When we say that something is travelling a certain speed, it's really travelling that speed relative to the Earth. When saying the speed of anything, it is, for the most part, relative to something else. That being said. If I have an object moving at half the speed of light, and another moving at just above half the speed of light in the opposite direction, would the second object be moving faster than the speed of light to the first one?

Note: I know this question is similar to some other questions like this one. However, with my limited physics knowledge (taking AP Physics class next year) I found the explanation a bit confusing. So, even though this might be a bit similar to other questions, I'm looking for a simpler explanation that could help me understand this and its foundation.

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    $\begingroup$ Velocities that are a significant fraction of the speed of light are not additative, so in your example the relative speed is still less than c. $\endgroup$ – Lewis Miller Mar 20 '16 at 2:02
  • $\begingroup$ @LewisMiller How do we know that they aren't additive? Why aren't they additive? $\endgroup$ – mdlp0716 Mar 20 '16 at 2:11
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    $\begingroup$ Why would they be additive? $\endgroup$ – CuriousOne Mar 20 '16 at 2:51
  • $\begingroup$ @CuriousOne I mean i would think so because velocities are relative. I don't know which situations there are where this isn't the case though. I'm open to being educated on this. $\endgroup$ – mdlp0716 Mar 20 '16 at 3:06
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    $\begingroup$ "Relative" means that everything is symmetric between any two observers, it doesn't tell us that the relations between three are additive. The reason for my question is, as you may have noticed, psychological, and I don't mean to hurt your feelings with that. Why is it so hard to let go of Galilean kinematics? Because we just spent so much time and effort on learning it! And then Prof. Einstein comes along and tells us that everything we had to learn in school is wrong... isn't that unfair? :-) Yes, it is, and trust me, absolutely everybody struggles with the "unlearning" process. $\endgroup$ – CuriousOne Mar 20 '16 at 3:14

In Special Relativity, we use Lorentz Transformations to add speed. The relevant formula here is $$u = \frac{u^{'}+v}{1+\frac{u^{'}v}{c^2}} $$ where $u^{'}$ and $v$ are the speeds of the objects and $u$ (what you are looking for) is the speed that the object at $u$ sees the object at $v$.

This embodies Einstein's postulate that no information can be transferred faster than the speed of light in vacuum. Now using this formula, we can put $u^{'} = 0.5c$ and v = $0.6c$ and still get that $u = \frac{1.1c}{1.3} = 0.85c$. Note that even if $u^{'}=v=c$ we get $u=c$, which tells us that the speed of light in vacuum is the same for all observers (which is really the more-precise text of Einstein's conjecture)

Note what you have learned that you can just add the two speeds up is only a good approximation when $v<<c$.

P.S. See e.g. here for a simple derivation of the formula, which we get from using the lorentz transformations of time and position. It is only after we realise that time and space can be a combination of each other that we arrive at this.

  • $\begingroup$ Thank you, this is along the lines of what i was looking for. I'm still confused though as for why this is the case. I'll probably understand it more next year when I take that physics class. $\endgroup$ – mdlp0716 Mar 20 '16 at 14:26
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    $\begingroup$ It is because space transforms as $x^{'}= \gamma (x-vt)$ and time transforms as $t^{'}= \gamma(t-\frac{xv}{c})$. and the speeds' relationship (called Einstein's velocity addition) then fall off from that. It is indeed non-intuitive and the class would surely help. $\endgroup$ – hsnee Mar 20 '16 at 14:37

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