Why does relativistic velocity subtraction produce a larger relative velocity than classically?

Question

I got a question for homework saying that there were two rockets on a parallel track, heading towards earth. Rocket A was in front of Rocket B. Rocket A was travelling at a velocity of $0.75c$ from the frame of reference (FOR) of earth and Rocket B was travelling at a velocity of $0.5c$ from the FOR of earth. The question asked to find the velocity of Rocket A from the FOR of Rocket B.

I got an answer of $0.4c$, $$v_{AB}=\frac{v_a-v_b}{1-v_av_b/c^2}=\frac{0.75c-0.5c}{1-0.75c\cdot0.5c/c^2}=0.4c,$$

which both my teachers said was right.

I am confused because this velocity is greater than what I got using the formula for classical velocity addition ($0.25c$). For every other question I have done, the relativistic velocity is less than the classical velocity. I am wondering why the relativistic velocity is larger in this case?

Answer 1

If the two move in the same direction, you are dividing the classical result by a number less than one: $1-v_1v_2/c^2$, thus the result will be always larger than the classical. If they were moving in opposite directions, then the sign changes and you are dividing by a number larger that one, and thus you will obtain a result smaller than the classical one.

Both results are intuitive. Imagine that both are moving in the same direction close to c. Classicaly, the difference to you will be very small (let us say 0.00000001c), but they could be moving relative to each other to a speed close to c. If they move, instead, in opposite directions, at speeds close to c, the classical result will be closer to 2c, but they canot see each other moving at a speed larger than c, so the result will be less than the classical.

Answer 2

My guess is just that the problem is inverted with respect to the problems you have seen, where the problems you have seen look like:

“Alice sees Bob move at velocity $u~\hat x$ and Bob sees Carol move at velocity $v~\hat x$, how fast does Alice see Carol move?”

The answer here is to construct the world line in Bob’s reference frame, $\text{Carol}=\{(ct, vt)_{\text{Bob}}, \text{ for all } t\}$, then boost it by velocity $-u\hat x$ into Alice’s reference frame and take the ratio of the space and time components (because the world line still goes through $(0,0)$), yielding $$v_{\text{Alice}}= c~\frac{\gamma_u~(vt+\beta_uct)}{\gamma_u~(ct+\beta_uvt)}=\frac{v+u}{1+uv/c^2}.$$

But now you are confronted instead with the problem,

“Alice sees Bob move at velocity $u~\hat x$ and Carol move at velocity $v~\hat x$, how fast does Bob see Carol move?”

Solving this problem is identical because in the earlier problem Bob also saw Alice moving with velocity $-u\hat x,$ so you have a complete description of the earlier calculation in precisely this format, just the names are different. If you will go through the derivation again you will see that the only difference is that you are boosting by velocity $+u\hat x$ hence the sign on $u$ has changed to give you, $$v_{\text{Bob}}= c~\frac{\gamma_u~(vt-\beta_uct)}{\gamma_u~(ct-\beta_uvt)}=\frac{v-u}{1-uv/c^2}.$$

Once you get that, it is not too hard to see that if the one speed (Carol seen by Alice) is slower (than Carol seen by Bob) and the directions stay the same, then the other speed (Carol seen by Bob) must be greater (than Carol seen by Alice). It's just the same numbers seen two different ways.

Answer 3

I don't think you're asking for the math to be checked, but for why the answer makes sense.

Think of the two boundary cases. First the easy one: You have two flashlights pointing in opposite directions. From the reference frame of one photon from flashlight A, photons in flashlight B are going the speed of light in the opposite direction even though the classical result would be that they are traveling from each other at 2c.

So here, relativistic travel in opposite directions leads to velocities smaller than classical result.

The one you're finding counter-intuitive is this thought experiment: You're on a train going 0.999c. Super fast. Then you switch on a flashlight and point it forward. From your reference frame, you KNOW that those photons have to be traveling at the speed of light, but that the classical result would be 0.001c.

In this example, relativistic travel in the same direction leads to velocities greater than the classical result.

Relativity isn't intuitive, we have to adjust our intuition to match its results. And simple gut check scenarios like the two I outlined above can aid with intuition.

Answer 4

Let a particle $\; a\;$ moving uniformly with velocity $\;\mathbf{v}\;$ with respect to an inertial system $\;\mathrm S$. A second particle $\;b\;$ is moving uniformly with velocity $\;\mathbf{u}\;$ with respect to particle $\;a$. An inertial system $\;\mathrm S_a\;$ is attached to particle $\;a\;$ in standard configuration to the inertial system $\;\mathrm S$. To find the velocity $\;\mathbf{w}\;$ of particle $\;b\;$ with respect to the inertial system $\;\mathrm S\;$ we must add the two vectors $\;\mathbf{v},\mathbf{u}\;$ addition non-relativistic or relativistic.

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A. The Non-Relativistic addition $\;\mathbf{w}_{_{\rm NR}}$

As shown in Figure-01 we have \begin{equation} \mathbf{w}_{_{\rm NR}}\boldsymbol{=}\left(\mathrm u\cos\phi\boldsymbol{+}\mathrm v \right)\mathbf{i}\boldsymbol{+}\left(\mathrm u\sin\phi\right)\mathbf{j} \tag{NR-01}\label{NR-01} \end{equation} so \begin{align} \mathrm w_{_{\rm NR}}^2 & \boldsymbol{=}\mathrm u^2\boldsymbol{+}\mathrm v^2\boldsymbol{+}2\,\mathrm u\,\mathrm v \cos\!\phi\vphantom{\dfrac{\mathrm u\,\sin\!\phi}{\mathrm u\,\cos\!\phi+\mathrm v}} \tag{NR-02.1}\label{NR-02.1}\\ \tan\!\theta_{_{\rm NR}} & \boldsymbol{=}\dfrac{\mathrm u\,\sin\!\phi}{\mathrm u\,\cos\!\phi\boldsymbol{+}\mathrm v} \tag{NR-02.2}\label{NR-02.2} \end{align}

Keeping the vector $\;\mathbf{v}\;$ and the magnitude $\;\mathrm u = \Vert\mathbf{u}\Vert\;$ constant the edge of $\;\mathbf{w}_{_{\rm NR}}\;$ is moving on a full circle as the angle $\;\phi\;$ is changing in $\;\left[0,2\pi\right]$.

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B. The Relativistic addition $\;\mathbf{w}_{_{\rm R}}$

In this case we have \begin{equation} \mathbf{w}_{_{\rm R}}\boldsymbol{=}\dfrac{\mathbf{u}\boldsymbol{+}\left(\gamma_{\mathrm v}\boldsymbol{-}1\right)\left(\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{v}}{\mathrm v^2}\right)\mathbf{v}\boldsymbol{+}\gamma_{\mathrm v}\mathbf{v}}{\gamma_{\mathrm v}\left(1\boldsymbol{+}\dfrac{\mathbf{u}\boldsymbol{\cdot}\mathbf{v}}{c^2}\right)}\,,\qquad \gamma_{\mathrm v}\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\mathrm v^2}{c^2}\right)^{\boldsymbol{-\frac12}} \tag{R-01}\label{R-01} \end{equation} From Figure-02 \begin{equation} \mathbf{w}_{_{\rm R}}\boldsymbol{=}\dfrac{\left(\mathrm u\cos\phi\boldsymbol{+}\mathrm v \right)\mathbf{i}\boldsymbol{+}\left(\sqrt{1\boldsymbol{-}(\mathrm v/c)^2}\,\mathrm u\sin\phi\right)\mathbf{j}\vphantom{\dfrac12}}{\left(1\boldsymbol{+}\dfrac{\mathrm{u}\,\mathrm{v}}{c^2}\cos\phi\right)} \tag{R-02}\label{R-02} \end{equation} so \begin{align} \left(\dfrac{\rm w_{_{\rm R}}}{c}\right)^{\!2}& \boldsymbol{=}1\!-\!\dfrac{\left[1\!-\!\left(\dfrac{\rm u}{c}\right)^{\!2}\right]\left[1\!-\!\left(\dfrac{\rm v}{c}\right)^{\!2}\right]}{\left(1+\dfrac{\rm u v}{c^2}\cos\phi\right)^{\!2}} \tag{R-03.1}\label{R-03.1}\\ \tan\!\theta_{_{\rm R}}& \boldsymbol{=}\sqrt{1\!-\!\left(\dfrac{\rm v}{c}\right)^{\!2}}\,\dfrac{\mathrm u\,\sin\!\phi}{\mathrm u\,\cos\!\phi+\mathrm v}=\sqrt{1\!-\!\left(\dfrac{\rm v}{c}\right)^{\!2}}\,\tan\!\theta_{_{\rm NR}} \tag{R-03.2}\label{R-03.2} \end{align}

Keeping the vector $\;\mathbf{v}\;$ and the magnitude $\;\mathrm u = \Vert\mathbf{u}\Vert\;$ constant the edge of $\;\mathbf{w}_{_{\rm R}}\;$ is moving on a closed ellipsis-like curve as the angle $\;\phi\;$ is changing in $\;\left[0,2\pi\right]$.

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Note that from equation \eqref{R-03.1} we have the well-known results when $\;\mathbf{u},\mathbf{v}\;$ are collinear \begin{align} \phi & \boldsymbol{=}0 \quad\boldsymbol{\Longrightarrow}\quad \cos\phi \boldsymbol{=+}1\quad\boldsymbol{\Longrightarrow}\quad \rm w_{_{\rm R}}\boldsymbol{=}\dfrac{\mathrm u \boldsymbol{+}\mathrm v }{1\boldsymbol{+}\dfrac{\rm u v}{c^2}} \tag{R-04.1}\label{R-04.1}\\ \phi & \boldsymbol{=}\pi \quad\boldsymbol{\Longrightarrow}\quad \cos\phi \boldsymbol{=-}1\quad\boldsymbol{\Longrightarrow}\quad \rm w_{_{\rm R}}\boldsymbol{=}\dfrac{\vert\mathrm u \boldsymbol{-}\mathrm v \vert}{1\boldsymbol{-}\dfrac{\rm u v}{c^2}}>\vert\mathrm u \boldsymbol{-}\mathrm v \vert \tag{R-04.2}\label{R-04.2} \end{align}

From \eqref{R-04.2} we conclude that for $\;\phi=\pi\;$ the magnitude $\;\rm w_{_{\rm R}}\;$ of the resulting relativistic sum is greater than the magnitude of the non-relativistic sum $\;\vert\mathrm u \boldsymbol{-}\mathrm v\vert\;$ for any values of
$\;\rm u,v\;$ less than $\;c$.

Answer 5

I don't see how you could get .25c with velocity addition. Clearly, you got that by subtracting .5 from .75. We have that .4c "plus".5c is .75c, which is less than what we would get with classical addition. If relativistic addition of $v_1$ and $v_2$ gives $v_3$, and $v_3$ is less than $v_1+v_2$, then clearly relativistic subtraction of $v_2$ from $v_3$ has to give something more than $v_3-v_2$. Hopefully, that's reasonably intuitive: subtraction is the opposite off addition, so the effect is opposite. If we add a velocity, and then subtract it, then we should end up with the velocity we started with. But if both addition and subtraction makes velocities smaller, then we would end up with a velocity smaller than the one we started with (for example, if we add $v$ to $u$, we end up with something less than $u+v$. If we then subtract $v$ from that, we would, by your way of thinking, get something smaller (something smaller than u+v)-v, which would be smaller than u). If $add(u+v)<u+v$ for all $u$, $v$, then $subtract(u,v)$ must be greater than $u-v$. That's because, by definition, $add(v+subtract(u,v))$ is equal to $u$ (if you subtract a number, then add it back, you end up with the number you started with), so if we have $add(v+subtract(u,v))<v+subtract(u,v)$, then we can substitute $u$ in for $add(v+subtract(u,v))$ and get $u<v+subtract(u,v)$, or $u-v<subtract(u,v)$, or $subtract(u,v)>u-v$.

So when you add two velocities, you get a number less than the classical sum, and when you subtract two velocities, you get a number greater than the classical difference. One thing you can compare it to is adding volumes and the resulting radius: if you have two spheres, and want a sphere with the volume of the sum of their volumes, the radius is going to be smaller than the sum of radii. If you want a sphere with the volume of the differences of volumes, the radius is going to be larger than the difference of radii.