Recognitions:
Gold Member

## using the force constant in equations

yesterday i took a bird feather out to an undeveloped canyon where there is this old peach tree which still blossoms
I tried to reach the blossoms to tickle them, the way a bee would,except that from a bee's point of view it might still be too cold ( sometimes there arent enough bees at this time of year)
the tree is between 50 and 100 years old and is half dead and has fallen over so the trunk is horizontal, out over a sharp dropoff, but the part that is still alive is still blossoming copiously every year. I guess the canyon used to be farm. but it is all overgrown now with brush and coyote bush and eucalyptus and bay. you should watch out for the poison oak.

I want to think about the natural unit of power---the rate of delivering energy, like the calories per unit time.
A "count" is about as fast as you can count outloud, say counting repeatedly up to 20. and a count is E42
So in one count, the natural unit power delivers E42 units of energy.

(each one can heat JB's hot tub, so E42 a huge amount of energy)

remember that one natural unit is E8 calories. so one can say that in one count it delivers E50 calories, if you like.

So as you count rapidly, with each number you say, the power brings E50 calories.

but how to visualize that much energy. For instance, how does it compare with mass-energy invested in the existence of the sun? Every particle of matter has some energy bound up in its very existence which is released if that particle experiences annihilation. By our standards it is quite a lot, even a small amount of mass (by our standards) when annihilated releases a large (by our standards) energy. If the sun went out of existence with a flash. What.

As you count, with each number you say, the power brings you the energy that would be needed to create how many suns?

I think it is 2000 suns, but i will have to check.

well that is about right, you are counting as fast as you can (222 counts a minute) and everytime you say a number
the natural power gives you 2000 suns

it gives you the energy that would be set free if 2000 suns suddenly went out of existence (not just the energy which they would produce in their lifetimes which is only a very small fraction of their total mass-energy)

so in not too long a time (maybe a couple hundred days) this power could deliver a galaxy-equivalent of energy.

I had better check that the number 2000 is right, or find out what it is more precisely just to be sure of not being too far off

-----
I did check it and it seems all right.
this time it actually seems simpler if you do NOT use calories or any of that, but stick with natural units. We already worked out that the mass of the sun is 4.6E38, and because c = 1 that means the ENERGY of the sun is the same number 4.6E38 units of energy. And at every count you get E42 units of energy. So how many suns-worth is that?

Easy just divide E42 by 4.6E38.

I get 2170 suns-worth. So about 2000.

in a minute (222 counts) you get about half a million suns.
That is enough about that for a while!

 Recognitions: Gold Member Science Advisor today's best estimate of the Hubble parameter is 71 km/second per Mpc. This is the reciprocal of a time, called the "Hubble time" which you sometimes see listed as 13.77 billion years, or some such. It is not the estimated age of the universe, although the figures are close, but just one over the Hubble parameter. in natural units, a year of 365.25 days is 1.1676 E50 I am keeping unnecessary precision to round off later. It seems handy that a year comes out close to E50 time units and it makes a lightyear also come out roughly E50 length units. this figure they have for the Hubble, of 71, translates to a Hubble time of 1.6080 E60 natural time units. (keeping spurious precision again for later rounding) if you divide that by the length of a year you do get 13.77 billion years, so it checks. I'm thinking that the reciprocal of 1.6E60 might actually be a very handy figure for the Hubble. I will use it to calculate the critical density---the density ( including dark energy) we theoretically must have in order for the universe to be spatially flat. I actually think that by the Friedmann equation it must be 3/(1.6E60)2------you just do 3H2 (other stuff is one) it comes out 1.16 E-120 By WMAP and other means they think they observe that as the actual density of the universe, and that dark energy is 73 percent of it. The 0.73 is another WMAP number. well what a nice surprise , 73 percent of 1.16E-120 is the figure we got for dark energy density earlier. 0.85 E-120 you may be disappointed that this is all so trivial, plz excuse, it's just me getting used to these units. I hadnt realized that a year was E50 and the Hubble time was of order E60 and the critical density was around E-120 (rho crit a little more, and rho Lambda a little less) A couple of years back when Baez opined, on SPR, that planck units would be better if 8piG = 1 they didnt bother to see how it would work out, they argued abstractly, and voices rallied to the flag of ancient custom. I think eventually he was persuaded to drop it. More important matters to discuss. Still many's the Loop Quantum Gravity paper I see that has a kappa in it standing for 8pi GN, and that kappa is not unlikely to be referred to as "the gravitational constant" at some point and set equal to one by a change of units.
 Recognitions: Gold Member Science Advisor some special numbers go with these units, most are pure math numbers and would be factors in the equation no matter what system of units, but here they sometimes jump out a little more clearly. 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 6 tells the density of a round planet. divide 6 by the square of the radian time in low orbit. 9 or thereabouts is the heat capacity of a molecule of water 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 (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. there's lots more but maybe this is enough for now
 Recognitions: Gold Member Science Advisor Back in post #71 I listed some rough sizes, including these force and power benchmarks. 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. The power output of that descending weight is E-49. To see that, 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 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 =================== In the "Force" system of natural units, the unit of power is of course E49 bulbs (because bulb was defined as E-49) and it is the power delivered by the unit Force pushing at the speed of light. this is a lot of power and if you count as fast as you probably can outloud, say 222 counts a minute, then WITH EVERY COUNT UNIT POWER DELIVERS ENOUGH ENERGY TO CREATE 2000 SUNS. We discussed this, it 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. As with conventional Planck units, these natural units are fundamentally Big Bang-scale. the temperature, the density, the pressure, ...and so on...are mostly at the level of big bang conditions. I guess that could be seen as reassuring. You can be sure ahead of time that you are not going to encounter any temperature less than zero or greater than one. the physical scales tend to be bounded between zero and one----like with speed too.
 Recognitions: Gold Member Science Advisor An explorer once visited three planets and went into low orbit around each in order to take pictures with his digital camera to put on his website. He finds each planet more delightful than the one before it and, while skimming around the third, he breaks down and calls you on the cellphone. Hello, says the explorer, on each planet it took a different length of time to travel one radian of the low orbit. In natural time units it took 7E45 4E45 and 3E45 on planets A, B, and C respectively. what are the densities of the three planets? the formula for the density is D = 6/T2 you just divide 6 by the square of the radian time, so the three densities are 6/(49E90) = 1.224E-91 6/(16E90) = 3.75E-91 and 6/(9E90) = 6.66E-91 the first, you tell the explorer, is virtually the same the density of water. planet B, on the other hand, is slightly over 3 times the density of water and is therefore comparable to many of the solar system's satellites including the earth's moon planet C, however, is 5.4 times the density of water, quite close to earth itself, which is 5.5! Indeed, says the explorer, that is within experimental error. I believe I am just passing over Sausalito. BTW E45 natural time units was listed a couple of posts back as lasting 4.5 minutes, so the radiantime for low orbit in the earth-like case, namely 3E45, becomes 3 x 4.5 = 13.5 minutes. that is for a hedgetop skimming orbit neglecting air--- not practical, of course, but raising it above the atmosphere does not make the orbit all that much slower. so it is a pretty good estimate. (multiply 13.5 minutes by 2pi to get the period)
 Recognitions: Gold Member Science Advisor Another time the explorer cruises by the nightsides of each of 3 planets in order to gauge the infrared heat brightness from each. He wants to know how warm or cold they are. He finds that the heatglow brightness of the three is as follows: 1.4E-117 1.9E-117 and 2.5E-117 the question is, what is the night-time surface temperature on each planet? Simply put, you just multiply each number by 6 and take the fourth root (press square root twice) although officially what you multiply each number by is 60/pi2. However, pi-square is almost the same as ten, and 60/10 is six, so it's almost the same either way. So let's multiply each planet's heatglow by 60/pi2 8.511 E-117 11.55 E-117 15.20 E-117 and press the squareroot button twice to get the temps 0.960 E-29 1.037 E-29 1.110 E-29 The first is below freezing, the second is room temperature, and the third is the perfect temperature for a hot tub! To help with interpreting these temperatures, remember all those we usually experience are close to E-29 and 1.000E-29 is our basic reference 49 Fahrenheit. 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. On the other hand, going from 1.000 up to 1.037 is equivalent to 37 of those steps which is 18-some Fahrenheit-degrees. Adding that which to 49 gets you 67 Fahrenheit, and what could be a more comfortable than that? So the moral is: go for the planet that glows 1.9 E-117 in the infrared.
 Recognitions: Gold Member Science Advisor two more benchmarks the temperature in the energyproducing core of the sun is 5E-25 (remember that solar surface temp is 2E-28 and avg earth surface temp is E-29 the way to remember solar surface is that E-28 is an eQ and green photons have energy 10eQ, so solar surface is going to be around E-28, and the temp there happens to be 2E-28. so from core out to surface, temp goes down by factor of 2500) and middle D on piano is (1/2)E-39, have to go, back later that means the D in the soprano/alto range, on the fourth line of the treble clef, is E-39 the natural unit frequency is E39 times higher than that note the sopranos in our chorus sing---maybe I could get it in falsetto. THE CAT ENGINES OF ORNISH I suspect Kea of liking cats, so I have devised a story about the wicked space pirates of Planet Ornish and how they propel their giant Bagel-shaped troopships. I hope this will scandalize Kea.
 Recognitions: Gold Member Science Advisor ------THE CAT ENGINE OF THE SHIPS OF ORNISH---- 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? ---answer--- "pound" is just a handle on E8 mass units so clearly 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 natural energy units. (the mass is the same number as the energy in natural units) The ship mass is 10 million pounds (E15), so the desired momentum change is E15 x E-3 = E12 momentum units. This requires discharging a photon pulse with E12 energy units, which consumes 1000 cats.
 Recognitions: Gold Member Science Advisor ---------FAST FOOD EXERCISE------ According to Defense Analyst Daniel Pinkwater, the earth is in danger of invasion by the Fat Men: a space-faring race which plunders other planets for their fast food. Their planet, planet Ornish, is deficient in basic resources like french fries, mayonaise, potato pancakes, Colonel Sanders fried chicken, and sour cream, which has forced them into a life of nomadic piracy. The Fat Men spacewar uniforms consist of loud plaid sports-jackets, green dacron slacks, loafers, and eyeglasses with heavy black plastic frames. They travel by the millions in a fleet of troopships shaped like enormous bagels. Professor Pinkwater fears that, before long, hordes of Fat Men will descend on earth and ravage our fast food outlets. The 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? ---answer--- 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 (E8 natural mass unit) pounds is 434 grams so it contains almost exactly 3000 Calories. Also to a reasonably close approximation, the natural unit of energy is 100 thousand food Calories. So a million pounds of mayonaise, with its 3 billion Calories, represents 30 thousand natural energy units.