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I've been looking for an answer to this question for a while now and I've seen the mathematical explanation

$$Vr=[V1+V2]/[1+V1*V2/c²]$$

I understand that the result cannot be greater than $c$ because that would break the laws of physics, but I'm stuck when it comes to how the speed can be calculated. Using that equation, for $v1$ and $v2$ being just under $c$, the result would be still under $c$ but just a bit closer. What space and time (relative to one of the particles) can be used to calculate that speed? (In a $d/t=Vr$ kind of way). How can I calculate those values? I've tried using the Lorentz factor but I've either used it wrong or miscalculated.

Any help will be appreciated, thanks!

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    $\begingroup$ I don't understand your question. It seems like you already have the formula for addition of velocities. Is there anything unclear about that formula? $\endgroup$
    – Allure
    Mar 28, 2022 at 9:27
  • $\begingroup$ @Allure what I'm trying to say is that I do know how to get the value of the speed, but I can't wrap my head around the concept. I can get the value of the speed of one of the particles (relative to the other particle), but what space over what time gives that value? Is it the time and space that one of the particles experiences? $\endgroup$ Mar 28, 2022 at 9:35
  • $\begingroup$ Both particles have a speed, both of them travel a certain distance in a given amount of time. One of them is slightly faster than the other, because it travels a slightly larger distance in the same amount of time. I still don't really understand your question, I'm afraid. $\endgroup$
    – Allure
    Mar 28, 2022 at 9:47
  • $\begingroup$ The particles velocities $ v_1$ and $v_2$ are measured in the same inertial frame, in the inertial frame's time and space coordinates. en.wikipedia.org/wiki/Inertial_frame_of_reference $\endgroup$
    – anna v
    Mar 28, 2022 at 10:01
  • $\begingroup$ @Allure I'm sorry, let me set an example. Let's say particle 1 and particle 2 both move at 0.99c. If we set the frame of reference to be at one of the particles (particle 1), we can see that the relative speed of the other (particle 2) is 0.999949c. If speed is space over time, what space and time gives 0.999949c? Is it the values that are relative to that particle? $\endgroup$ Mar 28, 2022 at 11:15

3 Answers 3

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Based on your comments, I think you are looking for a conceptual explanation, so try this.

Let us take the speed of light to be 300,000 kilometres per second.

Suppose I am moving away from you at 0.5c. After a second in your frame, I will be 150,000km further away from you. Suppose my friend is moving in the same direction as I am, at 0.5c relative to me. After a second in my frame, my friend will be 150,000km further away from me. However, after a second in your frame, my friend will only be 240,000km further away from you because his speed relative to you (according to the velocity addition formula) is only 0.8c.

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I'll work throughout with $c=1$.

Velocities are $3$-dimensional and in general not parallel, but you've asked about an already very informative special case in which they are parallel, so can be treated as $1$-dimensional and hence just numbers. If Bob has speed $\beta_1$ relative to Alice and Charlie has speed $\beta_2$ to Bob, and the aforementioned assumptions apply, Charlie's speed relative to Alice is $\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}$. But that formula may not be obvious, it's stay-under-speed-$1$(-$c$) consequences may not be obvious and may just look like a mathematical accident once deduced, and it feels like there should be something that adds the way we used to expect speeds would. I think that may what be troubling you.

Allow me to introduce rapidity. To go from Alice's reference frame to Bob's, use the $2$-dimensional Lorentz transformation $\left(\begin{array}{cc} \gamma_1 & \gamma_1\beta_1\\ \gamma_1\beta_1 & \gamma_1 \end{array}\right)=\left(\begin{array}{cc} \cosh\phi_1 & \sinh\phi_1\\ \sinh\phi_1 & \cosh\phi_1 \end{array}\right)$ with $\beta_1=\tanh\phi_1$. We call $\phi_1$ a rapidity. Rapidities add because$$\left(\begin{array}{cc} \cosh\phi_1 & \sinh\phi_1\\ \sinh\phi_1 & \cosh\phi_1 \end{array}\right)\left(\begin{array}{cc} \cosh\phi_2 & \sinh\phi_2\\ \sinh\phi_2 & \cosh\phi_2 \end{array}\right)=\left(\begin{array}{cc} \cosh(\phi_1+\phi_2) & \sinh(\phi_1+\phi_2)\\ \sinh(\phi_1+\phi_2) & \cosh(\phi_1+\phi_2) \end{array}\right)$$(proof is an exercise), in analogy with composing $2$-dimensional rotation matrices (indeed, you may see the above matrices called hyperbolic rotations). So the final speed is$$\tanh(\phi_1+\phi_2)=\frac{\tanh\phi_1+\tanh\phi_2}{1+\tanh\phi_1\tanh\phi_2}=\frac{\beta_1+\beta_2}{1+\beta_1\beta_2}.$$So

"adding" subluminal speeds gives subluminal speeds

can be rephrased as the much more obvious

literally adding finite rapidities gives finite rapidities.

As for more general cases where we have to bear in mind velocities' general directions, the computation of what happens is still done by multiplying Lorentz transformation matrices. So the fact that speeds remain subluminal just means such matrices are closed under multiplication, which follows from their defining equation $\Lambda^T\eta \Lambda=\eta$.

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$V1$ is the speed of one particle A relative to an observer O. $V2$ is the speed of the second particle B relative to the first particle A. $V_r$ is the speed of B relative to O. In Newtonian mechanics we would have

$$V_r = V1 + V2$$

and so if $V1$ and $V2$ were close to $c$ then $V_r$ could be greater than $c$. However, we know (from experiment) that $V_r$ never exceeds $c$ and that $V_r$ is in fact given by

$$\displaystyle V_r = \frac {V1 + V2}{1 + \frac {V1 \times V2}{c^2}}$$

If $V1$ and $V2$ are both close to $c$ then the second term in the denominator is close to $1$ and we have

$$\displaystyle V_r \approx \frac {V1+V2}{2}$$

However, if $V1$ and $V2$ are both small compared to $c$ then the second term in the denominator is very small and we have

$$V_r \approx V1 + V2$$

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