Scanning the brain activity

How much disk space we need to store 10 seconds of brain activity?

There are 1e11 neurons in the brain. If we can scan the action potential propagating from neuron to neuron real time, we can understand how the brain works.

Real time means every roughly 100e-9s, because that’s the time an action potential spends on a single neuron: dwell time of action potential on a single neuron = size of a neuron / action potential propagation velocity = 20um / 100 m/s = 200 ns.

Today’s SSD standard is 4TB = 4e12*8 = 3e13 bits. We need to store 1e11 bits every 200 ns, hence to record ONLY 10 seconds of brain activity we need x = 10 * 1e11 / 200e-9 = 5e18 bits (625000 TB, as contrasted for 4TB laptops and PCs today).

Moore’s law state that chip components roughly double every two years, hence from today's 3e13 (standard 4TB laptops and PC) bits to get to 5e18 bits we need 36 years.

But, we can do it even today, we can easily store 10 seconds of brain activity! To buy an SSD to store 5e18 bits today would costs (at $0.3/GB for SSD today) 5e18 / 8 / 1e9 * 0.3 = $187M.



To scan the brain in full capacity we need to know the position of all action potentials in time. The time it takes for an action potential to hop from one neuron to another is size of one neuron divided by the action potential propagation speed: 20e-6 m / 100 m/s = 200 ns. Hence, all 1e11 neurons must be scanned in 200ns, which mean we only have 200e-9 / 1e11 = 2e-18 s to scan each neuron.

Neurons are 10 cm deep from the skull. Light travels 10 cm in 1e-1/3e8 = 3e-10 s, which is >> 2e-18 s. Sounds travels 10 cm in 1e-1 / 1500 = 6e-5 s, which is >>> 2e-18 s. And we don't have fast enough electronic to record these signals. Presently impossible task.







FUN FACTS
Human brain has 1e11 neurons (1e15 including synapses) in 1000 cm3.One neuron occupies a volume of 1000/1e-6 / 1e11 = 1e-14 m3. The whole volume is formed by 3D matrix of 1e11^(1/3) =  5e3 x 5e3 x 5e3 neurons.
Diameter of one neuron is d =2*r =  2*(1e-14 / (4/3) / 3.14 )^(⅓) = 2*1e-5 m = 2*10um, where V = 4/3*pi*r^3, r = (V / (4/3) / pi)^⅓.
How many neurons occupy a typical nerve with diameter 1 mm2? X = 1e-6 m2 / 400e-12 m2 =  2.5e3 neurons. Only 2500 neurons! 2D matrix of a nerve cross section is 50 x 50 x 1 neurons.