Stars Red shift can be from a medium instead of a movement

1 The intensity of light should be inversely proportional to the distance between the source and the observer before considering how much intensity is lost by traveling through a medium

2 The intensity of light is lost as it travels through a medium by a different amount for each frequency or wavelength

3 The farther a light source is from an observer the more mass it will need to have as a source if made of the same material at the same temperature and density in order to have sufficient intensity of brightness to be seen by the observer and the more mass the light source has the greater gravitational red shift will occur and the greater the gravitational red shift that occurs the less energy per area per time the observer will experience from the light emitted by the source.  Light from some sources might not experience a red shift but the percent of those light sources that do not have enough mass to be visible would increase as their distance from the earth increases.  As the distance increases the minimum mass required and thus the minimal amount of red shift required for a stationary light source (of the same material, temperature, pressure and density) to be seen increases.  This is  assuming that gravitational red shifting and blue shifting really does occur in real life.

4 From a probabilistic viewpoint, the further the distance between the source and the observer the more likely at least one opaque object of a given size or greater will be between the source and the observer.  This might be similar to the relationship of the number of cracks on a section of road and the length of the segment of a road or the Poisson distribution.

Points 1 through 4 could account for red shifts from distant stars and a lack of unlimitted or infinite brightness experienced on earth by an individual in a universe that might be of unlimited or infinite volume and might have never experienced a big bang and in which some stars that were thought to be moving away from an observer due to a doppler shift might not be moving away from the observer in all cases that big bang believers have claimed them to be doing so.

Alleged shifts in light spectrums may occur from light sources that are stationary relative to the earth as the light from a stationary source traveles through the medium of outer space not necessarily through a doppler effect caused by a light source moving away from or towards the earth.

The intensity of light at one frequency or wavelength may be decreased by a different percent than the intensity at another frequency or wavelength for the same medium at the same thickness

Outer space is supposedly an imperfect vacuum which would mean it is a physical material.  If the physical material outer space is made out of reduces the intensity of blue light more strongly than it reduces the intensity of red light or absorbs blue light better than red light traveling through the same distance or thickness of material at the same temperature, pressure, chemical concentration and density then the further the distance between a stationary light source and an observer the more red shifted the light would appear to be to that observer.

By red shift in the context of the effects of light traveling through a medium, I mean a shift in the frequency and wavelength distributions of the light.  In this specific type of red shift the weighted average of frequency with respect to intensity would decrease and the weighted average of wavelength with respect to intensity would increase as the distance the light has to travel from the source through the medium to reach the observer increases.  A blue shift in this context would mean the weighted average of frequency with respect to intensity would increase and the weighted average of wavelength with respect to intensity would decrease.  The frequency of the individual photons would not change as the distance traveled through the same medium increases but the intensity would change for a given frequency as the distance traveled through the same medium increases.  The percent decrease in intensity would be different for blue light than red light since the medium the light is traveling through blocks or absorbs one wavelength or frequency a different amount than it does with another wavelength or frequency.  I mention traveling a further distance through the same medium because changing what medium a wave is in can also change it's wavelength intensity distribution and or frequency intensity distribution but that is not the source of the red shift I am talking about.  

One should also consider that if the path of a light source orbited around a observer in a perfect circular path at constant absolute value of acceleration and absolute value of velocity and maintained a constant distance from that observer what sort of doppler effect ( if any ) would happen even though the  light source is neither moving towards or away from the observer in terms of the absolute value of the distance between the source and the observer

Supposedly gravitation could also cause a shift in frequency that is separate from or in addition to the one allegedly caused by velocity.  The farther a star is from earth the more massive the star would usually have to be to be seen from earth because the intensity of light is inversely proportional to the square of the distance between the observer and the light source before intensity lost due to traveling through a medium is even considered.  This means a greater gravitational red shift would usually be experienced the farther the distance from the earth a star is observed from even if it was not moving away from the earth.

In the case of gravitational red shift I believe the claim would be that each photon actually individually increases it's wavelength and or decreases it's frequency.  It is possible for a wave to change it's frequency without changing it's wavelength or change it's wavelength without changing it's ftequency if the velocity or speed of the wave changes which can happen if it changes it's medium or if the medium changes some factor like temperature.

A universe of infinite or unlimited volume with an unlimited number of stars would not necessarily be unlimitedly or infinitely bright as viewed from the earth because objects between some of the stars and the earth would be opaque or partly opaque and block some of the light from other stars or objects in addition to light becoming more dim before it reaches or fails to reach earth through the inverse square law and the effect of traveling through a medium as well as allegedly shifting in frequency and or wavelength and reducing in energy per photon through gravitational red shifting it experiences on the way to earth from some objects near it's path exceeding any gravitational blue shifting it experiences on the way to earth from other objects.

Someone might argue that although I listed three different sources of red shifts ( something similar to Beer-Lambert Law, Doppler, Gravitation) we can still know the universe is expanding because each source of red shift effects things in a mathematically different way and we can know which one of each type contributes to how much of each and from there calculate the doppler shift and know how fast each of the stars are moving away from the earth based on the frequency distribution a star should emit based on it's chemical composition.  The problem with that is we do not know the chemical composition of stars.  Supposedly every element we have tested on earth emits a unique distribution of light frequencies or wavelengths and the distribution emitted by a star is the same as the distribution emitted by the elements or perhaps chemical compounds that make it up only shifted in a precise way such that there is only one possible combination of elements or compounds that could have emitted that light frequency or wavelength distribution with not a single one to possibly be added or removed from the list mathematically determinable for that star.  That however is simply untrue because we can only compare the distributions with distributions for materials found on earth and we have never flew a rocket to the sun and collected a sample of the matetial to know it is not made from something different, perhaps an undiscovered element that does not even fit on the periodic table with it's integer atomic numbers the way we currently understand it to work and maybe this element emits what looks like a shifted distribution of another element.  It sounds ridiculus but theoritical astronomers claim there is dark matter that is not like any material discovered on earth but somehow eliminates their massive calculation errors if you put the right amount of this dark matter in if I am being ridiculus for claiming the stars might be made of a material not discovered on earth then they are being ridiculus for claiming outer space has a material not discovered on earth that just so happens to exist in the quantities to reduce their calculation errors from the amount that disproves their theories to an amount that excuses their theories.  I am not saying the stars are made of a undiscovered or hypothetical material, I am saying they might be but to some people it is a definitive fact that 85% of the universe is made out of a hypothetical material.  Here on earth I have never known a construction worker to build a building out of a hypothetical matetial, they might build something out of a material they do not know the name of but it is not hypothetical they can actually touch it, unlike the sun which has never been touched by human hands if it is so hot you will burn up before you reach it as has been claimed and more importantly if it is so high in the sky I can not build a ladder or skyscraper or climb a mountain to reach it.  There is another problem than not knowing the material the stars are made out of and that is not knowing how much gravity they have which would mean not knowing how much of the red shift is from gravity vs how much is from doppler.  We can know how much acceleration is caused by local gravity on earth but we can not personally know how much acceleration is caused by local gravity on the moon unless we personally go to the moon which most people can not do and therefore can not personally confirm the moon data is not fake.  If the most people could go to the moon then most people could travel to the moon and take the acceleration measutements from there personally instead of going by the word of people who already have a track record of fabricating stories about dark matter.  We can not personally measure the Universal gravitational constant from the earth only the local gravitational constant because the universal gravitational constant is calculated by manipulating the position of objects in a physics laboratory and carefully measuring their masses but the person who takes the observation's mass and position is not measured but worse yet the mass and position of the walls, ceilings and floors of the building next to the objects is not measured.

We do not know

1 How much of the red shift is from gravity - because we do not know the universal gravitational constant - because they ignore the mass of the building but measure the mass of the equipment inside the building) - you can take a video camera and drop an object in front of a ruler with a stopwatch also running on video and record it and measure local acceleration due to gravity but that is different than proving a Universal Gravitational Constant even exists - And no you probably have not tested it on the moon if you are reading this

2 How much of the red shift is from traveling through a medium -  Because we do not know how outer space effects light over long distances involving something similar to the Beer-Lambert law - Because the common person can not fly up in a satelite and shoot several different types of monochromatic lasers through the medium outer space is made up of and take measurements.  Even if we could we have not taken those measurements to make sure outer space between the star and wherever the observers atmosphere becomes outer space is the same material

3 What the starting spectrum that the light frequency distribution was shifted from - Because we do not know what the spectrum of the star would have been based on it's material chemical composition - Since we do not know what material the stars are made out of since we can not assume they are made from chemicals that exist on earth because for all we know they could be 85% dark matter, why not 85% of everything else is except things we can observe on earth

4 Once we know those three things we do not know we can calculate the doppler shift of a star based on a extrapolation of other things that have never been tested in that data range  - If you do not believe in doppler you could set up a race track and take a video camera and could measure the location and time positions of the car and compare it with the doppler speed data for a wide variety of speeds less than 120 miles per hour that a commoner can afford a car to drive at but you can not drive your car at 3/4 the speed of light and take measurements.

 "The primary evidence for dark matter comes from calculations showing that many galaxies would behave quite differently if they did not contain a large amount of unseen matter"

That is not the primary evidence for dark matter that is the primary evidence of at least one of five things 

1 the "physics" assumptions that led to you doing those calculations are wrong

2 You made an expertimental error collecting the data

3 You made an error doing the calculations with the data you collected

4  It's not the theoritical astrophysisicts fault some other physicist calculated G wrong they just plugged the wrong number someone else calculated into the correct equations

Although a novice  would come to the instinctual conclusion that astronomers made a 85% error in how much mass is in the universe after further examination they would reslise it is only a negative 46 % error.  But a true expert would say there was no calculation error at all for the mass except we put in the wrong value for G which was clearly wrong because we ignored those massive walls every time we did a calculation to figure out G

We really need to multiply or perhaps divide what we thought G was by 3.4225 to get the true value of G.  Or perhaps G is another value altogether calculated a different way then how I suggested but it would not surprise me at all if G was pure fiction since g is measurable but no experiment to this day has used correct methodology to measure g that I know of taking into account the mass and position of the human being that looked into the equipment and the mass and position of any walls structures or buildings nearby.  Perhaps it could be measured out doors far away from massive buildings in a region where local g measurements are consistent at several locations near the test site to make sure there is not extra density mass in a section of the ground nearby and with no wind to shift the results.

1.85 * 1.85 = 3.4225

F = G*m1*m2 / (R^2)

3.4225*G*m1*m2  = 1.85*m1*1.85*m2

(G / 3.4225) * m1* m2= G* ( m1 / 1.85 ) * ( m2 / 1.85 )

It is theoritical that there is 85% more dark matter than expetimentally observed.

Traditional Percent Error formula based on assumption that the theory is correct and the experimental data is wrong

  experimental - theoritical / theoritical

Theoritical amount of mass in universe = 1.85 * Experimental

( E - 1.85E ) / 1.85 E= - 0.85 / 1.85 = - 0.459459459459

So I made up a theory and my observed data does not match my theory, I know reality is wrong not my theory the error is in the data not the theory

Keeping it real percent error based on that the experimental observations are correct and the theory that dark matter exists is wrong

theoritical - experimental / experimental

( 1.85E - E ) / E = 0.85

5 There was never any data and this is a practical joke and the common people can not ever look into one of these expensive telescopes to find out there never were any observations just fabricated observations used as an excuse to request tax funding for research and to brag about how smart you are but humbly still not educated enough until you get more money because there is a 85% unknown missing matter mystery that needs more funding.

Copyright Carl Janssen 2022

https://en.m.wikipedia.org/wiki/Beer%E2%80%93Lambert_law

http://web.archive.org/web/20220314215558/https://en.m.wikipedia.org/wiki/Beer%E2%80%93Lambert_law

https://en.m.wikipedia.org/wiki/Gravitational_redshift

http://web.archive.org/web/20211126104210/https://en.m.wikipedia.org/wiki/Gravitational_redshift

https://en.m.wikipedia.org/wiki/Inverse-square_law

http://web.archive.org/web/20220320133858/https://en.m.wikipedia.org/wiki/Inverse-square_law

https://en.m.wikipedia.org/wiki/Poisson_distribution

https://hubpages.com/@fatfist

http://web.archive.org/web/20220313060928/https://hubpages.com/@fatfist

https://discover.hubpages.com/education/OLBERS-PARADOX-A-Physical-Explanation-For-The-DARK-Night-Sky

http://web.archive.org/web/20210518201220/https://discover.hubpages.com/education/OLBERS-PARADOX-A-Physical-Explanation-For-The-DARK-Night-Sky

https://duckduckgo.com/?q=red+shift+not+caused+by+movement&ia=about

https://duckduckgo.com/?q=red+shift+beer+lambert&ia=web

The most absorbance is obtained when beam color is complementary to the color of solution.

https://colors-newyork.com/what-is-red-shift-what-does-it-indicate/

http://web.archive.org/web/20220323062447/https://colors-newyork.com/what-is-red-shift-what-does-it-indicate/

Dark matter is a hypothetical form of matter thought to account for approximately 85% of the matter in the universe

https://en.m.wikipedia.org/wiki/Dark_matter

http://web.archive.org/web/20220321143913/https://en.m.wikipedia.org/wiki/Dark_matter

 The primary evidence for dark matter comes from calculations showing that many galaxies would behave quite differently if they did not contain a large amount of unseen matter

http://web.archive.org/web/20220321143913/https://en.m.wikipedia.org/wiki/Dark_matter

14 out of 54 mentions of walls

That is to say, we now accept apples as having easily measurable and verifiable gravitational attractions, but we ignore the gravitational attractions of walls weighing thousands of pounds. 

http://milesmathis.com/caven.html

http://web.archive.org/web/20220130225907/http://milesmathis.com/caven.html

some scientists continue to work on models that might not require dark energy. Inhomogeneous cosmology falls into this class.

https://en.m.wikipedia.org/wiki/Inhomogeneous_cosmology

http://web.archive.org/web/20220322060158/https://en.m.wikipedia.org/wiki/Inhomogeneous_cosmology

https://en.m.wikipedia.org/wiki/Exotic_matter

http://web.archive.org/web/20220321143900/https://en.m.wikipedia.org/wiki/Exotic_matter

https://en.m.wikipedia.org/wiki/Exotic_atom

http://web.archive.org/web/20220319073945/https://en.m.wikipedia.org/wiki/Exotic_atom

Historical Origins Science is not real experimental science

 The only model consistent with science is an eternal universe not big bang or finite amount of time ago earth creation.  The assumption for the big bang is we can use input  data to extrapolate what happened in the past as an output and then put that output data in as a new input to figure out what happened even further in the past.  The only way to be consistent about it is to allow someone to extrapolate as far back in the past as they want, if you limit yourself to billions of years ago when the special event of the big bang occured instead of trillions or an unlimmitted amount of years big bangers are no better than the creationists who limit extrapolation to thousands of years in the past when the special event of creation occured or when the special event of a worldwide flood occured.  Both secular evolutionary big bang cosmology origin science and creationist origin science are not real experimental science.

Copyright Carl Janssen 2022 March 22

sine and cosine of sum of two angles

http://web.archive.org/web/20220314121223/https://upload.wikimedia.org/wikipedia/commons/thumb/f/f5/AngleAdditionDiagramSine.svg/338px-AngleAdditionDiagramSine.svg.png

http://web.archive.org/web/20201111211702/https://en.m.wikipedia.org/wiki/File:AngleAdditionDiagramSine.svg

http://web.archive.org/web/20150208110925/https://upload.wikimedia.org/wikipedia/commons/f/f5/AngleAdditionDiagramSine.svg

http://web.archive.org/web/20220319192920/https://en.m.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2013_November_24

http://web.archive.org/web/20220311205122/https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

https://en.m.wikipedia.org/wiki/Small-angle_approximation#Angle_sum_and_difference

http://web.archive.org/web/20210827113612/https://en.m.wikipedia.org/wiki/Small-angle_approximation

https://en.m.wikipedia.org/wiki/Proofs_of_trigonometric_identities#Angle_sum_identities

http://web.archive.org/web/20210506193820/https://en.m.wikipedia.org/wiki/Proofs_of_trigonometric_identities

https://ccssmathanswers.com/trigonometrical-ratios-of-90-degree-minus-theta/

http://web.archive.org/web/20210411120031/https://ccssmathanswers.com/trigonometrical-ratios-of-90-degree-minus-theta/

https://imgflip.com/i/69a0vj

https://i.imgflip.com/69a0vj.jpg

http://web.archive.org/web/20220319200030/https://imgflip.com/i/69a0vj

http://web.archive.org/web/20220319200030im_/https://i.imgflip.com/69a0vj.jpg

Square Roots of Complex Numbers without trigonometry or exponential functions

 

M, N, D and E do not equal 0 

D and E are real numbers

i * i = - 1

- i * - i = - 1

D + E i = ( M + N i ) ^ 2 = M ^ 2 - N ^ 2 + 2 M N i

D = M ^ 2 - N ^ 2

E = 2 M N

N = E / 2 M

D = M ^ 2 - ( E / 2 M ) ^ 2

You should not multiply or divide both sides of an equation by a variable unless you exclude cases of division by zero but that has already been done in the preconditions already set

D M ^ 2 = M ^ 4 - ( E ^ 2 ) / 4

0 = 4 M ^ 4 - 4 D M ^ 2 - E ^ 2

if a x ^ 2 + b x + c = 0 then x = ( - b +- [ b ^ 2 - 4 a c ] ^ 0.5 ) / ( 2 a )

x = M ^ 2

a  = 4

b = - 4 D

c = - E ^ 2

M ^ 2 = ( 4 D +- [ ( - 4 D ) ^ 2 - 4 * 4 * (- E ^ 2) ] ^ 0.5 ) / ( 2 * 4 )

M ^ 2 = ( 4 D +- [ 16 D ^ 2 + 16 E ^ 2 ] ^ 0.5 ) / 8

M = + - [ ( 4 D +- [ 16 D ^ 2 + 16 E ^ 2 ] ^ 0.5 ) / 8 ] ^ 0.5

N = +- 0.5 E / [ ( 4 D +- [ 16 D ^ 2 + 16 E ^ 2 ] ^ 0.5 ) / 8 ] ^ 0.5

Although there are eight possible solutions for N and four possible solutions for M based on what sign is chosen based on the plus or minus signs, only certain combinations of M and N values work and there should theoritically be exactly two solutions for the square root of D + Ei and not all combinations of solutions for M and N will work so you should check your solutions

if M is a pure imaginary number then N will also be a pure imaginary number and this would not prevent a solution

if M is a real number then N will also be a real number

Example

D + E i = 1 +  i

M =  + - [ ( 4 * 1 +- [ 16 + 16 ] ^ 0.5 ) / 8 ] ^ 0.5

Real solutions for M and N

In this case when M is positive then N is positive and when M is negative then N is negative because both D and E are positive

https://duckduckgo.com/?q=(+(+4+*+1+%2B+(+16+%2B+16+)+%5E+0.5+)+%2F+8+)+%5E+0.5&ia=calculator

https://duckduckgo.com/?q=0.5+%2F+(+(+(+4+*+1+%2B+(+16+%2B+16+)+%5E+0.5+)+%2F+8+)+%5E+0.5+)&ia=calculator

M = approximately +- 1.09868411347

N = +- 0.5 / 1.09868411347 approximately

N = approximately +- 0.455089860562

M ^ 2 - N ^ 2 = 1.09868411347 ^ 2 - 0.455089860562 ^ 2 = 1.00000000001 = D = = 1 approximately

2MN = 2*1.09868411347*0.455089860562 = 1 = E approximately

https://duckduckgo.com/?q=2*1.09868411347*0.455089860562&ia=calculator

https://duckduckgo.com/?q=(+1.09868411347+%5E+2+)+-+(+0.455089860562+%5E+2+)&ia=calculator

Pure imaginary solutions for M and N

Flip Flopped Solutions for the absolute values of M and N compared to real solutions

These have the same absolute values as before except M and N are switched and M and N are pure imaginary numbers instead of real numbers.  Only two out of these four solutions work based on choosing the sign correctly.  I recommend simply solving for the real solutions for M and N because figuring out the proper sign may be easier for most people that way.

M = +- 0.455089860562 i

N = +- 1.09868411347 i

https://duckduckgo.com/?q=(+(+4+*+1+-+(+16+%2B+16+)+%5E+0.5+)+%2F+8+)+%5E+0.5&ia=calculator

https://duckduckgo.com/?q=0.5+%2F+(+(+(+4+*+1+-+(+16+%2B+16+)+%5E+0.5+)+%2F+8+)+%5E+0.5+)&ia=calculator

Copyright Carl Janssen 2022 March 18

Skills, majors and careers to prepare you for medical school to become a doctor, physical therapist or pharmacist

The entry requirements for different career fields and college degrees may vary by time and region and the material in this article may not be correct or may no longer be correct at a future date 

Normally to attend college to get a doctoral degree to become a medical doctor or physical therapist or pharmacist one can get a Bachelors degree in anything as long as someone completes a minimum number of courses and gets a high enough grade point average.  

This however may not be best practice because there are certain hands on skills related to medicine someone can practice before completing their bachelors degree which may make them better equipped than other people who make it into the same medical school without learning these skills first. 

It is very common for people to major in biology or chemistry to try to become a medical doctor but I would suggest they could consider earning a bachelors degree where they actually learn how to practice medical skills first and use that with some biology and chemistry electives to get into medical school instead.  One could also get associate degrees or certificates that teach skills related to practicing medicine at a technical college and transfer those credits towards a bachelors degree to get in upon completing a Bachelors degree.  

Human dissection maybe traumatic for some individuals but maybe required as part of the coursework for Physical Therapy School or Medical School, more over someone who can not deal with human dissection of a dead individual might have even more difficulties as a surgeon in terms of either dexterity or psychologically handling gore and or fear of blood and germs involved with human dissection.  Some people physically or psychologically respond with pain upon seeing other people injured and this may also be a barrier to completing human dissection or doing surgery.  There is a high correlation between a career in surgery and the type of psychopathy or sociopathy defined as not experiencing pain upon witnessing other people experience pain, this type of psychopathy or sociopathy is not necessarily bad unless someone also enjoys other people being harmed or wants to harm other people.

Butcher - Better prepares someone for surgery and medical school human dissection labs.  Not all butchers kill animals some butchers simply dissect animals other people have killed, other butchers kill animals.  I am not endorsing killing animals by posting this.  Good preparation for becoming a Physical Therapist or Medical Doctor

Taxidermist - Better prepares someone for surgery and medical school human dissection labs.   Good preparation for becoming a Physical Therapist or Medical Doctor

Mortician - Better prepares someone for surgery and medical school human dissection labs.   Good preparation for becoming a Physical Therapist or Medical Doctor

Sewing - Practicing this skill as a hobby or career will make you better equipped to do surgery, not all medical students have the property dexterity to stitch patients and practicing sewing may help.   Good preparation for becoming a Medical Doctor

Nursing - You can get a nursing degree before going to medical school or there are some types of nursing that do not require a degree.   Good preparation for becoming a Physical Therapist, Pharmacist or Medical Doctor

Dietitian - A dietitian bachelors degree can meet the requirements to go to medical school if the right courses are taken.  Good preparation for becoming a Physical Therapist, Pharmacist or Medical Doctor

Athletic training submajor for a Kinesiology Bachelors degree - Some schools offer multiple Kinesiology Bachelor degree sub majors.  Although any major can get you in with the correct courses not all sub majors provide hands on skills.  Students who choose the correct sub major which may specifically be athletic training in the case of some Universities may be taught how to assess patients for certain injuries as a hands on skill where as if they choose the wrong submjaor for that specific university no hands on skills will be taught and they will simply only learn academic course material related to exercise.  Some of these skills would be the same skills in some cases as some of the same hands on assessment skills taught to complete a physical therapy doctoral degree in graduate school except that someone with this sub major can learn some of these skills ahead of time without completing a Bachelors degree first.  Good preparation for becoming a Physical Therapist or Medical Doctor 

Anatomy -  Anatomy classes often have been required to attend physical therapy graduate school but not medical school to become a medical doctor.  If I had to guess most physical therapists know skeletal muscle anatomy better than general practitioner medical doctors, but medical doctor specialists usually know anatomy in their specialty area better than physical therapists.  Even though Anatomy classes might not be required to enter medical school, taking anatomy in advance might make you better prepared.  Good preparation for becoming a Physical Therapist or Medical Doctor

Map Reading - Practicing using maps as a skill may help you learn anatomy easier.

Memory - There are memory championships and many world memory Champions wrote books on how to improve your memory but I recommend understanding the methods in the book "Use your perfect memory" by Tony Buzan.  Developing a good memory is good preparation for becoming a Physical Therapist, Pharmacist or Medical Doctor

Pharmacist Technician - You can get a pharmacist technician career or take pharmacy technician coursework without having to complete a bachelors degree first, to help you understand the pharmacological aspects of medicine before going to graduate school to become a medical doctor, physical therapist or pharmacist.  Good preparation for becoming a Physical Therapist, Pharmacist or Medical Doctor

Dental Hygenist - You can get a undergraduate degree and career as a dental hygenist before going to grafuate school to become a dentist

Physical Therapy/Therapist Assistant - You can sometimes take classes to become a physical therapist assistant at a technical college and transfer credits towards an associate degree rowards the bachelors degree required to go into medical school or Physical Therapy School.  Good preparation for becoming a Physical Therapist

Personal Trainer - This career can help you gain experience working with exercise hands on before enrolling in Physical Therapy school.  Do not call yourself a PT on your Physical Therapist college admission application but use the full term personal trainer if you are a personal trainer but not a physical therapist.  Some Physical Therapists get extremely offended by personal trainers calling themselves PT.  I have heard some people claim Physical Therapists have sued Personal Trainers for using the initials PT which Physical Therapists use claiming the initials PT are trademarked.   Personal Training is not a college degree but a career.  Good preparation for becoming a Physical Therapist

Massage Therapist / Massage Therapy School - Good preparation for becoming a Physical Therapist or Medical Doctor

Fire Fighter - Fire fighters often learn certain hands on medical skills 

Emergency Medical Technician

Paramedic

Bio-engineer Bachelors - You can use a bio engineering degree to transfer to medical school if you take any extra courses in Biochemistry or biology that maybe required to enter the school.  This maybe difficult however as engineering degrees usually do not have enough elective courses left over to meet course requirements for anything else in addition to the minimum degree requirements without attending college longer than it would take to complete an engineering degree alone.

Reading books ahead of time - There is nothing to stop you from reading textbooks like "Harrison's principles of internal medicine" before you ever attend medical school or even if you never attend medical school

Practicing math and or statistics ahead of time - You can get math and or statistics textbooks with practice problems and do those problems before taking any math courses at college that will later be required to get into other courses required to attend a medical school or required in and of them self to attend a medical school or that are required for a Bachelors degree major you are planning on working on to get into medical school.  Testing out of some math courses may save you a lot of time and money not taking as many classes at college.  And when you can not test out of a class but must take it the class will be much easier if you practiced the math ahead of time.  Unfortunately a large proportion of medical research is done with an emphasis on statistics  which is not predictable but "random" instead of math such as Algebra and Calculus without statistics that is better suited to real science which is testable, observable and predictable.  So someone in any medical field should be familiar with statistics enough in order to use valid reasoning and the correct choice of premises to show the flaws in any statistical study.

Advanced Placement courses and other means of testing out of college classes without taking them - This will give you more time to take elective college classes that teach hands on skills instead or to simply save time and money skipping classes covering material you already learned.

Actually learn concepts do not just memorize answers to complete tests - There was a Physics professor who was very angry that students wanting to become medical doctors who viewed a physics requirement as a barrier to admission into medical school would try to memorize test answers.  I am not talking about students who try to find out what problems were on a test ahead of time so they can know how to solve them, but actually memorizing a number with some units that was an answer to a past test question in a previous semester.  One of the teaching assistants might have tried to emphasize the level of stupidity involved by saying that they were not memorizing problems which would have been bad enough but instead memorizing answers to problems which was worse.  Some people could actually know how to solve a problem but cheat by trying to find out what the problems are going to be on the test before they are supposed to know and solve the specific problems on the test in advance and still put down work for solving the problem while doing the test pretending as if they never saw it before and were solving it live but these students just tried to memorize the answers whether it was as an attempt to cheat or it was where they genuinely thought it was not cheating and they simply thought the problem would occur again and the way to study was to memorize what the answer was to that problem and put it down again I do not know.  Someone can memorize problems they had done in the past without cheating in some circumstances such as memorizing the numbers for a practice problem assigned by the professor before the test and memorizing the steps to solve that practice problem and memorizing the solutions to that problem and understanding how the solution would change if the numbers change.  It might have been ok under certain circumstances of the non cheating variety if they knew how to answer the question if it was the same problem but with different numbers but if one number in the initial problem was changed the entire answer would be changed and they would not know how to get it by a methodology of problem solving, they would simply put down the number they memorized from what the correct answer was in a previous semester.  This might not be an exact description of the situation the Professor was experiencing with these students but it is a close enough description to convey the point I am trying to express.  Do not be like those bad students learn the methodology for how to get a proper answer instead of just memorizing answers.  Memorizing is important if used correctly but should not be misused that way.

Copyright Carl Janssen 2022 

https://duckduckgo.com/?q=surgery+sewing+and+dexterity&ia=web

Force on charges stacked in evenly spaced series at a single moment in time

 

Net force on point charges stacked in evenly spaced series at a single moment in time

It is at a single moment in time when all of the charges would be an equal distance from the next nearest charge as measured in a single dimension after that point of time some of the charges might move to a new position

In a one dimensional tube uncharged tube of length L going from left to right

The distance between the two farthest point charges from each other is L

N is an integer greater than or equal to two equal to the number of point charges

The total charge of all point charges is N

The charge of each point charge is Q/N

The distance between each point charge and the nearest point charge is D

D = L / ( N -1 )

Fi is the Individual force a point charge on the farthest right experiences from another point charge listed with an index of i where the charge farthest on the left has i = 0 the charge second farthest to the right has i = n - 2

Fi = [ L - i * L / ( n - 1 ) ] ^ 2

F = sum of all Fi from i = 0 to i = n - 2

Incomplete next steps

Look up notation I used in place of standard calculus symbols when letters and numbers and askey symbols are used instead of hand drawings from previous articles

Take limit as n approaches infinity to calculate F for a unlimmitted number of point charges using an integral

Compare that for all other variables having the same value but n equalling two

Do in two and then three dimensions next for the point charge in a single corner

Note :  trig functions might be useable to estimate distance between charge as a function of angle but might give wrong results if used as an integral of force with respect to angle from one angle to another angle to calculate net total force experienced by a charge in a corner of a square or cube, so it is probably better to use indexes i, j and k and do double or triple integral using caartesian coordinates instead of single or double integral using polar or spherical coordinates because the number of point charges in each angular direction maybe different 

Copyright Carl Janssen 2022 Jan 12

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

The tangent converts an angle into a slope

Take two points in polar coordinates of (1, A) and (1, B) on the unit circle

convert those points into caartesian coordinates and get (cosA, sinA) and (cosB, sinB)

A midpoint that is the average of their caartesian coordinates exists in between those two points having coordinates of 

( 0.5*cos(A)+0.5*cos(B) , 0.5*sin(A) + 0.5*sin(B) ) caartesian

It has a radius in polar coordinates of

( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

The midpoints angle in polar coordinates must be the average of the polar coordinates' angles of the two points it is in between in this specific case.  The midpoint of any two polar coordinates with the same radius away from a common origin ( pole 0 ) is located at the average of their angles but maybe with a different radius when expressed in polar coordinates.

This midpoint is located at

( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5, (A+B)/2 )    polar 

the slope from

 (0,0) to the midpoint 

(0,0) to (0.5*cos(A)+0.5*cos(B) , 0.5*sin(A) + 0.5*sin(B) ) caartesian

(0,0) to ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5, (A+B)/2 )    polar  

Is the same as the slope from (0,0) to a point on the unit circle with the same angle in polar coordinates as the midpoint

(0,0) to ( 1, (A+B)/2 ) in polar coordinates

N = 1 / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

The point on the unit circle (1, (A+B)/2 ) polar 

is the same as

( N*[0.5*cos(A)+0.5*cos(B)],  N*[0.5*sin(A) + 0.5*sin(B)] ) caartesian

Since that point is on the unit circle the following is therefore true

cos([A+B]/2) = N*[0.5*cos(A)+0.5*cos(B)] =

= [0.5*cos(A)+0.5*cos(B)] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

sin([A+B]/2) = N*[0.5*sin(A)+0.5*sin(B)] =

= [0.5*sin(A)+0.5*sin(B)] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

Since the tangent is the slope of an angle the following is therefore true

tan([A+B]/2) = N*[0.5*sin(A) + 0.5*sin(B)] / (N*[0.5*cos(A)+0.5*cos(B)])

tan([A+B]/2) = [0.5*sin(A) + 0.5*sin(B)] / [0.5*cos(A)+0.5*cos(B)]

tan([A+B]/2) = [sin(A) + sin(B)] / [cos(A) + cos(B)]

The following simplifications can be made for half and double angle trigonometric values

sin(0) = 0

cos(0) = 1

tan([A+0]/2) = [sin(A) + sin(0)] / [cos(A)+0.5*cos(0)]

tan(A/2) = sin(A) / [ 1 + cos(A) ]

[tan(A/2)]^2 = [ sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2

[sin(theta)]^2 = 1 - [cos(theta)]^2 = [1-cos(theta)]*[1+cos(theta)] 

[ sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2 = [1-cos(A)]*[1+cos(A)] / [1+cos(A)]^2

[ sin(A) ] ^2 / [ 1 + cos(A) ] ) ^2 = [1-cos(A)] / [1+cos(A)]

[tan(A/2)]^2 = [1-cos(A)] / [1+cos(A)]

tangent of half angle confirmed 

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae

let x = cos(A/2) and solve using quadratic equation

(1 - [cos(A/2)]^2 ) / [cos(A/2)]^2 = [1-cos(A)] / [1+cos(A)]

( 1 - x^2 ) / x^2 = [1-cos(A)] / [1+cos(A)]

1 - x ^2 = ( x^2 ) *  [1-cos(A)] / [1+cos(A)]

0 = ( x ^2 ) + ( x^2 ) * [1-cos(A)] / [1+cos(A)] - 1

0 = ( x ^2 ) * [1 + cos(A) + 1 - cos(A)] / [1+cos(A)]   - 1

0 =  ( x ^2 ) * 2 / [1+cos(A)]   - 1

a = 2 / 1 + cos(A) , b = 0 , c = -1

x= [ - b +- ( b^2 - 4*a*c) ] / ( 2 *a )

x = [- 0 +- ( 0 - 4 * 2 /  [ 1 + cos(A)] * -1) ^ 0.5 ] / 2 * [2 / 1 + cos(A)]

x = +- ( 8 / [1 + cos(A)] ^ 0.5 ) /  ( 4 / [ 1 + cos(A) ] )

cos(A/2) =  +- ( 8 ^ 0.5 ) * ( 1 / [ 1 + cos(A) ] ^ - 0.5 ) / 4

8 ^ 0.5 = 2 * 2 ^ 0.5

4 = 2 * 2 ^ 1

(8^0.5) / 4 = 2 ^ -0.5

cos(A/2) = +- ( [ 1 + cos(A) ] ^ 0.5 ) / ( 2 ^ 0.5)

[ sin(A/2) ] ^ 2 =  1 - [ cos (A/2) ] ^2

[ sin(A/2) ] ^ 2 = 1 - [ 1 + cos(A) ] / 2 = 0.5 - 0.5 * cos(A) = [ 1 - cos(A) ] / 2

sin(A/2) =  +- ( [ 1 - cos(A) ] / 2 ) ^ 0.5

confirmed

https://web.archive.org/web/20210912185657/https://www.intmath.com/analytic-trigonometry/4-half-angle-formulas.php

N = 1 / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

when B = 0  then N = 1 / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5

( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) = 0.25 * ( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 )

( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 ) = [ cos(A) ] ^ 2 + [ sin(A) ] ^ 2 + 2 * [ cos(A) ] + 1

( [1 + cos(A) ] ^ 2 + [ sin(A) ] ^2 ) = 2 + 2 * cos(A)

 ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) = 0.5 + 0.5 * cos(A) 

cos(A/2) = [ 0.5*cos(A) + 0.5*1 ] / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5  

[ cos(A/2) ] ^ 2 =  ( [ 0.5*cos(A) + 0.5* ] ^ 2 ) / [ 0.5 + 0.5 * cos(A) ]

[ cos(A/2) ] ^ 2 = 0.5 + 0.5 * cos(A) = [ 1 + cos(A) ] / 2

cos(A/2) =  +- ( [ 1 + cos(A) ] ^ 0.5 ) / ( 2 ^ 0.5)

[ cos(A/2) ] ^ 2 = [ 1 + cos(A) ] / 2

[ cos(A) ] ^ 2 = [ 1 + cos( 2 * A ) ] / 2

cos( 2 * A ) = ( 2 * [ cos(A) ] ^ 2 ) - 1 

confirmed

https://web.archive.org/web/20210421123600/https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php

cos([A+B]/2) = [ 0.5*cos(A)+0.5*cos(B) ] / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ^ 0.5

cos(A+B) = ( 2 * [ cos( [A+B] / 2) ] ^ 2 ) - 1

cos(A+B) = [ 2 * ( [ 0.5*cos(A)+0.5*cos(B) ] ^ 2 ) / ( [ 0.5*cos(A)+0.5*cos(B)  ] ^ 2+ [ 0.5*sin(A) + 0.5*sin(B) ] ^ 2) ] - 1

[ tan(A/2) ] ^ 2 = [ ( sin(A) ] ^ 2 / [ 1 + cos(A) ] ) ^2

[ tan(A) ] ^ 2 = [ ( sin( 2*A ) ] ^2 / [ 1 + cos( 2*A) ] ) ^2

[ sin( 2*A ) ] ^ 2 = [ tan(A) ] ^2 * [ 1 + cos( 2*A) ] ) ^2

[ sin( 2*A) ] ^ 2 = [ tan(A) ] ^ 2 * ( 1 + [ ( 2 * [ cos(A) ] ^ 2 ) - 1 ] ) ^2

[ sin( 2*A) ] ^ 2 = 4 * [ tan(A) ] ^ 2 * [cos(A)] ^ 4 

[ sin( 2*A) ] ^ 2=  4 * [ sin (A) ] ^ 2 * [ cos(A) ] ^ 2 

sin( 2 * A) = 2 * [ sin (A) ] * [ cos (A) ]

confirmed

https://web.archive.org/web/20210421123600/https://www.intmath.com/analytic-trigonometry/3-double-angle-formulas.php

sin(A/2) = [ 0.5 * sin(A) + 0.5*0 ] / ( [ 0.5*cos(A)+0.5*1  ] ^ 2+ [ 0.5*sin(A) + 0.5*0 ] ^ 2) ^ 0.5 

[ sin(A/2) ] ^ 2 =  ( [ 0.5 * sin(A) ] ^2 )  /  0.5 + 0.5 * cos(A) 

[ sin(A/2) ] ^ 2 = 0.5 * ( [ sin(A) ] ^2 ) / [ ( 1 + cos((A) ]

[ sin(theta) ] ^ 2 = 1 - [ cos(theta) ] ^ 2 = [ 1 - cos(theta) ] * [ 1 + cos(theta) ]

[ sin(A/2) ] ^ 2 = 0.5 *  [ 1 - cos(A/2) ]

sin(A/2) =  +- ( [ 1 - cos(A) ] / 2 ) ^ 0.5

[ cos(theta) ] ^ 2 = 1 - [ sin(theta) ] ^ 2

1 = ( 1 / [ cos(theta) ] ^ 2 ) - [ tan(theta) ] ^2

1 + [ tan(theta) ] ^2 = [ cos(theta) ] ^ -2

[ cos(theta) ] ^ 2 = 1 / (1 + [ tan(theta) ] ^2)

cos(theta) = +- 1 / (1 + [ tan(theta) ] ^2) ^ 0.5

[ sin(theta) ] ^ 2 = [ tan(theta) ] ^ 2 * [ cos(theta) ] ^2

[ sin(theta) ] ^ 2 = ( [ tan(theta) ] ^ 2 ) / (1 + [ tan(theta) ] ^2)

cos( [A+B] / 2 ) = +- 1 / ( 1 + [ tan( [A+B] / 2 ) ] ^2) ^ 0.5

cos( [A+B] / 2 ) = +- 1 / ( 1 + ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) ^2) ^ 0.5

[ sin( [A+B] / 2 ) ] ^ 2 =  [ tan( [A+B] / 2 ) ] * cos( [A+B] / 2 ) 

sin( [A+B] / 2 ) = +- ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) / ( 1 + ( [ sin(A) + sin(B) ] / [ cos(A) + cos(B) ] ) ^2) ^ 0.5

Using sin, cos and tan functions to calculate each other

[tan(theta)]^2 = [sin(theta)/cos(theta)]^2

[sin(theta)/cos(theta)]^2 = (1 - [cos(theta)]^2 ) / [cos(theta)]^2

[tan(theta)]^2 = (1 - [cos(theta)]^2 ) / [cos(theta)]^2

[tan(theta)]^2 = [sin(theta)]^2 / (1 - [sin(theta)]^2 )

Nothing in high school or college geometry is used to do this that has not been taught in prerequisite classes for high school or college geometry other than the definition of trigonometric functions the pythagorean theoreom and converting polar to caartesian coordinates or caartesian to polar coordinates all of which can be taught without prerequisite knowledge from high school or college geometry.  See my essay on why high school or college geometry should be removed and students should go straight to trigonometry without it which explains the cult like nature of high school or college geometry class.

http://teachingthenarrowway.blogspot.com/2021/12/why-you-should-be-able-to-skip-high.html

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2021/12/why-you-should-be-able-to-skip-high.html

Copyright Carl Janssen 2022

Incomplete work on the following

work on half and double angles, added angles and average angles for functions other than tan(0.5A + 0.5B)

I originally planned to get half an double angles for each trigonometric identity in terms of itself then planned to compare with results published elsewhere but apparently that is not commonly put that way in a table.  Such as for example 

sin of half an angle as a function of sin of a whole angle

cos of half an angle as a function of cos of a whole angle

tan of half an angle as a function of tan of a whole angle

"Each trigonometric function in terms of each of the other five" is not for half and double angles 

https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities

http://web.archive.org/web/20211227144328/https://en.m.wikipedia.org/wiki/List_of_trigonometric_identities

average with respect to time vs with respect to distance

mean average with respect to time vs with respect to distance

Notation

I am writing 2*(1second)^2 with a shorthand of 2 second ^ 2 that is why I am writing meter not meters and second not seconds.  I am doing this because it became very confusing writing out the units differently for plural and singular form such as 2 seconds but 1 second when the units were put to powers other than 1.

x = displacement

t = time

m = mass

f = force

a = acceleration

v = velocity

P = momentum

KE = Kinetic Energy

Int[y, g(y)] = the anti derivative of g(y) with respect to y

dg(y)/dy = the derivative of g(y) with respect to y

Dyg(y) = the derivative of g(y) with respect to y

Example 1

given x(t=0 seconds) = 0 meter

given v(t=0 seconds) = 0 meter second ^ -1

given a(t) = 1 meter second ^ -2

v(t) = t*(1 meter second ^ -2)

x(t) = 0.5 meter second ^ -2 * t^2

t^2 = x / (0.5 meters * seconds ^ -2 ) 

t^2 =  2*x / ( 1 meter per second ^ 2 )

t^2'= (2 meter^-1 second ^2) * x

t(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5)

v(x) = t(x) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ -0.5 second ^ 1) *(x^0.5) * (1 meter second ^ -2)

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

antiderivative of v(x) with respect to x shall be called Int[x, v(x)]

Dx(x^N) = N*x^(N-1)

Int(x, x^N) = (1/[N+1])*x^(N+1) + K0

Dx(1/[N+1])*x^(N+1) = (N+1)/(N+1)*x^(N+1-1)

Int[x, v(x)] = (1/1.5)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (2/3)*(x^1.5)*(2^0.5)*(1 meter ^ 0.5 second ^ -1) + K1

Int[x, v(x)] = (x^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

x(t) = 0.5 meter second ^ -2 * t^2

x (1 second) = 0.5 meter second ^ -2 * (1 second) ^ 2 = 0.5 meter

Int[x, v(x= 0.5 meter)] = (0.5 meter^1.5)*(2^1.5)*(1 meter ^ 0.5 second ^ -1) / 3 + K1

0.5 ^ 1.5 * 2 ^ 1.5 = 1

Int[x, v(x=0.5 meter)] = ( 1 meter ^2 second ^ -1 ) / 3 + K1

Int[x, v(x=0 meter)] = ( 0 meter ^2 second ^ -1 ) / 3 + K1

Average velocity with respect to displacement from a displacement of 0 meters to a position of 0.5 meter =

= ( Int[x, v(x= 0.5 meter)] - Int[x, v(x=0 meter)] ) / ( 0.5 meter - 0 meter) =

= ( 1 meter ^ 2 / 3 ) / 0.5 meter =

= 2 meter / 3

antiderivative of v(t) with respect to t shall be called Int[t, v(t)]

v(t) = t*(1 meter second ^ -2)

Int[t, v(t)] = 0.5 meter second ^ -2 * t^2 + K2

Int[t, v(t=1 second) = 0.5 meter second ^ -2 * (1 second )^2 + K2

Int[t, v(t=1 second) = 0.5 meter + K2

Int[t, v(t=0 second) = 0 meter + K2

Average velocity with respect to time from a time of 0 second to a time of 0.5 second =

= ( Int[x, v(t= 1 second)] - Int[x, v(t=0 second)] ) / ( 1 second - 0 second) =

= (0.5 meter-0meter) / (1 second-0second) = 0.5 meter second ^ -1

given mass = 1 kilogram

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

v(x) = (2^0.5)*(1 meter ^ 0.5 second ^ -1) *(x^0.5)

KE(x) = 1 kilogram meter second ^ -2 * x

F(x) = dKE/dx = 1 Kilogram meter second ^ -2

P(t) = 1 kilogram * v(t)

v(t) = t*(1 meter second ^ -2)

P(t) = t*(1 kilogram meter second ^ -2)

F(t) = dP(t)/dt = 1 kilogram meter second ^ -2

The average velocity with respect to time is not the same as the average velocity with respect to displacement for an object that starts stationary and experiences constant acceleration when averaged over a distance interval that is the same distance as it would have traveled in the time interval in which it's average velocity with respect to time is calculated.  However the average force it would experience over such an interval would be the same if either averaged by time or by displacement because the force is the same constant amount in either case.  Such would hold true for a falling object experiencing a constant force having displacement, time and velocity measured as zero at the initial starting condition of the fall.

Example 2

given a(t) = t^2 * 1 meter second ^ -4

given v(t=0 second) = 0 meter second ^ -1

given m = 1 kilogram

given x(t=0 second) = 0 meter

v(t) = 4*t^3 * 1 meter second ^ -4

x(t) = 12*t^4* 1 meter second ^ -4

[t(x)] ^ 4 = x / 12 meter second ^ -4

[t(x)] ^ 4 = x*(1/12)*1 meter ^ -1 second ^ 4

t(x) = x^(0.25) * (1/12)^0.25 * 1 meter^-0.25 second ^1

v(x) = 4 * 1 meter second ^ -4 * [t(x)]^3

v(x) = 4*1 meter second^-4*[x^(0.25)*(1/12)^0.25*1 meter^-0.25 second ^1]^3

v(x) = 4 * (1/12)^0.75 * x^0.75 * 1 meter ^0.25 second ^ -1

KE(x) = 0.5 * 1 kilogram * [v(x)]^2

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

Dx(x^1.5) = 1.5*x^0.5

dKE(x)/dx = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x) = dKE(x)/dx

int[x, f(x)] = KE(x) + K1

x(t) = 12*t^4* 1 meter second ^ -4

x(t = 1 second) = 12 meter second ^ -4 * (1 second)^4 = 12 meter

KE(x) = 8*(1/12)^1.5*x^1.5*1 kilogram meter ^0.5 second ^ -2

(1/12)^1.5*12^1.5=1

KE(x = 12 meter) = 8*1 Kilogram meter ^2 second ^ -2

v(t) = 4*t^3 * 1 meter second ^ -4

KE(t = 1 second) = 0.5 * 1 kilogram * [4 meter second ^ -1] ^ 2

KE(t = 1 second) = 8 kilogram meter ^ 2 second ^ -2

average force with respect to displacement from 0 meter to 12 meter =

=(int[x, f(x = 12 meter)] - int[x, f(x = 0)])/(12 meter - 0 meter) =

= 8 kilogram meter ^ 2 second ^ -2 / 12 meter =

= (2/3) * 1 kilogram meter second ^ -2

v(t) = 4 * t^3 * 1 meter second ^ -4

P(t) = 1 Kilogram * v(t)

P(t) = 4 * t^3 * 1 Kilogram meter second ^ -4

P(t = 1 second) = 4 * (1 second)^3 * 1 Kilogram meter second ^ -4

P(t = 1 second)  = 4 kilogram meter second ^ -1

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

int[t, P(t)] = P(t) + K2

average force with respect to time from 0 second to 1 second =

=(int[t, f(t= 1 second)] - int[t, f(t = 0)])/(1 second - 0 second) =

=  4 kilogram meter second ^ -1 / 1 second =

=  4 kilogram meter second ^ -2

f(x) = 12*(1/12)^1.5*x^0.5*1 kilogram meter ^0.5 second ^ -2

f(x=12meter)=12*(1/12)^1.5*(12meter)^0.5 *1kilogram meter ^0.5 second ^ -2

12*(1/12)^0.5*12^0.5=12

f(x = 12meter) = 12 kilogram meter ^ 2 second ^ -2

f(t) = dP(t)/dt = 12 * t^2 * 1 Kilogram meter second ^ -4

f(t = 1 second) = 12 kilogram meter second ^ -2

In this example the instantaneous force as a function of time is equal to the instantaneous force as a function of displacement at the displacement value that would occur for a given time value.  But the average force as a function of time is not the same as the average force as a function of displacement for the displacement interval that matches with the time interval.

Copyright Carl Janssen 2021 December 16

Estimating the gas laws based on collissions

Estimating the gas laws based on collissions

Starting with a simple one dimensional force calculation problem 

An object is traveling back and forth between two walls at a constant speed called travel speed except during collissions.  During collissions the object slows down to 0 then speeds up to it's original speed but in the opposite direction. The walls are parrellel to each other and a distance called travel length apart except during collissions where the wall hit by the object temporarily deforms then returns to it's original shape or is temporarily displaced then returns to it's original position after each collission ends.  The object moves in a path perpendicular to the walls.  The time the object is not in a collission is called travel time and the time the object is in a collission is called collision time.  The length a wall is temporarily displaced in the direction parrelel to the objects path of travel during a collission is called collission length.  The wall is moved or deformed during collission and the object is treated like a point mass with no deformation to make the math simpler.  

Mass refers to the mass of the object in one dimensional force calculations below.  Force below refers to a scalar being the absolute value of the force vectors magnitude and not to the direction of the force vector.  That is I am not counting forces going in opposite directions as having opposite values in the calculations below.  If forces were treated as vectors and no absolute value was taken there would be a zero average force when an equal number of complete collissions in each opposite direction occured.

The mean force the object would experience with respect to time using impulse = change in momentum = force * time

The absolute value of the change in momentum in a single collission is twice the absolute value of the objects momentum in this case because it switches to the same momentum in the opposite direction.

mean force with respect to time during travel = 0

mean force with respect to time during collission = 

= 2*mass*(travel speed) / (collission time)

mean force with respect to time = 

= 2*mass*(travel speed) / ([travel time] + [collission time])

(travel time) = (travel length) / (travel speed)

limit of mean force with respect to time as [(collision time) / (travel time)] approaches 0 is =

= 2*mass*(travel speed) / (travel time) = 

= 2*mass*([travel speed]^2) / (travel length)

The mean force the object would experience with respect to distance using work = force * distance

Work occurs during a collission to slow the object down to zero speed then again a second time to return it to original speed.  So the work done in a single collission is twice the traveling kinetic energy of the object in this case.

mean force with respect to distance during travel = 0

mean force with respect to distance during collission =

= 2*0.5*mass*([travel speed]^2) / (collission length)

mean force with respect to distance =

= 2*0.5*mass*([travel speed]^2) / ([collission length]+[travel length])

limit of mean force with respect to distance as [(collision length) / (travel length)] approaches 0 is =

= 2*0.5*mass*([travel speed]^2) / ([travel length])

= mass*([travel speed]^2) / ([travel length])

Start Side Note

the mean force with respect to distance is half the mean force with respect to time when you assume collission length and collision time are both insignificantly small compared to travel length and travel time.

See Article

http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2021/12/average-with-respect-to-time-vs-with.html

End Side Note

Now Let's instead assume there is a three dimensional object of a cube of length L and instead multiple particles collide into the walls in any one of six directions or three orthogonal / perpendicular directions and there opposites each of those six directions being perpendicular to one of the six faces of the cubes.  In this case we will assume the particles do not move any other direction than those six directions to make the math easier so that in each case it will work mathematically similar to the one dimensional case above.  That is we will assume the particles will only collide with the faces of the cube moving in a direction perpendicular to the face they hit making the collission calculatable as a one dimensional problem in each case.  Let us also assume the particles do not collide into each other or do not effect each other if they were to collide into another particle but "phase through" each other or act as though having "no clipping" with respect to each other but only collide with cube faces/walls or only experience collission forces through cube faces/walls and not other particles.  Additionally we shall assume all particles have the same speed except during collissions even though in reality their would be a distribution of different speeds at most temperatures people normally experience.

Temperature = Average Kinetic Energy per molecue

(Travel Speed)^2 = Kinetic Energy per molecue / mass per molecue =

= Temperature / (mass per molecue)

Limit as [(collision time) / (travel time)] approaches 0 of (Average Force per Molecue) with respect to time  = 

= 2* (mass per molecue) *([travel speed]^2) / (L)

= 2*(mass per molecue)*[Temperature / (mass per molecue)] / L

= 2*Temperature/L

From now on I shall assume collission time is negligible and refer to force as the force calculated using the calculus limit above and as measured as on average with respect to time

Total force from all molecues = 

= Force per molecue * Density * Volume / (mass per molecue)

 = Force per mole * Density * Volume / (mass per mole)

= (2*Temperature/L) * Density * Volume / (mass per mole)

Surface Area of a cube = 6*(L^2)

Volume of a cube = L^3

Pressure = Total Force / Surface Area =

(2*Temperature/L)*Density*(L^3)/[(mass per mole)*(6*[L^2])]

Pressure = Temperature*Density/[3*(mass per mole)]

N = Density*Volume/(mass per mole)

PV=NrT

P*V = r*Temperature*Density*V/(mass per mole)

P = r * Temperature*Density/(mass per mole)

Temperature*Density/[3*(mass per mole)] = r * Temperature*Density/(mass per mole)

r = 1 / 3

I remember getting  a different answer of r = 2 / 3 when I calculated this in high school.  I decide to search for the phrase unitless gas constant online and see if anyone actually calculated a unitless value or if I am the only person to ever do so.  Apparently someone else also got 2/3 which is different than the 1/3 (or 1/6 because I calculated force two different ways based on average force as a function of length vs average force as a function of time) value I calculated this time but the same as I remember calculating in high school.

https://duckduckgo.com/?q=unitless+gas+constant&ia=about

https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please

http://web.archive.org/web/20211215070129/https://en.m.wikipedia.org/wiki/Talk:Gas_constant#2%2F3_cut_-_ref_please

One might conclude that there should be different pressure depending on the shape of the container if it has a different volume to surface area ratio than that of a cube since the calculations were made using a cube shaped container but an assumption was made molecues do not experience force when they collide with each other only the container holding them in the calculations above.  If you remove the assumption of lack of molecular collission force except against the container walls the molecues would travel a much shorter distance between collissions.  You can imagine if there are N molecues in the container then imagine the container is made of N cubes,  the molecues would each inhabit a average volume of a cube shaped like the volume of the container / N and travel on average a length of the cube root of the (volume of the container divided by N) and have a average surface area of the cube shaped zone they move around in equal to 6 times the square of the cube root of the (volume of the container divided by N) you would end up with the same results as calculated above for the average pressure and force from these molecular collissions by partititioning a non cubed shape into many cubes.  

It might make more sense to assume the molecues typically inhabit a sphere shaped shape than a cube but a sphere which has a diameter equal to the length of a side of a cube has an equal surface area to volume ratio to that of a cube but would have a smaller surface area and volume than that cube requiring more spheres to fit in the same container if they each had a diameter equal to a side length of a cube as such it would result in a slightly different unitless constant number for R, especially if the molecues were considered to be able to move in more than six directions and a distribution of speeds rather than all moving at the same speed except when colliding.  

Pressure * Volume = Force/Area * Length/Area = Force * Length = Energy

Number of Moles * Temperature = Number of moles * Kinetic Energy per mole = Energy

Some number of Joules  = PV = rNT = r* some number of Joules

Some number of calories  = PV = rNT = r* some number of calories

r = some number without units

Even though r is technically unitless it is listed as a ratio of several units which cancel out each other to allow people to convert units from one type to another.

Surface Area / Volume for a cube of side length L

(6 L^2) / (L^3) = 6 / L

Surface Area / Volume for a Sphere 

(4*pi*r^2)/(4*pi*(r^3)/3) = 3 / r

(pi*D^2)/(0.5*pi*(D^3)/3) = 6 / D

Surface Area for a E*L by F*L by G*L rectangular solid

2*(EF+FG+EG)*(L^2)

Volume for a E*L by F*L by G*L rectangular prism

 E*F*G*(L^3)

Surface Area / Volume for a E*L by F*L by G*L rectangular prism

2*(EF+FG+EG)/(E*F*G*L)

Make a rectangular solid that is not a cube with the same surface area / volume as a cube

Let 2*(EF+FG+EG)/(E*F*G) = 6

When E = 2 and F = 1

2*(2+G+2G) / 2G = 6

12 G = 4 + 6G

6 G = 4

G = 2 / 3

2*(2+[2/3]+[4/3]) / [4/3] = 2*4 / [4/3] = 6

Let L = 1 inch

1 inches by 2 inches by 2/3 inch

volume = (4/3) inches^3

surface area = 2*([2*1]+[2*2/3]+[1*2/3]) inches^2 = 8 inches^2

8 inches^2 / (4/3) inches ^3 = 6 inches ^ -1

1 inch by 1 inch by 1 inch cube

6 inches ^ 2 / 1 inch ^ 3 = 6 inches ^ -1

Copyright Carl Janssen 2021

In some video games, noclip mode is a video game cheat command that prevents the first-person player character camera from being obstructed by other objects and permits the camera to move in any direction, allowing it to pass through such things as walls, props, and other players.

Noclipping can be used to cheat, avoid bugs (and help developers debug), find easter eggs, and view areas beyond a map's physical boundary.

http://web.archive.org/web/20170520115201/https://en.m.wikipedia.org/wiki/Noclip_mode

https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1

http://web.archive.org/web/20190515181746/https://www.khanacademy.org/math/ap-calculus-ab/ab-integration-new/ab-6-8b/v/antiderivative-of-x-1

https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4

http://web.archive.org/web/20211216034307/https://socratic.org/questions/what-is-the-antiderivative-of-ln-x-4

In the kinetic theory of gases, the mean free path of a particle, such as a molecule, is the average distance the particle travels between collisions with other moving particles. 

https://en.m.wikipedia.org/wiki/Mean_free_path

http://web.archive.org/web/20220629235820/https://en.m.wikipedia.org/wiki/Mean_free_path

https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases

http://web.archive.org/web/20220513084951/https://en.m.wikipedia.org/wiki/Kinetic_theory_of_gases

Why you should be able to skip high school geometry and go straight for trigonometry

 It was common in the late 20th century and early 21st century (AD/ACE) for high schools and colleges to teach pre algebra followed by alegebra then geometry, then trigonemtry (or Algebra 3 / Triginometry) then precalculus and finally a course equivelent to AB and BC AP calculus after that there are several calculus courses that all require a course equivelent to AB and BC calculus as prerequisites.

Pre algebra was an unnecesary course because it was a mixture of the math taught before Algebra in other math classes and Algebra that you would take a second time when you take the Algebra course.  Thus if you completed all the math classes that were prerequisites for Pre Algebra you could skip it and go straight to algebra and there would be nothing you would be taught in Pre Algebra that was not either taught in Algebra or the prerequisite courses for prealgebra that you would miss.

Precalculus was a combination of Algebra 3 trigonometry and AB calculus with the exception of the possible inclusion of matrixes and probability in Alegebra 3 and or Precalculus being included in one but not the other.   If matrixes and probability are included in both Algebra 3 Trigonometry and Precalculus then there is nothing someone would miss by skipping Precalculus and going straight to AB calculus from Algebra 3 Trigonometry.  If Matrixes and Probability were not included in both Algebra 3 Trigonometry and Precalculus then there is nothing someone would miss by skipping Precalculus and going straight to AB calculus from Algebra 3 Trigonometry.

The reason for being able to remove Pre Algebra and Pre Calculus courses and let someone skip them is overlapping and redundant teaching material with zero new teaching material added.  But there is a diffetent reason all together for the lack of necessity of the high school or college geometry course in that sequence in addition to any overlapping material.

The real life practical purpose of geometry is to measure distances or lengths of paths, distances between objects or lengths of objects, locations (coordinates) of objects, volumes of objects and cross sectional and surface areas of objects.  All that can be done using material usually taught in high school algebra, trigonometry and calculus classes without ever needing to take a high school geometry course.  If you understand caartesian coordinates taught in algebra class, polar coordinates taught in trigonometry and or calculus class and the meaning of the sin, cos, tan and cotangent functions and there inverses as well as what radians are and the pythagorean theoreom taught in trigonometry class you will be able to do every one of the practical purposes I have listed that you would be able to do taking a geometry class.  The only one of those things you might learn in calculus that you might not learn in trigonometry is polar coordinates which really should be taught in trigonometry before AB calculus because it is more of a trigonometry related thing.  Now there are some of those practical things I listed you might learn to do in calculus that you would not learn in high school trigonometry but you would not learn them in high school geometry either.  Examples would be learning how to calculate the surface area or cross sectional area or volume of certain shapes by integration or the distances of certain paths that are not a single straight line segment by integration.  Now there are some useful things involving geometry people might learn like names of shapes but people usually do not learn those in high school geometry but in other classes prior to high school geometry.  In Geometry class you might learn complicated arcane and obscure ways to measure those practical things I listed in very specific special cases that might not be taught in trigonometry but in trigonometry you will learn general wide reaching methods to do all those things you can do in geometry which will be easier to remember because you do not have to create a special case to figure out each thing but can simply figure out the coordinates of things using trig functions then use those coordinates to get the desired measurements.

* Start side note

For further reading about how to get trigonometric measurements by means of algebra with casrtesian and polar coordinates without geometry class you can read the following article in progress

Tangent of the average of two angles and other trigonometric identities derived from a combination of it and the pythagorean theoreom

http://web.archive.org/web/*/http://teachingthenarrowway.blogspot.com/2022/01/tangent-of-average-of-two-angles-and.html

http://teachingthenarrowway.blogspot.com/2022/01/tangent-of-average-of-two-angles-and.html

* End side note

But what about the philosophy of using axioms taught in high school geometry being useful to learn how to use logic to make practical decisions.  This so called philosophy taught in high school geometry is the most important reason it should be abolished except perhaps for those who wish to study how cultic thinking works.  High School Geometry is full of the bad kind of dogma that is presuppositions that one is not permitted to question, as opposed to stating presuppositions you use to come to a conclusion but acknowledging the possibility your presuppositions maybe wrong.  The starting point is there is a finite number of assumptions the student starts with called axioms and the claim is made that all geometric proofs in the geometry class can and will be taught either by using only those axioms or other proofs derived only through those axioms.  The student is given a list of those axioms at the beginning of the course and expected to derive all proofs assigned as homework only through those axioms or through other proofs they have derived eventually tracing back to a point of only through those starting axioms.  The fundamental problem with this is in reality assumptions are made to derive these proofs that are not in the list of initial axioms but the student is not permitted to admit that additional assumptions are required other than the axioms initially listed if you say another assumption is needed for the proof that is not listed in the initial list given at the beginning of the course that is classified as a thought crime and the problem is marked as wrong.  The type of erroneous methodology of proofs in high school geometry class influenced most participants (who mostly were not previously educated about cult psychology to develop a resistance to the undue influence) towards pseudomathematics, pseudoscience, pseudologic and magical thinking through the act of saying and or writing things in order to agree with the consensus of an authority figure even if those things are not true.  

In the Asch conformity experiment people were more likely to claim a untrue statement was true if someone else first claimed the same untrue statement was true.

The Asch conformity experiment was an experiment where volunteers were told a false answer about the length of a line segment then asked what length the line segment was.  It was found that when a unanimous group of people gave a false answer first people more frequently said the same measurement as the false answer they were previously told instead of the correct measurement than under other circumstances.

 Something similar to the Asch conformity experiment is replicated in geometry classes where the teachers assign what assumptions maybe used and students are not allowed to pick the assumptions they use for themselves and describe which assumptions they use and how and why they used them.  Even if the teachers claims about what assumptions are needed to prove a genuinely true statement is true are false by ommission or commission the students will (or more accurately have in the past implying others will in the future) more ftequently still give the same answer as the teacher than if they never heard the teachers claim.  False by ommission in this context would be where the teacher claims that such is the exact minimum list of assumptions required to validly prove a true claim when actually given the context on the list the teacher gives additional assumptions need to be added.  False by commission in this context would be to claim a assumption is needed to validly prove a true statement when it is not needed in the context of the other listed assumptions, that is if all the other assumptions on the list were included but that assumption was removed it would still be a valid proof of a true statement.  In addition a teacher may sometimes claim a false statement is true or a true statement is false and the students will more ftequently give the same answer as the teacher.  If the students did not first hear the teacher make a false claim or say a true claim is true based on logic that was not valid the students would be much less likely to make the same error when investigating if a statement is true independently in an ungraded environment without seeing someone else's example first.  Remember in geometry class people are not graded for right or wrong answers but they are graded for right or wrong reasoning.  

According to the way this type of geometry class is done the reasoning should be in agreement with the reasoning predefined in the curriculum before the student even started the class in order to not be marked as wrong or less than an A or 100% grade for a set of problems.  One of the fundamental problems is the assumption of minimum axioms that is the foundation for all reasoning in high school geometry classes of the type defined in this essay/article/manifesto is usually or maybe even always wrong based on a claim of being complete when more assumptions are actually needed in reality.  Say you claim the teacher or textbook is missing an assumption when they show how to do a geometric proof which was an assigned homework or test problem and that assumption is not in the axiom list or derived from the initial axiom list alone and you will get your problem marked as wrong.

If a list of five assumptions were given to reach a conclusion and you were asked if these five assumptions alone with nothing added or removed were valid to show the conclusion was true or untrue and nobody told you the answer beforehand and you were not given a hint what answer someone else came to and nobody else would see or grade your answer this would not be so similar to Asch conformity.  But in the case of geometry class the teacher starts by listing a certain number of assumptions and says they are sufficient.  According to the principle learned in the Asch conformity experiment by giving you the answer first people are more likely to verbally say the same answer they were first given by someone else even if the answer they were given is false.   Geometry classes of this sort do not start with the question of if those axioms are sufficient to prove the proofs but the statement that all the proofs you are required to prove in the class can be proved with those assumptions and no more excluding that which is derived from those assumptions alone.  Nobody in their right mind would believe no more assumptions will be needed based on independent thinking without someone else first telling them no more assumptions would be needed.  Those who are first told no more assumptions will be needed to prove the proofs in the book believe so (or have said they believe so) without even first reading the entire book more often than they should by which I mean more often than never.  Geometry classes of this sort are like religious instiuitions such as Churches or Mosques where they far too many members say they believe every word in a certain book such as a particular Bible or Quran or book of Mormon are true without ever having read the entire book.  Many self proclaimed Muslims can not read Arabic but claim every statement in the original Quran manuscript in Arabic is 100% true and many people who say they are Christian at Churches say every statement in the original Bible manuscripts of a particular Bible Canon are true without having even read an entire translation of it.  It is very different to have read a translation of the Bible and say you have found zero historical claims in it that you can confirm with great certainty to be false when correctly translated than to have never even read it and to claim every statement in it is true even statements you have not read.  Yet some geometry teachers will give their testimony to the axioms as being sufficient for all proofs in the assigned geometry book and far too many people will believe them without reading the entite book.  I am not saying to read an entire geometry book but do not give me your testimony of the sufficiency of the axioms to prove every proof in the book if you never read the entire book.

Copyright Carl Janssen 2021, 2022

https://en.m.wikipedia.org/wiki/Asch_conformity_experiments

http://web.archive.org/web/*/https://en.m.wikipedia.org/wiki/Asch_conformity_experiments

A 2003 effort (Meikle and Fleuriot) to formalize the Grundlagen with a computer, though, found that some of Hilbert's proofs appear to rely on diagrams and geometric intuition, and as such revealed some potential ambiguities and omissions in his definitions.[11]

[11] On page 334: "By formalizing the Grundlagen in Isabelle/Isar we showed that Hilbert's work glossed over subtle points of reasoning and relied heavily, in some cases, on diagrams which allowed implicit assumptions to be made. For this reason it can be argued that Hilbert interleaved his axioms with geometric intuition in order to prove many of his theorems."

http://web.archive.org/web/20210508113205/https://en.m.wikipedia.org/wiki/Hilbert%27s_axioms

Euclid's list of axioms in the Elements was not exhaustive, but represented the principles that seemed the most important. His proofs often invoke axiomatic notions that were not originally presented in his list of axioms.[23]

[23] Heath 1956, p. 62 (vol. I)

http://web.archive.org/web/20211209025416/https://en.wikipedia.org/wiki/Foundations_of_geometry

Absolute geometry is a geometry based on an axiom system for Euclidean geometry without the parallel postulate or any of its alternatives. Traditionally, this has meant using only the first four of Euclid's postulates, but since these are not sufficient as a basis of Euclidean geometry, other systems, such as Hilbert's axioms without the parallel axiom, are used.[1] The term was introduced by János Bolyai in 1832.[2] It is sometimes referred to as neutral geometry,[3] as it is neutral with respect to the parallel postulate.

[1] Faber 1983, pg. 131

[2] In "Appendix exhibiting the absolute science of space: independent of the truth or falsity of Euclid's Axiom XI (by no means previously decided)" (Faber 1983, pg. 161)

[3] Greenberg cites W. Prenowitz and M. Jordan (Greenberg, p. xvi) for having used the term neutral geometry to refer to that part of Euclidean geometry that does not depend on Euclid's parallel postulate. He says that the word absolute in absolute geometry misleadingly implies that all other geometries depend on it.

http://web.archive.org/web/20211209025836/https://en.m.wikipedia.org/wiki/Absolute_geometry

No one should be taxed to pay for child support

Mothers should not be required to pay child support to the father if he breaks up (or divorces) with her because he is not stuck taking care of the child and can allow the mother to take care of the child or if she is a unsafe caretaker or unwilling to be a caretaker he is still not stuck taking care of the child and can put the child up for adoption.  If he refuses to put the child for adoption and also refuses to allow the mother to take care of her own child then he should be required to pay.  Mothers should not be required to pay child support for a child they are not allowed to take care of.  Mothers should not be required to pay any child support for a child they are forbidden to see part of the time in the case of split custody

Fathers should not be required to pay child support to the mother if she breaks up (or divorces) with him because she is not stuck taking care of the child and can allow the father to take care of the child or if he is a unsafe caretaker or unwilling to be a caretaker she is still not stuck taking care of the child and can put the child up for adoption.  If she refuses to put the child for adoption and also refuses to allow the father to take care of his own child then she should be required to pay.  Fathers should not be required to pay child support for a child they are not allowed to take care of.  Fathers should not be required to pay any child support for a child they are forbidden to see part of the time in the case of split custody.

The government should not tax people to pay for taking care of children they are not allowed to take care of.  Single mothers and single fathers and adoptive parents should not get paid for choosing to be a parent through the taxation of people who are not allowed to take care of that child.

The person or people who take care of a child influence the child's thinking and worldview and behavior and only people who are responsible enough to plan how to economically (providing food, housing, etc. not necessarily money) support a child without taxing other people should be granted the privelege of taking care of a child.

Even if this sometimes means both biological parents lose access to funds and neither can take care of their own child economically after a break up or divorce and a third party adopts their children instead this is still an improvement over coerced child support payments, because it means more responsible people will raise the child in most cases, resulting in a child that makes better life decisions in most cases.

People should not pay taxes to send other people's children to public schools or daycare centers.  Responsible parents would have the time to home school if they are not required to work extra hours to pay for taxes to send other people's children to public schools.  Both parents working is not a valid excuse for public schools because at least one of two parents could choose not to be employed while the other one is working and still have enough resources to support the family if the parents were not collectively coerced to work extra hours through taxation to send other people's children to public school.

Although both biological parents might plan a means to be economically sufficient to support a child as a team, unforeseen circumstances such as being fired from a job for refusal to obey unethical orders of an employer may occur.  It is better therefore not to put ones primary means of economic sufficiency on an employer so that an employer can not threaten to take away your means to support your children if you do not follow unethical orders.  It is better to grow your own food and have your own means to produce clean fresh water.  In agricultural societies having more children usually meant more people that could help grow food.  None the less sometimes unforeseen factors occur even in agricultural societies where in spite of responsible economic planning of a couple before deciding to become biological parents disasters such as bad wrather, earthquakes or fire (not caused by negligence) may occur causing local famines or local crop damage and insufficient food and or drinkable water.  In such cases people may voluntarily provide charity or donations if they believe the people were responsible parents this is completely different than forcing other people to pay for child support.  Occassionally asking for donations to take care of your family in a time of unexpected need is very different than to plan to live off coerced child support payments as a regular and ongoing thing for the entire duration of the time you take care of your biological children.  In other cases a malicious third party may deliberately do actions that cause economic harm to biological parents such as burning their crops, in such cases it maybe reasonable to require them to pay for economic damage but this is not the same as requiring to pay for child support because they are not paying for the fact that other people had children but paying for the fact that they maliciously destroyed someone else's food supply.

If an adoptive or biological family is abusive a child should have a right not to be adopted by them anymore there are other people that can adopt them.  Nobody should be coerced to adopt other people's children but plenty of people are willing to adopt other people's children.

I am not here saying people should refuse to receive government subsidies to get extra resources to take care of children.  I am saying giving people subsidies to take care of children should not be used as an excuse to validate the moral legitamacy of coerced taxation or coerced childcare payments.  In cases where refusing to receive funds given by government agencies does not return the money to the taxpayers who were taxed with the excuse of raising money to pay for those funds the act of choosing to receive such funds on a individual (not policy) basis does not coerce people to pay for child support.  However if someone directly forces a ex spouse (who they initiated divorce against) to pay child support where as they could choose not to bill the ex spouse for child support this is a completely different matter.  If the ex spouse was divorced for malicious non defensive domestic violence and the violence was real the ex spouse should be in jail not out of jail employed and paying child support.

Copyright Carl Janssen 2021

https://grammarhow.com/peoples-or-peoples/

http://web.archive.org/web/20210118194052/https://grammarhow.com/peoples-or-peoples/