Using the force constant in equations

[marcus]

Quantum gravity research ties into 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

$$\mathbb{F} = \frac{c^4}{8\pi G}$$

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

 

[marcus]

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

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

$$G_{ab} = \frac{8\pi G}{c^4} T_{ab}$$

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.


$$\text{radius} = \frac{1}{4\pi} \frac{E}{\mathbb{F}}$$

$$\text{area} = \frac{1}{4\pi} \frac{E^2}{\mathbb{F}^2}$$

$$k\text{ temperature} = kT = \frac{\hbar \mathbb{F} c}{E}$$

$$\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2}$$

 

[marcus]

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

$$\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2}$$

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

$$\frac{5120 \pi G^2 M^3}{\hbar c^4}$$

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

 

[marcus]

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.

$$\hbar = \mathbb{E}\mathbb{T}$$

$$\mathbb{F}c = \frac{ \mathbb{E}}{\mathbb{T}}$$

$$\hbar \mathbb{F}c = \mathbb{E}^2$$

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

$$kT = \frac{\hbar \mathbb{F} c}{E} = \frac{\mathbb{E}^2}{E}$$

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).

$$\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2} = \frac{80}{\pi} \frac{E^3}{ \mathbb{E}^2 \mathbb{P}}$$

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.

 

[marcus]

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.

-------discussion-----

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

$$\text{area unit} = \mathbb{A} = \frac{\hbar c}{\mathbb{F}}$$

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

$$\text{pound} = 10^8 \mathbb{M}$$

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

$$\text{count} = 10^{42} \mathbb{T}$$

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.

 

[marcus]

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: don't turn the knob too much or things will go haywire inside the room. there won't 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.

WHAT IS THE FORCE BETWEEN THE TWO CHARGES?

----discussion----

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

$$\text{humanscale force unit} = \text{mark} = 10^{-43}\mathbb{F}}$$

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

$$\text{force between two electrons} = \frac{1}{138} \mathbb{F}$$

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

$$\text{force between two charges} = \frac{1}{138} 10^{-40} \mathbb{F}$$

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)

$$\text{force between two charges} = \frac{1000}{138} \text{ mark}$$

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

 

[marcus]

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.

$$\hbar = \mathbb{E}\mathbb{T}$$

$$\mathbb{F}c = \frac{ \mathbb{E}}{\mathbb{T}}$$

$$\hbar \mathbb{F}c = \mathbb{E}^2$$

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

$$kT = \frac{\hbar \mathbb{F} c}{E} = \frac{\mathbb{E}^2}{E}$$

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).

$$\text{evaporation time} = \frac{80}{\pi} \frac{E^3}{\hbar \mathbb{F}^2 c^2} = \frac{80}{\pi} \frac{E^3}{ \mathbb{E}^2 \mathbb{P}}$$

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.

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

$$\mathbb{L} = \sqrt{ \frac{\hbar c}{\mathbb{F}} }$$

unit time

$$\mathbb{T} = \sqrt{ \frac{\hbar}{\mathbb{F}c} }$$

unit force

$$\mathbb{F}$$

unit energy

$$\mathbb{E} = \sqrt{ \hbar \mathbb{F} c }$$

unit mass

$$\mathbb{M} = \sqrt{ \frac{\hbar\mathbb{F}}{c^3} }$$

unit acceleration

$$\frac{c}{ \mathbb{T}} = \sqrt{ \frac{\mathbb{F}c^3}{\hbar} }$$

unit power

$$\mathbb{P} = \mathbb{F} c$$

unit area

$$\mathbb{A} = \frac{\hbar c}{\mathbb{F} }$$

unit curvature (as in Einst. eqn.)

$$\mathbb{A}^{-1} = \frac{\mathbb{F}}{\hbar c} }$$

unit energy density (as in Einst. eqn.)

$$\frac{\mathbb{F}^2}{\hbar c}$$

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.

 

[marcus]

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

$$\mathbb{L} = \sqrt{ \frac{\hbar c}{\mathbb{F}} }$$

unit time: 10^-42 count

$$\mathbb{T} = \sqrt{ \frac{\hbar}{\mathbb{F}c} }$$

unit force: 10⁴³ mark

$$\mathbb{F}$$

unit mass: 10^-8 pound

$$\mathbb{M} = \sqrt{ \frac{\hbar\mathbb{F}}{c^3} }$$

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.

 

[marcus]

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

$$kT = \frac{\hbar c^3}{8 \pi G M_{hole}}$$


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

$$kT = \frac{\mathbb{E}^2}{E_{hole}}$$

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

 

[marcus]

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.

 

[marcus]

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 don't 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.

$$t_{evap} = \frac{5120 \pi G^2 {M_{hole}}^3}{\hbar c^4}$$


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.

$$t_{evap} = \frac {80}{\pi}\mathbb{T} \frac{E_{hole}^3}{\mathbb{E}^3}$$

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

 

[marcus]

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.

 

[marcus]

well here's a case where rewriting doesn't 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

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

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

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


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

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

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.

$$\text{radiant power per unit area} =\frac {\pi^2}{60}\frac{(kT)^4}{\mathbb{E}^3\mathbb{L}^2 \mathbb{T}}$$

 

[marcus]

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)

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

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

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

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

$$2.701 =\frac{3 \zeta(4)}{\zeta(3)}$$

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 don't 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.

 

[marcus]

I had better insert the natural unit of temperature in that earlier list a few posts back

...

unit length

$$\mathbb{L} = \sqrt{ \frac{\hbar c}{\mathbb{F}} }$$

unit time

$$\mathbb{T} = \sqrt{ \frac{\hbar}{\mathbb{F}c} }$$

unit force

$$\mathbb{F}$$

unit energy

$$\mathbb{E} = \sqrt{ \hbar \mathbb{F} c }$$

unit temperature

$$\mathcal{T} = \frac{1}{k}\sqrt{ \hbar \mathbb{F} c }$$

unit mass

$$\mathbb{M} = \sqrt{ \frac{\hbar\mathbb{F}}{c^3} }$$

unit acceleration

$$\frac{c}{ \mathbb{T}} = \sqrt{ \frac{\mathbb{F}c^3}{\hbar} }$$

unit power

$$\mathbb{P} = \mathbb{F} c$$

unit area

$$\mathbb{A} = \frac{\hbar c}{\mathbb{F} }$$

...

the notational problem with temperature is that T is for time as well as for temperature (making time lowercase t doesn't 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

 

[marcus]

continuing the comparison
how many photons per unit volue are in a room at temp T. ...
...
$$\text{number of photons per unit volume} = \frac{1}{2.701} \times \frac {\pi^2}{15}\frac{(kT)^3}{\mathbb{E}^3\mathbb{L}^3}$$

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

$$2.701 =\frac{3 \zeta(4)}{\zeta(3)}$$

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)

...
in rough terms tho, just to have an idea, it is still about a quarter trillion per pint.

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)

 

[marcus]

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 don't 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 won't after I get used to it.

 

[marcus]

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


$$\text{force per unit length} = \frac{1}{137} \mathbb{F}$$

this is where both currents are

$$\text{charge per unit time} = \frac{e}{ \mathbb{T}}$$

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 doesn't 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

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

where both currents are

$$\text{charge per unit time} = 10^{-22}\frac{e}{ \mathbb{T}}$$

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.

 

[marcus]

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 into 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
29/(2.6E18)

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

$$\text{speed of sound} = \sqrt{\frac{\frac{7}{5}kT}{\text{mass of molecule}}}$$

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

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

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

$$\text{speed of sound} = \sqrt{1.255 \times 10^{-12}}$$

$$\text{speed of sound} = 1.12 \times 10^{-6} c$$

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.

 

[Kea]

Thanks

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

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.

Regards
Kea

 

[marcus]

the solar constant

finding out how to use a natural system of units like this is like placing pitons in a rock wall

a nice bit of serendipity is that the surface temperature of the sun is
2E-28, and from what I see of various estimates based on color and brightness it is closer than that----more like 2.00E-28 of the natural unit of temp

that kind of pairs up with the fact that E-29 is a common temp at Earth surface (not far from global yearround average)------49 Fahrenheit on a familiar relative scale. You can see there is a factor of 20 between that and the sun.

well the angular size of the sun is 1/107.5 radian so the angular size of the sun's radius is 1/215---------in other words our distance from sun is 215 times larger than the sun's radius and equilibrium temps go as the SQUARE ROOT of the distance

so equilbrium temp of an unreflective flat surface facing sun, at this distance, is 2E-28 divided by sqrt 215

1.364 E-29

so far it's trivial
--------------------

now do you remember the solar constant in metric? it is the watts of sunlight power per square meter------or sometimes told as the calories per minute received by a square meter, in space at this distance from sun

anyway it is power per unit area. what is it in natural units?

we simply have to fourth the above number, 1.364E-29, and multiply by pi²/60

$$\text{solar constant} = \frac{\pi^2}{60} (1.364 \times 10^{-29})^4$$

it is going to look extremely small but anyway this measured against power on a universe natural scale---there is always the problem of assimilating the size after you get the answer. so let's get the answer and then interpret

0.57E-116

5.7E-117 of the power unit per unit area

that is the rate sunlight comes down on the Earth (apart from reflection of the tops of clouds back into space and attentuation by atmosphere etc.)

to interpret it I guess first I would multiply by E66 to get it on a square handbreadth (which is E66 of the natural unit of area)

5.7E-51 of the power unit per sq. hand

and I just happen to know that the natural power unit is E52 of a convenient humansize unit which is about 1/6 of a watt.

so our answer 57E-52 just needs to be multiplied by E52 to be 57 of these humanscale power units and then, if you like, divided by 6 to give it in watts. So the power of sunlight is 9-some watts per sq. handbreadth.

that is right, actually, because you can multiply by 152 to get metric, a square meter is 152 of these sq handbreadths, and it hits nice and close to the official value.

but I admit it is daunting to see the power of sunlight raw in natural unit terms, as 5.7E-117

the seeming disproportion is that nature's power unit is really vast. the natural unit of power in the system we are using would only need a few minutes to deliver enough energy to make a galaxy like Milkyway or andromeda. Our human scale power of 1/6 watt is related to it by this big E52 factor, which can be tough or boggling to assimilate

well, in terms of that humanscale power unit; sunlight
delivers 57 units of power on a square handbreadth area.

 

[marcus]

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.

Regards
Kea

Well aren't you nice to say so!
But frankly, as to what many people out there might think, I cannot say.
To me, assimilating a new system of natural units (even though similar to the Planck units) seems more and more like driving pitons in a bare rock face.

I will leave the pitons in, in case anyone else wants to climb (and guarantee that they are firm!) even though I do not know a single soul who is intrigued by this particular face of nature.

Maybe, if you are amused by this Kea, you would like to see a list of the pitons

 

[marcus]

nailing some pegs into the rock, for belaying to

reference quantities for the "Force" system of units

the system is defined by making hbar and c the units of ang.mom. and speed
and also by making the unit force be the coefficient in the Einst. eqn.

also one should make the charge unit be the elementary charge e,
and the unit of heat capacity be the Boltzmann k. but this is trivial
since it just says that kT and T are the same number.


then the mass unit will be 2.6E18 times the proton mass
and this number 2.6 quintillion will be like an Avogadro to us

the surface temperature of the sun will be 2E-28

the Earth's orbit speed will be (I guess everybody knows) E-4

the Earth's surface temp will average around E-29
(and this is about 49 Fahrenheit)

the force between unit charges separated by unit distance will be 1/137 of the Force unit

the average radius of the Earth is 7860 "halfmiles"
(a halfmile is E4 handbreadths which are E33 length units each, so one can
claw one's way up to 7.86E40 natural units of length)

a step of E-32 in temperature is about half a Fahrenheit step
and the CMB temperature is 9.6 of these steps
in other words 9.6E-32

the mass of the Earth is 1.38E33 natural mass units.

E8 mass units is roughly a pound

the power unit is awsome. if you take a humanscale power unit that is about a sixth of a watt then the natural unit is E52 of those.
it is equal to the Force multiplied by the speed of light.

E-24 of the natural current unit is about 0.6 amp

Air molecules average 29 proton weight and air temperature averages around E-29 natural. Normal atmospheric pressure is 1.4E-106.
As I guess anyone reading this realizes, E66 area units is a sq. palmwidth and E99 volume units is about a pint (generous pint actually, 532 cc)

maybe that is enough points of reference for now

 

[ohwilleke]

Interesting stuff. Of course, it doesn't change the substance of anything, but it is a useful way to look at things. It seems a bit like Kelvin temperature, useful to a certain class of scientists (physical chemists and physicists mostly in hte case of Kelvin, fundamental physicists in the case of these units), but otherwise not likely to be adopted any time soon by anyone else.

 

[marcus]

Interesting stuff. Of course, it doesn't change the substance of anything, but it is a useful way to look at things. ...

I am in essential agreement with willeke post above.
[edited to remove unnecessary detail]

 

[marcus]

the mass of the sun

I don't seem to have done the mass of the sun

well everybody should know the Earth's orbit speed is E-4
(a tenthousandth the speed of light) and everybody should know
that lite takes 8.3 minutes to get here
so multiply by 222 and E45 (dont mind me I'm just converting)
1.84E45

OK the basic data are circular orbit speed E-4
at a distance of 1.84E45, in natural units

that means GM, for the sun, is 1.84E37

but G, in these units is 1/8pi

so to get M by itself we have to divide by G, which means to multiply by 25.

that comes to 4.6E38 mass units, well that is what I get but I need to interpret it to get a notion of what it means. well E8 units is a pound (something like a conventional pound, 434 g).

So this answer means the mass of the sun is 4.6E30 pounds.

Hey that's right! if I multiply 4.6E30 by 0.434 I get 2.00E30 kilograms!

 

[Kea]

pitons

To me, assimilating a new system of natural units (even though similar to the Planck units) seems more and more like driving pitons in a bare rock face.

I will leave the pitons in, in case anyone else wants to climb (and guarantee that they are firm!) even though I do not know a single soul who is intrigued by this particular face of nature.

Its nice of you to make me feel so young - my friends don't use pitons these days - we like the new fangled pro.

And I really meant that this might be a useful resource...when we actually know how to recover Einstein's equations from the full theory.

 

[marcus]

...And I really meant that this might be a useful resource...when we actually know how to recover Einstein's equations from the full theory.

Maybe you are right!
that is, there would be an underlying theory in which one really did want hbar = c = 8piG = 1,

one only gets hints of this in bekenstein hawking temperature and such.

in all established quantum theory one sees that one should have hbar =c =1

and in Gen Rel one sees that one should have 8pi G = 1, that is, the Force coefficient = 1

but one does not yet really have a theory combining both, only hints of it

(this is from my not too well informed or sophisticated perspective)

anyway if that happened I would like that, and these units would not be a very important element but they would at least be in harmony with it

 

[marcus]

what do they use now to place anchors for rock climbing

I think pitons were destructive of the rock

do they use some kind of claw which does not have to penetrate the rock?

 

[Kea]

what do they use now to place anchors for rock climbing

Removable gear, as in

1. Wires: like nuts on wires in many sizes
2. little pieces of soft metal that can be hammered into small holes
3. Camming devices (such as 'friends'): these are very nice for putting in cracks
and come in many sizes

Of course, all this stuff is heinously expensive.

 

[nightcleaner]

Hi all.

Just want to add my vote in the Marcus popularity poll. I think this work is excellent, Marcus, and am very interested in seeing it in table form. Natural units are, in my opinion, likely to overtake and replace all our human based systems. It is very exciting for me to see them demonstrated so clearly.

Be well,

nc

 

[marcus]

Removable gear, as in

1. Wires: like nuts on wires in many sizes
2. little pieces of soft metal that can be hammered into small holes
3. Camming devices (such as 'friends'): these are very nice for putting in cracks
and come in many sizes

Of course, all this stuff is heinously expensive.

since you like mountaineering I want to try to calculate the dry air lapse rate. the rate that air gets colder with altitude.

you mentioned these cabins, one with a "freezer door", that hikers can use in the NewZealand mountains I guess to be more comfortable than in a tent. I guess there are some places where the height is great enough to be quite cold.

there is this beautiful temperature gradient which is the threshold for convection. if it doesn't cool off faster than this then the air can be stable, but if it cools off with height more abruptly than this then convection starts and wind, and mixing etc etc, which after a while redistributes the heat so that it cools off at only this rate, and then convection stops.

homeostasis is nice. Oh, Earth surface gravity is about 0.88E-50 natural.

$$\text{limiting temperature gradient} = \frac{\text{weight of air molecule}}{\frac{7}{2}k}$$

$$\text{limiting temperature gradient} = \frac{0.88E-50 \times 29 \div 2.6E18}{\frac{7}{2}k}$$

k is one, like the other constants. so the calculation is trivial and
comes to 28 of our "fahrenhalf degrees for each halfmile of altitude.

Well, in natural units it is 2.8E-68, but you know how these things go,
E37 natural length units is a half mile and E-32 on the temp scale is a fahrenhalf step so that is how it is coming out to be what I said.

whenever the wind blows you know that somewhere on Earth there must be convection which means that somewhere the gradient must have temporarily exceeded this 28 halfdegrees per halfmile. it is called the lapse rate. Kea knows this but I am saying it so as to be very explicit

BTW they used to sell a toy which was just a cube---a box---with a switch on the outside. You flip the switch and you hear the box whirr and it even shakes a little

and then a door flies open and a little hand sticks out

and flips off the switch

 

[marcus]

and then of course the little arm pulls back in and the door slams shut

and everything is the way it was before



hello nightcleaner that story of the box was partly for you
homeostasis is a good thing

 

[RingoKid]

I remember those boxes !

they were coffins and you had to put a coin on the switch then a withered hand would reach out and snatch it...

 

[marcus]

ringo, that could well be it. I am going on hearsay from someone I thought reliable, and partly on my imperfect memory. In my version there is no coin--the being in the box simply wants the switch restored to the off position.

Another piton or point of reference. Remembering the density of water. At ordinary swimming pool conditions it is around 1.225 pounds per pint----where a pound is E8 natural mass units and a pint is E99 volume units----so that translates to 1.225E-91. somehow it seems easier to remember 1.225 pounds per pint (which encode the powers of ten)

We had better rescue a baez-in-space story from a stalled thread in the Atoms Molecules forum. Kea got me in the frame of mind to tell this:

John Baez is exploring the galaxy in a small spacecraft and finds himself in low circular orbit around a glistening planet. His wristwatch tells time in natural units and after traveling one radian (1/2pi of full circle) he looks at his watch and sees it has taken

T_radian = 7E45 time units

What is the density of the planet?

$$\text{reciprocal density} = \frac{8 \pi G}{6} {T_{radian}}^2$$

remember that in the Force system 8 pi G = 1

$$\text{reciprocal density} = \frac{1}{6} (7\times 10^{45})^2$$

$$\text{reciprocal density} = \frac{49}{6} \times 10^{90}$$

$$\text{reciprocal density} = \frac{49}{60} \times 10^{91}$$

$$\text{planet's density} = \frac{60}{49} \times 10^{-91}$$

and 60/49 is about 1.2 so this planet is the density of water. in fact it is one large lucent drop. in the story Baez abandons the spacecraft and goes swimming. I suppose his orbit was low and slow enough to allow this.

The density of the Earth's moon is close to that of jade---the green semiprecious stone. I imagine that radian time for a planet made of jade would be just under 4E45

the radian time of a round planet depends only on the density and not on the size----big or small, the same radian time.

for the Earth I rather think the radian time is 3E45, but I haven't checked.