativity that all observers will measure the same speed of light irrespective of their relative motion.

At first it might seem impossible, and that's why you might think that there is a mistake in your chain of thought - but a consistent theory was developed by Einstein including the postulate.

Apparent contradictions are resolved due to time passing at different rates for different observers, changes to our notion of simultaneity, length contraction etc...

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  • $\begingroup$ But why is it happening? It is clear that if you are stationary then it will take less time for somebody to overtake you. But if you are walking, then that somebody will overtake you a little later. And if you are running pretty fast, that person will overtake you even yet later. Why this is not the case with light? Why even if you travel fast, the light overtakes you the same speed as if you were stationary? $\endgroup$
    – pro100tom
    Dec 8, 2021 at 14:16
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    $\begingroup$ @ pro100tom it's a good question, ("Why this is not the case with light?") We (as humans) have evolved experiencing mainly slow speed events. Sometimes it's best (also in quantum theory) to accept that nature doesn't behave as we would expect from our limited experience. Also we have to realise that there was no justifiable way to say $B$ and $C$ were in motion and $A$ was stationary, from $B$ and $C$'s point of view $A$ is moving and they are stationary (that's more or less the other postulate of relativity), so maybe it's natural that they would measure the speed of light to be the same. $\endgroup$ Dec 8, 2021 at 14:47
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    $\begingroup$ To your commented question, the light does not overtake you in the same time nor does it travel the same distance, when viewed by different observers. But you do measure the same speed of light $\endgroup$
    – RC_23
    Dec 8, 2021 at 15:22
  • $\begingroup$ This doesn't expalin the inverse situation. Can you explain for when the carriage moves towards the source of light? In this case, the way you all explained, the clock should tick fatser. And this oppostie from Einetins theory $\endgroup$
    – user316791
    Dec 10, 2021 at 9:11
  • $\begingroup$ @Aresenal Creation is your comment about the answer or about one of the comments above? $\endgroup$ Dec 10, 2021 at 16:29
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I think that it might be helpful to take a step back from the math, and re-evaluate the question itself.

There is no why or how to the constancy of the speed of light: it is an empirical observation. The job of physics is to construct a theory of motion that fits observations of the real world as best as it can. For hundreds of years, the best theory was Newtonian mechanics, which you describe in your question. In the 19th century, as more precise measurements were made, it became clear that, contrary to the predictions of Newtonian mechanics, all measurements of the speed of light turned up the same number, regardless of the relative velocity of the origin and observer.

Einstein started with the assumption that the speed of light is constant to all observers, and worked backwards to figure out what laws of motion would lead to such a phenomenon while not contradicting other observations about the world. This requires discarding the straight-forward, intuitive model of how a bullet fired from a moving train behaves, and starting over: masses in motion do not have an "objective" velocity, you can't just add relative velocities together, and not even the passage of time is consistent from observer to observer.

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But why is it happening?

I suggest that you look at Einstein's paper On the Electrodynamics of Moving Bodies. Part I of the paper is quite accessible. It is the key to much of 20th century physics, because it introduced the idea that you need to measure something before you can really talk about it. If you want to measure velocity, you need to measure distances and times. If you set out from somewhere, say a specific door of the Empire State Building, we define that place to have coordinates (0,0,0), and the time that you set out is T=0. Easy. Now when you get to somewhere else, we measure the time at that place, but what does that actually mean? Einstein developed an ingenious method for synchronizing clocks at different places using light signals: On the Electrodynamics of Moving Bodies is about the consequences of following this method. NB: Einstein doesn't attempt to answer your question, though; he makes the constancy of the velocity of light into a postulate; we don't ask why all right angles are equal in Euclidean geometry, and we don't ask why the velocity of light is constant. How do we know that special relativity is a correct description of the world? It is the simplest theory that is consistent with experiment. Fast forward 20 years, and Heisenberg invented quantum mechanics by analogy with Einstein's 1905 paper: if you can't measure it (even in principle), you have no business talking about it.

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Okay, look at it this way. Imagine a train that is 300,000 km long and there are clocks located at the front and rear. They have been synchronized.

Now the train whips by the train station at 260,000 km per second.

Those at the train station will notice that the clock at the rear of the train is ahead of the clock at the front by 0.866 of a second, and they will see that the length of the train at this speed has shrunk down to 150,000 km, and they also will notice that the clocks on board the train are ticking at half speed.

Those on board the train still think that the two clocks are still synchronized, and they still think that the spatial length of the train is 300,000 km long.

And so if a burst of light is sent from the rear to the front of the train, the observers at the train station will see that it takes about 3.73 seconds for the light to reach the front of the train.

This is because the light is only going roughly 40,000 km per second faster than the train. Thus 150,000 km divided by 40,000 km per second, equals 3.73 seconds.

But for those on board the train, 3.73 seconds becomes 1.866 seconds since their clocks are ticking at half speed. But due to the clock offset of 0.866, they will measure 1.866 - 0.866 which equals 1.000 second. Now since they still think that the spatial length of the train is still 300,000 km, they think that the light had crossed 300,000 km in one second, hence the speed of light.

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You just stumbled upon one of the ultimate questions of out universe, that is, why do massive observers see massless particles (like photons that make up EM waves) propagate always at the same ultimate speed (that we call the speed of light) regardless of the motion of the observer?

the speed of light was created by Nature to be one, the number whose multiplication influences nothing. But the primitive people who lived in spacetime and moved by speeds much smaller than c=1 - along small angles in the spacetime - were not able to see that their speeds were particular fractions of the maximum speed.

The origin of the value of speed of light in vacuum

Intuitively it is very hard to understand, if you are standing still on Earth, and shoot a light beam, or you are in a train that is moving (in your example) and shoot a light beam, why will both beams seem to propagate at the same ultimate speed, as seen from both inertial frames (standing on Earth or moving with the train)?

If you want to intuitively understand why this is so, there are two very important things:

  1. massive objects always move at a certain speed that is a relative fraction of the ultimate speed of massless particles

Originally in the photon epoch, there were only massless particles and they were all zipping around at the same and only one ultimate speed, that we call now the speed of light. Certain particles gained rest mass, slowed down in space, and built us up, and now as we massive observers look around, see all massless particles zip around the this same ultimate speed still. Regardless how fast a massive observer is moving, its speed will always be relative to the ultimate speed, the observer will always move at a certain fraction of the ultimate speed.

  1. Massless particles/fields propagate independently of the source

Massless particles that build up for example EM waves, will always propagate independently of the source, this is very important, because no matter the speed of the massive object that emitted the massless particles, the radiation will propagate away from the source at the ultimate speed regardless of the original speed of the emitting object at the moment of emission.

Intuitively, you would think that when you put your hand into the pond, the waves will propagate (away from your hand) at a speed that depends on the original speed of your hand when you disturbed the water.

Though, with EM waves, it is different, because the speed of the radiation (wave) that spreads from the source (charge) that disturbed the fields is independent from the original speed of the source. The radiation will always propagate at the only one ultimate speed, the speed of light.

So there are two things to bear in mind:

  1. a massive observer is always moving at a speed that is a relative fraction of the speed of light, and will always see massless particles move at this ultimate speed

  2. EM waves (and all massless particles) always move independently of the source, at the speed of light, regardless of the original speed of the object that emitted them

The ultimate message of relativity is that speed (of massive objects) is relative, but the speed of light is absolute, and the massive (inertial) observer will always see the massless particles propagate at the speed of light.

This is the ultimate answer to your question and we do know that relativity is the correct framework because all our observations and experiments are best described by it.

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I will present a diagram (called a spacetime diagram) and explain what it shows. You will have to trust that I have constructed the diagram correctly. My aim is to show how it can be that two observers in motion relative to one another along a line can nevertheless find that something else (here a light pulse) moving along that same line can have the same speed as observed by both of them.

enter image description here

The diagram shows a region of spacetime with two clocks and a light pulse. Clock 2 is moving to the right relative to clock 1. Clock 1 is moving to the left relative to clock 2. The light signal is moving to the right relative to both clocks. At event A clock 1 registers 1 second since the light went past both clocks. At event C clock 2 registers 1 second since the light went past both clocks.

Event B is simultaneous with A according to an observer travelling with clock 1.

Event D is simultaneous with C according to an observer travelling with clock 2.

The main thing to note is that the lines AB and CD have the same length. This means that the two observers agree about the distance travelled by the light signal during 1 second.

That's it! That's how two observers in relative motion can find that a light signal has the same speed for both of them. The reason it is surprising to our intuition is that we are not used to simultaneity being different for observers in different states of motion. Newtonian physics would say that the lines AB and CD ought to be both horizontal on the diagram, and then they would have different lengths. That Newtonian way of constructing a notion of simultaneity does not match with what is observed however.

Note that for this result it is not necessary to mention either time dilation or Lorentz contraction. It is only necessary to understand how simultaneity is modified.

In constructing the above diagram I have used a symmetric set-up, where the two clocks have equal and opposite velocities relative to any observer whose worldline is vertical on the diagram. This symmetry is why one can claim that equal distances up the two clock worldlines must represent equal amounts of elapsed time for each. This is an example of the principle of relativity. Owing to this same symmetry the distance scales along the two lines of simultaneity in the diagram must agree. A question in the comments asks how distance is measured along these lines. Since they both have the same scale you can simply use a ruler. The line AB answers the question "how far away is the photon after 1 second, for observer with clock 1?" The line CD answers the question "how far away is the photon after 1 second, for observer with clock 2?"

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  • $\begingroup$ Andrew Steane: "I will present a spacetime diagram and explain what it shows. You will have to trust that I have constructed it correctly." -- That's fair warning but ... "[...] The main thing to note is that the lines AB and CD have the same length." --... Better to specify how to measure "length" at all, of lines in your diagramm, as well as in general. Also, the OP introduced capital letters "A", "B" ... as names of "(material) points". We should stick to this convention. (In order to identify and denote individual events, we can name their respectve coinciding participants.) $\endgroup$
    – user12262
    Dec 10, 2021 at 8:08
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    $\begingroup$ @user12262 I added some remarks to address this. $\endgroup$ Dec 10, 2021 at 23:57
  • $\begingroup$ Andrew Steane: "I added some remarks [...]" -- Well -- too bad you didn't change your figure: to add representations of additional material points (in order to point out pairs which are at rest wrt. each other, to which a specific value of distance could be attributed) and to denote all of them by capital letters (following the OP convention). (And a real shame that the available MathJax version doesn't seem to support "pstricks"-commands ...) To be fair, I should post here an answer of my own; and I intend to, hopefully within the next few days (although likely without any figure). $\endgroup$
    – user12262
    Dec 12, 2021 at 22:58
  • $\begingroup$ I really like your explanation, however I curious why the event D that is simultaneous to the event C is so "ahead" in comparison to the the A and B pair? Does it mean that the smaller the angle between the direction of the clock and the beam, the more "ahead" the event will be perceived as simultaneous? Or is rather something to do with the speeds at which clocks move? $\endgroup$
    – pro100tom
    Dec 14, 2021 at 11:45
  • $\begingroup$ Also what does "opposite velocities relative to any observer whose worldline is vertical on the diagram" mean? Geometrically I can see this. But what would be the velocity of the observer with the vertical worldline? And what does "opposite velocities" mean in reality? $\endgroup$
    – pro100tom
    Dec 14, 2021 at 11:54
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To me, the key to understand special relativity was to view it as co-ordinate transformation, together with the notion that space and time aren't distinct things, but closely related.

Bear with me, take yourself a piece of paper, and follow me.

(To the experts: I'll first use a model that ignores the imaginary-number property of the Lorentz transformation)

Take your piece of paper and draw a co-ordinate system with t (time) horizontally to the right, and x (positions) vertically upwards. That's the "reference frame" that you'd like to call "static", non-moving.

As a non-moving observer at the zero position, you stay at the same position all over time, so your "track" is along the t axis.

Now a second person is moving with a speed of e.g. 0.1c in positive direction, then his track is a sloped straight line with increasing x for increasing time. If you scale the axes so that 300000km are as long as 1sec, then you get a slope of 5.71 degrees.

Special relativity says that this second person can't feel that he's moving, so he'd think that his track is the "correct" t axis, from his point of view. So he lives in a different co-ordinate system, one with its t axis rotated by 5.71 degrees against your "static" one.

Let's have a look at the other axis. Your x axis doesn't form a 90 degrees angle with his t axis any more. So, to get a valid "moving" co-ordinate system, we have to rotate the x axis as well.

What does that mean? The x axis is all the events (spacetime points) that happen at the same time. As you and the other person now disagree about the x axis, it means that things that happen "at the same time" for you, have a time difference for him, if they happen at different x locations.

Now, if you shoot a ray of light from the zero point and observe it at some distance, then you and the other person will not only disagree about the distance travelled, but also about the time it took. And it happens for the speed of light that these differences in both distance and time compensate one another, so that we always observe the same quotient.

(And here come the unintuitive parts of the transformation)

One question remains:

  • What is the correct scaling factor between space and time? The one that explains the experiments showing a constant speed of light, no matter how fast the observers move?

It turned out that no real number, used as scaling factor, did the trick, but a complex number did (speed of light, multiplied with the imaginary unit). With this factor, the co-ordinate transformations got the special property that the 45-degrees speed-of-light diagonal on our sheet of paper always stays a diagonal (okay, that is counter-intuitive, but if it were intuitive, it wouldn't have taken an Einstein to discover it).

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  • $\begingroup$ This is a great answer and I think I understand it! I've got a question regarding the "45-degrees speed-of-light diagonal". What do you mean "it always stays a diagonal"? Do you mean, it means no matter what is the frame of reference, after the calculations the diagonal remains at 45 degrees? $\endgroup$
    – pro100tom
    Dec 14, 2021 at 12:06
  • $\begingroup$ Also, what do you mean by "scaling factor between space and time"? Do you mean the factor by which they dilate depending on the "tilt"? And finally (but not really important). That 5.71 degree angle; how did you define the scaling of the axis so that it gets to the 5.71. One can mark distance whatever one likes and the slope angle would change then as well (on the paper) $\endgroup$
    – pro100tom
    Dec 14, 2021 at 12:09
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    $\begingroup$ The scaling factor is basically the scaling of the axes: do we draw 1 second with the same length as 1 meter, 1 kilometer, or 300000 km? If you use a scaling where 300000km as the same drawing length as 1 second, then the speed of light is a 45-degrees diagonal. $\endgroup$ Dec 15, 2021 at 9:44
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A and B start in the same inertial frame of reference.

Observer A has a meter stick and synchronized clock.

Observer B has a meter stick and synchronized clock.

A and B observe a distant source of light or electromagnetic radiation. Both specify the same wavelength and frequency for the source. Speed is wavelength times frequency so A and B specify the same speed of light.

Taking their respective measurement instruments, which agree in the frame of reference, A travels toward the source at 0.5 the speed of light, and B travels away from the source at 0.5 the speed of light. A and B will both measure the speed of light as constant, neither will think the meter stick or clock have changed length or tick period in their respective frame of reference, but their respective measures of length and time are no longer equal to each other as they were when A and B were making observations in the same frame of reference.

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The intuition can be obtained if you understand the retarded time. For a stationary body producing light and another body moving close to C , you are evaluating the field at the retarded time in a sphere around the emmiter thus the moving body is moving out of the horizon of the retarded time bubble, thus I measure that it takes longer for the field to be updated for the moving body. But from the moving bodies perspective When the emmiter produces an acceleration in his perspective, he is stationary and the emmiter is moving close to c so at the instant it produces light, the retarded time bubbles move out from that initial position . in the stationary frame it takes longer for light to reach the moving body as he is moving out of the horizon of the retarded time.bubble, but in the moving frame, he is stationary relative to that initial retarded time bubble. thus we can also conclude that the time has slowed down for the moving body as e.g light takes 1 second to reach him, but 15 seconds for the stationary observer

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I've always found it helpful to remember that the observed photons may have the same velocity c, but different energies/momentums. Hence blue/red shifting.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Dec 10, 2021 at 8:07

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