Using the force constant in equations

marcus · Feb 5, 2005

remember that the solar constant (sunlight power per unit area) is
5.7E-117
in the South China Sea let us suppose that 5E-117 is available during the hours of peak demand. comparing this to 5E-48 we see that the area must be E69. My goodness that is ten square paces (a square pace area is (E34)²)

marcus · Feb 5, 2005

... at standard conditions of temperature and pressure (E-29 and E-106) the mass of air is
about 1 pound per cubic pace. I calculate it's 1.115 E8 mass units.

the number of molecules in (E34)³ volume

PV/T = n
E-106 x E102/E-29 = E25

the mass of E25 molecules, each 29/(2.6E18).
E25 x 29/(2.6E18) = 1.115E8, to humanize it, call it 1.1 "pound"

--------------
It was a crisp Fall day, the temperature was E-29.
The hills of rural Vermont had turned bright colors.
A dog and a goat wished to take a ride in a balloon, so they went
to the goat's barn and got out his hot-air balloon.

the mass of gear and passengers is 400 pounds----that is 4E10 natural.

the goat asked the dog, who was a physicist,
how much they would have to heat the air to get lift-off.

What is the volume of the balloon? said the dog.
8000 cubic paces, replied the goat, naturally that is 8E105.

Well, said the dog, who enjoyed off-the-cuff order-of-magnitude calculation, the mass of air at ambient conditions is about 8000 pounds.
We have to heat it by about 5 percent, to lighten it by 5 percent, which is the weight of us and our equipage.

Excellent said the goat, as he opened the propane valve and pressed the igniter. We will raise the temperature in the bag from 1.00E-29 to 1.05E-29. Being a Vermonter, I call that 25 Fahrenheit degrees----shouldn't take too long!

an obvious followup concerns how much propane they are going to burn in the initial heating, to raise the temperature in the 8000 cubic pace bag by 0.05E-29---or 5E-31
we just calculated that the number of molecules in a cubic pace is E25
so we are talking about some 8000E25---or 8E28---molecules.

In natural units, heat capacities are generally easy because k=1, so for most metals it is around 3k per atom, for water 9k per molecule,...I will drop the k since it is one... for monatomic gasses at constant pressure 5/2
for biatomic gasses 7/2, per molecule always.

so the energy to heat the air in that balloon is about
(7/2) x 8E28 x 5E-31 = 140E-3 = 0.14 energy unit.

we just multiplied the delta-tee by the number of molecules by the heat capacity per molecule. In cases like this there is nothing to look up.

How much propane does the goat require, to get lift-off?

marcus · Feb 5, 2005

In answer to this inquiry, the dog recited a poem:

"whether it's to burn or eat
the Oh-Two count will tell the heat
on the average each Oh-Two
releases 17 eekyoo."

Now a propane is C3H8, and it uses 5 Oh-Twooz when it burns,
so it releases 85 eekyoo which is 85E-28 natural energy unit.
Moreover a propane is 36+8 = 44 baryons, and therefore a pound (mass E8) is 2.6E18 x E8 baryons. Accordingly the dog calculated that a pound is (2.6E26)/44 propanes and must therefore supply 85E-28 x 2.6E26 x (1/44) = 0.05 energy units.

To get off the ground, said the dog, we will need 0.14 unit of heat.
Behold! said he, we will burn 3 pounds of propane for lift-off!

That is fine, said the goat, who had experience in these matters: after that we will not need so much, because we will mostly just be keeping the air in the bag warm. The propane tank will be ample for our trip.

marcus · Feb 5, 2005

The captain of a million-pound scout vessel of the Ornish fleet is searching for planets rich in junk food for his men to plunder. As the ship comes out of warp, he discovers that it is bearing directly down on Atlantic City New Jersey at a speed of E-4.
With no time to spare he must order a photon pulse to cancel the ships momentum.

How many standard 10 pound cats will be consumed?

----
answer: a pound is E8 so the ship mass is E14 and the speed (30 km/second) is E-4, so the momentum to be canceled is E10 natural momentum units.
The light pulse with this momentum, directed at Atlantic City so as to avoid collision, delivers E10 natural energy units.

This will of course vaporize the famed vacation spot and some of the surrounded ocean. Of interest to the captain, however, is how many cats need to be converted to supply the energy.

the mass of a 10-pound cat is E9 and therefore, when converted in accordance to the emcee-square rule, the cat will yield E9 units of energy. Therefore 10 cats are needed from the ships fuel reserves to accomplish this maneuver.

marcus · Feb 5, 2005

It might be of interest to judge the effects of the scout ship's maneuver on the environs of Atlantic City. The boardwalk and casino Mecca is located on a narrow sandbar with water on both sides. For simplicity let us assume that the pulse of light misses the ciity and is entirely absorbed by the adjacent ocean.

the handbook figure for the latent heat of vaporization of water (2260 joules per gram) translates into one natural unit of energy vaporizing 399 pounds of water, let us say 400 pounds for round numbers. If we allow for some of the energy to go into preliminary heating then we can estimate that one unit is sufficient to vaporize 300 plus pounds.

We may estimate that ten billion units from the Ornish ship, on being absorbed, then vaporizes 3E12 pounds of water. It is clear why junk food pirates are generally looked on with disfavor.

nightcleaner · Feb 5, 2005

Hi Marcus.

I am very much enjoying your tour de force. But I wonder if you would address a question about natural units which has been bothering me.

The Planck mass, as I recall, is the mass that would be required, if compacted somehow into the size of a proton, for the creation of a mini-black hole. Wikipedia suggests this mass is about the mass of a small flea.

Wouldn't it be more in line with the other natural units if the basis of mass were made to be the amount of mass required, if compacted to form a black hole, into the volume of the Planck space? Then, one Emass would be the mass of one Evolume at the birth of a nascent black hole.

Just wondering what you would think.

Thanks for all this,
as well as for tickling the peach blossums,

nc

marcus · Feb 5, 2005

nightcleaner said:

... mass were made to be the amount of mass required, if compacted to form a black hole, into the volume of the Planck space? Then, one Emass would be the mass of one Evolume at the birth of a nascent black hole.

Hi nc, good to hear from you. what you what to be the case IS almost the way things are with conventional Planck units except for a factor of 2. It is normal for physicists not to worry too much about "factors of order one" (that is to say like 2 or 1/2pi and suchlike smallish numerical factors) especially in these extreme situations.

So one can say that if the Planck mass were compressed down to a ball with radius equal to the Planck length then its own gravity (which increases the closer you get to centerpoint) would be so strong that it would take charge and collapse the mass to a black hole.

Actually, putting in the factor of two, one can say that as soon as Planck mass is compressed to a ball with radius TWICE Planck length then a black hole will result.

It might be only a fanatical perfectionist would want to change the Planck length to get rid of that factor of 2. If everybody could be satisfied with the extent that we already have harmony between Planck units and black holes, it would avoid unnecessary difficulty of trying to eliminate the last factors of pi and 2pi and 2 and so on.

In the variant of Planck units I am working with, the Schwarzschild radius of the mass unit is 1/(4pi) of the length unit. this factor of one over 4pi does not bother me at all. I am very happy with the other neat things that happen with this version of Planck units. Like the clean form of the Einstein equation

marcus · Feb 6, 2005

The goat was flying his balloon over central Vermont, with his friend the dog. the two admired the Fall colors.

At our house, said the dog pensively, when someone goes to put air in the tires it is called "weighing the family car".

This is because, the dog continued, passing the binoculars over to the goat, we know the footprint of a properly inflated tire is 3E66.

(here the dog held out his paws to show a square about 3 human palms in area) and the combined footprint on the pavement of all four tires is 12E66.
Moreover it is our practice to inflate the tires to 2.8E-106 natural pressure, according to the gauge.

Therefore, declared the dog, the weight of the car is discovered by multiplying these two numbers
2.8E-106 x 12E66 = 34E-40

the dog did not explain, but this force is the weight of a 3400 pound mass in E-50 gravity. That mass is 34E10 natural and one multiplies it by "gee" or E-50 to find the weight.

After that it was deemed proper to unpack the sandwiches: liverwurst for the dog and cucumber for the goat.

marcus · Feb 6, 2005

At our house, said the goat when each of the friends had finished his sandwich, it is our custom to drink a gin-and-tonic on hot afternoons, and the preparation of a gin-and-tonic is called "measuring the height of the clouds".

this is because we believe that a proper drink of that sort should be just cold enough to make the glass sweat. We add just enough ice to make that happen. The person making the drinks can always measure Delta-T, the difference between ambient and making the glass sweat.

So you know how cold the air must get for condensation to start, observed the dog, and of course you also know the rate temperature falls off with altitude.

Yes, agreed the goat, ballooning has at least taught me that. In these parts on days like this lapse rate is about 3E-68. So we just divide Delta-T by the lapse rate and it tells you how high you'd have to go for the air to be as cold as a gin-and-tonic. That is where the clouds form.

marcus · Feb 6, 2005

by a Planck accident, lapse rate 3E-68 = 30E-69 = 30E-32/E37 is thirty halfdegrees per half mile, and in humanly familiar terms that is simply thirty degrees per mile. The goat remembers this so that he always knows what sweaters and down jackets to take with him on a balloon flight. If the lapse rate is 3E-68, he knows it will be 30 degrees colder one mile up.

On a day when the gin-and-tonic temperature differential Delta-T is 30 degrees the clouds are a mile high.

To be pedantic about it, if Delta-T is 6E-31
and the lapse rate is 3E-68 we just divide
Delta-T/lapse rate = 6E-31/3E-68 = 2E37 = two halfmiles = height of clouds

there are some unstated assumptions, like convection is occurring and cumulus clouds are forming locally (not just drifting in) but these conditions are not uncommon in Vermont on warm summer days.

marcus · Feb 7, 2005

the gypsy's keepsake

Artem Starodubtsev had a brief passionate love affair with a gypsy girl while her tribe was visiting the solar system.
On the day she left, she gave Artem a memento to remember their time together:
it was a black hole with the same mass as the earth.
How wide was the black hole?

Imagine Artem holding it in his palm and compare the width of the black hole (in it's protective jacket) with the breadth of his palm. The hole's diameter is:
A. 1/20 of his palm
B. 1/5 of his palm
C. 1/2 the width of his palm
D. exactly the width of his palm

marcus · Feb 7, 2005

the Au Pair Girl business

Besides raiding planets for their junk food, the Men of Ornish run an au pair girl business.

The have arranged an exchange between Earth and a planet inhabited by air-breathing giant squid.

An Ornish troop transport has been diverted from raiding in order to transfer 1000 young Republican women from Iowa to the Squid planet, where they will care for cephalopod children and attend high school.

On its return the ship will bring 1000 girl squid to look after Earth children.

The ship is currently in geosynchronous orbit around earth, and the young Republicans have been beamed on board.

How fast is the ship going relative to Earth?

marcus · Feb 7, 2005

the cube of geosynchronous orbit speed

there are several ways to calculate the speed in synchronous orbit from the planet's mass (1.38E33) and rotational period (3.19E47)
to be brief, one way is the cube of the speed is the planet's mass divided by 4 times the period, assuming things are expressed in natural units

v³ = mass/(4 period) = 1.38E33/(1.28E48) = 1.08E-15

the cube root of that is about E-5, more precisely 1.02E-5

so the Ornish craft was going one 100 thousandth of the speed of light.

that is the same speed that communication satellites in synchronous orbit go as well, so as to remain over one spot on the equator.

The captain prepares the ship to clear Earth's gravitational field and enter warp. The soon-to-become au pair girls are happily discussing impending bankruptcy of national healthcare and consequent opportuntities for private enterprise.

marcus · Feb 7, 2005

the spherical mirror in Artem's hand

the gypsy has placed in Artem's hand a ball with a perfect mirror coating. this is the tight-fitting protective jacket surrounding a black hole with the mass of the earth

they have chosen a place on the outskirts of the solar system for their last meeting, but the protective jacket isolates the holes gravity and inertia from the outside world so that in any case planet orbit would not be perturbed by her gift. (love is like this, so momentous that it should turn planets from their tracks and yet does not)

expressed in natural terms, the diameter of a black hole is simply equal to the mass divided by 2pi.

the Earth's mass is 1.38E33 and dividing that by 2pi gives 0.2E33
or about one fifth the width of Artem's hand.

for a moment, they watch their two small reflections mirrored on the ball

marcus · Feb 7, 2005

how the giant squid heat their hot tubs

It is widely known that the proper temperature for the hot tub is 1.11E-29.

expressed like this in natural units, it is an eternal number, and will no doubt be remembered (and respected by hot tubbers) long after the metric system and other arbitrary human constructs are forgotten

what is not so well known is how the giant airbreathing squid heat their tubwater to this very temperature

they have a bed of encapslated black holes each of which is at hawking temperature 1.11E-29
the tight-fitting reflective jacket surrounding each small hole protects the outside world from its gravity and inertia, but let's the nice warmth of the hawkingradiation escape into the water. this brings the water to exactly the right temperature

what is the mass and diameter of a black hole whose temperature is thus?

marcus · Feb 7, 2005

although the formula for the hawking temperature of a mass M hole is rather complicated and hard to remember when written in conventional metric format, it is quite simple in natural terms.

the temperature is simply the reciprocal of the mass

so if one wants the temperature of the hole to be 1.11E-29 then one makes the mass of the hole be the reciprocal of 1.11E-29, which is
0.9E29 natural mass units.

to humanize this, since E8 mass units is a pound, this is 0.9E21 pounds.
which is why the squid encapsulate these small black holes in a protective jacket, isolating the effects of their gravity and inertia.

Remember that the diameter of a hole is 1/2pi times its mass, when we are using this system, as appeared in the story of Artem and the gypsy.
To find the diamter of each tiny micro-capsule, we divide 9E28 by 2pi and get some 1.4E28 natural length units.

(this is dust particle size, about one micron)

marcus · Feb 8, 2005

you are orbiting a small planet at a steady speed of 67 miles per hour, and after you've circled the planet 4 times the clock says you have been in orbit 7 and an half hours.

what is the mass of the planet?

marcus · Feb 8, 2005

4 x orbit period = 4 P = E47
speed = 67 mph = E-7

mass M = 4P v³ = E47 (E-7)³ = E(47-21) = E26

to put the planet mass E26 in more familiar terms call it E18 pounds, a quintillion pound planet.
Or since the Earth mass is around E33, perhaps think of it as E-7 (one ten millionth) of the mass of earth.
to get a handle on the time E47, think E45 = 4.5 minutes, so 100E45 is 100x4.5 = 450 minutes = 7 and 1/2 hours. this is just footnotes.
the main thing is that with any circular orbit you have

mass = 4 x period x speed³

marcus · Feb 8, 2005

Imagine a planet of so little mass that you can orbit close to the surface at only 6.7 miles per hour

Perhaps the exceptionally benign atmosphere offers no air resistance but is good to breathe and of a comfortable temperature. In that case you can orbit without using any kind of spacecraft ---in your street clothes.

Let us suppose you can smell plumblossom and magnolia as you orbit (just grazing the hilltops) seeing everything on the planet at the speed of a run.

You are invited to calculate the mass of this planet, using one additional piece of information: In an orbit with a steady speed of 6.7 mph it takes
one and 7/8 hours to go full circle around the planet.
Calculate the mass any units you please. I've stated it in common units so it should make no difference.

marcus · Feb 9, 2005

about this thread, and natural units

I am trying out these natural units----like ordinary Planck but with |F|=1 instead of the more usual |G|=1

They do seem to work better than conventional Planck and this is consistent with what I notice in Quantum Gravity research papers. Often I see a kappa ("gravitational constant") which is 8piG, which can be set to equal one to further simplify the equations.

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.

Here are some rough sizes of familiar things expressed in the units

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
tesla E-53
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
D on treble clef E-39
one "gee" acceleration E-50
weight of 50 kg sack of cement E-40
power of 160 watt bulb E-49

some constants (approx.):

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
sun core temp 5E-25
CMB temperature 9.6E-32
earth surface air pressure 1.4E-106
earth surface gravity 0.88E-50
fuel energy released by one O₂ 17E-28
density of water 1.225 E8/E99

timescale:

1/222 of a minute E42
4.5 minutes E45
As a handle on the natural timescale, imagine counting out loud rapidly at the rate of 222 counts a minute, each count is E42 natural time units. A thousand counts is 4 and 1/2 minutes. It just happens that one year is roughly E8 counts, or E50 natural.

marcus · Feb 9, 2005

now what I want to condense into a post or so is a sampling of how the formulas look, which this thread has been illustrating

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 don't 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 water, per molecule
For the liquid, it is 9
So making some liquid water's temperature increase by E-30 takes an amount of energy equal to (the number of molecules) x 9E-30. The latent heat of vaporization is 1.7E-28 per molecule.

for a metallic solid, heat capacity is about 3 per atom
for a biatomic gas like air, 5/2 per molecule at constant volume, 7/2 per molecule at constant pressure

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.

marcus · Feb 9, 2005

A couple of posts ago, post #120, there's a list of useful constants including the electron mass 2.1E-22. I was reminded by listening to the online
http://www.vega.org.uk/series/lectures/feynman/
Feynman QED lectures that one of the big triumphs of QED (which he talks about several times) is predicting the ratio of electron magnetic moment to the Bohr magneton.

mu_e/mu_B = 1.001159...

defined in our units as e hbar/2m_ec, the Bohr magneton
numerator is 1 and the denominator is about 4.2E-22.
So Bohr magneton in natural units is about
mu_B = 0.24E22 = 2.4E21

The actual magnetic moment of an electron is very close to this---the ratio is only a tenth of a percent different from one---and the ratio was predicted by QED out to many decimal places. Like for starters look at
1 + alpha/2pi +...
that is already not bad, something like 1.001161...

marcus · Feb 10, 2005

talking about magnetic moment means having some handle
on the natural unit of elec./mag. fields (it is the same unit in this system, unit force per unit charge) by coincidence to a reasonable approximation:

magnetic field E-57 = gauss
magnetic field E-53 = tesla

to give an idea how close:
1 Tesla = 0.9974 E-53 natural = 1.00E-53
1 gauss = 0.9974 E-57 natural = 1.00E-57
to two decimal accuracy the relevant factor is just one!
It is not the same unit, because of two different forms of the Lorentz force equation, but a magnetic field that registers as 1 Tesla on a metric gauge will read 1.00E-53 on the natural scale.

I haven't been bothering to show precision in this thread since we rarely if ever need it, but some additional accuracy is available
natural energy unit = 3.9018E8 joule
natural charge = electron charge = 1.602176E-19 coulomb
1 conventional volt = 4.1062E-28 natural voltage units
1 meter = 1.2342E34 natural length units.

The handbook's value of 0.58 gauss for the Earth magnetic field at the north pole converts directly to 0.58E-57 natural.

a propos Bohr magneton, it and the electron mag. moment are both about 930E-26 joule per tesla (e.g. given in metric by NIST) and a tesla is E-53
so we are talking 930E27 joules per natural field unit. dividing that by
3.9E8 gives 2.4E21. Just a check. It agrees with what we got directly from the definition.

marcus · Feb 10, 2005

the kinetic energy of the solar wind is measured on the temperature scale
If I remember correctly it is on the order of E-25
that is hot compared to the surface of the sun which is temperature 2E-28

(I find 2E-28 makes a lot of sense because the energies of visible photons are around 10E-28, something that temperature would make a bunch photons those energies. I can almost SEE that sun surface temp is 2E-28. But grasping that solar wind is E-25, or even E-26, is harder.)

but let us think of that E-25 as just a way of describing the speed of a proton. then what actually is the proton's speed?

well energy at rest is 1/(2.6E18)
(1/2)m v² =v²/(5.2E18) = E-25
v² = 5.2E-7
v = 7E-4

so from the speed point of view it is no big deal, the Earth orbit speed is E-4, so what sounds like a terribly hot wind is just some protons going a few times faster than earth
(I may have misremembered the wind's temperature, it may be closer to E-26. but that would only reduce the speed a little, to like 2E-4. qualitatitively similar)

marcus · Feb 10, 2005

Length of organ pipes for various pitches

for definiteness let's call the D right next to "middle C" on the piano
"middle D" (as some people do anyway)

I'm always using angular format for freq., wvlength, etc. because more convenient to stick to one format consistently. Here are some frequencies of pitches in the human voice range

Code:

D above middle D     E-39
middle D             0.5E-39
D below middle D     0.25E-39

the length of organ pipe you want to make a particular pitch depends on the speed of sound and the speed depends on the temperature of the air.
Commonly the speed is around 1.1E-6
but for simplicity I'm going to use the cold air speed of E-6 (a millionth of the speed of light.

organ pipes are either open at both ends or open at just one end
the both-open kind has to be pi x wvlength
the half-open kind has to be (pi/2) x wvlength (they get to be shorter and have been used in some very nice-sounding small compact pipe organs)
but both-ends-open kind is more common.

Code:

musical pitch        freq         wvlngth        pi x wvlngth = pipelength
D above middle D     E-39         E33            piE33
middle D             0.5E-39      2E33           2piE33
D below middle D     0.25E-39     4E33          4piE33

this reference length E33 is the width of your hand, or 8.1 cm., or 3.2 inches. So when it says D below middle needs a pipe 4pi that, then
it means on the order of a yard long---some 40 inches.

The wavelengths are gotten by dividing: speed of sound E-6 divided by frequency (like E-39) gives wavelength (like E33). If the speed were 10 percent faster then the pipes wd hv to be 10 percent longer. But this gives a rough idea.

What I'm thinking is it isn't hard to use natural units in a way that embraces college-level general physics. If anybody has any favorite problems please type them in and I will see if they translate nicely into natural units terms.

marcus · Feb 10, 2005

the weight of the 100-pound monkey

standard Earth gravity is actually 0.88E-50 but I tend to think of the "round number planet" situation where a gee is simply E-50

so if a monkey's mass is E10 then his weight is E-40

(Hundred pounds is 100 E8 = E10, and that is the mass.
Always multiply the mass by gee to get the weight force: E10 E-50 = E-40)

the monkey is hanging on a rope and he begins to climb up the rope
but first the professor draws a picture: the rope goes up to a pulley and down to a 100 pound sack-----the same mass as the monkey.

when the monkey isn't climbing his weight exactly balances the other weight and nothing moves. (BTW this is one of these massless frictionless idealized pulleys that one often hears about from physics teachers)

now the professor grins gleefully and says "the monkey starts to climb the rope, what happens?"

marcus · Feb 10, 2005

Electrified dwarves

Once there were 7 dwarves who all lived together and operated a bed and breakfast. The dwarves house was a handsome old high-ceilinged Victorian: it was 4E34 from floor to ceiling, with dumbwaiters the dwarves could ride in, and bannisters to slide down and all that.

The dwarves were all unusually tolerant of static electricity and liked to give each other shocks. There were thick rugs in all the rooms and they were always shuffling around on the rugs getting charged up and zapping each other.

One day a dwarf whose name was Stinky got charged up to E16
(this is a huge charge, in metric terms it would be 1/600 of a coulomb!)
and to play a trick on him the other dwarves electrified the ceiling of his room with a huge voltage of 2E-22
(remember that E-28 is a quartervolt so this voltage was 2 million quartervolts, half a million conventional volts.)

Well, when his housemates did this terrible thing, Stinky floated! He became weightless and turned slow cartwheels and somersaults in midair (cursing shrilly all the time) until the others took pity on him and put the voltage back to normal.

How much did Stinky weigh?

marcus · Feb 10, 2005

Count Rumford and the Genii

Count Rumford, born in Massachusetts 1753, was a yankee inventor who made a lot of improvements in Bavaria and was appointed a Count of the holy roman empire and did some basic physics experiments too.
In summer Rumford liked to take baths in a four-footed sheet-metal bathtub he had placed in the palace garden. Rumford wasnt his real name either, he just liked the sound of it.

On his travels to Arabia Rumford had obtained a Genii Lamp, which he kept around to rub in case he needed the Genii to do something. This Genii could do fantastic things but he absolutely refused to violate Conservation Laws.

One summer afternoon Rumford had the servants heat enough water from ambient 1.04E-29 up to a good hot 1.1E-29 to fill his bath and he was sitting in the tub scrubbing his back with a large oak-handled brush and enjoying the hot water. It was a nice bright day and at that moment he conceived a desire to be up in the sky, so he rubbed the lamp. "What is your will, master of the Lamp?" said the Genii. "Lift me and my tub into the sky so I can enjoy the view of the Bavarian countryside while I bathe," said Rumford.

The Genii did this and for a moment the Count was in bliss. Suddenly he realized he was sitting in disgustingly cold water, like 69 Fahrenheit, which is 1.04E-29 natural. "Yow," said the Count, "this water is freezing!" It wasnt, but that is how room temperature water feels.

"Right," said the Genii, "energy cannot be created or destroyed. I changed the energy that went into heating the water into gravitational potential energy." The Genii had no compunction about violating stuff like the 2nd Law of Thermodynamics, which is routine proceedure for competent Genii, but he drew the line at conservation laws.

how high up was the tub?

(neglect the mass of the count and the sheetmetal tub. the main thing is the water.)

marcus · Feb 11, 2005

any reader is welcome to work the problem in metric units, if desired.
in metric terms, the servants raised the temperature of the water by 17 Celsius and the question is: how high would you need to raise some water so as to endow it with grav. pot. energy equal to what you have to put into it as heat to raise it 17 Celsius?

nightcleaner · Feb 11, 2005

Marcus I am dreadful at calculation but I will have a go at it anyway.

The monkey and the weight both rise half as much as the length of rope the monkey pulls.

I am not sure how much the dwarf weighs. Maybe if I start the problem you will help me finish it?

Well, first the attractive force between the dwarf and the cieling is set equal to the weight of the dwarf when he is in midair...say halfway to the cieling, so the electrostatic force equals the gravitational force when the dwarf is at 2E34.

The gravitational force is the weight of the dwarf, so we just have to calculate the electrostatic force. Now I am in trouble, but here goes.

The electrostatic force falls off according to the inverse of the square of the distance, so we will have to square half the distance and invert it...so .25E-68. Then the volage acts to attract the charge, so E16 x 2E22 is 2E38. Then the attraction by the inverse square is .5E30. That seems like too many fleas. A pound is 10E8, and that makes Stinky on the order of E22 pounds, way too big for a dwarf.

I am sure this is the wrong answer, but will have to go review electrostatics to make any headway. Meanwhile put a cone on my head and sit me in the corner.

nc

I see that the force between two equal charges is q^2/r^2. The dwarf is hanging from the cieling just as the monkey hangs from the pulley. So if I have to square the charge, I think I get 4E76, times the inverse square makes E8, makes poor Stinky weigh a tenth of pound, closer, but no banana for the monkey. I'll go have another think.

nc

Looking at the coefficients again, I get Stinky up to a quarter pound. I am using weight=(qV)^2/r^2 where r is half the room height, q is the charge on Stinky, and V is the voltage on the cieling. I'll have to find a better method.

nc

Actually, it seems to me this problem is very like Millikin's oil drop experiment, in which he determined the charge on an electron by holding a drop of oil with a single charge stable in an electric field by varying the voltage. Can't find the reference. Still looking.

nc

Ok that was no help. Millikin had three forces, the weight, the electric field, and the buoyancy. Charge in Millikin's experiment was found by setting q equal to buoyancy by volume by g over the electrostatic force.

I guess we can ignore buoyancy of dwarves in air.

q=g/E? Then back to the problem, how to find E at that voltage and distance?

I don't know. F should just equal q/r^2. Then what's volts got to do with it? argh.

Well I used up my time and got nowhere.

I have to get some sleep, work again tomorrow night, maybe have time to play some in the afternoon. Sorry to be a dissappointing student.

nc out.

marcus · Feb 11, 2005

nightcleaner said:

The monkey and the weight both rise half as much as the length of rope the monkey pulls.

I am not sure how much the dwarf weighs. Maybe if I start the problem you will help me finish it?

You are absolutely right about the monkey.
About the dwarf you have made a brave attempt!

BTW you correctly pointed I was neglecting the buoyancy of dwarves in air.
It hadn't even crossed my mind. I believe it is a small consideration which we may continue to ignore.

Perhaps selfAdjoint will kindly add some explanation here that will (as often does when he comments) make it easier to understand.
what I can say, for starters at least, is that the electric field tells you the force per unit charge

and in this case the electric field is equal to the voltage gradient
that is, it is not simply equal to the voltage, because if the ceiling were very far away it would be felt less.
the electric field is equal to the rate the voltage changes with distance
that is called the voltage "gradient" and it is what determines the force on a unit charge

now let's work it in metric because I think this is more familiar to everybody, and then work it in new units:

the ceilingheight is 3.25 meters and there is a halfmillionvolt difference betw. floor and ceiling
this means about 150 thousand volts per meter (gradient)
this means a metric UNIT charge (a "coulomb") would experience 150 thousand Newtons
but Stinky is charged up to 1/600 coulomb
so we divide the 150,000 Newtons that a unit charge would experience by 600 and we get the force on Stinky
which is 250 Newtons.
this 250 newt is the force of his weight.
(we can estimate his mass at 25 kilograms or so but that doesn't matter if all we want to know is the force of his weight)

here I am not trying to be especially accurate, just want to get approximate idea of his weight

now I will do the same in new units and get approx. the same answer.
his mass will come out to about 50 "pounds" that is 50E8 of these tiny natural mass units-----which more or less checks with the 25 kilos, so it will be OK

marcus · Feb 11, 2005

marcus said:

Once there were 7 dwarves who all lived together and operated a bed and breakfast. The dwarves house was a handsome old high-ceilinged Victorian: it was 4E34 from floor to ceiling, ...
...
One day a dwarf whose name was Stinky got charged up to E16
(this is a huge charge, in metric terms it would be 1/600 of a coulomb!)
and to play a trick on him the other dwarves electrified the ceiling of his room with a huge voltage of 2E-22

... Stinky floated! He became weightless and turned slow cartwheels and somersaults in midair (cursing shrilly all the time) ...

How much did Stinky weigh?

The voltage gradient is 2E-22 divided by the distance 4E34
(the total voltage difference divided by the distance over which the voltage changes)
2E-22/4E34 = 0.5E-56 = 5E-57

that means each electron (each unit charge) feels a force of 5E-57.

that is a small force, but Stinky has E16 extra electrons on him!
So the total force on Stinky is E16 times 5E-57, which is 5E-41

just to get a rough idea, earlier I was saying that the force E-40 natural was similar to the weight of a 50 kg sack of cement, and this is 0.5E-40, or half that. So very crudely his weight is like the weight of 25 kg. which is what we got before using metric.

nc, thanks for trying the problem out. having some dialog adds considerably to the pleasure of posting the problems. Hoping you find some others of interest. Will consider posting other monkey and dwarf problems.

nightcleaner · Feb 11, 2005

Marcus your approach here is beautiful and entertaining, and I feel like I am learning more using fundamental units that I was able to learn using metrics. I very much enjoy your stories of squids and gypsies and I find your sense of humor very much in line with my own.

I am not so sure, personally, about the cats. It isn't that I am worried about throwing them into the mass conversion generator, altho on general principles I would have to object to that procedure, if asked, but that using cats as energy units instead of just using the natural mass unit is, for me, an added bit of information which I would rather not have to remember.

To me, it seems more natural to learn to use the fundamental units as they are and then to remember, if it seems necessary in some problem of interest, what my own anthropometric values are.

Well, I have a few questions. One, why does the voltage placed on the cieling fall off to zero as it reaches the floor? I mean, if we are going to distribute the voltage, shouldn't it be distributed to infinity? That is unless the floor is specially made to be highly reflective to EM or something. Or, should the problem state that the voltage difference between the floor and the cieling is 2E-22? I am learning something already. Shouldn't we always have to say that the voltage placed on a surface has to be compared to the voltage on some other surface? Voltage is by its nature a difference, correct?

Please do continue to post problems, monkeys or young Republicans or whatever comes to mind. Perhaps it would be a good idea to start a parallel thread or two, one with the solutions, another for discussion with gratefully eager students?

Thanks, Marcus.

BTW, I thought you might like to know that the buds on the birch trees on the shores of Lake Superior are beginning to swell. We have 29 inches of snow on the ground, melting fast in unseasonably warm temperatures. I may go out today to taste the birch buds, which are bitter but have a faint aroma of wintergreen. I may taste the aspen buds as well, which are mostly just bitter, but they do contain some salicylic acids, good for easing the headaches I get from trying to force my poor brain to compute.

Be well, Marcus

thanks for being here,

Richard

marcus · Feb 11, 2005

marcus said:

Count Rumford, born in Massachusetts 1753, was a yankee inventor who made a lot of improvements in Bavaria and was appointed a Count of the holy roman empire and did some basic physics experiments too.
In summer Rumford liked to take baths in a four-footed sheet-metal bathtub he had placed in the palace garden. Rumford wasnt his real name either, he just liked the sound of it.

On his travels to Arabia Rumford had obtained a Genii Lamp, which he kept around to rub in case he needed the Genii to do something. This Genii could do fantastic things but he absolutely refused to violate Conservation Laws.

One summer afternoon Rumford had the servants heat enough water from ambient 1.04E-29 up to a good hot 1.1E-29 to fill his bath and he was sitting in the tub scrubbing his back with a large oak-handled brush and enjoying the hot water. It was a nice bright day and at that moment he conceived a desire to be up in the sky, so he rubbed the lamp. "What is your will, master of the Lamp?" said the Genii. "Lift me and my tub into the sky so I can enjoy the view of the Bavarian countryside while I bathe," said Rumford.

The Genii did this and for a moment the Count was in bliss. Suddenly he realized he was sitting in disgustingly cold water, like 69 Fahrenheit, which is 1.04E-29 natural. "Yow," said the Count, "this water is freezing!" It wasnt, but that is how room temperature water feels.

"Right," said the Genii, "energy cannot be created or destroyed. I changed the energy that went into heating the water into gravitational potential energy." The Genii had no compunction about violating stuff like the 2nd Law of Thermodynamics, which is routine proceedure for competent Genii, but he drew the line at conservation laws.

how high up was the tub?

(neglect the mass of the count and the sheetmetal tub. the main thing is the water.)

heat capacities are interesting, in a lot of materials the heat capacity is (in our units) about 3 per atom.
that is actually how it works out in liquid water! so it is 9 per molecule

the bath water temp was raised by the servants from 1.04E-29 to 1.10E-29, so its temp went up 0.06E-29 = 6E-31
and this means that each water molecule should have been supplied on average 54E-31 energy unit.

How high would you have to raise a water molecule to endow it with that much energy as gravitational potential?

well the mass of the thing is 18/(2.6E18) and "gee" is about E-50. Let's use the more accurate figure 0.88E-50 for gee. Multiplying the mass by gee gives 6.1E-68 for the weight-force. so we can just solve for the height you raise it (pushing against the force of its weight)----that has to give the energy:

height x 6.1E-68 = 54E-31
height = 8.9E37

to get a familiar perspective on it, E37 is half a mile, so the Genii lifted the Count up some 9 halfmiles------4-some ordinary miles into the air.
Rumford teeth must be chattering, so hopefully the Genii got him back down right away and restored the heat to his bath.

marcus · Feb 12, 2005

a nice example of planet equilibrium temp

in another thread saltydog and Mean-Hippy were trying to find the equilibrium temp of a round ball at 1 AU from a star that is 20 percent more luminous than the sun
https://www.physicsforums.com/showthread.php?p=460118&posted=1#post460118

Mean-Hippy got the answer 476510.66 K .

Saltydog got the answer 117.5 Kelvin.

the right answer is about 283 Kelvin.
If you work it in natural it is pretty simple and you get the equilib. temp is E-29
this is the same as 283 Kelvin (if you like kelvin) or 49 Fahrenheit (if you like Fahrenheit) but I just think of it as E-29. It is a nice temp for a planet and not very different from global avg. temp on Earth surface.

How to get it. intensity of sunlite at Earth dist from sun is 5.7E-117
20 percent more is 6.84E-117
surface area of ball is 4 times cross section area
so divide by 4
1.7E-117

that is how much surface of ball must radiate in order to get rid of same amount of energy that the ball is intercepting from it's "sun"

stef-boltz says

power per unit area = (pi²/60) T⁴

so to solve for T (the surface temp the ball must have to radiate fast enough to stay in balance)
we just have to multiply 1.7E-117 by (60/pi²)
which gives E-116

and then take fourth root
which gives E-29

that is the nice 49 Fahrenheit temp.

it is a comfortable example. Thanks to mean-hippy. he has some particular extrasolar planet in mind around some particular star. here is a link to his thread

marcus · Feb 12, 2005

the force F = c⁴/(8piG) is the main constant in Gen Rel, the prevailing theory of gravity since 1915. The constant in the Einstein equation is not Newton's G, but rather F. In Quantum Gravity one often uses units in which |F| = 1
(this can come about by stipulating that |8piG|=1, since normally one already has adjusted the units so |c|=1)

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.

I am trying out this version of natural units to see how they work. In order to try out the units one must keep a list of rough sizes of things handy----to use the units for practical purposes one must have a sense of scale. Here are some rough sizes of familiar things expressed in the units.
I periodically bring this list forward to keep it handy.

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
tesla E-53
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
D on treble clef E-39
one "gee" acceleration E-50
weight of 50 kg sack of cement E-40
power of 160 watt bulb E-49

some constants (approx.):

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
sun core temp 5E-25
CMB temperature 9.6E-32
earth surface air pressure 1.4E-106
earth surface gravity 0.88E-50
fuel energy released by one O₂ 17E-28
density of water 1.225 E8/E99

timescale:

1/222 of a minute E42
4.5 minutes E45
As a handle on the natural timescale, imagine counting out loud rapidly at the rate of 222 counts a minute, each count is E42 natural time units. A thousand counts is 4 and 1/2 minutes. It just happens that one year is roughly E8 counts, or E50 natural.

marcus · Feb 12, 2005

In Astronomy forum was a thread by MeanHippy about the temperature of a planet and it had a figure for the luminosity of the sun. Now i will see how to find this out from the list of numbers in natural units that we already have. I'd like to exercise the data we have instead of going all the time to the handbook and converting from metric. To get the sun's power output we just need to know the solar constant and our distance from it.

The solar constant is 5.7E-117
(brightness of sunlight, very basic number) and how far are we from sun?
Well year is 1.17E50 and orbit speed is E-4 so circumference is
1.17E46. Divide by 2pi and we have the orbit radius. 1.86E45.
now we just multiply 5.7E-117 by the AREA of a ball with that radius and that tells us the total power of the sun.
By (4pi/3)R² the area is 14.5E90 and multiply by 5.7E-117 gives 8E-26. so that is the luminosity of the sun: the energy units output per unit time. It looks about right.

MeanHippy was interested in a star with 20 percent bigger luminosity.
That would have a power of right around E-25. so that is a nice example star for natural units! Apparently a planet has been detected circling such a star at radius 1 AU. This is also a good example. the equilibrium (black ball) temperature turns out to be E-29.

marcus · Feb 13, 2005

Frog was out driving his vintage Morgan.
This car is great! said Frog. It can really take the curves.
Suddenly, coming around a bend in the road, he saw a sign BRIDGE OUT.

Frog jammed on the brakes and locked the wheels. The sporty vehicle skidded to a stop.

Toad emerged from the bushes beside the road and told Frog to wait while he measured the skidmarks. If you were speeding, said Toad, I will give you a ticket.

Toad paced out the skidmarks. They were 50 paces long (50E34).
He applied the formula that assumes a friction coefficient of one for rubber on pavement:
v² = 2gL = 2 x E-50 x 50E34 = E-14
v = E-7

You were going right at the speed limit, said Toad, E-7 is 67 miles per hour. I will not write you a ticket. You may proceed on your merry way!

But the bridge is out, said Frog.

No, said Toad, the sign is just part of our Emergency Preparedness program, in case terrorists blow up the bridge. We are testing the sign to see if it works. The bridge is passable. Do not make me tell you again to proceed on your merry way.

Frog drove the Morgan along the winding country roads. As dusk approached, a thin crescent moon appeared in the west. Ah, said Frog, the moon is curved just like the bends in the road

marcus · Feb 13, 2005

Giant chickens have invaded from outer space and are living in a castle.
They are holding Robin Hood's girlfriend captive.

Robin sneaks into the castle and surprises the chickens by swinging from a chandelier. The chickens are alarmed and flee to their ships. Maid Marian is free!

The chandelier was hanging 9 paces down from the stone gothic-arch ceiling of the grand dining hall of the chickens. that is 9E34 of course.
What was the period of the pendulumswing?

period = 2pi sqrt(length/gee) = 2pi sqrt( 9E34/E-50) = 2pi sqrt(9E84)

period = 2pi x 3E42 = about 19E42

footnote this is about as long as it takes to count rapidly to 19 out loud.
More precisely it is 19/222 of a minute which you can work out in seconds if you like: it comes to a bit over five seconds.

marcus · Feb 13, 2005

It was good weather for bikeriding yesterday. I rode up to the ridge, stopped at some friends house, coasted (mostly) back.

the weight of me and bike is around 2.2E-40
our frontal area is about 6E67
air density is 1.5E-94
drag coefficient is around one so can be ignored

coasting down a 5 percent grade without using brakes, I would get up to what speed?

----------
answer: 5 percent of our weight is 1.1E-41
we get going fast enough so the drag force balances that 1.1E-41
drag force = density x area x v²/2
= 1.5E-94 x 6E67 x v²/2

Set that equal 1.1E-41 and solve for v and you get 5E-8

(it is around 33 mph, actually I use the brakes some going down that hill,
no speed maniac)

nightcleaner · Feb 13, 2005

marcus said:

Frog was out driving his vintage Morgan.
This car is great! said Frog. It can really take the curves.
Suddenly, coming around a bend in the road, he saw a sign BRIDGE OUT.

Frog jammed on the brakes and locked the wheels. The sporty vehicle skidded to a stop.

Toad emerged from the bushes beside the road and told Frog to wait while he measured the skidmarks. If you were speeding, said Toad, I will give you a ticket.

Toad paced out the skidmarks. They were 50 paces long (50E34).
He applied the formula that assumes a friction coefficient of one for rubber on pavement:
v² = 2gL = 2 x E-50 x 50E34 = E-14
v = E-7

You were going right at the speed limit, said Toad, E-7 is 67 miles per hour. I will not write you a ticket. You may proceed on your merry way!

But the bridge is out, said Frog.

No, said Toad, the sign is just part of our Emergency Preparedness program, in case terrorists blow up the bridge. We are testing the sign to see if it works. The bridge is passable. Do not make me tell you again to proceed on your merry way.

Frog drove the Morgan along the winding country roads. As dusk approached, a thin crescent moon appeared in the west. Ah, said Frog, the moon is curved just like the bends in the road

What about the mass of the Morgan? What if Froggy had a ton of gold bars in the bonnet?

marcus · Feb 13, 2005

Hi nc,

some odds and ends. You know all thru this thread I've been using the number 2.6E18 (for masses and weights of atoms and molecules)

well I sometimes think of that number as "Wilczek's number" because he wrote a series of 3 articles in "physics today" about that number, how interesting it is and how to explain its large size (or the small size of its reciprocal)

today Wilczek's Nobel acceptance speech posted on arxiv
http://arxiv.org/abs/hep-ph/0502113

this speech is NOT about the number 2.6E18, it is about "asymptotic freedom" and quarks etc. that Wilczek and two others got prize for.
but probably in the end the size of that number will be explained out of the same fundamental ground as QCD-----the reciprocal is the proton mass expressed in natural units.
Wilczek is a good writer so i think anybody might like to try reading his speech, even tho about difficult stuff.

About that gold in the Morgan
putting gold bars in Morgan would have no net effect (at least to first approx.)
because it has two opposite effects that cancel
heavier makes the car have more traction
more massive makes it have more momentum (at given speed)
and so it needs more force to make it stop

double the mass and you double the stopping force of tires on pavement, but also double the need for stopping force, so it cancels and get same result.
therefore the simple formula they teach which relates speed to length of skidmarks does not have the mass of the car in it as a factor.
(at least this is how I remember, please correct me if you have better info)

marcus · Feb 13, 2005

...today Wilczek's Nobel acceptance speech posted on arxiv
http://arxiv.org/abs/hep-ph/0502113
this speech is NOT about the number 2.6E18,...

well he does mention it after all (a bit vaguely using more popularly recognizable h instead of hbar and leaving out 8pi) on page 21,

nontechnically, but mentions it anyway-----it is the paper's equation #2
which says that mass of proton is approx 10^-18 of Planck mass.

nightcleaner · Feb 13, 2005

marcus said:

Hi nc,
double the mass and you double the stopping force of tires on pavement, but also double the need for stopping force, so it cancels and get same result.
therefore the simple formula they teach which relates speed to length of skidmarks does not have the mass of the car in it as a factor.
(at least this is how I remember, please correct me if you have better info)

Hi Marcus

I don't have better info, just curiosity and a vague memory from a defensive driving course many years ago. The instructor said you have to watch out for motorcycles because they can accelerate very fast, and also can stop very fast, due to their light weight. Give them extra following space, was his advice, but he was no physicist.

Even more vague, it takes something like half a mile of screeching to stop a freight train.

Here, near The Lake, it has begun to snow. I am on my way to work in a few minutes. The forcast was rain, freezing rain, then snow, so I expect it to be slippery. There is zero traffic on the highway, so I suspect the forcast was correct, and I will have to be very careful walking up the little hill to the truck. When it is icy, sensible people drive very slowly or better, not at all. You can be very surprised how long it takes to stop. I once skidded half a block after rounding a corner onto an icy side street. I didn't think I was going fast at all.

I think you are right about the traction thing, though, but there are probably mass effects on the coefficient of friction that are neglected in the formula. I used to drive a semi truck and I do know it takes a lot longer to stop a loaded truck than an empty one, but I never had to lock up the brakes. Tapping the brakes repeatedly is better than jamming them on, because once the tires break free of the pavement, as in a skid, the friction goes way down.

Be safe,

nc

marcus · Feb 13, 2005

nightcleaner said:

Here, near The Lake, it has begun to snow. I am on my way to work in a few minutes. The forcast was rain, freezing rain, then snow, so I expect it to be slippery. There is zero traffic on the highway, so I suspect the forcast was correct, and I will have to be very careful walking up the little hill to the truck. When it is icy, sensible people drive very slowly or better, not at all. You can be very surprised how long it takes to stop. I once skidded half a block after rounding a corner onto an icy side street. I didn't think I was going fast at all.

I think you are right about the traction thing, though, but there are probably mass effects on the coefficient of friction that are neglected in the formula. I used to drive a semi truck and I do know it takes a lot longer to stop a loaded truck than an empty one, but I never had to lock up the brakes. Tapping the brakes repeatedly is better than jamming them on, because once the tires break free of the pavement, as in a skid, the friction goes way down.

I think you are right, pumping the brakes is better. I do that whenever I think there is a danger of tires breaking free and have avoided skids thank heaven so far, except (as you mention) on ice.
one time near NYC the hutchinson river parkway was iced and we and all the other cars were going slow. I remember turning around 360 degrees and sort of waltzing down a gentle slope. it was better than turning around 180 degrees and coasting down backwards

roads can ice suddenly and nobody have chains. at times we had to push on that trip, fortunately there were several of us in the car.

However there is this classic freshman physics problem where the driver (Frog in this case) actually does lock the wheels and does the whole skid thing, EVEN THOUGH it is the wrong way to stop.
and then the textbook usually assumes some value like one for the kinetic friction of rubber on pavement. It may not be altogether realistic (especially modern cars have automatic anti-locking so you can't do it even if you want) but firstyear physics problems often have a slight unreality anyway.

BTW Wilczek says in that nobel acceptance talk that 90 percent of the mass of a proton is due to the kinetic energy of the quarks buzzing around inside it. Curious thought. the protons mass is owing to the speed of its constituent parts which sitting still would not weigh much.

marcus · Feb 15, 2005

marcus · Feb 15, 2005

The cyclotron frequency of the cat

On a planet famous for its potato pancakes, they have special cats which are highly tolerant of electric charge. It is possible to charge one of these cats up to E19 a truly amazing charge equal, in conventional terms, to 1.6 metric Coulombs!

A large magnetic room has been built to discover the cycloton frequency of fully charged cats.

A cat of charge E19 and mass E9 (about 10 pounds) is launched into a uniform vertical field of strength E-53 (in conventional terms one Tesla). The cat is observed to travel in a circle with a fixed constant angular frequency determined by its mass m, charge q, and the strength B of the field

frequency = qB/m = E19 x E-53/E9 = E10 x E-53 = E-43.

This constant frequency is called the "cyclotron frequency of the cat"

A modest vertical electric field cancels the cat's weight, so it is effectively in zero gravity. It follows a large horizontal circular path which eventually spirals inwards as the animal slows down (due to aerodynamic drag).
The reason for spiraling is that it has to keep circling with constant frequency E-43, so as it slows down it must go in smaller and smaller circles to maintain the same frequency.

question If the cat is launched into the room at a speed of E-8 (in familiar terms 6.7 mph), what is the radius of its circular path?

answer We have shown that the cat takes time E43 to go one radian, if it is traveling at speed E-8 then the distance traveled in that time is E-8 x E43 = E35. this is the radius of the cat's path. In traditional terms it is 10E34 or 10 paces.

marcus · Feb 16, 2005

Quantum Hall Resistance

the quantum hall resistance (also called "von Klitzing's constant") is a certain definite ratio of voltage to current built into nature

in terms of these natural units it is simply 2pi

if you want to know the metric value (adopted in 1990 to standardize current measurement) you probably have to look it up:

the NIST website gives it as exactly 25,812.807 volts per amp, in other words as 25,812.807 Ohms.

we encountered this kind of thing earlier with the StefanBoltzmann radiation law constant which in natural units is pi²/60
but if you want to know it in metric you look it up and it is about
5.6704 x 10^-8 watts per sq meter per Kelvin⁴

--------------
I found some stuff about QHE (quantum hall effect) on google. I'm not an expert about this---just know that the effect exists and is used to base the current standard on at various countries' national labs
http://www.warwick.ac.uk/~phsbm/qhe.htm
-----------------

so you have a cold horizontal rectangle placed in a vertical magnetic field and you send a certain current down the length of the rectangle and a crossways voltage is induced across the width, which depends on the current in a fixed ratio (you can even vary the magnetic field some and as long as you don't change the field too much the ratio of voltage to current stays the same). See for example the picture and graph at that link.

In natural terms the ratio is 2pi. So like if you make the current E-28 then the transverse voltage across the width of the thing will be 2piE-28

(in familiar terms voltage E-28 is a "quartervolt" so we are talking 6.28 quartervolts or about one and half volt)

At national labs they use the Quantum Hall effect to standardize current measurement. Voltage can be measured using the atomic clock and the Josephson effect, and then once one has a voltage standard (based on atomic clock) then one standardizes current using the ratio 25,812.807 volt per amp adopted by the CIPM in 1990.

Richard, I went googling for QHE references in response to your questions in next post. Will edit in some links

I tried Wikipedia but it had no pictures or diagrams.
this JohnHopkinsUniv. page has pictures but also a huge amount of formulas so you have to go searching for the pictures:
http://www.pha.jhu.edu/~qiuym/qhe/qhe.html
It does give the dimensions and materials of an actual QHE device.

nightcleaner · Feb 16, 2005

Hi Marcus.

This is very interesting, and gets at a point which I have always found confusing. I hope you will entertain some questions here.

First the magnetic field. We know the field lines by looking at iron filings near a magnet, which line up in the familiar way, joining north to south poles in loops. But you can shove the iron filings over to the right or left a bit with no problem, until you come to the limit where the filings need to touch each other. The "lines" therefore are artifacts brought about by the presense of the iron filings. The field lines are not really lines through empty space at all, but need the presense of material particles for definition, correct? In the absense of magnetizable particulate matter the field is continuous, undivided, perhaps more like a density cloud than a lattice.

We might imagine that there is a density cloud of photons or even of virtual particles which exists even in non-material space, and that these particles line up in their own scale in a manner analogous to the way the iron filings behave. So the "lines" may be present down to the Planck scale, even in a vacuum. Then when a charged or magnetized particle enters the region affected by the magnet, it encounters these microscopic particles, which are arranged in lines due to the effect of the magnet, and so is deflected in predictable ways.

Something about this field idea bothers me.

Then there is the rectangular conductor. If one places an electrode in one place and an annode in another place, does the current go directly from one diode to the other, in a straight line, or does it fill the conductor? Or at least, does it cover the surface of the conductor? Or do we need to apply a path integral approach and say that the current is everywhere on the conductor but cancels itself out mostly except for along the line between the diodes?

Cats! I am out of time. As you can see I still have more points of confusion to consider. I hope to return to this tonight. Meanwhile, chores.

Be well,

nc

marcus · Feb 16, 2005

hi Richard,
the best I found in a quick search was
http://www.warwick.ac.uk/~phsbm/qhe.htm

and this shows that in our units the QHE "resistance" i.e. the voltage/current ratio can take on several values. It has several plateaux it can be on depending on the strength of the field.

2pi, but for a weaker field 2pi/2, and for even weaker field 2pi/3, and I also see a plateau at 2pi/4,...etc.

So if you put in a current of E-28 (roughly a tenthousandth of an amp)
then it will create a transverse voltage (helped by magetic field) of
either 2piE-28, or (2pi/2)E-28, or (2pi/3)E-28, and so on...
You can see the plateaux in the diagram.

the strongest voltage, 2piE-28, is when the field is strongest. According to the diagram it happens when the field is 10 Tesla, or more. Say 11 Tesla to be safe.
In our units a tesla is 1.00E-53, by accident it was really close. So I can
translate that to say the field has to be 11E-53 or more.

Using the force constant in equations

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