Puny Earth doomed?

Calculus Level 4

A race of super intelligent aliens have very large perfectly spherical heads. Each kilogram of their brain operates at the maximum possible computational speed ( Bremermann's limit ) which is

c 2 h 1.359 × 1 0 50 bits s 1 kg 1 . \frac { { c }^{ 2 } }{ h } \approx 1.359\times10^{50}\text{ bits}\cdot\text{s}^{-1}\cdot\text{kg}^{-1}.

The radius of an alien's head is 0.5 m . 0.5\text{ m}. Since the density of its brain is non-uniform, it is given by

ρ ( r ) = 0.5 r 1 0 3 kg/m 3 , \rho (r)=\frac { 0.5-r }{ 10^{ 3 } }\text{ kg/m}^3,

where r r is the distance (in meters) from the center.

A single human brain is capable of computing 400 × 1 0 9 400\times10^9 bits per second. In terms of computing power, how many human brains is a single alien brain equivalent to?

Details and Assumptions

  • For this problem the given approximation of Bremermann's limit will suffice.

  • Take a look at Micheal Steven's cool YouTube video " How many things are there? " where he uses Bremermann's limit to count everything .


The answer is 8.895E+34.

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3 solutions

d V = 4 π r d r dV=4\pi r dr

d m = ρ ( r ) d V = 0.5 r 1 0 3 4 π r d r = 0.5 r r 2 1 0 3 4 π d r dm=\rho(r)dV=\frac { 0.5-r }{ 10^{ 3 } }4\pi r dr=\frac { 0.5r-r^2 }{ 10^{ 3 } }4\pi dr

m = 0 m d m = 0 1 2 0.5 r r 2 1 0 3 4 π d r 261 1 0 6 \displaystyle m=\int_{0}^{m}{dm}=\int_{0}^{\frac{1}{2}}{\frac { 0.5r-r^2 }{ 10^{ 3 } }4\pi dr}\approx 261\cdot10^{-6}

C o m p u t a t i o n a l s p e e d o f a l i e n = 1.359 1 0 50 261 1 0 6 = 3.547 1 0 46 Computational speed of alien=1.359\cdot10^{50}\cdot261\cdot10^{-6}=3.547\cdot10^{46}

R a t i o = 3.547 1 0 46 400 1 0 9 = 8.895 1 0 34 Ratio=\frac{3.547\cdot10^{46}}{400\cdot10^9}=8.895\cdot10^{34}

A K
May 3, 2014

Average density of the alien brain is given by:

0 0.5 0.5 r 1 0 3 d r = 1.25 × 1 0 4 k g m 3 \int_0^{0.5} \frac{0.5-r}{10^{3}}\,\mathrm{d}r = 1.25{\times}10^{-4}kgm^{-3}

Since the alien's head is a sphere, and assuming that the brain takes up all of its head volume:

M a s s = 4 π 3 ( 0. 5 3 ) × 1.25 × 1 0 4 = 6.545 × 1 0 5 k g Mass = \frac{4\pi}{3}(0.5^{3}) \times 1.25{\times}10^{-4} = 6.545{\times}10^{-5}kg

The total computational power of the alien brain is therefore:

6.545 × 1 0 5 × 1.359 × 1 0 50 = 8.895 × 1 0 45 b i t s / s e c o n d 6.545{\times}10^{-5} \times 1.359{\times}10^{50} = 8.895{\times}10^{45} bits/second

R a t i o = 8.895 × 1 0 45 400 × 1 0 9 = 2.224 × 1 0 34 Ratio = \frac{8.895{\times}10^{45}}{400{\times}10^{9}} = \boxed{2.224{\times}10^{34}}

Moderator note:

This solution is incorrect. Do you know why?

The calculation is correct, but the density factor in the problem is wrong. The density of water is 1000 kg/m^3 , so a credible brain density would be (0.5 - r) 10^3 kg/m^3 ; the problem uses a density much less than air! So the ratio should be 10^6 larger, which just makes things all the worse for humankind!

Gregory Ruffa - 7 years, 1 month ago

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You are right,it was a simple error when I was converting units.Unfortunately I couldn't change the answer once I posted..Even though it is implausible I guess it serves to demonstrate the power of the Bremmerman limit.

Thaddeus Abiy - 7 years, 1 month ago

This solution is incorrect. Do you know why?

Calvin Lin Staff - 5 years, 7 months ago

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May I ask why?

A K - 5 years, 5 months ago

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Hint: Review Volume of Revolution - Disc Method and contrast it against Volume of Revolution - Shell Method .

Calvin Lin Staff - 5 years, 5 months ago

Wouldn't you get the mass of it by having the integrand be 4 pi r^2 * (.5-r)/1000 ? I get pi/4800 for the mass when I did it, which makes the final answer 3.9E+33

Alex Fischer - 7 years, 1 month ago

Nice solution.

Vaibhav Chaturvedi - 7 years, 1 month ago
Zakaria Salameh
May 2, 2014

mass of alien's head = integral(rho .dr)[from 0 --->0.5] * v(volume of his head)= 0.125 10^-3 * (0.5)^3 pi (4/3) computational speed of alien=m (c^2/h)=8.8946 10^45 bets/s ratio between alien and human minds c.p =2.223658 10^34 (approx.)

Moderator note:

This solution is incorrect. Do you know why?

This solution is incorrect. Do you know why?

Calvin Lin Staff - 5 years, 7 months ago

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