Relativistic flies

Classical Mechanics Level 5

Consider a group of $N=100$ relativistic houseflies, flying along the x-axis. Each fly $F_{k}$ sees the next fly $F_{k+1}$ moving away at $u=0.01 c$ (c is the speed of light). What is the relative speed $v_{100}$ in meters per second between the leftmost fly, $F_{1}$ , and the last fly, $F_{100}$ ?

Details and assumptions

Assume $c=3\times 10^{8}~\text{m}/\text{s}.$

The answer is 2.27E+8.

2 solutions

Đinh Ngọc Hải
Dec 23, 2013

The speeds of flies $F_k$ and $F_{k+1}$ in the frame of $F_1$ are connected by the equation: $v_{k+1}=\frac{v_k+u}{1+\frac{uv_k}{c^2}}$ .

Therefore, we can calculate $v_{100}=0.757c=2.27 \times 10^8 m/s$ .

The simplest way to solve this is to use the hyperbolic tangent tanh function.

If w is the sum of two relativistic velocity xc and yc, we have w = c(x + y)/(1 + xy)

Let w/c = z, then z = (x + y)/(1 + xy)

If we let x = tanh(p) and y = tanh(q), then we get z = [tanh(p) + tanh(q)]/[1 + tanh(p)*tanh(q)] which is the identity for tanh(p+q)

In this question, the relative velocity of the flies are 0.01c. Therefore, x=y=0.01, and p=q=arctanh(0.01)

In other words, adding the relative velocities of two flies gives us c tanh(2p), and this can be inductively shown that adding the relative velocities of n flies gives us the sum of relative velocities of c tanh(np).

This in problem, there are 100 flies and therefore 99 relative velocities to consider. Therefore the relative velocity between the fly 1 and fly 100 is c*tanh(99p) where p is arctanh(0.01).

Try it. This method allows you to solve it without brute force of punching numbers repeatedly into your calculator or spreadsheets.

Tong Choo - 7 years, 4 months ago

Beautiful problem!

Sam Thompson - 7 years, 4 months ago

Hi Dinh!

How do you get that formula?

Pranav Arora - 7 years, 5 months ago

Pranav, it is a well known formula from special relativity. Here is a link to more info.

Jonathan Schirmer - 7 years, 5 months ago

Thanks Jonathan!

I am not yet introduced to special relativity so I did not know about the formula.

Pranav Arora - 7 years, 5 months ago

Yes, as Jonathan said, it's a well known formula from special relativity. I don't know the words in English but in my country, it's called formula to transfer the velocity in two frames:

$v_{31}=\frac{v_{32}+v_{21}}{1+\frac{v_{32}v{21}}{c^2}}$ .

Đinh Ngọc Hải - 7 years, 5 months ago

Hey Dinh,

How did you calculate v100 without going through each frame one by one?

Samuel Jackson - 7 years, 5 months ago

Yes, I did go through each frame one by one. In my calculator, I pressed 0.01 and =, then pressed (0.01+Ans)/(1+0.01*Ans) and pressed = for 98 times. I thought 98 times are not too much, it took about 1min.

Đinh Ngọc Hải - 7 years, 5 months ago

Chew-Seong Cheong
Aug 1, 2014

According to the Theory of Specific Relativity, the relative speed $u'$ in a frame $S'$ moving at a speed $v$ in a frame at rest $S$ , when measured in frame $S$ is given by: $u = \cfrac{u'+v}{1+\cfrac{u'v}{c^2}}$ where $c$ is the speed of light.

For our case, the frame at rest is that of $F_1$ , the relative speeds of $F_1$ and $F_2$ to $F_1$ are:

$v_1=0$

$v_2=0.01c$

Since the relative speed of $F_3$ to $F_2$ is $0.01c$ and that to $F_1$ is given by: $v_3 = \cfrac{0.01c + v_2}{1+\cfrac{0.01v_2}{c}}=0.019998c$ Similarly, $v_k = \cfrac{0.01c + v_{k-1}}{1+\cfrac{0.01v_{k-1}}{c}}$ Using the above equation, we find $v_3, v_4.... v_{100}$ , and $v_{100} = 0.757376396c$ $= \boxed{2.272 \times 10^8}$ $m/s^2$ .

Relativistic flies

The answer is 2.27E+8.

2 solutions

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