Q: How do velocities add? If I’m riding a beam of light and I throw a ball, why doesn’t the ball go faster than light?

Physicist: Everything in special relativity, no matter how weird, eventually boils down to the speed of light being the same to everyone.  It’s just not immediately obvious how.  The very short answer is: if v and w are pointing in the same direction, then their sum, V, is V =  (v+w)\left(1+\frac{vw}{c^2}\right)^{-1}.

A fanatical adherence to this principle (that the speed of light is the same to everyone) allows relativistic effects to change the distance and time between events, slow the passage of (other) people’s time, even (under the right conditions) rearrange the order of events.  However, what it can’t change is that an event happens, and (oddly enough) it can’t change the ordering of speeds in any particular direction.

Say you take one direction, and you have a variety of things traveling in that direction going fast, faster, and fastest.  While different perspectives may disagree on exactly how fast each one is going, they will all agree on the same fast-faster-fastest ordering.  And light is always the same, and always the fastest.

Imagine pitching great Walter Johnson throws a baseball at 0.3c (30% of light speed, or about 200 million mph).  Clearly Walt (being stationary) is moving slower than the ball.  In turn, the ball is clearly moving slower than a passing beam of light.

Top: for the pitcher's point of view he's sitting still (zero velocity), the ball is moving at 0.3c, and light is moving at c. Bottom: the same situation, but moving to the right at 0.8c. This is how you "add" 0.3c and 0.8c.

Now imagine that, unbeknownst to Walt, the entire baseball field was actually on some kind of super train tearing through the countryside at 0.8c.  “Intuitively”, if the usual addition of velocities held up, the ball should be traveling at 1.1c (0.8c+0.3c).  But relativity, being neither kind nor reasonable, elects instead to be stubborn about the whole “speed of light thing”.

Even on the train, Mr. Johnson is certainly still traveling to the right slower than the ball, which is still traveling slower than the light.  Moving to a different frame doesn’t change the order, it just scrunches all the velocities up.

If light obeyed the normal addition of velocities, then the speed of the light beam that pitching great Walter Johnson saw should now be 1.8c as seen by someone watching from beside the train tracks (the c Walter saw, plus the 0.8c of the train moving the whole shebang), and there should be no scrunching.  But it isn’t 1.8c!  That’s where the fact that the speed of light is the same in all reference frames comes up.

In Walter’s frame (top) the light is (naturally) traveling at light speed, and the ball is traveling slower than that.  In the frame where the whole thing is passing by on a train (bottom), the same light is still moving at c, and the ball (although traveling at a different speed) is still slower.

The classic question “what happens if I’m already on a beam of light and…” sadly doesn’t make sense from a physicsy standpoint.  No material can travel that fast.  In fact, nothing that experiences time can move that fast.  You’re always restricted by the fact that, from your own point of view, you’re sitting still and light is traveling at light speed (try it!).

If you’d like to find out why in the example above the ball ended up traveling at 0.88c instead of something else, or what would happen if Walter “The Big Train” Johnson had thrown the ball sideways off of the train, you’ll find it milling about in the answer gravy.

By the way, here’s the same example as seen from the baseball’s point of view:

On the off chance that the question would come up, and just to emphasize the symmetry of it: this is the same example from the baseball's point of view.


Answer gravy: When you go from one frame to another (changing from moving at one speed to moving at another) all you’re really doing is rearranging coordinates.  This rearrangement is called a “Lorentz boost” or a “Lorentz transformation“, and it’s surprisingly similar to a rotation (which is also just a rearrangement of coordinates).  Adding velocities amounts to applying a series of Lorentz boosts.

When doing a “coordinate transform” the weapon of choice is matrix multiplication.  It’s easy stuff.  If you can add and multiply you can do matrix multiplication.  Although I won’t lay out the technicalities here, I’ll try to be gruesomely slow algebraically.

Also, when dealing with 4-velocities (velocities that have 3 space components and one time component) it happens to be useful to not consider the time dimension, “t”, by itself, but rather “ct”.  You can think of this as natural units, or whatever.  It just tends to make everything work out beautifully.

A general 4-velocity looks like this: \left( \begin{array}{c} \frac{d(ct)}{d(c\tau)}\\\\\frac{dx}{d (c\tau)}\\\\\frac{dy}{d (c\tau)}\\\\\frac{dz}{d (c\tau)}\end{array}\right) where \tau is the on-board time of the moving thing.  This allows you to keep track of both the velocity in space (\frac{dx}{d(c\tau)}, \frac{dy}{d(c\tau)}, and \frac{dz}{d(c\tau)}) as well as time (\frac{d(ct)}{d(c\tau)}).

An object at rest has a velocity of \left( \begin{array}{c} 1\\\\0\\\\0\\\\0\end{array}\right), that is: it’s traveling through time at one second per second (one second of your time per second of on board time).

The matrix of the Lorentz transformation to a frame moving at velocity v in the x-direction is given by: \left( \begin{array}{ccccccc} \gamma && \gamma \beta&&0&&0\\\\\gamma\beta&&\gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right), where \beta = v/c and \gamma = \frac{1}{\sqrt{1-\beta^2}}.  Gamma (as regular readers are probably sick of regularly reading) is a measure of, among other things, how much time slows down at high speeds, and is often used as a short-hand way of describing those speeds, as in “moving at a gamma of 3″.  Beta (\beta), on the other hand, is just a more reasonable way of talking about speed.  For example, when you say “half of light speed” what you’re saying is “\beta=0.5“.

Matrix aficionados may notice that this looks almost like a rotation in the time and x directions.  Just a note, those 1′s mean that the y and z components are unaffected.

So when the pitcher throws the stationary ball he “boosts” it into a new frame (by making it move).  Mathematically you can do this with matrix multiplication:

\left( \begin{array}{ccccccc} \gamma && \gamma  \beta&&0&&0\\\\\gamma\beta&&\gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right) \left( \begin{array}{c} 1\\\\0\\\\0\\\\0\end{array}\right) = \left( \begin{array}{c} \gamma\cdot 1+\gamma\beta\cdot 0+0\cdot 0 +0\cdot 0\\\\\gamma\beta\cdot 1+\gamma\cdot 0+0\cdot 0 +0\cdot 0\\\\0\cdot 1+0\cdot 0+1\cdot 0+0\cdot 0\\\\0\cdot 1+0\cdot 0+0\cdot 0+1\cdot 0\end{array}\right) = \left( \begin{array}{c} \gamma\\\\\gamma\beta\\\\0\\\\0\end{array}\right)

The top \gamma is time dilation.  \frac{d(ct)}{d(c\tau)} = \frac{dt}{d\tau} = \gamma means that for every second that passes for the baseball (\tau) \gamma seconds pass for you.  Also, you can get rid of the on-board time dependence and recover “normal” velocity (dx/dt, no “\tau‘s”) with a little algebra.  Normal velocity (in the x-direction) is: \frac{dx}{dt} = c\frac{dx}{d(ct)} = c\frac{dx}{d(c\tau)}\frac{d(c\tau)}{d(ct)} = c\left( \gamma\beta \right) \left(\frac{1}{\gamma} \right) = c\beta = c\left(\frac{v}{c}\right) = v.

So, the baseball is moving with ordinary velocity v (like you’d hope) and 4-velocity (\gamma, \gamma\beta, 0,0).  To see what happens when we boost again, by moving to a new frame where the ball’s velocity is being added to the velocity of the train, just do another matrix multiply.  This time with: \left( \begin{array}{ccccccc} \Gamma && \Gamma  \alpha&&0&&0\\\\\Gamma\alpha&&\Gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right)  It’s the same thing, just with a different speed from last time.  This time \alpha = w/c, and \Gamma=\frac{1}{\sqrt{1-\alpha^2}}.

\left( \begin{array}{ccccccc} \Gamma && \Gamma   \alpha&&0&&0\\\\\Gamma\alpha&&\Gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right) \left( \begin{array}{c} \gamma\\\\\gamma\beta\\\\0\\\\0\end{array}\right) =\left( \begin{array}{c} \Gamma\cdot \gamma+\Gamma\alpha\cdot \gamma\beta+0\cdot 0  +0\cdot 0\\\\\Gamma\alpha\cdot \gamma+\Gamma\cdot \gamma\beta+0\cdot 0 +0\cdot  0\\\\0\cdot 1+0\cdot 0+1\cdot 0+0\cdot 0\\\\0\cdot 1+0\cdot 0+0\cdot  0+1\cdot 0\end{array}\right) =\left( \begin{array}{c}  \Gamma\gamma(1+\alpha\beta)\\\\\Gamma\gamma (\alpha+\beta)\\\\0\\\\0\end{array}\right)

So, what’s the normal velocity after adding the speeds v and w?  What is the final velocity “V”?  Algebra!:

\begin{array}{ll} V = \frac{dx}{dt}\\= c\frac{dx}{d(ct)}\\= c\frac{dx}{d(c\tau)}\frac{d(c\tau)}{d(ct)}\\= c\Gamma\gamma (\alpha+\beta)\left(\frac{1}{\Gamma\gamma(1+\alpha\beta)}\right)\\= c\frac{\Gamma\gamma (\alpha+\beta)}{\Gamma\gamma(1+\alpha\beta)}\\= c\frac{\alpha+\beta}{1+\alpha\beta}\\= c\frac{w/c+v/c}{1+(w/c)(v/c)}\\= \frac{w+v}{1+(w/c)(v/c)}\\= \frac{w+v}{1+\frac{wv}{c^2}}\end{array}

Holy crap!  The equation we normally think is right, V = v+w, doesn’t work.  But stir in a little special relativity and you get: V=(v+w)\left(1+\frac{vw}{c^2}\right)^{-1}.

So, in the example from (way) above the speeds were v=0.3c and w=0.8c.  So V=\frac{0.3c+0.8c}{1+(0.3)(0.8)} \approx 0.887097c

Seems like a long trip for \left(1+\frac{vw}{c^2}\right)^{-1}, but there it is.  This little term keeps the speed of light in the right place, and adjusts every other velocity to stay in the right place.  At low (compared to light) speeds vw is overwhelmed by c2 and 1+\frac{vw}{c^2} is very close to 1.  In other words, when you’re not cruising about at crazy speeds you get the usual rules.

Finally, if you choose to add velocities that aren’t in the same direction, you just have to rotate one of the Lorentz matrices.  For example, while all of the examples so far have been in the x direction, a boost in the y direction would be given by: \left( \begin{array}{ccccccc} \gamma && 0&&\gamma   \beta&&0\\\\0&&1&&0&&0\\\\\gamma\beta&&0&&\gamma&&0\\\\0&&0&&0&&1\end{array}\right)

If Walter Johnson had decided to throw the ball in the y direction on the train, and everything else stayed the same, you’d figure out the end 4-velocity of the ball by doing the following boosts:

\left( \begin{array}{ccccccc} \Gamma && \Gamma     \alpha&&0&&0\\\\\Gamma\alpha&&\Gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right) \left( \begin{array}{ccccccc} \gamma && 0&&\gamma    \beta&&0\\\\0&&1&&0&&0\\\\\gamma\beta&&0&&\gamma&&0\\\\0&&0&&0&&1\end{array}\right) \left( \begin{array}{c} 1\\\\0\\\\0\\\\0\end{array}\right)= \left( \begin{array}{ccccccc} \Gamma && \Gamma    \alpha&&0&&0\\\\\Gamma\alpha&&\Gamma&&0&&0\\\\0&&0&&1&&0\\\\0&&0&&0&&1\end{array}\right)  \left( \begin{array}{c}  \gamma\\\\0\\\\\gamma\beta\\\\0\end{array}\right) =\left( \begin{array}{c}  \Gamma\gamma\\\\\Gamma\gamma\alpha\\\\\gamma\beta\\\\0\end{array}\right)

It seems a little spooky that when the velocities aren’t parallel, velocity addition is no longer commutative (notice that the alphas and betas aren’t on equal footing), and the order in which you do the boosts becomes important.  Of course, it may be spooky that that isn’t always the case.

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18 Responses to Q: How do velocities add? If I’m riding a beam of light and I throw a ball, why doesn’t the ball go faster than light?

  1. Justin Hilyard says:

    “A fanatical adherence to this principle (that the speed of light is the same to everyone) allows relativistic effects to change the distance and time between events, slow the passage of (other) people’s time, even rearrange the order of events.”

    Wait, rearrange the order of events? I thought that was impossible even with relativity, because it would violate the principle of causality. I know that if you’ve got two events A and B, you could have A and B simultaneous in one reference frame and A following B in another, but I was under the impression that if there exists one reference frame in which A follows B, then in all reference frames the time of A is >= the time of B.

  2. The Physicist The Physicist says:

    I should have been more precise and less showy.
    Spacelike separated” events A and B can happen at the same time, A then B, or B then A. However, causality is maintained since they’re too far away in space, and too close together in time to send any kind of signal to the other.
    I’ll put a disclaimer in the post.

  3. Neal says:

    Recommendation for further reading: The Geometry of Spacetime by Callahan. Only requires understanding of basic linear algebra and calculus.

  4. Terry Hayward says:

    So put simply, the ball will travel in time ?

  5. The Physicist The Physicist says:

    Not quite that simply. Everything (with mass) travels in time, it’s just a matter of how exactly.

  6. Terry Hayward says:

    Since time is a function of light speed, would you experience time if you could travel at light speed, and if you could, then would the thrown ball effectively travel into the future?
    To quote Douglas Adams, “this is of course, impossible” :)

  7. The Physicist The Physicist says:

    Nopers! If you could travel at light speed you’d find that time wouldn’t pass at all. Any journey, no matter how long (distance-wise), would be instantaneous from your perspective.
    Keeping with the “ordering of speeds” idea of the post; since the baseball needs to travel between your speed (the speed of light) and the speed of light, the only speed it has the option of going is also the speed of light.
    Of course, not being able to move at the speed of light, this is a non-issue.

  8. Terry Hayward says:

    What you seem to be saying is that the faster you travel, the slower you are able to move, so that, at light speed, you would not be able to move the ball in the direction of travel.
    This seems to imply that were you able to travel faster than light, you would travel into the past.

  9. Sabeeh says:

    Hi, there has recently been an experiment at Cern with results indicating faster than light travel occured. I am interested on your thoughts on this. I studied physics at degree level some years ago and will never forget one event. Our lecturer stated that the faster speed possible is c. He then went on to’proove’ this with 4 pages of mathematics with the answer being zero. I queried that i did not follow WHY c is the max speed. Lo and behold he went through the maths again. He could not explain in physical terms why.
    I believe science is guilty of over indulgence in theories without experimental backup and use of assumptions. We must use observation as the key and then try to explain what we see with theories. Otherwise it all gets ‘fantastical’.
    I know there is currently a variable speed of light theory gaining ground. And why not ! Makes more sense than forcing a constant c then having to deal with bending space time, worm holes, parallel universes etc…. If we are honest with ourselves we know we do not really fully understand relativity (special but especially general) until we can EXPLAIN it to the lay person.
    One little thought experiment if I may. Imagine you have a piece of string the length of 299km I.e. the length that light would travel in 1 second. I am at one end and you at the other. Ignoring the fact that it would be heavy and assuming it is rigid, if I pull it you should notice it moved almost instantly. So hasn’t information just travelled faster than c ?

  10. The Physicist The Physicist says:

    You’re absolutely right! Science in general is an observational practice, and almost all practicing scientists take that very seriously (not including mathematicians, of course).
    We’ve done a couple of posts on this stuff:
    Q: Why is the speed of light the fastest speed? Why is light so special?
    Q: CERN’s faster than light neutrino thing: WTF?
    I’ve never heard that the constantness of C requires worm holes and parallel universes. Do you have a reference for that?
    The perfectly rigid rope or pole thought experiment is a classic. But it turns out, it doesn’t work. The information that passes from one atom to another down the length of an object that says “move” moves at no faster than light speed.

  11. Neutrino says:

    You say that the maximum attainable speed permissible by the relativity is the light speed. What about the “inflation”? We have enough evidence that it occurred soon after the “big bang”. Inflation expanded the universe at a rate faster than the speed of light. How?

  12. The Physicist The Physicist says:

    “Inflating faster than the speed of light” is a phrase that shows up in a lot of science documentaries, that really shouldn’t. The inflation refers to space itself, whereas “speed” refers to movement through space. In fact, the rate of inflation is described in terms of “speed per distance”, as opposed to just “speed”.
    You can think of inflation as the expansion of a balloon, and movement through space as things moving across the surface of that balloon. If the balloon expands quickly, that doesn’t mean that dots drawn on the surface are moving fast. In fact, it doesn’t say anything about the speed of things on the surface.

  13. Anthony Rose says:

    This is the BEST explanation of this topic I have ever seen! Thank you!

  14. Albertie says:

    If two photons are moving in opposite directions, what is their relative velocity?
    IS it c or 2c?

  15. Albertie says:

    how can it be 2c when any speed cannot exceed the speed of light?

  16. The Physicist The Physicist says:

    Depends on your perspective.
    In between the two photons you’d say that their relative velocities are 2C.
    However, neither of them sees the other moving faster than light (or in fact at all), and you of course see them both moving at exactly light speed.
    The important thing is that nothing is moving faster than light with respect to anything else.

  17. Sheldon says:

    I had asked a physicist a very similar question and got a beautiful answer. He was nice enough to take the time to make charts and everything. I had asked “if a spaceship leaves space station A toward space station B at 0.6c and another spaceship leaves B toward A also at 0.6 then I sit at poin C in the middle to witness this 1.2c collision what would happen? The actual question was would each spaceship see the other spaceship as moving at 1.2c toward them? The answer was no. I forget the exact speed from the math but it basically turned out where each ship would see me traveling at 0.6c toward them and the other ship at some number > 0.6c and < 1.0c.

    One of the ways I break it down in my lay mind is that to the Newtonian relativity we experience intuitively, space and time are constants that all frames agree on. They are rigid. Well once the speed of light was realized to be rigid… Then everything else has to flex. So it's most intuitive to preface any discussion of relativistic speeds as saying "on a scale of 0 to c how fast were things going" and that's where you will get two reference frames that agree … Would draw a scale that looks similar in order and proportion but the actual numbers in the middle would not be the same. They would correlate though in relation to everything else. And the idea of breaking the speed limit by simply going faster does the same weird things when you try to exceed an infinite limit in any system. Merely going faster than c is like merely going colder than absolute 0. As you approach the limit and your curve on the graph gets closer its like you can't get there you just zoom in on the gap more.

    It's like if you spend twice as much energy you only go a fraction faster or a fraction colder. 0.99 becomes 0.999 becomes 0.9999 or 0.01 becomes 0.001 becomes 0.0001. It's like the way extents work.

  18. JoooBlooo says:

    If light can’t move at different speeds relative to a perspective, how do you account for red shift? The baseball thought process is archaic and repeated so many times as fundamental, it hasn’t been challenged in modern times. Of course! a baseball being thrown is practically zero compared to the speed of light, therefore has no bearing in calculations or reality for that matter. What physical experiments prove this?… It’s easy to write equations based on illusions, and to consider them truth when you are out of the realm of perception.

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