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Let's say I fire a bus through space at (almost) the speed of light in vacuum. If I'm inside the bus (sitting on the back seat) and I run up the aisle of the bus toward the front, does that mean I'm traveling faster than the speed of light? (Relative to Earth that I just took off from.)

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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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    $\begingroup$ holy crap, awesome. $\endgroup$ – ed209 Mar 23 '11 at 13:54
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    $\begingroup$ Great answer. I second the recommendation of Spacetime Physics. $\endgroup$ – Ted Bunn Mar 23 '11 at 14:07
  • $\begingroup$ Also in case you're confused how that can work (being much faster than the bus from the bus' viewpoint but only little faster viewed from earth), there is additionally the effect of time dilatation, i.e. time "runs" differently when you're fast $\endgroup$ – Tobias Kienzler Mar 23 '11 at 15:06
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    $\begingroup$ 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) $\endgroup$ – ed209 Mar 24 '11 at 14:13
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    $\begingroup$ 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. $\endgroup$ – user566 Mar 24 '11 at 14:59
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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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    $\begingroup$ Doesnt the bus then have insane mass density? Why doesnt it collapse into a black hole from the observers reference? $\endgroup$ – Skúli Mar 16 '14 at 22:29
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    $\begingroup$ I wonder how much Cherenkov radiation that would cause. $\endgroup$ – Cees Timmerman May 7 '14 at 8:59
  • $\begingroup$ @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. $\endgroup$ – user82794 Jun 10 '15 at 19:58
  • $\begingroup$ Sir will answer of question changes frame of reference that is ground also moving with speed of light $\endgroup$ – yuvraj singh Aug 19 at 7:12
  • $\begingroup$ Sir shrinkness you refer in your answer is govern by distance=speedxtime since speed is constant distance decreases time tends to infinity , $\endgroup$ – yuvraj singh Aug 19 at 7:18
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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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    $\begingroup$ Why did you think this would be closed? (In general, if a question should be closed, flag it, don't answer it) $\endgroup$ – David Z Mar 23 '11 at 16:47
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    $\begingroup$ Then what about slow light? $\endgroup$ – Cees Timmerman May 7 '14 at 8:41
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    $\begingroup$ @Lawrence, You didn't actually answer the question... $\endgroup$ – Pacerier Jul 10 '14 at 16:15
  • $\begingroup$ @CeesTimmerman: You are confusing the phase velocity and the group velocity. Slow light has just the group velocity (speed of propagation of the wave envelope) slow. $\endgroup$ – pabouk Nov 13 '15 at 9:06
  • $\begingroup$ @DavidZ many things will be closed which should not be $\endgroup$ – amara Mar 20 '18 at 5:01
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Hmm.

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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    $\begingroup$ 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. $\endgroup$ – ed209 Mar 23 '11 at 13:51
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    $\begingroup$ The gas from your rocket will hit the back wall of the bus and transfer the momentum. $\endgroup$ – anna v Mar 23 '11 at 15:11
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    $\begingroup$ @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. $\endgroup$ – Cees Timmerman May 7 '14 at 8:50
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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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    $\begingroup$ I deleted a slightly inappropriate comment. diracpaul, please don't make trivial edits to your post. $\endgroup$ – David Z Jun 22 '15 at 12:54
  • $\begingroup$ Sir will answer of question changes frame of reference that is ground also moving with speed of light $\endgroup$ – yuvraj singh Aug 19 at 7:13
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As mentioned by physicist Brian Greene in his book The Elegant Universe, (See the_Elegant_Universe-B.Greene.pdf, pages 26 and 27, titled as "Motion through Spacetime"), all objects are constantly on the move within space-time, and that they do so at a speed that is identical to the speed of light. For a view of Brian Greene's mathematical representation of this ongoing constant "c" motion within space-time, click on image below. http://www.outersecrets.com/real/image/brian00.png Despite the ongoing constant "c" motion of all objects located within space-time, a change in the direction of travel is still possible. The more you direct your travel across space, rather than across time, the closer to the speed of light you will be moving across space, and, concerning motion across time, the closer you will be to being at a standstill.

Thus unless a way is discovered to elevate the speed of your constant motion across space-time, the speed of light will remain as the limit. Thus no matter how fast from your point of view you move from the rear to the front of the bus, you will still not be able to surpass the speed of light.

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  • $\begingroup$ To see the outcome of this constant "c" motion across space-time, see goo.gl/fz4R0I . Make sure that you watch the entirety. $\endgroup$ – Sean Oct 28 '15 at 3:49
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According to special relativity, information can't ever travel faster than light for the reason described at https://www.quora.com/How-does-relativity-work/answer/Timothy-Bahry. The universe does obey special relativity in the absense of a gravitational field. Diamond has the highest speed of sound of any substance on Earth which is 12 km/s so if you have a super long diamond rod in outer space and you twist one end, the twisting will travel to the other end at only 12 km/s, far slower than the speed of light. There might actually be a substance whose bulk modulus is larger than its density times c^2 so the wave equation would predict that a longitudinal wave in it will travel faster than light, which is what a neutron star is made of. The reason I'm wondering that is because I read that some neutron stars have a photon sphere. Obviously, even if that is the case, if you bombard its surface, the shockwave in it will not travel faster than light because then it would be travelling backwards in time in another frame of reference and therefore it doesn't obey the wave equation. It can still be destined to obey the wave equation for any analytic initial state without information travelling faster than light. That's because given all derivatives of velocity with respect to position at one point and the fact the the function is analytic, it can be fully determined what the velocities at all the other points are. For that reason, it wouldn't break any laws for a sinusodial wave to travel through a neutron star faster than light.

Since the universe doesn't follow special relativity, it might be possible to travel faster than light. Accoriding to http://munews.missouri.edu/news-releases/2007/0601-gravitomagnetic-field.php, gravitomagnetism has been detected so we're pretty sure the universe follows general relativity at the macroscopic level. Objects actually do get dragged faster than light by moving space beyond the event horizon of a black hole. See https://www.quora.com/What-is-a-black-hole-How-can-we-understand-it/answer/Timothy-Bahry. Also, galaxies will receed from us faster than light because of the nonzero cosmological constant and just like if they were falling into a black hole, they will appear to approach the certain finite distance away from us without ever reaching it while getting ever more red shifted. However, since our universe follows general relativity, I don't think its laws predict that anything can go from a point in space-time back to the same point without going faster than light within the space itself but I'm not absolutely sure. If an object does do so, its path is called a closed timelike curve. After all, there probably does exist a deterministic set of laws a universe could have that doesn't let information travel faster than light within the space and yet for some initial states, gives a contradictory prediction of what it will evolve into because it predicts the that a closed timelike curve will exist. There's no way a space ship could travel faster than light by bending space around it according to those laws because then the information telling the space to warp would have to travel through the unwarped space far enough ahead of the ship faster than light. Those laws do however permit you to build a track through space to faster than light where time is contracted by a factor of 21 within the track enabling things to go 20 times faster than light within the track as observed from outside. After that, those laws would then also permit another track to be built right beside it that contracts time by a factor of 21 in another frame of reference enabling a spaceship to travel into its own past light cone. If general relativity just like that other set of laws really does allow for closed time-like curves, here's one possible solution. The moment a closed light-like curve gets created, it will nucleate the disappearance of space at the speed of light before it gets a chance to evolve into a closed time-like curve. Such a nucleation may have already occurred in a black hole but never got to us destroying us because light can't escape from a black hole.

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