Recently, relativistic physics has sparked interest in me. I read in one of my textbooks that the classical addition of velocities does not apply to light but the explanation given in the book is pretty incomprehensible. Somebody please explain it doesn't apply to light in layman's terms.
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1Without knowing what the book has to say on the matter we can't really clarify that for you, it would be better to include the books information (like title and author) and a quote of the explanation given. – Triatticus Feb 19 '22 at 20:19
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10It doesn't apply to anything. The closer speeds get to the speed of light, the more noticeable this becomes. – Connor Behan Feb 19 '22 at 20:33
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1It is fundamentally because there is no reference frame in which light is stationary. The specific situation that the velocity addition formula is valid for to can not exist for light! – m4r35n357 Feb 19 '22 at 20:38
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1What Connor said. It comes down to spacetime geometry. As I said here, velocity is the spacetime slope of a worldline. But I guess that's not very helpful when you're just starting to learn about relativity. – PM 2Ring Feb 19 '22 at 20:51
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2@PM2Ring Velocity is the slope on a position-vs-time graph, which should be familiar to anyone who has taken introductory physics. – robphy Feb 19 '22 at 21:09
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Actually, the normal rules of vector addition apply when calculating relative velocity as long as you know that motion is discrete by nature. That's exactly what I did in this preprint: https://osf.io/4msb3/ – Agona Apr 06 '23 at 22:16
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the classical addition of velocities does not apply to light
Actually, if that is what the book said, then it did not go far enough. It should have said that “the classical addition of velocities does not apply”. It is true that it doesn’t apply to light, but it doesn’t apply to anything else either.
The relativistic velocity addition formula is $$v’=\frac{u+v}{1+uv/c^2}$$ This is the correct formula that applies for velocity addition in general. As you can see, if $uv/c^2$ is negligible then this formula is approximately equal to the classical one. But whenever $uv/c^2$ is big enough to notice then this is the correct formula, even if neither $u$ nor $v$ is equal $c$
Dale
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Despite the votes, I don't believe this answers the real question. The question is why they are not additive, the relativistic formula is a consequence of them not being additive. – Alexander Sep 18 '22 at 02:19
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The speed of light has to be the same for everyone. If you travel with a speed near to the speed of light away from me, and I send a light beam in your direction, you won't see light moving with a smaller velocity towards you. You don't subtract your speed from the light speed. Instead, you see light travelling towards you at the speed of light too. You will see a different frequency though. Can you imagine a light beam standing still or moving slowly? Indeed it doesn't. It's the same for everyone and that's the base of special relativity. Velocities add in a different way in special relativity, and space and time are transformed between different frames moving with a constant relative velocity. So, $c+c=c$.
Arc
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In special relativity velocities do not add up as in classical mechanics. Consider for example two objects moving towards each other with speeds $v_1$ and $v_2$ in the observer frame of reference. If the observer travels along with object 1, that is, in the rest frame of object 1, classical mechanics gives the speed of object 2 is $$v_{rel}= v_1 + v_2 \,.$$ In relativistic mechanics this speed is $$v_{rel}= \frac{v_1 + v_2}{1+v_1 v_2/c^2} \,.$$ This formula does not permit $v_{rel}$ to exceed $c$.
my2cts
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"This formula does not permit vrel to exceed c." but you can't tell it to tachyon :( – Billy Istiak Feb 20 '22 at 07:18
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@BillyIstiak Even tachyon do not travel faster than light. "In the 1967 paper that coined the term, Gerald Feinberg proposed that tachyonic particles could be made from excitations of a quantum field with imaginary mass. However, it was soon realized that Feinberg's model did not in fact allow for faster-than-light particles or signals and that tachyonic fields merely give rise to instabilities, not causality violations. Nevertheless, in modern physics the term tachyon often refers to imaginary mass fields rather than to faster-than-light particles." https://en.m.wikipedia.org/wiki/Tachyon – my2cts Feb 20 '22 at 08:24
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What kind of velocity would you measure for yourself if you are moving along with yourself? I only understand what $v_1$ and $v_2$ would be if the observer is at a different location, if the observer travels along with object 1 then $v_1$ would be zero? – Emil Feb 20 '22 at 08:32
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I think you should look at it from the other direction: what is "addition of velocities" and why would it ever make sense? You can add many things, but the result isn't necessarily sensible or useful.
You can make this argument for taking sums of velocities: if you have two trains traveling toward each other with speeds $v$ and $w$, then after a time $Δt$, the distance between them has decreased by $vΔt$ on one end and $wΔt$ on the other, for a total of $(v+w)Δt$, which gives a clear meaning to $v+w$. If they're a distance $d$ apart at time $0$ then they collide at time $d/(v+w)$.
As long as you use a consistent definition of distance and time, this argument is valid even in special relativity. It continues to be valid if you replace the trains by beams of light. If they start a distance $d$ apart, they meet at time $d/(2c)$.
Where this doesn't work is where you have more than one standard of distance and time – where you have clocks and metersticks attached to the ground and also clocks and metersticks attached to the trains, and you use both to measure velocity. The argument from before fails to go through because, e.g., it assumes the existence of a "time $Δt$", but we no longer have a single time standard. It turns out that the conclusion of the argument is also wrong, and when you have velocities measured to different standards, the relationship between them is more complicated. As Dale's answer said, the relationship is different for all speeds, not just for light speed.
benrg
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If the velocity of light were additive with that of its source, then there would exist a frame of reference in which we would see effects precede their causes, which feature our universe does not exhibit.
For example, imagine a speeding fire truck coming straight towards you down the road with all its warning lights on. Between you and the truck, there is a slow-moving car approaching the road on a perpendicular side street and it is on a collision course with the fire truck i.e., they will occupy the intersection at the same moment.
Now imagine that the velocity of light is added to that of its source.
The driver of the truck swerves to miss the car in the intersection, but because the speed of the truck adds to that of the light coming from its lights, you will see the light from the speeding truck before you see the light from the car, and you will therefore see the truck swerve before the car has entered the intersection.
niels nielsen
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Note : speed of light is always constant. And nothing can ever travel faster than the speed of light.
If you can remember it then you will think that classical addition rule isn’t obeying the postulates relativity. Let's take a look at math.
Assume that you are in a train and that train is traveling at $0.98c$ and you are also running in the train with velocity $0.55c$(It's nonsense. But just suppose). So your total velocity will be(from classical addition rule or using galilean tranformations). $$v_{2nd}=0.98c+0.55c$$ $$=1.53c$$ If someone outside from the train look at you then he should think that you are traveling faster than the speed of light. So you aren’t obeying the postules of relativity.
nothing can travel faster than the speed of light.
But if you plug your velocity in relativity's addition rule of velocity. Then you will find: $$v_{2nd}=\dfrac{v_1+v_2}{1+\dfrac{v_1v_2}{c^2}}$$ $$=0.9941....c$$ You will tend the speed of light but never reach it.
OK! Let's see if it agrees for light. Again, assume that there's light beam in a moving train and velocity of that moving train is $0.99c$
$$v_{2nd}=\dfrac{1c+0.99c}{1+\dfrac{0.99c\cdot 1c}{c^2}}$$ $$=1c$$ Is that magic? Nope, that's science. All observer will see that light always travel at $c$. For verification, I wanted to do another nonsensical thought experiment. In this case, I took a light-like train. Means, train is traveling at the speed of light. $$v_{2nd}=\dfrac{1c+1c}{1+\dfrac{1c\cdot 1c}{c^2}}$$ $$=c$$
Billy Istiak
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0.98c according to whom? This is always what stops me from understanding these kinds of answers... rarely stated who measured that the train moved at 0.98c and how it was measured... – Emil Feb 20 '22 at 08:23
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@Emil If you see 0.98c in speedometer of train then what will you say? – Billy Istiak Feb 20 '22 at 11:05
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Depends on how the speedometer works. Does it work by calling the rest api of your external observer friend which asks him to use his external laser speedometer? – Emil Feb 20 '22 at 12:58
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@Emil Have you ever seen speedometer in a bus or car? Do you know how does that work? :| If you can't come to conclusion by answering these question. Than I am not interested in further discussion. – Billy Istiak Feb 20 '22 at 13:03
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Or does it work by counting wheelturns and measuring the wheel diameter inside the train frame? (Because of length dilation the wheels won't be the same diameter for an outside observer, so need to specify by whom they are measured) – Emil Feb 20 '22 at 13:05
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Most speedometers probably assume the wheel diameter is the same as when it was built in the rest frame, so the regular speedometer is probably just showing nonsense. – Emil Feb 20 '22 at 13:12