## Planck Units and Other Units in Physics

Copyright (C) 2006 by Lew Paxton Price

Information added as acquired.

## Main Menu - More About the Electron - Planck's Constant - Photons and Red Shift

This is an update to this website which is being added on February 11, 2006. Other units may be added later on as time is available and the need arises.

Planck units are largely based upon three fundamental units,

h, G, and c. If any one of these fundamental units is wrong, the units that are based upon them are also wrong. Whether by intent or by accident, there are numerous websites which have the wrong units listed. This addition to my website is designed to shed some light on the fundamental units and to aid in corrected the problem.

The gravitational constant is given the symbol "

G". It is a measured value used in the force equation for gravity (see below).F = Gm

_{1}m_{2}/ r^{2}The

munits are masses of two bodies which are separated by a distancer. By rearranging the equation, we haveG(see below).G = Fr

^{2}/ m_{1}m_{2}The gravitational force,

F, was measured between two masses to arrive atG. At various times, the laboratory equipment and methods were improved to arrive at more accurate values forG. The physics texts were not usually updated for the new values because (1) the changes were not great enough to justify the added expense to the texts, and (2) the changes were happening frequently enough to make each text obsolete before it arrived in the hands of students. Consequently, there are numerous variations ofGto be found, but the differences between them are slight.There are many systems of weights and measures used in physics. Fundamental constants such as

Goften have two values according to the measuring system used. When two such units are mixed to arrive at subsidiary units such as the planck length, the result is a completely erroneous value along with units of measure which do not apply. Consequently, it is important to convert the various units of the fundamental constants used to the same system of units. In may instances of late, this has not been done.Finally, there is human error involved in copying from an old text to create a new next. This means that it is wise to check various texts to see if they all agree (they usually don't), and decide what is correct and what is not correct.

For

G, the following was discovered.

G = 6.670x10^{-11}newton meter^{2}/kilogram^{2}

From a text created by the Department of Physics at the U.S. Air Force Academy in about 1955,Formulas and Tables.

G = 6.673x10^{-11}newton meter^{2}/kilogram^{2}

From a textbook last copyrighted in 1972,Elements of Physics.

G = 6.6742x10^{-11}meter^{3}/kilogram second^{2}

From a science publication announcing an improved value based upon data from a recent experiment 1994. Note that the the text mentioned below, probably written prior to 1994, did not include the change.

G = 6.67259x10^{-11}newton meter^{2}/kilogram^{2}

From a textbook last copyrighted in 1997,Fundamentals of Physics Extended.

Note that one of the systems of units is

newton meterand one is^{2}/kilogram^{2}meter.^{3}/kilogram second^{2}1 newton = 1 kilogram of force = 1 kilogram of mass x 1 meter/second

^{2}= 100,000 dynesIf we substitute

kilogram meter/secondfor the newton in the newton equation, we have^{2}meter. The units are based on a conversion factor of one, so the numbers do not change.^{3}/kilogram second^{2}The newton is defined as

the force that will impart to one kilogram of mass an acceleration of one meter per second squared.The dyne is defined as

the force that will impart to one gram of mass an acceleration of one centimeter per second squared.In the metric system, the kilogram may be either that of force or that of mass, which means that the units are different for the two. This leads to confusion.

Planck's constant is given the symbol "

h". It was discovered by Max Planck by using temperature. Some of the details of his experiment are given elsewhere on this site. In nether theory, when it is divided by one second of time, it is the energy in one wave of light.For

h, the following was discovered.

h = 6.6252x10or^{-34}joule second4.134x10^{-15}electron volt second

From a text created by the Department of Physics at the U.S. Air Force Academy in about 1955,Formulas and Tables.

h = 6.624x10^{-27}erg second

From a two-volume dictionary copyrighted in 1960.

h = 6.62618x10or^{-34}joule second4.1357x10^{-15}electron volt second

From a textbook last copyrighted in 1972,Elements of Physics.

h = 6.6260755x10^{-34}joule second

From a textbook last copyrighted in 1997,Fundamentals of Physics Extended.

Note that all of the units published for

hare incompatible with the units published forG. My wife and I have often wondered why. Perhaps those who write the texts do not want the students to have it too easy. Or perhaps they don't know the difference between systems of measure. In any case,joule secondsmust be converted to work with eithernewton meteror^{2}/kilogram^{2}meterif they are to be useful for calculating other planck units - or the units for^{3}/kilogram second^{2}Gmust be converted to work withjoule seconds. Actually, the conversion requires going to units which work with those shown rather than literal equivalent units.The joule is defined as

the unit of work or energy equivalent to work done or heat generated in one second by an electric current of one ampere against a resistance of one ohm - or raising the potential of a coulomb by one volt.1 joule = 10,000,000 ergs = .737324 (one book gave .7376) foot pounds = 1 watt second = .1020 kilogram meter

The erg is defined as

the unit of work and of energy, being the work done in moving a body one centimeter against a force of one dyne.1 erg = one centimeter dyne = 980.7 centimeter grams = 10

^{7}joules = 10^{7}watt secondsThe electron volt (sometimes called the equivalent volt) is defined as

the unit of energy equal to that acquired by an electron passing through a potential of one volt.Once the conversion for

his completed, it can be used along with the compatible units forG. I recommend the less-seen versions ofGwhich is inmeterbecause you will need to have both^{3}/kilogram second^{2}handGin those units if you are to use them with the units forclater on. Once again, many have wondered whyhandGwere not published in the most usable units, but apparently someone was against allowing students to work problems or to understand how certain things were derived. So the usable version ofhfollows along with a usable version of the planck unit of actionwhich is merelyhh/2pi.Converted from

h = 6.6261x10^{-34}joule second

h = 6.7565x10^{-35}kilogram meter^{2}/second

h= 1.0753x10^{-35}kilogram meter^{2}/second

The speed of light in a "vacuum" is usually given as "

c". Apparently, the units for the speed of light were too simple for someone to get them wrong, so they have been published in a useful form as long as the units for the constants that must be used with them are converted properly.

c = 2.997929x10^{8}meters/second

From a text created by the Department of Physics at the U.S. Air Force Academy in about 1955,Formulas and Tables.

c = 2.9979x10^{8}meters/second

From a textbook last copyrighted in 1972,Elements of Physics.

c = 2.99792458x10^{8}meters/second

From a textbook last copyrighted in 1997,Fundamentals of Physics Extended.

Using the planck area as an example of how to calculate from the three constants given, we will be using the following values and units. All of these units are compatible with one another and this compatibility allows us to arrive at the correct answer within the limits of values given.

h= 1.0753x10^{-35}kilogram meter^{2}/second

G = 6.6742x10^{-11}meter^{3}/kilogram second^{2}

c = 2.9979x10^{8}meters/second

We will give the planck area the symbol

A. It is defined as shown below in mathematical form. We will abbreviate_{p}kilogramaskg,meterasmtr, andsecondassec.

A_{p}=hG/c^{3}

A_{p}= (1.0753x10^{-35}kg mtr^{2}/sec) (6.6742x10^{-11}mtr^{3}/kg sec^{2}) / (2.9979x10^{8}mtr/sec)^{3}Taking one step at a time, we first simplify the units in the numerator and arrive with what is shown below.

A_{p}= (1.0753x10^{-35})(6.6742x10^{-11}) mtr^{5}/sec^{3}/ (2.9979x10^{8}mtr/sec)^{3}Now we simplify the units in denominator.

A_{p}= (1.0753x10^{-35})(6.6742x10^{-11}) mtr^{5}/sec^{3}/ (2.9979x10^{8})^{3}mtr^{3}/sec^{3}We now simplify the units using both the numerator and the denominator.

A_{p}= [(1.0753x10^{-35})(6.6742x10^{-11}) / (2.9979x10^{8}^{3}] mtr^{2}At this point we can see that we do indeed have an area because the units have become meters squared. Now we begin the calculating by working first with the numbers on top.

A_{p}= [(7.1767672)(10^{-35})(10^{-11}) / (2.9979x10^{8}^{3}] mtr^{2}Now we work with the numbers at the bottom by calculating their cube.

A_{p}= [(7.1767672)(10^{-35})(10^{-11}) / (26.943339)(10^{8})^{3})] mtr^{2}Next we add the exponents in the numerator algebraically and that gives us the exponent for the number in the numerator. Remember that -35 plus -11 equals -46.

A_{p}= [(7.1767672x10^{-46}) / (26.943339)(10^{8})^{3})] mtr^{2}The exponent at the bottom must be multiplied by 3 which is the same as adding it to itself twice. This gives us an exponent of 24.

A_{p}= [(7.1767672x10^{-46}) / (26.943339x10^{24})] mtr^{2}Next we move the bottom powers of ten to the top by changing the sign of the exponent. So 24 becomes -24.

A_{p}= [(7.1767672x10^{-46})(10^{-24}) / (26.943339)] mtr^{2}Now we algebraically add the exponent at the top. So -46 plus -24 equals -70.

A_{p}= [(7.1767672x10^{-70}) / (26.943339)] mtr^{2}It is time to do the divide the numerator by the denominator.

A_{p}= .2663651x10^{-70}mtr^{2}Next we move the decimal point over one place which is the equivalent of multiplying by ten.

To compensate, we change the -70 to -71 which is the equivalent of dividing by ten.

A_{p}= 2.663651x10^{-71}mtr^{2}

The planck length "

Lp" is the square root of the planck area. Now would be good time to do find its value, so we find a way to take the square root of the plank area. We cannot take the correct square root as it is shown above because the power of ten is an odd number and cannot be divided evenly by two. So first we move the decimal one more place to the right which is the equivalent of multiplying by ten, and change the exponent

from -71 to -72 which is the equivalent of dividing by ten.

A_{p}= 26.63651x10^{-72}mtr^{2}Now we can take the square root of 26.63651 and divide the exponent by two to arrive at the planck length. The square root of "meters squared" is a "meter", so we take the exponent off of "mtr".

L_{p}= 5.161057x10^{-36}mtrThe last thing we do is round off all the numbers to four places to the right of the decimal which is the extreme limit of our accuracy.

A_{p}= 2.6637x10^{-71}mtr^{2}

L_{p}= 5.1611x10^{-36}mtrYou have seen how this was done in careful steps which you can use in other instances to check values you see on the web and discover which are correct. Good luck!

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