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In this question the accepted answer says:

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.

What is "fast" in this answer? Is there a certain cutoff for when it stops being $u+v$ and becomes $\frac{u+v}{1+ uv/c^2}$?

Community
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    Note that when we write "24 km/s or so", we already imply that our tolerance for input errors is several %. "24 km/s + 5 km/s = 29 km/s", with the error margins dominated by the input error regardless of the formula used to add speeds. – MSalters May 21 '15 at 10:35
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    You got good answers here, but I really hope that what you took away from it is the way that you could have answered the question yourself. The formula is in your post, and whatever device you used to post the question certainly has enough computing power to compute and graph the difference between the two formulae :) – hobbs May 23 '15 at 04:25
  • I could, but going into this I didn't know that the relativistic velocity addition formula worked for every speed, I thought it just worked for speed above a certain threshold. –  May 23 '15 at 11:37

6 Answers6

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For simplicity, consider the case $u=v$. The "slow" formula is then $2u$ and the "fast" formula is $\frac{2u}{1+(u/c)^2}$. In the plot you can see these results in units of $c$. The "slow" formula (red/dashed) is always wrong for $u\ne0$, but it is good enough [close enough to the "fast" formula (blue/solid)] for small $u/c$. The cutoff you choose depends on the accuracy required. When $u<c/10$ then the difference is only likely to be important for high precision work. enter image description here

A series expansion about $u=v=0$ shows the "slow" formula as the first term and that the corrections are small for $uv \ll c^2$:

$$ \frac{u + v}{1+uv/c^2} = (u + v)\left[1-\frac{uv}{c^2} + \left(\frac{uv}{c^2}\right)^2 + O\left(\frac{uv}{c^2}\right)^3\right] $$

Ramashalanka
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    In other words, reality only has one level - relativity. It's just that you can safely use approximations if the error is small enough for your use case. That differs between calculating the speed of a car and, say, relativistic particles hitting the atmosphere :) – Luaan May 21 '15 at 12:24
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    you could say we don't know that relativity is reality either; however it's a better approximation than the Newtonian at high speeds, and at least equally good elsewhere. – user_1818839 May 21 '15 at 13:38
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    I cannot tell which of those lines is red and which is not. I don't know if it's because I'm colorblind (seriously, I am) or if it's because of the format/being on my screen. I know which one is which because I know what the functions look like, but for others who may not, could you remake the plot with either much thicker lines so the different colors are obvious or with a dashed line to make it very clear which one is which? – tpg2114 May 21 '15 at 14:56
  • The upper linear equation is the red line. The lower curve is the blue line. – Jesan Fafon May 21 '15 at 15:59
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    @tpg2114 I'm only color weak and I can't find a red line, either--but I can tell the blue line, the other must be red by elimination. – Loren Pechtel May 21 '15 at 17:03
  • Maybe it is just me, but I also like seeing a Taylor series with error terms. It should be easy to show that the high-order components are the "slow" equation, and the bounded error on the tail gives you how far off your "slow" equation is to the accurate one. @LorenPechtel it is a dark red that is 1 or less pixels wide and anti-aliased. Ick! – Yakk May 21 '15 at 18:29
  • @LorenPechtel color strong here, red is top. feels good, man. – easymoden00b May 21 '15 at 19:11
  • @BrianDrummond: I agree. I was careful to call the "slow" formula wrong, rather than the "fast" one correct (though it is implicit). – Ramashalanka May 22 '15 at 02:23
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    @tpg2114 and `@Loren: I annotated the graph and made the "slow" formula dashed which should make it more readable, sorry for not thinking about that before. – Ramashalanka May 22 '15 at 02:28
  • @Yakk: I added the first terms of a Taylor series below the graph. I experimented with adding the better approximations to the graph, but I felt they cluttered it without adding much. – Ramashalanka May 22 '15 at 02:28
  • @Ramashalanka much better, thanks for adapting to those of us who struggle :) – tpg2114 May 22 '15 at 02:44
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    @tpg2114 as a sidemark that I'll delete later... you opened my eyes to the importance of producing graphics considering some audience that I'm not used to. Thanks! – woliveirajr May 22 '15 at 13:04
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    @Ramashalanka Notice that $1/(1+x) = 1 - x + x^2 - x^3 + ...$ and for $0 \leq x \leq 1$ this is between $1$ and $1-x$, as the remaining terms alternate in sign and monotonically decrease in magnitude. So a simple bound on the error from the Newtonian approximation is a factor of $1-\frac{uv}{c^2}$. Which is funny, because a simpler (and more accurate) to derive bound is a factor of $1+\frac{uv}{c^2}$. ;) – Yakk May 22 '15 at 17:34
  • @Luaan we need to count in relativistic effects when we deal with satellite navigation so I'd rather say your statement is oversimplification of what world we really live in, – shabunc Feb 06 '23 at 08:20
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I'm usually a little more specific with my students. Let's consider a car traveling down the highway at $\rm 30\,m/s \approx 60\,mph$. Then the denominator in the relativistic formula is something like $$ 1 + \left(\frac vc\right)^2 = 1 + \left( \frac{30\rm\,m/s}{3\times10^8\rm\,m/s} \right)^2 = 1 + 10^{-14} $$ In your car, then, the difference between Galilean and relativistic velocity addition starts in the fourteenth decimal place. Do you know the speed of your car to fourteen decimal places? This is where they laugh with me.

It's common in experiments to ignore percent-level errors — that is, to trust a number to about three decimal places. Errors this size start to appear between classical and relativistic dynamics when you have $v/c \approx 0.1$, so that's a common cutoff for "fast."

Note that if you can't ignore percent-level corrections, your definition of "fast" changes. For example, a satellite in low-Earth orbit has a speed of $$ v = \frac{2\pi R_\oplus}{90\rm\,minutes} \approx 7500\,\mathrm{m/s} \approx 2\times10^{-5}c $$ and so has relativistic corrections $(v/c)^2 \approx 10^{-10}$ beginning roughly in the tenth decimal place. In the Global Positioning System (GPS), the total relativistic correction of about $38\rm\,\mu s/day \approx 5\times10^{-10}$ is about at this level as well, as is the centimeter-scale precision $$ \frac{1\rm\,cm}{26\,000\rm\,km} \approx 4\times10^{-10} $$ which is occasionally claimed for military-grade GPS hardware.

rob
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    I've heard the centimeter-level precision claim too, though I can't fathom how they deal with the varying light travel time though a dynamical atmosphere to the point where GR becomes relevant. –  May 22 '15 at 01:18
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    AIUI it's done by having another station very close by, but fixed and at a carefully surveyed location. Assuming that because they are close by the signals traversed similar parts of the atmosphere and looking at the differences, they can model the atmospheric effects and subtract them. I think they also average over comparatively long times, too, so they don't get that kind of position accuracy when the receiver is moving in the ECEF frame. It's still hard to believe though. – Doug McClean May 23 '15 at 03:11
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    @ChrisWhite DGPS works by assuming that the ionospheric delay is similar between two nearby stations, as Doug McClean says. WAAS retransmits a "map" of ionospheric delays measured from ground stations through a satellite. And military receivers use two different frequency bands which are subject to different delays (due to dispersion), allowing the receiver to autonomously estimate the delay from the difference between the two bands, even if the other options aren't available. It's pretty cool stuff. – hobbs May 23 '15 at 04:30
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Actually, regardless of the velocity of the objects in concern, the 'velocity addition' formula is always $\frac{u+v}{1+uv/c^2}$.

There is no transition point where the formula changes from $u+v$ to the special relativity one. Its just that the difference you get in both the formulas at 'low velocities' is very very negligible. $c^2$, in $m^2/s^2$, is $9*10^{16}$. The average speed of a bus is $3.6$ $m/s$. Even for much larger $u,v$ values than that, $uv/c^2$ is negligible. This is why the formula is not of much use in low velocity cases. The significance is observed when $uv/c^2$ is a significant portion of $1$.

Hritik Narayan
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The "fast" formula is always the correct one.

A "fast" speed is one that is comparable to the speed of light. However, when both the speeds involved are much smaller than the speed of light, the "slow" formula is a very good approximation.

Buzz
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When does a object go fast [enough to require the use of $\frac{u+v}{1+ uv/c^2}$]?

When you are approaching light speed.

How fast do you need to go to be "approaching" light speed?

That depends on the precision.

The key is that $\frac{u+v}{1+ uv/c^2}$ always works. However, it's not exactly simple, so generally we like to use an approximation: $u+v$. This approximation is rather accurate, unless you start getting close to the speed of light. Unless you require high precision, you should be fine using $u+v$ as long as $u < 0.1 * c$ and $v < 0.1 * c$.

Disclaimer: I think everything hd already been said, I just tried to reword it so it more accurately addresses the question.

Jasper
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Other answers have effectively, or actually, said when you're dealing with $0.1 c$ or above is when you should be considering relativistic effects.

You can see this by just slightly rearranging the $\frac{u+v}{1+ uv/c^2}$ formula:

$$\frac{u+v}{1+ uv/c^2}=\frac{u+v}{1+ (\frac{u}{c})(\frac{v}{c})}$$

And from there you can see while both $u$ and $v$ are $\lt 0.1 c$ the variance is less than $1\%$.

Of course, you can also use this to see $(\frac{u}{c})^2$ is the relativistic correction mentioned in Rob's answer.

Mark Hurd
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