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Let's say I fire a bus through space at (almost) the speed of light. If I'm inside the bus (sitting on the back seat) and I run up the aisle of the bus will I in fact be traveling faster than the speed of light? Relative to earth that I just took off from.

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Related, but not quite a duplicate (I think...): physics.stackexchange.com/q/1557 –  David Z Mar 23 '11 at 16:47

5 Answers 5

up vote 63 down vote accepted

Your question has to do with addition of velocities in special relativity. For objects moving at low speeds, your intuition is correct: say the bus move at speed $v$ relative to earth, and you run at speed $u$ on the bus, then the combined speed is simply $u+v$.

But, when objects start to move fast, this is not quite the way things work. The reason is that time measurements start depending on the observer as well, so the way you measure time is just a bit different from the way it is measured on the bus, or on earth. Taking this into account, your speed compared to the earth will be $\frac{u+v}{1+ uv/c^2}$. where $c$ is the speed of light. This formula is derived from special relativity.

Some comments on this formula provide direct answer to your question:

  1. If both speeds are small compared with the speed of light, they approximately add up as your intuition tells you.

  2. If one of the speeds is the speed of light $c$, you can see that adding any other speed to it does not in fact change it: the speed of light is the same in all reference frames.

  3. If you add up any two speeds below $c$, you end up still below the speed of light. So, any material object which has a mass (unlike light, which doesn't), moves at a speed less than $c$. Adding to it according to the correct rule makes it closer to the speed of light, but you can never exceed it, or in fact not even reach it.

I'd recommend Wheeler and Taylor's "Spacetime Physics" to read about this. Unlike the reputation of the subject it is actually pretty intuitive (I learned that formula in high school).

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holy crap, awesome. –  ed209 Mar 23 '11 at 13:54
Great answer. I second the recommendation of Spacetime Physics. –  Ted Bunn Mar 23 '11 at 14:07
I heard that space can expand faster than the speed of light. so what about if I am on a planet and you are on another, at opposite ends of the universe. We'd be waving goodbye to each other and disappearing into the distance at faster than the speed of light (although we'd never actually see each other as the light is not fast enough to reach each other) –  ed209 Mar 24 '11 at 14:13
That's true: once space is stretching, instead of material things moving, more general things can happen. This is the subject of general relativity, which combines the finite speed of light (but only "locally") with gravity. There are still rules, they are just not as simple. –  user566 Mar 24 '11 at 14:59
+1 Nice answer here –  Sklivvz Apr 3 '11 at 16:53

No. Relative to Earth your bus will have (almost) zero length, so moving from back to the front of the bus will contribute nothing to your speed relative to Earth.

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Doesnt the bus then have insane mass density? Why doesnt it collapse into a black hole from the observers reference? –  Skúli Mar 16 '14 at 22:29
I wonder how much Cherenkov radiation that would cause. –  Cees Timmerman May 7 '14 at 8:59
@Anixx If the aisle of the bus is normal to its rectilinear travel path on which it moves there is no length contraction. Length contraction undergoes the projection (component) of a vector parallel to the direction of motion while the length of its projection (component) normal to the line of motion remains unchanged. Think the bus as a system of coordinates and its aisle as a line oblique to the line of motion of this system to realize that your answer has no sense and is wrong. –  diracpaul Jun 10 at 19:58

I will have to answer this quickly, for I suspect this question will be closed. However, this thought experiment is similar to what Einstein thought about 10 years before he published his paper on special relativity. The problem is this. If you were on a reference frame moving at the speed of light you would observe that light, or any electromagnetic wave, as a wave of oscillating electric and magnetic fields. However, this would be stationary, which contradicts the Maxwell equations for the propagation of electromagnetic radiation. Einstein worked to fix this contradiction, which lead to special relativity. The conclusion is that you can’t place yourself on a frame where light is observed to have any velocity other than the speed of light $c~\simeq~300,000km/sec$.

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Why did you think this would be closed? (In general, if a question should be closed, flag it, don't answer it) –  David Z Mar 23 '11 at 16:47
Then what about slow light? –  Cees Timmerman May 7 '14 at 8:41
@Lawrence, You didn't actually answer the question... –  Pacerier Jul 10 '14 at 16:15


Assume that your bus is approaching the speed of light, because if it had reached it, its mass would be infinite and the question becomes metaphysical as far as the contents and passengers.

Generally, momentum conservation insures that the bus would drop back from the speed it has to compensate for your momentum,as long as you are airborn but when you hit the front glass, it will gain it back.

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so if I'm floating inside the bus and I have a rocket pack on my back and fire it, how would the bus's momentum be affected? I'm not touching the bus. –  ed209 Mar 23 '11 at 13:51
The gas from your rocket will hit the back wall of the bus and transfer the momentum. –  anna v Mar 23 '11 at 15:11
@annav You could float next to the bus and fire your rocket to pass it. I'd be worried about obstacles and oncoming traffic, though. –  Cees Timmerman May 7 '14 at 8:50

This answer here covers the general case where the aisle of the bus you run up, so your velocity, is not necessarily collinear (parallel) to the velocity of the bus relative to Earth. Let define the velocities : \begin{equation} \text{(1) Velocity of bus relative to Earth : } \qquad \mathbf{v}=\upsilon \;\mathbf{n} \tag{A-01a} \end{equation} \begin{equation} \text{(2) Your Velocity relative to bus aisle :} \quad \mathbf{u}=u \;\mathbf{k} \tag{A-01b} \end{equation} \begin{equation} \text{(3) Your Velocity relative to earth : } \quad \mathbf{u}^{\prime}=u^{\prime} \;\mathbf{k}^{\prime} \tag{A-01c} \end{equation} Note that in equations (A-01) the vectors $\:\mathbf{n},\:\mathbf{k}\:,\:\mathbf{k}^{\prime}\:$ are unit vectors and $\:\upsilon,\:u\:,\:u^{\prime}\:$ are real numbers in the interval $\:\left(-c,+c\right)\:$ and not the non-negative numbers representing the norms of these velocity vectors.

Now, the relativistic equation connecting the velocities $\:\mathbf{v},\:\mathbf{u}\:,\:\mathbf{u}^{\prime}\:$ is : \begin{equation} \bbox[#FFFF88,12px]{\mathbf{u}^{\prime} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\circ \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\circ \mathbf{u}}{c^{2}}\Biggr)} } \tag{A-02a} \end{equation} where \begin{equation} \gamma\ \stackrel{\text{def}}{\equiv} \ \left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}}=\dfrac{1}{\sqrt{1-\dfrac{\upsilon^2}{c^{2}}}}\\ \tag{A-02b} \end{equation} The symbol "$\:\circ\:$" refers to the usual inner product in $\:\mathbb{R}^{3}:$ \begin{equation} \mathbf{a}\circ\mathbf{b}=\mathrm{a}_{1}\mathrm{b}_{1}+\mathrm{a}_{2}\mathrm{b}_{2}+\mathrm{a}_{3}\mathrm{b}_{3} \tag{A-03} \end{equation}

For $\:c\rightarrow\infty\:$ equations (A-02) yield the well known non-relativistic composition of velocities:
\begin{equation} \lim_{c\rightarrow\infty}\mathbf{u}^{\prime} = \dfrac{\mathbf{u}+(\gamma-1)(\mathbf{n}\circ \mathbf{u})\mathbf{n}+\gamma \mathbf{v}}{\gamma \Biggl(1+\dfrac{\mathbf{v}\circ \mathbf{u}}{c^{2}}\Biggr)}= \mathbf{u}+\mathbf{v} \tag{A-04a} \end{equation} since \begin{equation} \lim_{c\rightarrow\infty} \gamma = \lim_{c\rightarrow\infty} \left(1-\frac{\upsilon^2}{c^{2}}\right)^{-\frac{1}{2}}= 1 \tag{A-04b} \end{equation} If the velocities $\:\mathbf{v},\:\mathbf{u}\:$ are collinear then $\:\mathbf{k}=\mathbf{n}\:$ and (A-02a) yields \begin{equation} \mathbf{u}^{\prime}=u^{\prime}\mathbf{k}^{\prime}=\left(\dfrac{u+\upsilon}{1+\dfrac{u\upsilon}{c^{2}}}\right)\mathbf{n} \tag{A-05} \end{equation} So with the choice $\:\mathbf{k}^{\prime}=\mathbf{n}\:$ we have \begin{equation} u^{\prime}=\dfrac{u+\upsilon}{1+\dfrac{u\upsilon}{c^{2}}} \tag{A-06} \end{equation} In case the bus speed is approaching the speed of light we have $\:\gamma^{-1}\longrightarrow 0\:$

\begin{equation} \lim_{\upsilon\rightarrow c}\mathbf{u}^{\prime} = \lim_{\upsilon\rightarrow c}\dfrac{\gamma^{-1}\mathbf{u}+\left(1-\gamma^{-1}\right)(\mathbf{n}\circ \mathbf{u})\mathbf{n}+\mathbf{v}}{\Biggl(1+\dfrac{\mathbf{v}\circ \mathbf{u}}{c^{2}}\Biggr)}=\lim_{\upsilon\rightarrow c}\left[\dfrac{(\mathbf{n}\circ \mathbf{k})u+\upsilon}{1+\dfrac{(\mathbf{n}\circ \mathbf{k})u\upsilon}{c^{2}}}\right]\mathbf{n} \tag{A-07} \end{equation} so \begin{equation} \lim_{\upsilon\rightarrow c}\mathbf{u}^{\prime} =c\;\mathbf{n} \tag{A-08} \end{equation}

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I deleted a slightly inappropriate comment. diracpaul, please don't make trivial edits to your post. –  David Z Jun 22 at 12:54
@David Z Many many thanks. I'll try to comply with your instructions. –  diracpaul Jun 22 at 15:27

protected by David Z Dec 17 '11 at 22:15

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