Yes, I Am Now Convinced The Brain Ages In Stages

 

                Tracking how brain connections change using MRT imagery

The notion that our brains age in distinct stages is one I have seriously considered, especially over the past quarter century as I've found my own mental agility and processing rate diminish. (Of course, a lot of the recent decline can be attributed to having to start androgen deprivation therapy to try to contain metastatic prostate cancer.)  But even before commencing ADT with Firmagon I have noted brain changes over decades  - especially in the time since graduating with my M.Phil. in Physics. To wit, my most recent attendance at a scientific conference last year:

Looking Back On Last Week's Solar Dynamics Workshop - One Of The Best Meetings I've Ever Attended

Showed me clearly that the time for delivering scientific papers, far less attending conferences, is basically over. It was all I could do to basically keep up and absorb the material and research presented in the SDO 2025 workshop. (And there was no way in hell I was staying for the final day's Python coding workshop!)

But how and why do these distinct brain 'ages' occur and is there any evidence for them? It turns out there is, now finally,  as neuroscientists have been able too use advanced imagery to track brain connections, development. (See top graphic).  Basically, this research has shown different ages at which the connections in our brains shift. The average ages turn out to be: 9, 32, 66 and 83, Well I am now 13 years past 66 and only 3 1/2 to 83 - and it shows. (Including in the much longer time to write a blog post)

According to a December, 2025 WSJ article, ('Your Brain Changes In Distinct Stages Research Shows): 

 "The adolescence phase lasts until age 32. The brain then enters a period of stability until early aging begins at age 66."

I can vouch for this as when I hit 66 was when I first confronted my mortality after being given a prostate cancer diagnosis,

Verdict Is: Prostate CANCER......So Now What?

And basically, for all intents, things haven't been the same since. This is given each succeeding year has resulted in more tests, more treatments, and sped up aging especially after the radiation in Sept., 2012. Add in the effects of the ADT and I feel I aged 25 years in the actually elapsed 13. (And each forthcoming PSA test is looked on with dread.)

But let's bear in mind the WSJ report is based healthy aging for brains, and first published in November, 2025 in the journal, Nature Communications. We find therein that the study researchers examined results from about 4,000 brain scans taken from people in the U.S. and UK.  These scans ranged from those of a newborn baby to a nonagenarian.

They basically showed how white matter -  a fatty substance insulating nerve fibers connecting different brain regions - enabled the researchers to see the changing connections over time. (Using an MRI).  Such mapping also enabled them to create an "average brain" for each year of life.  Machine learning then helped to pinpoint phases of significant change defined by the data.

For example, during the period from birth to age 9, the brain undergoes massive pruning of connections which aren't used - given the amount of excess wiring we're born with. But during the period from age 9 to 32  the brain does more with less wiring, hence becoming more efficient.

When does the brain finally become 'mature'?  Likely when it becomes most efficient in terms of the wiring, at age 32.  There is then the blessed period of 'consistency' and 'stability' (age 32 to 66) as opposed to endless emotional upheavals, convulsions and petty neuroses. (75% of mental health conditions begin in the early 20s we are informed).  This period between ages 32-66 also aligns with a plateau in our intelligence.

But the bad news begins at just past 65 when brain shrinkage occurs - increasing year by year.  This diminishes the integrity of white matter and also begins the decline in cognitive function. Past the age of 83 the brain must "rely on a small  number of highly used interregional pathways."

Bottom line: If you want to keep that brain functioning then once you're past 66 you better use it - or risk losing it. Forget the Tik Tok and other smart phone obsession - stick to chess and reading adult books, like Jean-Paul Sartre's Being and Nothingness - or at least Stephen Hawking's 'A Brief History of Time'.

See Also:

Topological turning points across the human lifespan | Nature Communications

Solutions To Algebra II Simple Machine Application Problems

1) Show, with reasons, that the pulley system shown in Fig. 3(b) has a mechanical advantage of 2. If the weight w is lifted 1m, how much distance must the force F cover?

Solution:

This is a single moveable pulley as with 3(a) but with just a small modification: adding another fixed pulley. Since M.A. = s/d = 2 then F = wd/ s and the machine moves a distance d while the applied force moves a distance s. Then: F = ½w. Then if the weight w is lifted 1m, the force must be applied a distance 2d = 2(1m) = 2m.

2) An inclined plane has an angle of Θ = π/6. The coefficient of static friction m_s = 0.3.How much force, parallel to the incline, is needed to push a 100 N object up the incline at constant speed?

Solution:

Note: Θ = π/6 = 30 degrees (e.g. 180 deg/ 6). We have w = mg = 100N and m _s  = 0.3. Then the force F' required is:  F' = mg(sin Θ + m_s cos Θ).

F' = (100 N) [sin 30 + 0.3 cos 30] = 100N[½ + 0.3(0.866)]

F' = 100N(0.50 + 0.25)= 100N (0.75) = 75N

3) The work done in sliding a 50 kg mass up an incline 1 m high and 5m long is 640 Joules. What is the frictional force in N? Compare the ideal and the actual mechanical advantage. (Take g = 10 N/kg)

Solution:

 W= mg = 50 kg (10 N/kg) =    500 N

Work done against gravity (Useful work) = mgh =    500 N (1 m) = 500 J

The total work done is: W_T   = 640 J

Work done against friction is the difference: 

W_T -  W_u   =  640 J - 500 J=  140 J

Work done by friction =  WF  = 140 J = frictional force x distance (5m)

So frictional force = F   =  140 N-m/ 5m = 28 N    (Rem: 1J = 1 N-m)

Ideal M.A. = L/h =  5m/ 1m = 5

Actual M.A. =  load/ effort =  total work output/ work input 

work in =  W_T  / s =    640 N-m/ 5m  =   128 N

work out =  mg = 500 N-m/1m =500 N

Actual M.A. = 500 N/ 128 N = 3.9

4) In the wheel and axle device (Fig. 4 (A)) the radius r = 1 cm and R = 23 cm. Find the mechanical advantage and the applied force needed to lift a load of 80 N.

Solution:

We have r = 1 cm and R = 23 cm for the wheel and axle (Fig. 4A). The distance the load (w= mg) will move is just d = 2πr. The distance the force moves will be s = 2πR. If we take the mechanical advantage:

M.A. = s/d = (2πR)/ (2πr) = R/r = 23 cm/ 1 cm = 23

mg/ F = w/F = R/r and so: F = (r/R)w

Now, if w = 80N then F = (1/23) 80N = 3.4 N

5) In the grouped pulley system depicted in Fig. 4 (C) the force applied F will move 6 times as far as the load w. If the load has a mass of 40 kg, and assuming g = 9.80 m/s^2, find the applied force. Thence or otherwise, obtain the mechanical advantage of the system. If the force F is applied through 10 m what is the work done?

Solution:

If F will move 6 times as far as the load w, then the M.A. = 6. So: F = w/6. And w = 40 kg(9.8 m/s^2) = 39.2 N. Then: F = 39.2N/6 = 6.5 N.

The work done is: W = F s = (6.5N) (10m) = 65 N-m = 65J.

6)A man raises a uniform plank 12' long and of weight 40 lbs. until it is horizontal. His left hand is on one end of the plank and his right hand is 3' from the same end. Assuming both hands exert vertical forces, find the forces exerted by each hand to support the plank.

Solution:

The diagram for Problem. #6. is shown above..

For #6 we see where the man's right and left arms are. The point is that the combined vertical forces must balance a total clockwise torque or moment of 40 lbs. x 6 ft. = 240 lbs.-ft. (By the law of levers) The right arm exerts a counterclockwise torque at its position of 3' x F1 = 240 ft-lb. So that: F1 = (240 ft.-lbs.)/ 3 ft. = 80 lbs.

The left arm exerts a second counterclockwise moment at its position (6' from the load, at the center of gravity of the plank- since it's uniform) of 6' x F2. Therefore: F2 = (240 ft.-lbs.)/(6 ft.) = 40 lbs. is the upward force exerted by his left hand.

7) In the sample lever problem it is feasible to reduce the work done to only 125 J by re-arranging the lever distances (effort and load distance). Using a sketch show how could this could be done and give the new applied force in this scenario.

Solution:

The diagram for Problem. #7. is shown above.

This is done by lengthening the effort distance to 4 m (from 3.5 m) and shortening the load distance to 1 m (from 1.5m). Then the new force exerted needed to lift the block is found from the law of levers: F x 4m = 500N x (1m) = 500 N-m.

Or: F = 500 N-m/ 4m = 125 N. (Compared to the original force exerted of 214 N). The new work done is: Fs = (1/4) 500N x 1.0 m = 125 J.

A Deeper Dive Into The Basic Principles Of Radio Astronomy

 

    Examining the basic principles of radio astronomy means first recognizing we are talking about a window of observation that involves the ionosphere as opposed to the atmosphere.  In the first case we have a terrestrial layer transparent to radio waves, in the second case we have a layer transparent to light waves  - for optical observation.   The basic electromagnetic (EM)  spectrum showing the relation of visible light to radio waves is shown below:

Note here that radio waves are at much longer wavelengths.  The earliest radio wave observations were made at kilohertz (kHz) frequencies and then gradually were able to also detect sources at megahertz and beyond.   All radio astronomy - at whatever frequency- begins with knowing the basic radiation characteristics.  To this end we wish to know:

i) The direction it comes from, 

ii) The emergent flux,  

iii) The polarization 

 In respect of the flux, consider a small element of surface ds: Then to find the amount of energy which flows per unit solid angle we use the graphic below:

From this we can obtain:

dE _v  = I cos  q ds dw dt dv

Where:

I = f(u,f,q, x, y, z, t)

  is defined as the specific intensity or brightness.  The units we are making use of here are:

[ergs / cm ²   steradian sec  cps]  

Integrating the specific intensity over the solid angle:

FLUX/  F _v  =    ò I _v   (q,  f) cos  q dw

And over the total solid angle:  

F _v  =   Total flux  =  ò _4p  I (q, f) dwf

 F _v   Units  [ergs / cm ²  sec  cps] 

  Note: The infinitesimal sold angle dw  is equivalent to:   sin q dq df

 Now we examine the specific situation for a radio telescope antenna. In the most basic kind of radio telescope system we will have:

Which shows a parabolic dish with the receiver at the prime focus.  Note that a parabolic dish has the property whereby radio waves reflected off it come to a focus at the same time.  This is critical given if the waves arrived at different times the telescope would be essentially useless.  The signal obtained would be incoherent which the reader might be able to deduce by the end of this post.  

Note that  the antenna is essentially some sort of conducting wire or metal rod which is the 'heart' of the telescope.  Its basic function is to capture incoming electromagnetic waves and convert them into an electric current in the wire.  For maximum efficiency the dimension of the antenna wire or probe should be one fourth the size of the radio wave it is intended to intercept.  Thus for the 21 cm line of hydrogen the probe should be 5 cm in length.   

There is still the matter of the parameters of the source in relation to the antenna and the solid angle formed. In the sketch below I show a radio source of area A, at a distance D from an antenna of dimension (C) and we wish to find the radio flux reaching the antenna from the source.

                     Comparison of solid angles, source & antenna

We have in the first instance:

dE _v  = I _v cos  q ds dw dE _v  =   

Then the flux reaching C from the radio source is:

F _v  =  dE _v  /  C dv dt  = I _v  A W  /  C  

 Now, if   b is the solid angle:  A/  D ²

And W  is the solid angle     C/  D ²

  Then:  A/ C   =    b / W

The  flux then received becomes:  

F ' _v  =    I _v  b

We  must use:  F ' _v   = P _v    /  C D u

Since the radio telescope measures Power:  P _v    = dE _v  / dt 

The next important feature when considering the basic principles of radio astronomy is the polarization of the radio source. This is given that the information is considerable and every non-thermal process emits some random polarization which can be described by two components, and electric (E) vector and a magnetic (H) vector:

1) E(y, t) = E ₀ sin 2 p u (t – y/c )

2) H(x, t) = H ₀ sin 2 p u (t –x/c )

Electromagnetic waves with the above vectors are often represented as in the diagram below:

Where S denotes the Poynting vector: S = E X H.  The EM- waves are  polarized when their E- field  components are preferentially oriented in a particular direction.  Basically, the up and down motion of the EM radio waves is converted to the up and down motion of electrons in the antenna wire which gives rise to the current detected at the telescope.

Other forms of polarization include:

Linearly or horizontally polarized: I.e. the E- vector is confined to one  (horizontal) plane

---------à E

Vertically polarized: I.e. the E- vector is confined to one  (vertical) plane

^ E

!
!
!

Circular: The E-vector rotates through 360 degrees

Elliptic: Any polarization not circular or plane, but note in most texts circular is regarded as a limiting case of elliptic polarization.

The equivalent circuit for a basic radio telescope system may be represented as shown below:

This contains a motor producing an oscillating potential difference (V)  and two resistors:

R _A  (Antenna resistance)   and

R _B       (transmission line and receiver resistance)

Further:  R _A   =  R _r    +   R ℓ

  

Where R ℓ  is the radiation loss resistance, and R _r   is the fictitious resistance due to oscillating electrons which radiate. Since there is an oscillating voltage there is an oscillating current, i.e.

I   =  V /  (  R _r    +  R ℓ    +  R _B )

   

The Power is then:  P   =   I ²  R _B 

The maximum power is obtained under the condition:

dP/ dR_B  =  0   (N.B.  In general, R ℓ  ≈  0,  and  R ℓ << R _r 

dP/ dR _B  =   d/ dR _B [ R _B V²  /  (R _r + R _B) ² 

=  V² { [(R _r + R _B) ²  -   R _B   2(R _r + R _B)] /(R _r + R _B)⁴   

dP/ dR _B  =  R _r ² -  R _B ² /(R _r + R _B) ⁴

Then the available power is: 

P_A   = V² /4R _B   =  V² /4R _r 

The diffuse power is:  P _r  =  I R _r ²

Let the total area of an antenna be A _T  then the total power:

P _T  =  A _T  F

Where F is the energy flux of the incoming wave . 

The Gain of an antenna:

 For an isotropic lossless antenna the flux emitted is:

F  =  (1/4p )   1/ d ²

^{(Where P is the applied Power )}

For an anisotropic antenna:

F  =  g (1/4p )   1/ d ²

 Where the gain  g is subject to the constraint:

  ò _4p  g  dw   =    4p

If:

r  =   ò   g  dw /4p    

For all secondary lobes then the beam efficiency of the antenna is:  1 -  r

The relation between the gain of the antenna and its effective aperture (A') such that:

g( (q ,  φ )  =   4p  A' (q ,  φ)/ l² 

Now, every radio telescope receives waves that come from fairly restricted areas of  the sky and this is defined by what's called the beam.  Every antenna - and by extension radio telescope- has a main lobe called the "main beam" and also side lobes off to the side.  The side lobes have different causes but most result from diffraction (Consult the astronomy archives for past answers pertaining to diffraction in optical telescopes).   The pattern of a radio telescope beam and typical side lobes are shown below:

                              Main lobe and directions for side lobes

Note the side lobes are weak but can still represent a significant fraction of the power a radio telescope receives. Generally 80-90 percent of the signal entering a receiver from a typical dish comes through the main beam, the other 10-20 percent through the side lobes.  In the case of the initial radio telescope diagram, try to imagine the wave fronts coming in at an oblique angle instead of directly., e.g.

Here, the arrival of the left side of the wave is later than the right, because it happens to travel an extra distance.  If one finds this distance d is such that:

 d sin  q     =     l/  2

Then we have the left -radio wave fully cancelling the right and no signal gets through. Obviously one wants to avoid this condition.  Ideally then, we need: 

 d sin  q     =  0.

Again, every radio telescope has maximum reception in location and this falls off according to the beam pattern.  A convenient way to specify this pattern is by using the beam width and one half the peak (power) height.   

From this one can deduce the optimum beam width in terms of the resolution. Thus, if two objects are closer than the beam width then they will blur into one, and the astronomer will be unable to distinguish the sources.  This conveys the importance of a narrow beam width, which just means the signal falls off rapidly as one shifts from one side of the source to the other.The narrower the beam width the better the resolution of the radio telescope. In general:

Beam width =  Resolution =   Wavelength /   diameter    =     l/  D 

Hence, if the wavelength is 21 cm (say for the hydrogen line) and the diameter D  of the telescope is 10 m (1000 cm) we have:   21cm /  1000 cm  = 0.021 arcseconds

See Also:

Brane Space: All Experts Redux: Basics of Radio Telescopes