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Using the force constant in equations

  1. Jan 7, 2005 #1


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    Quantum gravity research ties in to Planck units and it is possible to have variations on that theme.
    At one point (I think a couple of years ago) john baez was advocating that the units should be adjusted so that the central coefficient in the Einstein eqn have value one.

    that is the main equation of Gen Rel, and the coefficient that relates the left and the righthand sides is a force

    [tex]\mathbb{F} = \frac{c^4}{8\pi G}[/tex]

    I want to see what equations it makes look cleaner to use that constant rather than the more traditional G. Maybe nature would like us to use that constant
    Last edited: Jan 7, 2005
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  3. Jan 7, 2005 #2


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    First of all it obviously makes the main eqn of GR cleaner because you just have
    [tex]G_{ab} = \frac{1}{\mathbb{F}} T_{ab} [/tex]

    instead of the more messy way the einstein eqn is usually written
    (when the author wants to show the dimensions involved)

    [tex]G_{ab} = \frac{8\pi G}{c^4} T_{ab} [/tex]

    But what about some other equations like the specs of a black hole.

    Instead of indexing black holes by mass M, I am going to index them by
    their energy E = M c2
    so I will talk about a black hole with rest energy E, that should be all right

    then I will write the formulas for things like Schw. radius, area, BekensteinHawking temperature, evaporation time.

    [tex]\text{radius} = \frac{1}{4\pi} \frac{E}{\mathbb{F}}[/tex]

    [tex]\text{area} = \frac{1}{4\pi} \frac{E^2}{\mathbb{F}^2}[/tex]

    [tex]k\text{ temperature} = kT = \frac{\hbar \mathbb{F} c}{E}[/tex]

    [tex]\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2}[/tex]
  4. Jan 7, 2005 #3


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    I guess there are two things to say about what happens when you do this.
    the first thing is that the formulas simplify
    they have considerably fewer symbols in them in some cases

    like the evaporation time, as we write it, is

    [tex]\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2}[/tex]

    but WIKI writes in the conventional way and it is more mess

    [tex]\frac{5120 \pi G^2 M^3}{\hbar c^4}[/tex]

    but there is another point here which is that not only are there fewer symbols but it is dimensionally more transparent

    For example when you read the formula for the radius you see a ratio
    of E/F and energy over a force
    well energy IS force X distance
    so clearly dividing energy by force gives you a length

    to me, that is more primitive than parsing the conventional formula for the radius, which is 2GM/c^2. From a nooby viewpoint, why should dividing GM by the square of the speed of light give anybody a length?

    To take another example, for any force F, multiplying by a speed gives a power. Force F pushing at speed v delivers power Fv.
    That applies to Fc, where F and c are universal constants of force and speed, you get a power quantity.
    And hbar is energy X time, so clearly (hbar F c) is the square of an energy.

    It makes the Hawking temperature formula a no-brainer in a certain sense because it just says the the kT energy of the whole is equal
    to the square of a natural energy constant divided by the energy of the hole.

    If you look at it right, it is immediately clear that the formula for kT does in fact give an energy, as it should
    Last edited: Jan 7, 2005
  5. Jan 7, 2005 #4


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    Let's take another look.

    If we work in a certain VARIANT of planck units in which hbar and c are made units but (instead of working with G) we make this F into the unit force and do everything with that, then it we wll get variant planck energy and time units which I will write the obvious way analgous to the force unit.

    [tex] \hbar = \mathbb{E}\mathbb{T}[/tex]

    [tex] \mathbb{F}c = \frac{ \mathbb{E}}{\mathbb{T}}[/tex]

    [tex] \hbar \mathbb{F}c = \mathbb{E}^2[/tex]

    So the Hawking temperature formula is saying something extremely primitive. It says that the kT for the hole is simply the square of the natural energy unit divided by the energy of the hole

    [tex] kT = \frac{\hbar \mathbb{F} c}{E} = \frac{\mathbb{E}^2}{E}[/tex]

    and in the formula for the evaporation time we also recognize the square of the energy unit, as before: (hbar F c) , and the natural unit of power, unit force x unit speed, namely (Fc).

    [tex]\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2} = \frac{80}{\pi} \frac{E^3}{ \mathbb{E}^2 \mathbb{P}}[/tex]

    So it is again dimensionally transparent, you come in with the black hole's energy cubed and you divide it twice by the unit energy and once by the unit power, and you get a time.
    Oh, there is also that numerical factor of 80/pi.
    Last edited: Jan 7, 2005
  6. Jan 7, 2005 #5


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    telling John Baez stories

    one can use these variant Planck units to tell Baez stories, like this

    How the Gypsies Stole the Moon

    Some gypsies were roaming this part of the galaxy and on their way thru the solar system they stole the moon
    and replaced it by a black hole of the same mass

    Dont worry, said the gypsies (when the people complained) you will still have tides and everything will be the same because the black hole we put in has the same mass----1.7E23 pounds---as the moon.

    the people appointed John Baez the noted explorer to negotiate.

    Give us back our moon, Baez told the gypsies, this black hole you gave us will eventually evaporate. It is not a fair exchange, you have cheated us.

    the gypsies thought about it some. then they said "All right, we will give you one chance. If you can tell us how long the black hole will take to evaporate we will swap the moon in for the black hole and take our black hole away with us!"

    John baez cubed 1.7E23 to get 4.9E69

    then he multiplied that by (80/pi)E-18 to get 125E51

    he knew gypsies like to use units based on the Force, so he said "125 x 1051 counts"

    Very good, said the gypsies, but you are from Earth and measure time in minutes. Tell us the time in minutes so we can be sure you really understand.

    john baez took 1250 x 1050 counts and divided by 222 and said
    "5.6 x 1050 minutes"

    All right said the gypsies and they swapped the familiar moon back into its orbit.


    people might be worried because they think the units are not clearly enough defined. they are all based on this hbar and c and this F that was introduced in this thread (the coef. in the Einst. eqn.)

    that is because hbar X c is equal to a force X area. It is the standard of coupling for whatever interaction, as one sees in the definition of the fine structure alpha.

    and so, if nature gives us a force unit F, we immediately have an area unit

    [tex]\text{area unit} = \mathbb{A} = \frac{\hbar c}{\mathbb{F}}[/tex]

    and then the sqrt of the area is unit length, and unit time comes from c, as the time for light to travel unit length, and so on.

    these are just like the usual Planck units except differing by factors of 5 and sometimes 25, where by 25 I really mean 8 pi. you have to be a little watchful because 8 pi and the sqrt of 8 pi are getting into things.

    And so, well, what the gypsies call a pound is just a practicalscale unit that is exactly ten million universal mass units

    [tex]\text{pound} = 10^8 \mathbb{M}[/tex]

    and what they call a count is 10^42 (sextillion sextillion) of the time units.

    [tex]\text{count} = 10^{42} \mathbb{T}[/tex]

    I already said how to define all the units just starting from the Force, so these are just named power-of-ten multiples that are convenient scale.
    in conventional terms a pound is 434 grams and counts are 222 to the minute.
    Last edited: Jan 7, 2005
  7. Jan 7, 2005 #6


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    John Baez and the Mad Inventor

    John Baez and the Mad Inventor

    John Baez was driving around the galaxy in a small spacecraft and he came to the planet where the Mad Inventor lives.

    The Mad Inventor has constructed a room in which Alpha, the fine structure constant, can be different. There is a knob by the door.

    Alpha is an important number in our universe which is usually equal to 1/137.036 and which we nevertheless write 1/137. And if you turn the knob by the door you can make it 1/136 instead. or 1/138

    Some friendly advice: dont turn the knob too much or things will go haywire inside the room. there wont be the same elements on the periodic table and chemistry will be different and some things you didnt expect might be radioactive and so on. Just turn it a little bit.

    the Inventor gives John Baez two insulated balls with a gauge to measure the force between. There are 1013 extra electrons in each ball.

    The balls are anchored a handbreadth apart, center to center (i.e. 8.1 centimeters, this is 1033 of the Force-based length unit we were talking about)

    The mad inventor smiles and turns the knob to 1/138.



    instead of expressing it in terms of the huge force constant,
    we can use a convenient power of ten fraction

    [tex]\text{humanscale force unit} = \text{mark} = 10^{-43}\mathbb{F}}[/tex]

    all the relevant constants are powers of ten and they mostly cancel
    so in terms of the mark force unit it is
    1000/138 which comes to about 7.25

    a mark force is a around half a newton. this 7.25 answer comes to some 3.5 newtons or (if you know conventional ounce force) a little over 12 ounce.

    -----how did we figure it out?---
    alpha tells you the force between two idealized unit point charges, in terms of the natural unit, if they could be placed at unit distance.
    In that idealized situation the

    [tex]\text{force between two electrons} = \frac{1}{138} \mathbb{F}[/tex]

    But because each ball has not one but 1013 extra electrons, the force on that account will be 1026 stronger
    and because the separation between is not unit distance but 1033 longer, we have to square the distance and the force will be on that account 1066 weaker. So the net effect is

    [tex]\text{force between two charges} = \frac{1}{138} 10^{-40} \mathbb{F}[/tex]

    but our humanscale unit of force is 10-43 of the natural unit, so when we rewrite the above in marks (of roughly half a newton)

    [tex]\text{force between two charges} = \frac{1000}{138} \text{ mark}[/tex]

    Here the 1/138 was arbitrary (depending on the knob) and could have been 1/136 or a more realistic 1/137.036....

    it would have been still more convenient to work the exercise entirely in practicalsize units (avoiding all big numbers!) if we were more used to them. In the practical units the coulomb constant is equal to alpha multiplied by a power of ten
    Last edited: Jan 7, 2005
  8. Jan 8, 2005 #7


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    OK I think Baez was right 2 or 3 years ago or whenever he was advocating this. here goes. I am redefining planck units. the variant works better
    if you know the conventional definitions of pl. units you will see these are simpler (and they often make equations neater and more transparent, but here you will see that even the definitions are simpler)

    unit length

    [tex] \mathbb{L} = \sqrt{ \frac{\hbar c}{\mathbb{F}} }[/tex]

    unit time

    [tex] \mathbb{T} = \sqrt{ \frac{\hbar}{\mathbb{F}c} }[/tex]

    unit force

    [tex] \mathbb{F} [/tex]

    unit energy

    [tex] \mathbb{E} = \sqrt{ \hbar \mathbb{F} c }[/tex]

    unit mass

    [tex] \mathbb{M} = \sqrt{ \frac{\hbar\mathbb{F}}{c^3} }[/tex]

    unit acceleration

    [tex] \frac{c}{ \mathbb{T}} = \sqrt{ \frac{\mathbb{F}c^3}{\hbar} }[/tex]

    unit power

    [tex] \mathbb{P} = \mathbb{F} c [/tex]

    unit area

    [tex] \mathbb{A} = \frac{\hbar c}{\mathbb{F} }[/tex]

    unit curvature (as in Einst. eqn.)

    [tex] \mathbb{A}^{-1} = \frac{\mathbb{F}}{\hbar c} }[/tex]

    unit energy density (as in Einst. eqn.)

    [tex]\frac{\mathbb{F}^2}{\hbar c} [/tex]

    the point being that dimensionally curvature and energy density only differ by a factor of the force constant F. And that is the proportion in the Einstein equation which relates the LHS (curvature) to the RHS (energy density). the force is the proportion by which space is bent.
    Last edited: Jan 8, 2005
  9. Jan 8, 2005 #8


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    One way to grasp the sizes of these natural units is to first assimilate the size of some convenient humanscale power-of-ten multiples of them
    handbreadth (8.1 cm)
    count (222 to the minute, 0.28 second)
    mark force (half a newton, an ounce and 3/4)
    "pound" mass (434 grams)

    unit length: 10-33 hand

    [tex] \mathbb{L} = \sqrt{ \frac{\hbar c}{\mathbb{F}} }[/tex]

    unit time: 10-42 count

    [tex] \mathbb{T} = \sqrt{ \frac{\hbar}{\mathbb{F}c} }[/tex]

    unit force: 1043 mark

    [tex] \mathbb{F} [/tex]

    unit mass: 10-8 pound

    [tex] \mathbb{M} = \sqrt{ \frac{\hbar\mathbb{F}}{c^3} }[/tex]

    in other words, instead of memorizing strings of digits, visualize a handbreadth and then think of the natural length unit as a decimal power of that. whatever works for you.
  10. Jan 8, 2005 #9


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    Something I found kind of persuasive was when i realized that in conventional format the Hawking temperature was

    [tex] kT = \frac{\hbar c^3}{8 \pi G M_{hole}}[/tex]

    but in this format, where it depend on the energy E sub hole

    [tex] kT = \frac{\mathbb{E}^2}{E_{hole}}[/tex]

    it is simply the square of the natural energy unit divided by the energy of the hole
    that made me think favorably of the set of units
  11. Jan 8, 2005 #10


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    what this says is that in terms of these units

    the Hawking temperature's kT is simply the reciprocal of the massenergy of the hole

    one takes the mass energy of the hole, and expresses it in the natural energy unit and gets a number

    one simply inverts that number

    and that is how many natural units kT is.
  12. Jan 8, 2005 #11


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    this is the kind of thing that sophisticates like to poopoo because at a certain level it is selfevident. Like----"well if you set all the constants equal to one then you dont see them and of course it's simpler!"

    But I am calling attention to this anyway, look at how ugly the Evaporation Time of a black hole is , in the usual formulation.

    [tex] t_{evap} = \frac{5120 \pi G^2 {M_{hole}}^3}{\hbar c^4}[/tex]

    but if you are working in natural units all you have to do is cube the hole's energy and remember to multiply by 80/pi

    thats all. if the energy is expressed in natural energy units then the time will come out in terms of the natural time unit.

    [tex] t_{evap} = \frac {80}{\pi}\mathbb{T} \frac{E_{hole}^3}{\mathbb{E}^3}[/tex]

    and in natural units the energy and the mass of the hole are the same number so one can say the same thing differently. working in these units you just need to do is cube the hole's mass and multiply by 80/pi
    Last edited: Jan 8, 2005
  13. Jan 8, 2005 #12


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    so here's an example
    Europa is about 1023 pounds

    (a pound is a convenient name for 108 of thenatural mass units)

    suppose we have a black hole with the mass of europa, what is the evaporation time?

    well in natural units obviously the mass is 1031
    and you cube that: 1093
    and multiply by 80/pi

    that gives 25.5 x 1093 and the only problem is the time unit is so small, but remember a count is 1042 of the natural time units, so the evaporation time is 25.5 x 1051 counts
    and you can go from there: 255 x 1050 counts is
    about 1050 minutes, counts being 222 to the minute.

    The evaporation time for a black hole with Europa's mass is about
    1050 minutes.

    so it was not too bad after all.

    The main thing is we didnt have to go thru all that complicated formula that you would with metric units, namely 5120 pi G-square divide by hbar and c-to-fourth etc etc etc, which you need to look up esp when raising to powers and so on---what a mess.
    Here it is exact, cube the mass and multiply by 80/pi.
  14. Jan 8, 2005 #13


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    well here's a case where rewriting doesnt make the equation simpler
    it is the thermal energy density in a room at temp T. Not surprisingly it depends on the fourth power of the energy kT.

    here's how a metric user might calculate it

    [tex]\text{energy per unit volume} = \frac {\pi^2}{15}\frac{(kT)^4}{\hbar^3 c^3}}[/tex]

    here's a version using natural units of energy and length
    the two are the same because
    [tex]\hbar c =\mathbb{E}\mathbb{L}[/tex]

    [tex]\text{energy per unit volume} =\frac {\pi^2}{15}\frac{(kT)^4}{\mathbb{E}^3\mathbb{L}^3}[/tex]

    here's a similar case, the StefanBoltzmann 4thpower radiation law.
    the conventional form is this

    [tex]\text{radiant power per unit area} = \frac {\pi^2}{60}\frac{(kT)^4}{\hbar^3 c^2}}[/tex]

    at least one alternative way of writing the same equation, while there may be some gain in transparency because one sees the "per area" and the energy radiated "per time" explicitly, is more cluttered.

    [tex]\text{radiant power per unit area} =\frac {\pi^2}{60}\frac{(kT)^4}{\mathbb{E}^3\mathbb{L}^2 \mathbb{T}}[/tex]
    Last edited: Jan 8, 2005
  15. Jan 9, 2005 #14


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    continuing the comparison
    how many photons per unit volue are in a room at temp T. Not surprisingly it depends on the cube of the energy kT.

    here's how a metric user might calculate it (in fact dextercioby just did it this way in another thread)

    [tex]\text{number of photons per unit volume} = \frac{1}{2.701} \times \frac {\pi^2}{15}\frac{(kT)^3}{\hbar^3 c^3}}[/tex]

    here's a version using natural units of energy and length.
    the two are the same because
    [tex]\hbar c =\mathbb{E}\mathbb{L}[/tex]

    [tex]\text{number of photons per unit volume} = \frac{1}{2.701} \times \frac {\pi^2}{15}\frac{(kT)^3}{\mathbb{E}^3\mathbb{L}^3}[/tex]

    the reason for the 2.701 is because in thermal radiation at temperature T, the average photon energy is 2.701 kT.
    2.701 comes from the Riemann zeta function

    [tex]2.701 =\frac{3 \zeta(4)}{\zeta(3)}[/tex]

    what the E and the L in the denominator are telling is is that if you work just in terms of the natural units then you simply have to cube the kT and do the stuff with pi-square over 15 and 2.701 ( factors that nature insists be there)

    So we can do an example. A common (Fahren. 49) temperature at the earth's surface is E-29, put in natural terms, cube that: E-87.
    and the numerical factors come to 0.24

    so the number of photons per unit volume in a space at this common temperature is
    2.4 x 10-88

    To interpret that it is good to remember that a PINTsize volume---the cubic handbreadth thing---is 1099 of the natural volume units. So at a more familiar scale our result is just

    2.4 x 1011 photons per pint.

    roughly a quarter of a trillion per pint.
    the problem with using natural units is always AFTER you get the answer you have to know some familiar multiples of the natural units so that you can visualize the answer in humanscale or practical terms

    getting the answer tends to be expedited by the main constants being ONE so you dont bother with them, but then afterwards you need to take a moment to assimilate or interpret the answer.

    Conventional room temperature, on the natural temp scale, is a bit more than E-29, in fact it is 1.04E-29 (that is 69 or 70 Fahrenheit) so if you are curious about the photons per volume at room temp you would cube that
    and the answer would be larger by a factor of 1.12----so 12 percent larger number of photons.

    in rough terms tho, just to have an idea, it is still about a quarter trillion per pint.
    Last edited: Jan 9, 2005
  16. Jan 9, 2005 #15


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    I had better insert the natural unit of temperature in that earlier list a few posts back
    the notational problem with temperature is that T is for time as well as for temperature (making time lowercase t doesnt always take care of things)
    so here I am using a script-T for the natural unit temperature.

    a familiar size degree (happens to be half the Fahrenheit step) is
    10-32 of this script-T
    Last edited: Jan 9, 2005
  17. Jan 9, 2005 #16


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    I happen to know the temperature of the CMB in natural units, it is
    9.6 x 10-32------more precisely 9.64 IIRC.
    It is not too hard to remember if you know that 10-32 of the natural temperature unit is about half a Fahrenheit sized step.

    So I might wonder HOW MANY CMB PHOTONS per unit volume are there in the universe? we we just did a problem like this and the numerical factor out front came to 0.24

    (that is what the pi-square divided by 15 and by 2.701 comes to.)

    working in natural units all I need to do is cube the temperature, cube
    9.6 x 10-32 that is, and multiply by 0.24 and that's the answer.
    the hard part will be interpreting it (because natural units are unfamiliar and not human scale)

    2.1 x 10-94 photons per unit volume

    the thing to remember is that a handbreadth is 1033 of the natural length unit and that makes a pint 1099 of the volume units, so to make a number per unit volume, like that above, understandable you multiply it by 1099 and change it to

    2.1 x 105 photons per PINT.

    So for every pintsize cube, you can make one a hand's breadth width and height, there are 210,000 photons of the cosmic microwave background.

    as far as we know this is true throughout all space at this epoch. they used to be more crowded and they have been dispersing out as space expands. so this number 210,000 is a kind of indicator of the age of the universe and an handle describing the present moment. or if you prefer, that number is just an alias of the CMB temperature.

    the secret of getting along with natural units is that it is very easy to calculate a lot of stuff because so many of the constants are ONE and you just ignore them----but then after getting the answer you need to be able to interpret it humanscale

    so it helps to know 1033 length units is a handbreadth and
    1099 volume units is a pint and
    10-32 of the temperature is a Fahrenhalf (more exactly it's
    0.2826 kelvin if you need the additional accuracy but having a rough idea of the sizes is what seems to matter)
    Last edited: Jan 10, 2005
  18. Jan 10, 2005 #17


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    So what do you think of this set of natural units as compared with conventional Planck?

    Conventional Planck units are established to some extent. One sees them frequently in the literature. So this may moot all the issues. An variant system may stand no chance.

    But I think I will continue trying out these.
    Like calculating speed in low earth orbit.

    In natural units earth mass is 1.38E33
    earth radius is 7.86E40
    We have to do sqrt (GM/R) and the catch is that G value is 1/8pi
    this is what putting the force constant equal one gets us. All the other main constants are one, and the force, but G is peculiar and has to be 1/8pi.

    So divide 1.38E33 by 7.86E40 AND by 8pi, and you get 7E-10
    and sqrt of that is 2.64E-5

    what speed is that? It is 26.4 millionths of the speed of light, because c is the natural unit of speed.

    a millionth of the speed of light is roughly the speed of sound (in a substantial part of the earth's atmosphere) so this is like "Mach 26"

    As always, getting the answer is easy but then you have to interpret it so your mind can assimilate it

    What speed is 2.6E-5? Mach 26. 26 times the speed of sound
    and that is an upperbound on circular orbit speed, a kind of zero altitude low earth orbit. Escape velocity from the surface is sqrt(2) times that.
    Which comes to sqrt(14E-10)

    How can anyone remember that the earth mass is 1.38E33 and the average
    earth radius is 7.86E40? I really dont have a good answer. Should I compartmentalize my brain and never use natural units for macroscopic calculation----be a metric-only thinker---and use natural/Planck for microscopic? Probably not, seems like a dead-end. maybe it would help to see the mass of the earth as 1.38E25 pounds and the radius as
    7.86E7 hands. I know that 10,000 hands is half a mile, so that makes the radius 7860 halfmiles. Seems about right. OK, 7860 halfmiles is something i can kind of remember, and that is 7.86E7 hands and that is 7.86E40 natural.
    Seems a tough stretch, maybe it wont after I get used to it.
  19. Jan 10, 2005 #18


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    Force between parallel wires

    In an idealized picture where you have two long straight currents in the same direction
    and the half-separation is one unit
    and the each current is one electron per unit time
    then the attractive force (per unit length) between the wires is

    [tex]\text{force per unit length} = \frac{1}{137} \mathbb{F}[/tex]

    this is where both currents are

    [tex]\text{charge per unit time} = \frac{e}{ \mathbb{T}}[/tex]

    I am writing 1/137 for 1/137.036..., out of laziness.

    the unit of length can be anything here. If the half-separation if a handbreadth then we are calculating the force per handbreadth of conductor. Or it could be the natural length unit. It doesnt matter because
    it's proportional. It could be a centimeter----the separation is 2 centimeters and the force is the force per centimeter on the wire.

    the main thing is to adjust for the current not being that huge current of one electron per natural unit of time.

    A humanscale charge unit is 1018 e, which is a sixth of a metric coulomb, and a practical timeunit is 1042 of the natural one, which comes to 0.27 second-----222 to the minute. These give a practical current unit which is 0.6 conventional amperes and 10-24 of the natural current unit. So let each wire carry 100 of these

    So in natural units terms, each wire carries 10-22 current.
    and the square (since there are two currents) is 10-44 and
    THAT is what we have to multiply the force we got earlier for the huge unit current.

    The attractive force (per unit length) between the wires is

    [tex]\text{force on segment equal to half-separation} = \frac{1}{137}\times 10^{-44} \mathbb{F}[/tex]

    where both currents are

    [tex]\text{charge per unit time} = 10^{-22}\frac{e}{ \mathbb{T}}[/tex]

    finally, to interpret the answer recall that 10-43 of the natural force unit is half a newton.

    It may seem harder than it is because unfamiliar. I guess the point is that if you know that E-22 is a reasonable current and that E-43 is a reasonable force, then all you need besides that is the fine structure constant 1/137.

    Implicit here is taking the elementary charge e as the unit.
    Last edited: Jan 10, 2005
  20. Jan 10, 2005 #19


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    Speed of sound in air

    I am trying out a variation on planck units that I guess you could call "Force" system of units because the system gives value one
    to the Force which is the coefficient in the Einstein equation.

    this is the proportion relating energy density to curvature---our main equation about gravity. Setting the Force equal to one has the side effect of giving Newton's G the value 1/8pi.

    the conventional planck units are fairly well established and used by a considerable number of people, but I think it worth while to test out
    this variant (which I first saw proposed on SPR by John Baez but which I also see implicitly used in some Quantum Gravity papers where they are often setting 8piG equal to one.)

    Anyway, if you use the "Force" system then you need to know the number 2.6E18, that is 2.6 quintillion, or if you like metric prefixes for number it is 2.6 "exa".

    This is the reciprocal of the proton mass.

    Or if you like it is the (angular format) Compton wavelength of the proton.

    It gets in to hundreds of calculations and formulas for other important physical constants and stuff like that.

    As a Dada or Surrealist gesture I will use this number to calculate the speed of sound in air. Recall that a normal temperature for air at earth surface is E-29

    also the average molecular weight of air is 29
    this is a weighted average of 32 for oxygen O2 and 28 for nitrogen N2
    this means that in natural mass unit the average molecule mass is

    but in any system of units a standard speed of sound formula is

    [tex]\text{speed of sound} = \sqrt{\frac{\frac{7}{5}kT}{\text{mass of molecule}}}[/tex]

    So if we are talking about a usual surface temp of around 49 Fahrenheit which in natural terms meand E-29 then this formula is simply

    [tex]\text{speed of sound} = \sqrt{\frac{\frac{7}{5}\times 10^{-29}}{29/(2.6E18)}}[/tex]

    [tex]\text{speed of sound} = \sqrt{\frac{7}{5}\times 10^{-29}\times 2.6 \times 10^{18} \div 29}[/tex]

    [tex]\text{speed of sound} = \sqrt{1.255 \times 10^{-12}}[/tex]

    [tex]\text{speed of sound} = 1.12 \times 10^{-6} c[/tex]

    it comes out 1.12 millionths, and that is millionths of the speed of light because c is the natural speed unit

    and that is right, it is the speed of sound in air at Fahrenheit 49

    we could work it for other bi-atomic gasses, if we are told the molecular weight, and for other temperatures, but that is one example

    this number 2.6E18 is kind of like the Avogadro number in our context, it is very basic and into a lot of things.
    Last edited: Jan 10, 2005
  21. Jan 10, 2005 #20


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    Hi Marcus

    So that you don't feel too alone....I want to thank you for all this marvellous stuff. I'm sure many people out there will think of it as a great resource.

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