# Enigmatic Code

Programming Enigma Puzzles

## Enigma 218: Relatively speaking

From New Scientist #1364, 30th June 1983 [link]

The professor during his lecture on relativity asked: “If I am in a spacecraft travelling at half the speed of light and pass another craft travelling in the opposite direction at a quarter of the speed of light, what is our relative velocity?”

“Three quarters of the speed of light,” replied one student.

“You weren’t paying attention at my last lecture,” said the professor. “We proved that, according to the special theory of relativity, when two velocities are to be added then the result is not their sum but this,” he broke off to write $(v_1 + v_2) / (1 + v_1 v_2 / c^2)$ on the board then continued, “where c is the velocity of light — 300 000 kilometres per second.

“Is it possible for two equal rational velocities to be added so that the result is an integral number of 1000 kilometres (we shall say megametres) per second?”

“No professor,” answered a bright student. “But if the speed of light is decreased by an integral number of megametres per second then it is possible.”

“But you can’t reduce the speed of light! — It is constant,” protested the professor.

“But we can imagine it to be less,” persisted the student.

The professor then suggested the amount his student had taken as the velocity of light.

“I took it as more than that, professor.”

“In that case I calculate what you took as the speed of light and all the possible sums of the equal velocities.”

Can you?

Note: I am waiting for a phone line to be connected at my new house, so I only have sporadic access to the internet at the moment.

[enigma218]

### 4 responses to “Enigma 218: Relatively speaking”

1. Jim Randell 30 August 2014 at 11:27 am

This puzzle appears to be based on a flawed premise.

This program looks for possible integral values of c (in megametres/s) and integral values of m (in megametres/s) such that the the relative sum of velocities v, vv = m implies that v is rational.

```from fractions import Fraction as F
from enigma import irange, is_square, printf

# speed of light in megametres/s
for c in irange(1, 300):
# sum of velocities (less than c)
for m in irange(1, c - 1):
# compute possible velocities
r = is_square(c * c - m * m)
if r is None: continue
for s in (+1, -1):
v = F(c * (c + s * r), m)
if not(v > c):
# check the computed sum
assert m == F(2 * v * c * c, c * c + v * v)
printf("c={c}, v={v} ({f}), m={m}", f=float(v))
```

And it finds lots. Including the following values for c = 300.

c = 300, v=300/7, m=84.
c = 300, v=100, m=180.
c = 300, v=150, m=240.
c = 300, v=225, m=288.

These can be verified by computing the relative sum vv for each value of v:

2×300/7 / (1 + (300/(7×300))²) = 600/7 / (1 + (1/7)²) = 600 / (7 × (1 + 1/49)) = 600 / (7 × 50/49) = 600 / (50/7) = 12 × 7 = 84.

2×100 / (1 + (100/300)²) = 200 / (1 + (1/3)²) = 200 / (1 + 1/9) = 200 / (10/9) = 20 × 9 = 180.

2×150 / (1 + (150/300)²) = 300 / (1 + (1/2)²) = 300 / (1 + 1/4) = 300 / (5/4) = 60 × 4 = 240.

2×225 / (1 + (225/300)²) = 450 / (1 + (3/4)²) = 450 / (1 + 9/16) = 450 / (25/16) = 18 × 16 = 288.

So, in fact, there are several possible rational (and even integral) velocities, such that the relative sum of two equal velocities is an integral number of megametres/s without reducing the speed of light from 300 megametres/s.

The next lowest integral number of megametres/s that works is 299, and this gives rise to the following two solutions:

c = 299, v = 299/5, m=115.
c = 299, v = 598/3, m=276.

• Jim Randell 30 August 2014 at 11:28 am

The published solution is as follows:

Let integral V be the sum of the two equal rational velocities v.

Then V = 2v(1 + v²/c²) = 2vc²/(c² + v²).

So Vv² – 2c²v + Vc² = 0;

so v = (c² ± c√(c² – V²))/V;

so that v is rational 300² – V² = U² (say), so 300² is the sum of two
squares which is possible only if 300 can be expressed as a² + b²
(where a and b are integers). This is possible. 299 equals a² + b² is
also possible.

Therefore 299² = 115² + 276², so V = 115 or 276 and the bright student
took 299,000 km per second as the speed of light.

Oddly this answer points out that 300² = a² + b² is possible:

300² = 84² + 288² = 180² + 240², and this gives rise to the 4 solutions found above.

But if we ignore these solutions and look for the next lowest solution we get:

299² = 115² + 276²

And this gives rise to the solutions:

c = 299, v = 299/5, m=115.
c = 299, v = 598/3, m=276.

Which is the published solution:

Solution: The student took 299,000 km/s as the speed of light. The possible sums of the equal velocities are: 115,000 km/s (when the velocity is 59,800 km/s), and 276,000 km/s (when the velocity is 199,333 ⅓ km/s).

2. arthurvause 31 August 2014 at 12:16 pm

A very strange Enigma. Just for fun, I have written a variation, that hopefully makes sense. I will wait a few days before adding a solution:

“The speed of light in a vacuum is exactly 299 792 458 m/s” stated the professor, who then asked:
“Is it possible for two equal rational velocities to be added relativistically so that the result is an integral number of metres per second?”

“Yes professor,” answered a bright student. “But if the speed of light is decreased by an whole number of metres per second, by a medium such as air for example, then sometimes it is not possible.”

What is the largest speed of light, with an integral value in m/s, for which there are no equal rational velocities that add relativistically to an integral number of metres per second?

• arthurvause 31 August 2014 at 1:59 pm

I think relativistic addition of velocities always uses the speed of light in a vacuum, so in order to make my variation work, the assumption has to be that the speed of light in a vacuum is decreased by a whole number of m/s.

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