Athletic Conundrum......

[ToraToraTora]

If I run a lap of the track in 2 minutes, how fast must I run a second lap so that my average speed for both laps is twice that of my first lap?

[ubasses]

6 minutes. 6+2 =8 / 2 = 4 (twice first lap)
unless it is a trick question

[New Judge]

You can't do it.

Your first lap was run at a rate of 0.5 laps per minute. You want to run 2 laps at 1.0 laps per minute - which would take 2 minutes. You've already taken 2 minutes to do the first lap.

[ClareTG0ld]

I agree with NJ.

[ToraToraTora]

yes judge/CTG correct. It's one of those things that sounds like it should be a) doable and b) simple to work out but when you try you reach a mathematical impasse.

[ClareTG0ld]

One reason it might sound doable is because NJ was talking about the "time-averaged speed", which is the usual definition: average speed = total distance travelled divided by time taken, with (in this case) a formula

s_avg = (s1*t1 + s2*t2)/(t1 + t2) .

On the other hand, you could formulate a different measure of averaging speed, the "distance-averaged speed". In that case, the "average" would be written

s_d = (s1*d1 + s2*d2)/(d1 + d2).

Since d1 = d2 = 1 (one lap), then this reduces in our case to the more familiar average, s_d = (s1 + s2)/2, and you can get s_d = 2 s_1 as long as s_2 = 3 s _1.

I'm not clear on how useful, if at all, this second notion of "average speed" would be in practice, and as far as I remember the only place I came across it previously was in some weird exam question. But you clearly *can* define this, and call it an average, even if it's not the usual (and most sensible, in this case) notion of average speed.

I was going to put all this in my previous answer, but decided to hold off. And, besides, even considering all the above, I still agree with NJ.

[ToraToraTora]

ok thanks for the effort CTG.

[ClareTG0ld]

On reflection, it's simply related to the link between harmonic and arithmetic means.

Let d1 = d2 = 1. Then the formula for distance-averaged speed is, as I quoted, (s1 + s2)/2 (arithmetic mean). Meanwhile, the formula for time-averaged speed is

s_avg = (s1*t1 + s2*t2)/(t1 + t2) = (d1 + d2)/ (d1/s1 + s2/s2) = 2/(1/s1 + 1/s2)

or:

2/s_avg = 1/s1 + s/s2

which is the definition of harmonic mean.

[Old_Geezer]

Doesn't time stop for you as you reach light speed ? If so an instant burst of infinite acceleration up to C may be all you need.

[ClareTG0ld]

Whilst I expect that OG was joking, it's worth taking seriously that suggestion for long enough to explain why it wouldn't work.

Firstly, even in theory you can never reach the speed of light, c, assuming that you weren't already travelling at c. You can get arbitrarily close, it is true, and we do have to be aware of the limiting behaviour, but you'd always measure your speed as less than c.

Secondly, it's misleading to say that time stops at the speed c anyhow. What equals zero, when moving at c, is instead the "proper time", the difference defined as (ct)^2 - x^2, where t is the time taken to go between two points a distance x apart. For a given object moving from A to B, then they'd always measure the same value of (ct)^2 - x^2 even when they measure a different value of x and t separately. For the case x/t smaller than c (that is, for any physical object), this allows you to define a proper time, which basically records how long you'd note on your own clock as you move around; but when x/t = c then the concept of proper time, rather than equalling zero, no longer makes sense. This is often referred to as "time stopping", but it is not that, because the other aspect of proper time is that it's "time in your own rest frame", but light is never at rest in any frame.

Anyway, that's a long-winded way of saying that no, time doesn't stop when you move at speed c. Still, let's ignore this for a second and consider a practical case of accelerating arbitrarily often -- as noted, you never reach c, but if you accelerate at a constant rate, then you can work out what time is recorded using the fairly simple formulas described on, for example, this page:

https://en.wikipedia.org/wiki/Acceleration_(special_relativity)#Curved_world_lines

Specifically, equations (6a). In this notation, x and t are the length and time as measured by the audience, and τ is the proper time recorded by our superhuman athlete, while α is the "proper acceleration", the value of acceleration on which all observers can agree.

As a matter of pedantry, in this set-up our athlete is "convinced" that they aren't actually moving, so that the proper distance moved is always zero. However, x is still the relevant distance -- it's the length of the track the athlete perceives before starting the race.

Just to get our bearings, the current 400m world record is 43.03 seconds, and this turns out to be equivalent to a proper acceleration of around 0.415 m/s^2 (NB in practice, athletes don't accelerate constantly through the race, but we will ignore this piffling little detail). Even if we suppose that our athlete was capable of accelerating at, say, 10^10 m/s^2 (this is actually physical despite, on the surface, implying that the athlete can exceed the speed of light -- ask about this if you care) -- for the entire race, then the athlete would still record a time of 0.000282842 seconds for the entire lap -- and, importantly, the audience would see a slightly longer time of 0.000282846 seconds. This last matters because it's evidently good practice to record the time taken to do a lap according to somebody watching, to avoid bias (and also because of the physics!).

Finally, if the athlete did indeed accelerate infinitely rapidly, then the time recorded by the athlete *does* approach zero. But, the time recorded by the audience -- or, for that matter, the referee -- is still going to be equal to just 400/c = 1.33*10^(-6) seconds. You can accelerate however fast you like, but you're never going to beat the speed of light.

tldr, NJ is still correct :)