J*di Units: the F*rce is One

[Martin]

is that 1.7E23 pounds the imperial unit for force or is it a jedi unit for mass?

 

Gypsies always prefer to use jedi units

the moon's mass is 1.7E23 jedi mass units

[Martin]

backing off for a moment

the whole reason to play around with Jedi or "The Force" units is

because there is this force constant in nature

 

and when you use it to write formulas they often turn out cleaner or more streamlined

 

and then if you adjust the units so this force AND also hbar and c (also important universal constants) have powr of ten values then a lot of

the arithmetic goes away as well

 

(essentially you are going more decimal than the metric system so whatever is good about metric makes this even better, and it is interesting to see what happens)

 

so i want a symbol for the Force

 

I was calling it einstein force or F_Ein

but it looks klunky always having the subscript

that might turn out best but I want to try some other symbols first

 

like lets see if this works

 

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

[Martin]

Looks like LaTex is down for the moment but i will write these equations in, using the constant Force

 

First of all it obviously makes the main eqn of GR cleaner because you just have

[math]G_{ab} = \frac{1}{\mathbb{F}} T_{ab} [/math]

 

instead of the more messy way the einstein eqn is usually written

(when the author wants to show the dimensions involved)

 

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

 

Here are some sample other equations involving the specs of a black hole.

 

Instead of indexing black holes by mass M, I am going to rank them by

their energy E = M c²

so instead of a hole of mass M, I will talk about a black hole with rest energy E

 

Here are the formulas for things like Schw. radius, area, BekensteinHawking temperature, evaporation time.

 

 

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

 

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

 

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

 

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

[ydoaPs]

afaik, all our measurement systems but one are arbitrary. so, the only one that isn't arbitrary is radians.

i was thinking of a system that is not random and all i could come up with is making the plank length and plank time both = 1, which would make c=1, which makes calculation easier. also, electron volt=1.

the only thing i can find wrong is that for normal calculation, the numbers will be really big. i guess we could use REALLY BIG prefixes on the units.

[Martin]

...

i was thinking of a system that is not random and all i could come up with is making the plank length and plank time both = 1' date=' which would make c=1, which makes calculation easier...[/quote']

 

well, why dont you try using it? lots of physicists use units in which c = 1, and hbar = 1.

 

lots of journal articles get written using Planck units.

 

try getting used to it.

 

the planck units of Mass, Length, Time, and Temperature are listed at the government's website. the best current estimates.

 

so what is your mass in Planck mass units?

 

what is your age in Planck time units?

 

http://physics.nist.gov/cuu/Constants/

[Martin]

Time to pick up where we left off.

 

this thread is about a system of natural units that you get when you take some natural quantities to be basic units (like the speed of light and Planck's hbar, and the force constant in the Einstein equation) and derive the rest from them.

 

it's a variant of conventional planck units

 

to get acquainted with the basic scales, try writing down your age, height, weight.

 

we dont need accuracy, just order-of-magnitude, because what matters is familiarity with the rough sizes.

 

there are some named power-of-ten multiples to help.

E33 natural length units is a handbreadth 8.1 centimeters or 3.2 inches

E34 is a pace of 81 cm or 32 inches

E37 is a halfmile

E50 is a lightyear (actually a lightyear is 1.1676 E50 length units but here the rough size is all that matters)

 

so if your height is 22 handbreadths, call it 22 E33.

 

as for time, E42 is what I've been calling a "count". there are 222 to the minute and 222 per minute is about as fast as I can count out loud

E50 is about a year (more exactly a year is 1.1676 E50 but we are only doing rough sizes)

 

So if your age is 21 years, call it 21 E50

 

as for mass, I have been calling E8 a "pound"----it's 434 grams which is less than the traditional 454 but who cares--it is just a bridge to the natural unit

 

so if your weight is 180 pound, call it 180 E8

 

Yeah, why would anyone be interested, or want to do this, but at this point anyone who wants can write down their age height weight in natural units.

[Martin]

I am exploring what it's like to use these "Force" natural units----like ordinary Planck but with |F|=1

instead of the more usual |G|=1

 

In conventional Planck you set G = 1 but the coefficient in the Einstein equation (the real gravity equation, not Newton's) is a force which is related to G this way

 

F = c⁴/(8piG)

 

so if you set that equal one, as people often do to get the cleanest form of einstein eqn., then since c always = 1 that makes 8pG = 1.

 

so these are streamlined Planck units, which i see used more and more in quantum gravity papers. Increasingly I notice a kappa ("gravitational constant") which is 8piG. And which can be set to equal one to further simplify the equations. The newtonian G is yielding a little---not yet a secondary constant but not as predominant.

 

the theory of gravity is the theory of the (geometry of) spacetime

when there is finally a good Quantum Gravity theory it will provide physics with a new model spacetime and physics will be rebuilt on that new basis.

 

so I really want to check out these units because they are the intrinsic ones that are naturally being used in QG, they are the units of physics in the future.

 

the moment one sets

|F|= |c|=|hbar|=|k|=|e|=1

one has a fairly universal set of units and it is interesting to see what some familiar quantities come out to be.

 

the first thing to ask about is the rough sizes. After that one can ask about basic physics and astronomy constants like Hubble parameter, proton mass, cosmological constant.

 

First the rough sizes:

 

pound E8

year E50

handbreadth E33

pace (32 inch) E34

halfmile E37

lightyear E50

food Calorie E-5

lab calorie E-8

quartervolt E-28

green photon energy 10E-28

average earth surface temp E-29

2/3 mph E-9

67 mph E-7

cold air speed of sound E-6

earth orbit speed E-4

one "gee" acceleration E-50

weight of 50 kg bag of cement E-40

power of a 160 watt lightbulb E-49

 

Now the approximate values of some useful constants:

 

reciprocal proton mass 2.6E18

electron mass 2.1E-22

Hubble time 1.6E60

Lambda 0.85 E-120

rho-Lambda 0.85 E-120

rho-crit (critical density) 1.16 E-120

more exact earth year 1.1676 E50

more exact lightyear 1.1676 E50

avg earth orbit speed E-4

earth mass 1.38 E33

earth radius 7.86 E40

sun mass 4.6 E38

solar surface temp 2.0E-28

CMB temperature 9.6E-32

earth surface pressure 1.4E-106

earth surface gravity 0.88E-50

common fuel energy released by one O₂ 17E-28

 

 

the way I remember earth surface air pressure is to think of

the weight of a bag of cement (E-40) on a sq. handwidth area (E66)

which gives me an idea of the pressure E-106, and it is 1.4 of those.

I remember the earth radius as 7860 halfmiles (a halfmile being E37)

instead of 7.86E40.

 

Named powers of ten, like saying "halfmile" for E37 natural length units, is a way of bridging between humanscale and nature's intrinsic scale.

[Martin]

there are some universal constants and pure math numbers that are factors in various physics equations no matter what system of units, but here they sometimes jump out a little more clearly.

 

Basically, i am going to run down some formulas, partly to show the simplicity. If you know how to use metric units---say to calculate something like the Black Hole temperature or evaporation time---then you know that metric formulas are full of garbage. There can be half a dozen different things you have to look up or remember and multiply in to get the answer. By contrast here, for example, you just cube the mass and multiply by 80/pi.

 

as this list illustrates, what may be a cluttered formula in metric or some other system, tends to boil down to just one number (which is there in the metric formula already, but you tend not to see it because it is drowned out by other details). So here is, like, a list of some essential numbers in nature and what they tell you.

 

80/pi this tells the evap time of a BH. cube the mass and multiply by 80/pi

 

pi²/15 tells the per-volume radiant energy density at some temp. quart the temp (raise to fourth) and multiply by pi²/15

 

pi²/60 tells the brightness at some temp (power radiated per unit area). quart the temp (raise to fourth) and multiply by pi²/60

 

3zeta(4)/zeta(3) = 2.701 tells the average photon energy at some temp.

just multiply the temperature by 2.701. Since sun temp is 2E-28, the average sunlight photon has energy 5.402E-28---anyway that's the idea.

 

1 tells the bekenhawking temperature of a BH. just take 1 over the mass.

 

1/4pi tells the Schw. radius of a BH. just take that times the mass.

 

1/4pi tells the area of the BH. take that times the square of the mass.

 

3 tells the critical density of the universe. just multiply 3 by the square of the hubble parameter

 

29 is the molecular weight of air. It is handy to know.

(atomic and molecular weights generally are)

 

Oh, they tell us that the density of the universe is at or very close to the critical value. So 3 also tells the actual density of the universe.

 

1/137 (more exactly 1/137.036...) is the coulomb constant. it tells the force between two charges separated by a distance. just multiply the charges by 1/137 and divide by the square of the distance.

 

1/137 also tells the force between parallel currents (measured on a test segment with length equal half the separation). just multiply the currents by 1/137

 

in each case i am assuming that the calculation is done in natural units terms, so that I don't have to specify the units each time I say something.

[Martin]

in post #57 I had these two rough-size comparisons

 

weight of 50 kg sack of cement E-40

power of a 160 watt lightbulb E-49

 

I am thinking of the force E-40 as a "sack" force benchmark

and imagine a 50 kg weight on a pulley descending at speed E-9

(which is 2/3 mph, or a billonth of the speed of light)

and as it descends it does work, like turning a spindle, maybe even

generating electricity.

 

I say that the power output of that descending weight is E-49

you just have to multiply the force E-40 by the speed E-9

and you get the power. of course if you are generating electricity there

will be some loss because of friction and suchlike inefficiencies.

but basically this force exerted at that speed delivers that much power.

and I'm going to call that level of power a BULB of power.

 

this is a drastic solution to the problem of remembering the brightness of sunlight. the solar constant at this distance from the sun---the power per unit area delivered by direct unattenuated sunlight----is 5.7 BULBS PER SQUARE PACE.

 

In natural unit terms, a pace (81 cm) is E34 and a square pace is E68 and a bulb of power is E-49. So a bulb of power spread over a square pace is

E-49/E68 = E-117

I am saying that the brightness of sunlight is 5.7 times that.

It is like about SIX of those 160 watt litebulbs set in a pace-wide square.

 

In natural units, 5.7E-117 is what the handbook value of the solar constant actually turns out to be. but I dont find that so easy to remember. So I visualize it as 5.7 bulbs per sq. pace.

 

A pace is just one of my steps----around 32 inches----so I can easily pace out a square that size on the flagstones in the garden. It is an easy area for me to visualize. and the litebulbs are easy to visualize. so I have a visual handle on this 5.7E-117

===================

 

the power unit in J*di, or "Force", units is E49 bulbs and it is the power delivered by the Force pushing at the speed of light.

 

this is a lot of power and if you count as fast as you can outloud, say 222 counts a minute, then WITH EVERY COUNT THE POWER DELIVERS ENOUGH ENERGY TO CREATE 2000 SUNS.

 

this is enough power to create a galaxy in something on the order of 100 days. or if you wanted to produce such a power by annihilating stars and converting their whole mass into energy then you would have to annihilate about 2000 stars like the sun with every count.

 

the natural units are fundamentally Big Bang-scale units. the temperature, the density, the pressure, ...and so on...are mostly at the level of big bang conditions. however that does not seem to disqualify them from ordinary mundane calculations either!

[ed84c]

are you enjoying yourself Martin? I certainly am reading.

[Martin]

are you enjoying yourself Martin? I certainly am reading.

 

Hi ed84, great to hear from you.

Yeah, I find these units absolutely fascinating!

many formulas get streamlined down to almost nothing

as if nature is trying to tell us these are the right ones to use

 

thanks for reading too, knowing someone else might be interested is part of what motivates me to write it down

[Martin]

This quote contains the essentials of Force units.

I am exploring what it's like to use these "Force" natural units----like ordinary Planck but with the coefficient of the Einstein equation |F|=1

instead of the more usual |G|=1

 

F = c⁴/(8piG)

 

If you set that equal one, as people often do to get the cleanest form of Einstein eqn., then since c always = 1 that makes 8pG = 1

I want to check these units out because they are the intrinsic ones that are getting used in Quantum Gravity, and may be those of physics in the future.

 

The moment one sets

|F|= |c|=|hbar|=|k|=|e|=1

one has a fairly universal set of units and it is interesting to see what some familiar quantities come out to be.

 

the first thing to ask about is the rough sizes. After that one can ask about basic physics and astronomy constants like Hubble parameter, proton mass, cosmological constant.

 

First the rough sizes:

 

pound E8

year E50

handbreadth E33

pace (32 inch) E34

halfmile E37

lightyear E50

food Calorie E-5

lab calorie E-8

quartervolt E-28

green photon energy 10E-28

average earth surface temp E-29

2/3 mph E-9

67 mph E-7

cold air speed of sound E-6

earth orbit speed E-4

one "gee" acceleration E-50

weight of 50 kg sack of cement E-40

power of a 160 watt lightbulb E-49

 

Now the approximate values of some useful constants:

 

reciprocal proton mass 2.6E18

electron mass 2.1E-22

Hubble time 1.6E60

Lambda 0.85 E-120

rho-Lambda 0.85 E-120

rho-crit (critical density) 1.16 E-120

more exact earth year 1.1676 E50

more exact lightyear 1.1676 E50

avg earth orbit speed E-4

earth mass 1.38 E33

earth radius 7.86 E40

sun mass 4.6 E38

solar surface temp 2.0E-28

CMB temperature 9.6E-32

earth surface pressure 1.4E-106

earth surface gravity 0.88E-50

common fuel energy released by one O₂ 17E-28

 

 

the way I remember earth surface air pressure is to think of

the weight of a sack of cement (E-40) on a sq. handwidth area (E66)

which gives me an idea of the pressure E-106, and it is 1.4 of those.

I remember the earth radius as 7860 halfmiles (a halfmile being E37)

instead of 7.86E40. Named powers of ten, like saying "halfmile" for E37 natural length units, is a way of bridging between humanscale and nature's intrinsic scale.

 

I'll continue working exercises and examples using this "Force" or "J*di" system.

 

Once in the course of his explorations John Baez came to a giant planet with three moons, each being a different temperature.

 

Now by virtue of much meditation and the teachings of Yoda, he had achieved the level of Taoist Monk grade Two, which means that he could see in the infrared. Accordingly Baez was able to judge the brightness of the thermal radiation from each moon and tell how warm it was on the ground.

 

Just for comparison sake, we note that the solar constant (the brightness of sunlight at earth's distance from the sun) is 5.7E-117 natural units of power per unit area. As a crutch, i think of this as 5.7 lightbulbs per square pace. A 160 watt bulb being E-49 of the natural unit of power, a square pace area being E68.

 

Baez cruised by the nightsides of each moon in order to gauge the infrared heat radiation from each in the absence of reflected light. And he found that the brightnesses were as follows:

 

1.4E-117

1.9E-117

and 2.5E-117

 

using my crutch I am picturing 1.4, and 1.9, and 2.5 of those 160 watt bulbs. and a square pace of area. the power is all in the infrared, not in the visible spectrum, but Baez can see it because of the mental teachings of Yoda.

 

the question is, what is the temperature of each of the three moons? If possible to picture it in familiar terms (is it freezing, is it boiling, is it the perfect temperature for a hot tub?) and say which moon you would choose to land on.

[Martin]

Baez cruised by the nightsides of each moon in order to gauge the infrared heat radiation from each in the absence of reflected light. And he found that the brightnesses were as follows:

 

1.4E-117

1.9E-117

and 2.5E-117

 

the question is, what is the temperature of each of the three moons?...

 

you just multiply each number by 6 and take the fourth root (press square root twice)

 

officially what you multiply each number by is 60/pi²

but pi-square is almost the same as ten, and 60/10 is six.

 

so let's be real official and multiply each moon's brightness by 60/pi²

 

8.511 E-117

11.55 E-117

15.20 E-117

 

then we press the squareroot button twice and get the temps

 

0.960 E-29

1.037 E-29

1.110 E-29

 

I am used to this scale so i can interpret: the first is below freezing, the second is room temperature, and the third is the perfect temperature for a hot tub!

 

Obviously you dont go to that one because it would be like being in a hot tub all the time. the crutch for interpreting these temperatures is to say all the temps we experience are right around E-29 (that small fraction of the primal big bang natural Unit temperature) and 1.000E-29 is 49 Fahrenheit. a kind of average common earth surface temperature. going up from 1.000 to 1.110 is equivalent to going up 110 halfFahrenheit steps, that is 55 F-degrees, which if you add it to 49 gets you 104 Fahrenheit.

and going from 1.000 up to 1.037 is going up 37 of those steps which is 18-some F-degrees which if you add it to 49 gets you 67 Fahrenheit. what could be a more comfortable moon than that? Clothing optional.

 

So the moral is that if you ever see a moon glow 1.9 E-117 in the infrared, that's the one.

[Martin]

You have traveled to a distant planet to study with Yoda.

The brightness of sunlight there is the same 5.7E-117 as at earth so much the same wildlife has evolved to grow there. In fact everything looks pretty similar to Marin County in Northern California.

 

Yoda teaches you to see better, not only in normal colors but also in different wavelengths, like infrared and ultraviolet, and to be in harmony with the Force and stuff like that. The usual guru business.

 

One night you have a dream of being on a strange planet where the sun appeared very small in the sky-----the angular diameter was only 1/350 radian, and the radius was of course half that: 1/700. After morning meditation, you tell Yoda about it and he asks what color the sunlight was, in the dream.

"The average photon is green," you tell him, because the average sunlight photon energy was 10E-28 in your dream. Yoda's teaching has enabled you to judge the average energy and wavelength, just by looking at the light, which is pretty amazing and almost worth the trip on its own account.

 

Then Yoda asks about the brightness of the sunlight on the planet in your dream, was it brighter or dimmer than the 5.7E-117 here?

 

You remember perfectly how things looked in the dream but you want to calculate as a check, so you divide the color 10E-28 by 2.701 to get the temperature of the star's surface 3.7E-28 square twice multiply by the usual pi²/60, divide by 700²

It comes out 6.3E-117. So it is a little brighter there than what we have here, and on Yoda's planet, with our 5.7E-117 sunlight

[Martin]

The characters of J.R.Tolkien were furious with their author for never putting any good sex scenes in his books, so to get revenge they decided to hang him.

 

"He has us do all that dumb stuff and screw around with rings and that," said one Hobbit who was wiser than the rest, "hanging is too good for that phony medieval windbag."

 

The Hobbit search team found the fugitive writer on a small planet where the air had no resistance and you could go into orbit skimming over the surface at 67 miles per hour. When they arrested him, he was modifying an old Chevy pickup truck to go into orbit.

 

J. R. Tolkien stood bravely on the gallows with the noose around his neck.

Please don't hang me, he said, I will write a new Hobbit book with plenty of wonderful sex scenes.

 

The Hobbits considered his offer.

 

And you get twenty percent of the movie rights, conceded the famous novelist.

 

All right, said the wisest Hobbit, if you can tell us how much this planet would weigh in its own gravity, we will let you go. We will reconsider your case after the film release.

 

==========

How would you weigh a planet in its own gravity anyway? Dig up a chunk which you think is a millionth of the planet, weigh that with giant scales sitting on the surface of the planet, and multiply by a million.

 

or if a millionth is still too much, use a smaller fraction and multiply by more. Like weigh a billionth part of it, and then multiply by a billion.

 

How much would the moon weigh in its own gravity? How much would Mars weigh, in its?

 

Does Tolkien escape the Hobbit's clutches?

[Martin]

It's easy, in natural units you just take the speed, square it twice, and multiply by 8pi.

 

Anytime you know the speed of the lowest orbit around a planet (same as the top speed for circular orbits because the lower they are the faster they go) you can tell how much the planet weighs on its own terms. that doesnt say how much it would weigh on earth! we're talking about its weight in its own gravity.

 

67 mph is the skimming orbit speed in this case and in natural terms that speed is E-7. So raise to the fourth: E-28, and multiply by 8 pi, which for practical purposes is 25.

 

the planet's weight is 25 E-28

 

does anyone know the low-orbit speed at the earth's moon? might be interesting to take that to the fourth power.

I believe it is right around 5.6E-6 of the speed of light. If you were in circular orbit near the lunar surface that is how fast you'd be going.

 

BTW how to interpret the force E-28 in humanscale terms? It is like the normal weight of a trillion sacks of cement. Because the natural force unit is E40 sacks, and this is E-28 of that, so it is E12 sack-weights.

(the earlier post with all the comparisons listed E-40 of the force unit as the weight in earth gravity of a 50kg sack of cement)

[Martin]

According to Daniel P, the earth is in danger of invasion by the Fat Men.

these are a space-faring race of creatures

who plunder other planets for their fast food.

 

their planet, planet Ornish, is deficient in basic resources like french fries, mayonaise, potato pancakes, donuts, Colonel Sanders fried chicken, and sour cream, which has forced them into a life of nomadic piracy.

 

the Fat Men wear spacewar uniforms of loud plaid sports-jackets, loafers, and eyeglasses with heavy black plastic frames, they travel by the millions in troopships shaped like enormous saucers.

 

Professor Pinkwater fears that, before long, swarms of Fat Men will descend on earth and ravage our fast food outlets.

 

A defense analyst has calculated that one raid by the Fat Men could deplete the earth of a million pounds of its mayonaise. How much energy does this represent?

[Martin]

this is a very easy problem because if you just go into the kitchen and look on a jar of Best Foods Real Mayonaise it will say that a 13 gram serving has 90 food Calories.

One of our Jedi pounds is 434 grams so it contains almost exactly 3000 Calories.

 

But to a reasonably close approximation, the natural unit of energy which is built into the universe (derived by setting the speed of light and other basic constants equal to one) is 100 million Calories.

 

So a million pounds of mayonaise, with 3 billion Calories, represents 30 energy units on the natural scale.

[Martin]

The ships of Ornish are driven by Cat Motors

which consume cats as fuel by converting each cat entirely into energy.

 

The captain of a ten-million-pound troopship wishes to achieve a speed of E-3 (one thousandth of the speed of light, about 300 km/second) in order to depart a plundered system. Once in the clear he will enter warp and rendezvous with the rest of the pirate band. The initial change of velocity is accomplished by the ship's efficient photon drive.

 

How many standard 10 pound cats must be converted?

 

*********

we obviously the mass of each cat is one billion mass units (10 pounds is 10 E8 units) so since c = 1 each cat yields one billion energy units.

The ship mass is 10 million pounds (E15), so the momentum change is E15 x E-3 = E12 momentum units. This requires discharging a photon pulse with E12 energy units, which consumes 1000 cats.

[Martin]

in natural units the temperature at the core of the sun is

5E-25

 

the energy in our food is born in the core of the sun (that is where the fusion reaction happens that gives the sun its energy) and it gradually percolates out to the surface

 

by the time it gets to the surface the energy is not so concentrated as in the core, and the temperature at the surface is 2E-28

 

the average temp at the earth's surface is E-29

 

so it goes down by a factor of 2500, getting out to the surface, and then by another factor of 20 between there and us.

 

I am trying to think of a physics exercise involving the temperature 5E-25 at the core of the sun. Anyone have an idea for one?

[Martin]

a nice inflammatory note in the female range is the D on the treble clef

and the frequency of that note is E-39 on the natural frequency scale.

 

that means E-39 radians per natural time unit, or E39 time units to go one radian (2pi E39 to go a full cycle)----radian is the measure of phase used with natural units whether they are Planck or a variant thereof.

 

male response to voices in the frequency range around E-39 is latently animal---albeit buried under layers of civilization---and one of the reasons we enjoy singing in mixed chorus.

 

ON THE OTHER HAND the temperature of the sun's surface is 2E-28 which means that a fairly typical energy for a photon of sunlight is 2E-28

and such a photon has frequency 2E-28

(on the natural scales it is always the same number)

 

HOW MANY OCTAVES are there between the soprano and sunlight?

 

that is, about how many doublings between the two typical frequencies

E-39 and 2E-28?

[Martin]

A few days ago I calculated what the temperature gradient in the Titan atmosphere should be (with the simplifying assumption that it is pure nitrogen, a few percent methane would make it less---a more gradual cooling off with higher altitude)

 

I found out today that my estimate was roughly right, according to what Huygens measured on the way down.

 

I calculated a maximum gradient of 3.7E-69 (less if there is methane vapor) and Huygens measured 1.4E-69

the measured value being lower can be accounted for by the presence of methane and the fact that what I calculated was a theoretical maximum (the critical or threshhold gradient for conduction and mixing)

 

the way you calculate the maximum gradient uses the body's gravity (0.12E-50 in Titan case) and the mass of a typical molecule in the atmosphere (28/(2.6E18) for N2 molecule in Titan case).

Multiplying these together gives the force which the molecule weighs.

Multiplying that by 2/7 gives the socalled "lapse rate" or maximum gradient. Any gradient above that is not stable because it sets up mixing by convection.

 

To refer the Titan gradient back to human scale remember temperature step E-32 is a Fahrenhalf degree and E37 is a half mile. So the measured gradient which extended up into the atmosphere for 60 halfmiles, was

1.4 Fahrenhalf degrees per half mile. At 60 halfmiles it was about 85 Fahrenhalf degrees colder. That height corresponds, in the picture I am using, to the Earth's "tropopause" which is the highest convection and clouds extend up. Passenger aircraft typically fly above the clouds and turbulence that is above the tropopause. Because of the earth's stronger gravity, this comes at a much lower altitude.

 

Above the tropopause the atmosphere actually does not continue to get colder----an odd fact you may have noticed if you'v checked out a temperature profile of the atmosphere.

[Guest StartingOver]

How long does it take to get the email with instructions on how to activate my username? I have a posting for Abstract Algebra and am indesperate need of some help!

[Martin]

How long does it take to get the email with instructions on how to activate my username? I have a posting for Abstract Algebra and am indesperate need of some help!

 

I'm not a tech and I dont know anything about how they run the SFN board, but I am guessing your account is fully active and that you can post your request for help wherever. So why don't you try?

[Martin]

Here's a sample of how the formulas look in natural units. I've been illustrating some of these with examples in this thread.

 

1. for a satellite in circular orbit

mass = 4 x period x speed³

 

e.g. a planet's year is E50 and its speed is E-4 (both very like Earth's)

how massive is its star?

4 E(50-12) = 4E38

 

e.g. a planet's mass is E33, its year is E50 and the speed of a synchronous satellite circling it is E-5 (similar to Earth case as well)

how many of this planet's days to a year?

4 period E-15 = E33, 4 period =E48, 400 days in a year.

 

e.g. you are orbiting a small planet at the speed of a run, 6.7 mph, and find that full circuit takes 1 and 7/8 hours. What is the planet's mass?

speed = E-8, 4 x period = 450 minutes = E47, E47 E-24 = E23

 

2. for black hole radius, area, temperature, evaporation time

 

radius = (1/4pi) mass

area = (1/4pi) mass²

temp = mass^-1

evaporation time = (80/pi) mass³

 

3. radiant energy density and brightness

(energy per unit volume, power per unit area)

 

energy density = (pi²/15) temp⁴

brightness = (pi²/60) temp⁴

 

4. average photon energy

3zeta(4)/zeta(3) = 2.701 tells the average thermal photon energy at some temp. Multiply the temperature by 2.701.

 

avg photon energy = 2.701 temp

 

Since sun surface temp is 2E-28, the average sunlight photon has energy 5.402E-28.

Sun core temp is 5E-25, so the average core photon has energy 13.5E-25.

Room temperature is 1.04E-29, so the average energy of a photon in the room with you right now is 2.8E-29

 

5. critical density of universe

(just multiply the square of the hubble parameter by 3)

H = (5/8)E-60

H² = (25/64)E-120

critical density = 3(25/64)E-120 = (75/64)E-120

It's the overall concentration of energy needed in the universe so that it can be spatially flat---too little makes negative curvature and too much makes positive curvature, either way triangles dont add up to 180 degrees--- and since it looks flat, folks think the actual density is at or close to critical.

This is where "0.83 joules per cubic km" comes from. It is just a metric translation of 1.2E-120

 

 

6. radian time in low orbit.

(time to go one radian, that is 1/2pi of full circle, lowest possible orbit)

radiantime² = 6/density

 

e.g. if density of planet is E-91 (similar to water) then square of radiantime is 6E91 = 60E90, so radiantime roughly 8E45 = 8 x 4.5 minutes.

 

e.g. if density of planet is 6E-91 (similar to Earth) then square of radiantime is E91 = 10E90, so radiantime roughly 3E45 = 3 x 4.5 minutes.

 

7. the heat capacity of a molecule of water

It is 9

So making water temperature increase by E-30 takes an amount of energy equal to (number of molecules) x E-30.

(often the number of molecules is not so hard to tell)

 

8. some 1/137 stuff

 

1/137 (more exactly 1/137.036...) is the coulomb constant. it tells the force between two charges separated by a distance. just multiply the charges by 1/137 and divide by the square of the distance.

 

1/137 also tells the force between parallel currents (measured on a test segment with length equal half the separation). just multiply the currents by 1/137

 

(1/137)² tells the energy needed to ionize a hydrogen atom. multiply the rest energy of an electron (2.1E-22) by it and you get a quantity of energy called the Hartree----which is twice the ionization energy (so you still need to divide by two)

 

in each case i am assuming that the calculation is done in natural units terms, so that I don't have to specify the units each time I say something.

 

I need to be able to write a small Greek omega [math]^{\omega}[/math] in the middle of ordinary text so let's see how this works