Using the force constant in equations

marcus · Feb 19, 2005

Astronuc just posted about a bright flash witnessed in December 2004
http://www.everything-science.com/index.php?option=com_content&task=view&id=111&Itemid=2
see his post "brightest explosion ever observed"
https://www.physicsforums.com/showthread.php?t=64274
...

Here's a NASA link that goes with that. It has an animation of the flare and some photographs of the aftermath taken in the optical range.
http://www.nasa.gov/vision/universe/watchtheskies/swift_nsu_0205.html

It says that for a tenth of a second the power of the explosion was E40 watts.
the natural unit of power is 1.444E51 watts, so roughly 100 billion time brighter still.
In case anyone is curious and wants to calculate the natural power unit for themselves it is really easy, you just need to know c and G
the power unit is just c⁵/(8piG)

so you put in 299792458, raise it to the fifth,
then you divide by 8pi,
then you divide by 6.6742E-11 (what NIST gives for G)
and that's it

at this point I believe you have 1.444E51 watts.

this is the number that I somehow goofed with and got 1.6E51
in a rough calculation a couple of weeks ago. too much in a hurry
or just careless

nightcleaner · Feb 19, 2005

Hi Marcus

Goat boy's gramma wonders if the bright flash could have had anything to do with the Christmas Tsunami? I told her, probably not, but then got to thinking about it. The CT event was several days before the flash, and the flash traveled at speed of light, so no particle or energy could have preceded the flash.

However, the flash was on the far side of the galaxy, and the galaxy curves space, so the light flash had to travel in a curve to get here. In fact, if the galaxy is roughly a hundred thousand lightyears across, it took a hundred thousand years to get here. What if spacetime distortions, gravitational waves if you like, take a straighter path? Maybe gravitational waves could outpace the light flash by a few days in a hundred thousand years?

Don't tell gramma about this. She has enough on her mind as it is.

Richard

marcus · Feb 19, 2005

Once there was a Taoist sage who had a special tooth that allowed him to listen to programs on the airwaves. His dentist had accidentally made the filling in his first bicuspid act like a crystal radio set. The sage could receive two stations, one when he was wearing his spectacles and one when he was not.

That would have been wonderful enough but something even more amazing was true. The sage could test batteries by LISTENING TO THE VOLTAGE. If the pitch was high enough it meant that the battery had enough voltage.

The way it worked was that his upper and lower bicuspids served as contacts and when he bit the battery just right he could hear a kind of humming or whistling and this tone, which he could hear by virtue of senses heightened by years of spiritual discipline, was exactly a trillion times lower than Josephson frequency for the same voltage

For instance if he heard the pitch "A440" when he bit the battery (that's the note orchestras often tune by) then it meant the battery was only putting out 0.91 conventional volts.

But if the battery was a fresh Dragon class Alkaline Power-cell delivering a perky 1.53 volts, the he would hear the Soprano High G. This is the G right above the treble staff. It is only a few notes down from the High C which is sometimes said to fracture fine crystal.

Question: My rechargeables put out 1.23 conventional volt which is around 5 quartervolts. We've been using "quartervolts" in this thread to stand for E-28 on the natural voltage scale. What pitch would correspond to 5E-28 natural?

marcus · Feb 19, 2005

...
Question: My rechargeables put out 1.23 conventional volt which is around 5 quartervolts. We've been using "quartervolts" in this thread to stand for E-28 on the natural voltage scale. What pitch would correspond to 5E-28 natural?

Answer: to convert any voltage to its Josephson frequency (see the post about the AC josephson effect) you simply multiply by 2.

Since the voltage is 5E-28, the Jo frequency must be 2 x 5E-28 = E-27.

But by virtue of the tooth, what he hears is a trillionfold (E12) lower frequency instead:
E-27 x E-12 = E-39
And E-39 is the D on the treble staff.

marcus · Feb 19, 2005

we just went for a walk up the hill and looked at the lights of the bay area

they are mostly sodium pinkish orange. apparently sodium lights make especially efficient use of electricity

the basic sodium yellow is actually a pair of frequencies
8.64E-28
and 8.63E-28
these both come from the transition 3p down to 3s
most of the light comes from the 8.64E-28 (the higher of the two frequencies----the shorter of the two wavelengths) but both are quite strong.

you may recall that 10E-28 is green and over 10 gets into blue and violet
8E-28 on the other hand is red. So it makes sense that 8.6E-28 should be orange or yellow.

Here's a web source about sodium that shows the electron transition and stuff. It shows how you can plug a salty pickle into housecurrent and it will glow yellow from the sodium lines
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/sodium.html

[edit: we went up the hill again tonight and I asked my wife what color she thought the East Bay streetlights were and she said gold. That is right. It is a gold with a slightly reddish tinge to it]

marcus · Feb 20, 2005

anyone who wants to can convert back and forth between conventional visible wavelengths in nanometers and frequencies in natural.

you just divide 5091 by the conventional wavelength and that gives the frequency in E-28 terms, or likewise the photon energy.
so if you know that conventional wavelength 650 is red, 500 is green, 430 is blue then you calculate the frequencies as 5091/650 for red,
and 5091/500 for green, and 5091/430 for blue (and multiply each by E-28)

but that is a bit cumbersome so I have just gone ahead and learned the visible range in natural frequency terms is from 7 to 13, with 10 in the middle for green. I mean from 7E-28 to 13E-28 with 10E-28 in the middle.

Now this whole planet full of life survives mostly by eating red and blue photons.
the fact is that "Chlorophyl A" the main thing that eats sunlight here, can only absorb and utilize red photons around 7.6-7.7
and blue photons around 11.6-11.8.

Chlorophyl A is a very picky eater. It gets some help from "Chlorophyl B" which can absorb some little bit more of the blue and red and transfer energy to Type A. but there is a big gap from 8.2 to 10.4 which hardly get used at all.

7.6-7.7 YUM! delicious red photons *smack* *slurp*

8.2-10.4 YUK! nasty yellow and green, spit it out!

11.6-11.8 CHOMP! SLURP! yummy blue photons

what impresses me is that I just went out to the kitchen and got a bowl of granola and sliced a banana onto it and all that energy that I just ate came from a narrow range of red and a narrow range of blue photons
Red and blue light is my phood ----- I just eat the energy second or third hand.

[edit afterthought: it's strange that life hasnt evolved to make better use of photons in the frequency/energy range 8E-28 to 11E-28.
If it did then leaves would be black. Common knowledge that the reason leaves are green is they don't absorb and utilize much light in that central 8-to-11 section of the visible spectrum, so it gets reflected back to us.]

marcus · Feb 20, 2005

Markers in the visible range 7-13 eekyoo

The numbers stand both for photon energies and for (angular format) frequencies. Think of them all multiplied by E-28, which I'm sometimes to lazy to write.
Below 7E-28 light is infrared, above 13E-28 it is ultraviolet.
By good luck it turns out that a typical green frequency 10E-28 is right in the middle of the visible range. So the three main landmarks are 7--10--13.

Low end of visible 7

Chlorophyl red peak 7.7

Sodium street lights ("D-lines") 8.64

Middle green 10

Chlorophyl blue peak 11.7

High end of visible 13

Suppose you needed to know the conventional (cyclic format) wavelength in nanometers for some reason. Just divide 5091 by any of these numbers. Like 5091/10 = 509.1 nm is the conventional wavelength for the "middle green" I'm using as a landmark color.
My source on the endpoints of the human visible range is Campbell's Biology, a modern textbook. He gives two decimal places compared with only one in the Handbook of Chemistry and Physics. Presumably the boundaries vary from person to person and are only approximately known.

marcus · Feb 20, 2005

Batman in Zero Gee

Batman is at Burger King enjoying a tasty batburger.
The plate glass beside him shatters and a giant hand reaches in.
It is King Kong again!

King Kong's arm is 10 paces long----E35 in natural length units.
He does a circular windmill-style pitcher's windup, preparing to throw Batman a long distance.

At the top of Kong's windup arc, Batman feels no gravity! It must be that the planet's normal E-50 gravity is canceled by "centrifugal force"!

What is Batman's speed?

marcus · Feb 20, 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 appears suddenly in the grand dining hall.
He surprises the chickens by singing the Queen of the Night Aria from Mozart's Magic Flute.

This is one of the great coloratura arias of all time and it goes up to the high E-flat above the soprano's high C. Even professionals can often only make a kind of indeterminate squeak for this E-flat.

The chickens are alarmed and flee to their ships. Maid Marian is free!

What is the frequency of Robin Hood's high note on the natural scale?

Listen to a midifile of the Queen of the Night aria:
http://www.geocities.com/Vienna/Strasse/2915/flute2.mid
the very high passage starts when the aria is about 1/3 of the way thru

marcus · Feb 20, 2005

The celebrated opera singer Kiri Te Kanewa once went to a masquerade party as Minny Mouse. Her friends persuaded her to an make an impromptu performance and she sang Donna Elvira's aria Ah! Chi mi dice mai.

Some hydrogen-breathing aliens from outer space were at the party and the prince of a distant planet fell in love with her. He therefore commissioned skilled doll-makers to create a copy of the famous Diva for him to take home. The doll was to be life size, able to sing several Mozart arias, and satisfy the desires of the prince.

When the masterful reproduction of the Diva was delivered to the prince's hotel, however, it was rejected scornfully. The prince declared that they had put the wrong head on the singer by mistake. She was supposed to have large round black ears, he said, and a pointed nose.

Indeed he had fallen in love with Kiri while she wore a mouse mask and had no idea she had any other face.

This oversight was quickly remedied and the prince departed happily accompanied by his vocally talented paramour with the long eyelashes and big ears.

When they arrived at the prince's home planet, trouble immediately developed. It was found that the doll would now only sing Mozart's arias two octaves too high. It was like fingernails scraping on slate and gave everyone a very bad impression. What caused this to happen?

Kea · Feb 20, 2005

hello

marcus said:

When they arrived at the prince's home planet, trouble immediately developed. It was found that the doll would now only sing Mozart's arias two octaves too high. It was like fingernails scraping on slate and gave everyone a very bad impression. What caused this to happen?

Perhaps when they changed the heads, the mouse thought it was really
supposed to be singing arias from the The Tales of Hoffmann?

Wouldn't the mouse have trouble breathing hydrogen?

marcus · Feb 20, 2005

Kea said:

...Wouldn't the mouse have trouble breathing hydrogen?

I was thinking of Tales of Hoffmann too

It was a doll, like the kind that runs on NiCad batteries, or you have to wind up, so it didnt have any metabolic trouble with the planet's hydrogen atmosphere.
But the doll's vocal apparatus was analogous to flutes and woodwinds, so she had the same problem that a flute would have if someone tried to play it in a hydrogen atmosphere. The speed of sound is almost four times faster.

I think you thought of that

and that is why you mentioned breathing hydrogen.

Kea I would really love it if you would deign to work some of the simple problems in this variant of Planck units. Like the Batman problem I just posted, or the one about the chickens.

marcus · Feb 20, 2005

There were seven dwarves who lived in a old Victorian in San Francisco and operated a bed-and-breakfast. For extra cash flow these dwarves had a mailing business where they pack and ship things for you. They had all the materials ready--the right size crates and boxes, the plastic bubbles-- and it saved time to let them do it.

One time the Stanford Board of Trustees got tired of Leonard Susskind and they decided to have him shipped to an alternative universe, so they called Dwarves We-Send-It and the dwarves came and got Susskind and packed him up. What a lot of fussing he made about it!

Then the most disagreeable dwarf, whose name was Stinky, put the address on. He wrote

Code:

Deliver to:

alpha = 1/138
Lambda = 0.9E-120

At last they wheeled the crate out onto the loading dock and it disappeared.

The reason Stinky wrote that address was because of the Anthropic Principle, he wanted to send Leonard Susskind where he could be happy because the basic parameters were close to ours. We have alpha = 1/137.036... and this leads to a nice big periodic chart of elements, rich chemistry, stable long-lived stars etc. So Stinky made the alpha be near ours, but slightly different just to make sure they really got rid of Susskind.

Again just to be on the safe side, he made the cosmological constant Lambda be slightly different from our Lambda, which is 0.85E-120.

If its Lambda is too small a universe will short-lived and collapse, but if Lambda is too big it will be rather vacuous: space will expand so fast that galaxies and stuff never have a chance to collect and condense.
Stinky judged that having a wee bit more vacuousness would suit Susskind which was why he made it 0.9.

Kea · Feb 20, 2005

Batman and King Kong

marcus said:

King Kong's arm is 10 paces long----E35 in natural length units. He does a circular windmill-style pitcher's windup, preparing to throw Batman a long distance. At the top of Kong's windup arc, Batman feels no gravity! It must be that the planet's normal E-50 gravity is canceled by "centrifugal force"!

What is Batman's speed?

About 3E-8 in natural units. In other words, speed squared = gravity times armlength = E-15.

Assuming a pace is roughly 0.25 metres, in metric the speed is 5 m/s.

Batman was visiting a Wellington Burger King, so he screamed out to Peter Jackson, who was scoffing burgers at the next table, to ask King Kong to let him down, which King Kong did so that there could be a next episode.

Kea · Feb 20, 2005

Giant Chickens

marcus said:

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 appears suddenly in the grand dining hall. He surprises the chickens by singing the Queen of the Night Aria from Mozart's Magic Flute.
This is one of the great coloratura arias of all time and it goes up to the high E-flat above the soprano's high C. Even professionals can often only make a kind of indeterminate squeak for this E-flat. The chickens are alarmed and flee to their ships. Maid Marian is free!

What is the frequency of Robin Hood's high note on the natural scale?

High E-flat is about 1.25 KHz. Using a speed of sound of 343 m/s, the wavelength of high E-flat is about 0.27 metres. Converting to natural units, the frequency is

[tex]f = \frac{1250}{3.7 E42} = 3.4[/tex]E-40

Fortunately this is quite a bit smaller than the visible frequencies above.
Analogously to the visible case, we talk in terms of E-40 for musical notes.
In these terms the conversion factor [itex]k[/itex] in

[tex]\frac{k}{\lambda} = \nu[/tex]

for wavelength in metres and frequency in natural units E-40, is [itex]k = 0.92[/itex].

The chickens weren't happy because Maid Marion had been treating them to a few arias by Handel, who died only 3 years after Mozart was born, and the chickens immediately thought Mozart was a postmodern anarchist, and they left in disgust to join Susskind in another universe.

marcus · Feb 20, 2005

Kea said:

About 3E-8 in natural units. In other words, speed squared = gravity times armlength = E-15.

Assuming a pace is roughly 0.25 metres, in metric the speed is 5 m/s.

Batman was visiting a Wellington Burger King, so he screamed out to Peter Jackson, who was scoffing burgers at the next table, to ask King Kong to let him down, which King Kong did so that there could be a next episode.

Yes yes yes! THANK you Kea, you are very kind. that is exactly right for the speed. It is sqrt(E-15) = about 3E-8

I am glad that you have arranged to keep Batman around. We all admire his style and his relentless battle against the wicked

as for metric conversions the speed unit in this system is c, so that E-8 is 3 meters/second, sqrt(E-15) is around 9.5 m/s
the pace is E34 times the (variant) Planck length or around 81 cm.
but the metric conversion part is totally non-essential, what is a balm to my spirit is when someone works the problem companionably with me in natural units and gets the answer in those terms

marcus · Feb 20, 2005

Kea said:

... a postmodern anarchist, and they left in disgust to join Susskind in another universe.

is right!

marcus · Feb 20, 2005

Kea said:

... we talk in terms of E-40 for musical notes.

Exactly! Let me clarify: the units are defined by |8piG| =|hbar| = |c| = 1
and that has the outcome that the D right above middle C (near the top of my range) is 0.50E-39 = 5.0E-40

The frequency E-39 is an octave up so it is the note D on the treble staff (up in soprano/alto range)

So this E-flat the Queen of the Night sings, that is one halfstep up from a still higher D: 2E-39.

The D which is an octave above the D on the treble staff is 2E-39

and since the dreaded E-flat is a halfstep up from there, I have to multiply by 1.06------and it is 2.12E-39

(I should mention that I am using angular format for wavelengths and frequencies unless stated otherwise, as in the case with conventional cyclic wavelengths in nanometers for light)

marcus · Feb 20, 2005

Kea said:

[tex] \frac{...}{3.7 E42} [/tex]

I am going to focus on just one thing in Kea's post #190, the number
3.7E42 that she divided by in one of her equations

this is the natural unit of angular frequency expressed as radians per second. (in these units we have |hbar| = 1 so it makes sense to work consistently in angular format for freq. and wvlength.)

It is a really good number to remember. Also it has latent (undeclared) accuracy.
According to current estimates of the fundamental constants you could say that the fundamental frequency, or natural frequency unit, is
actually 3.700E42.

So we can copy Kea's approach and get frequency equivalents like this.

The A that orchestras use to tune by is 880pi radian/second.
All we need do is divide by 3.7E42 and we will know the frequency of A in natural terms!

[tex]\omega_A = \frac{880 \pi }{3.7 E42} = 7.472E\text{-}40[/tex]

So the A they tune by is 7.472E-40
The D on the treble staff is 5 musical halfsteps above that so one just needs to multiply the frequency by 2^5/12 = 1.335.

So the D on the treble staff is 1.335 x 7.472E-40 = 1.00E-39

this is so close that one could tune by setting D = E-39 and not know the difference.

So using this number 3.7E42 (natural unit of ang. freq.) that Kea supplied is a good way to explain why, for instance, the D right above the soprano's high C has frequency 2E-39

and the E-flat a halfstep up from that high D has frequency 2.12E-39.

we could call this the "Kea number" and go back and define the whole set of natural units on the basis of that exact frequency of

[tex]\omega_{kea} = \sqrt{\frac{c^5}{8\pi G \hbar}} = 3.7 E42 \text{ per second (exact)} [/tex]

If there were some reason to do it over it might not be such a bad way to proceed.

marcus · Feb 21, 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 appears suddenly in the grand dining hall
with half of a Kentucky Fried Chicken in a paper bag. He consumes this delicious fast food at the Invaders' victory banquet, thoughtfully sucking each bone clean before tossing it into their punch bowl.

The chickens are sickened with disgust and flee to their ships moaning and barfing. Maid Marian is free!

What fraction of a natural unit of energy did Robin consume at the banquet?

Anne Collins lists the Calories in Kentucky Fried Chicken piece by piece here:
http://www.annecollins.com/calories/calories-fried-chicken.htm
WING 135
BREAST 370
DRUMSTICK 140
THIGH 360
One can easily see that half a bird is 1000 Calories.

marcus · Feb 21, 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 appears suddenly in the grand dining hall
with half of a Kentucky Fried Chicken in a paper bag. He consumes this delicious fast food at the Invaders' victory banquet, thoughtfully sucking each bone clean before tossing it into their punch bowl.

The chickens are sickened with disgust and flee to their ships, barfing and moaning. Maid Marian is free!

What fraction of a natural unit of energy did Robin consume at the banquet?

Anne Collins lists the Calories in Kentucky Fried Chicken piece by piece here:
http://www.annecollins.com/calories/calories-fried-chicken.htm
WING 135
BREAST 370
DRUMSTICK 140
THIGH 360
One can easily see that half a bird is 1000 Calories.

with other food problems in this thread we've been using a crude approximation that one natural energy unit is 100,000 food Calories.
So clearly Robin consumes 1/100 of a natural unit.

But just for a change let's be real precise and use Kea's value for the natural frequency expressed in conventional radian-per-second terms.

[tex]\text{energy unit = hbar x frequency unit} = \hbar \times \omega_{kea} = 1.05457E\text{-}34 \times 3.7 E42 = 3.90E8 \text{ joules} \approx E5 \text{ food Calories} [/tex]

marcus · Feb 21, 2005

This summary needs to be brought forward periodically.
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.

Another way (suggested by a Kea post) to define the same units is to make the unit angular frequency be 3.7E42 per second. Call it omega_nat and set its value equal one, instead of F. Same result.

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 144 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
solar constant 6.2E-117
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:
3.700E42 rad per sec 1 (the unit frequency)
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.

Kea · Feb 21, 2005

Marcus

Where is everybody?

marcus · Feb 21, 2005

Kea said:

Marcus

Where is everybody?

Hi Kea, I went to a friends house this afternoon to practice singing Brams Liebeslieder Waltzes. they are fun. bouncy. and with nice harmony.

I just got back. I don't notice fewer or more people than usual. I hope you check in soon and tell me who is missing.

Kea · Feb 21, 2005

Hi Marcus

It's great that you sing. I guess you're right about it being as busy as
usual. I have been talking to Urs on the Coffee Table lately.

Kea

marcus · Feb 21, 2005

Kea said:

I have been talking to Urs on the Coffee Table...

fraid I can't compete with Urs for conversation but if you would like i will make you up a physics problem. just tell me what you would like it to be about (anything as long as involves only basic physics and is not too risque)

marcus · Feb 22, 2005

a physics P-uhh-zzz-L for Kea

Kea suggested, or got me to look at, the frequency which is exactly
3.7E42 per second. I am going to call that frequency [tex]\mathbb{Z}[/tex]

[tex]\mathbb{Z} = 3.7E42 \text{ per second}[/tex]

In return, here's a P

:zzz: L for Kea. The idea is in the case of each quantity, evaluate it in metric terms.

[tex]\text{Can I put you on hold for just a } \frac{1}{\mathbb{Z}}\text{?}[/tex]

[tex]\text{Don't yield a } \frac{c}{\mathbb{Z}}![/tex]

[tex]\text{A good breakfast gives you } \hbar\mathbb{Z}\text{ all day long.}[/tex]

[tex] \hbar\mathbb{Z}^2 \text{ corrupts, absolute } \hbar\mathbb{Z}^2 \text{ corrupts absolutely.}[/tex]

[tex]\text{Early wins build campaign } \frac{\hbar \mathbb{Z}}{c}[/tex]

[tex]\frac{\hbar \mathbb{Z}}{c^2} \text{ mind is made by } \frac{\hbar \mathbb{Z}}{c^2}\text{ media. }[/tex]

marcus · Feb 22, 2005

marcus said:

I am going to focus on just one thing in Kea's post #190, the number
3.7E42 that she divided by in one of her equations
...this is the natural unit of angular frequency expressed as radians per second. ... has latent (undeclared) accuracy.
... could say that the fundamental frequency, or natural frequency unit, is
actually 3.700E42..we could call this the "Kea number" and go back and define the whole set of natural units on the basis of...If there were some reason to do it over it might not be such a bad way to proceed.

[tex]\sqrt{\frac{c^5}{8\pi G \hbar}} \approx \mathbb{Z} = 3.7 E42 \text{ per second } [/tex]

usual symbol for (angular format) frequency in physics----the "radians per second" description of rotation, vibration, waves etc.---is lowercase Greek omega. But omega still confuses some people and, well, Z and Omega are the last letters in their respective alphabets. Saying "alpha to omega" is about the same as saying "A to Z". So...

And maybe the above fizix p-uhh-zzz-L is dumb. Yeah it probably is dumb. It is just another way of getting the very same set of natural units that we have used all thru this thread---a set I see used implicitly in Quantum Gravity papers a lot these days.

But dumb or not, since no one else here has, I will calculate using Z = 3.7E42 sec^-1 what all the natural units are in metric terms...or at least have a cup of coffee and think about doing it

marcus · Feb 22, 2005

Kea suggested, or got me to look at, the frequency which is exactly 3.7E42 per second. I am going to call that frequency [tex]\mathbb{Z}[/tex]

[tex]\mathbb{Z} = 3.7E42 \text{ per second}[/tex]

[tex]\sqrt{\frac{c^5}{8\pi G \hbar}} \approx \mathbb{Z} = 3.7 E42 \text{ per second } [/tex]

[tex]\text{time unit } = \mathbb{T} = \frac{1}{\mathbb{Z}} = \frac{1}{3.7E42 s^{-1}} \approx 0.27027E\text{-}42 s [/tex]

[tex]\text{length unit } = \mathbb{L} = \frac{c}{\mathbb{Z}} = \frac{299792458 ms^{-1}}{3.7E42 s^{-1}} \approx 0.081026 E\text{-}33 m [/tex]

[tex]\text{energy unit } = \mathbb{E} = \hbar\mathbb{Z} = 1.0545715E\text{-}34 Js 3.7E42 s^{-1} \approx 0.39018E9J [/tex]

[tex]\text{power unit } = \mathbb{P} = \hbar\mathbb{Z}^2 = 1.0545715E\text{-}34 Js (3.7E42 s^{-1})^2 \approx 144.4E49 watts [/tex]

[tex]\text{momentum unit } = \mathbb{M}c = \frac{\mathbb{E}}{c} = \frac{\hbar\mathbb{Z}}{c} = \frac{1.0545715E\text{-}34 Js 3.7E42 s^{-1}}{299792458ms^{-1}} \approx 1.3015kgms^{-1} [/tex]

[tex]\text{mass unit } = \mathbb{M} = \frac{\mathbb{E}}{c^2} = \frac{\hbar\mathbb{Z}}{c^2} =\frac{ 1.0545715E\text{-}34 Js 3.7E42 s^{-1}}{(299792458ms^{-1})^2} \approx 0.434E\text{-}8 kg [/tex]

marcus · Feb 22, 2005

marcus · Feb 22, 2005

carousel

On Sundays Toad operated the merrygoround in Golden Gate Park.
The painted animals went up and down. There was a bright-colored parrot you could ride on, and a unicorn with a blue saddle.

This is a big merrygoround, said Toad, the outside mounts are 10 paces from center. He turned on the calliope, which played Skater's Waltz and Stars and Stripes Forever. It wheezed and snorted and occasionally beat its drum and jangled bells.

Frog got on. He went to the outer circle of beasts, where he climbed onto a giant wooden chicken . Make it turn faster! said Frog. I am having a good time but if it turns faster I will have an even better time.

Cant, said Toad, it turns exactly E-43 of universal standard. It's a constant and won't change.

What was Frog's constant acceleration towards the center?

marcus · Feb 23, 2005

Cavendish

my purpose in this thread is to "test drive" a set of unit that crop up in QG a lot (but afaik are mainly restricted to QG topics like hawking temp, black hole, unruh radiation, LQG, spinfoam, LQ cosmology etc.). I want to see what happens if you breach that restriction and apply this set of natural units outside QG.

so let's do an unlikely thing. let's cast the Cavendish 1798 experiment in these terms.

we have a delicate torsion beam and a couple of M = E10 (i.e. hundredpound) masses.

we want to see how much the masses deflect the torsion beam.

the torsion beam is a stick with knobs at the end, suspended by fiber
the mass of the two knobs totals m (unspecified) and each is a distance R from the centerpoint of the beam. R = 2E33 (i.e. twice 8.1 cm, two palmwidth). the torsion beam moment of inertia mR²

can also call this delicate "dumbell" suspended by thread or wire or fiber by the name "torsion pendulum". it does not swing back and forth like usual pendulum, it rotates one way until the thread develops a twist torque and then it rotates the other way...and so on.

Now the 18th century English eccentric Cavendish contemplates the torsion beam slowly twisting this way and that in a glass cabinet and he TIMES the oscillation (often an astute thing to do) and this gentleman finds that its (angular of course) frequency is E-44.

By this I mean that it is E-44 of the "Z" frequency defined as exactly 3.7E42 per second (I picked this up from Kea). This is something Cavendish might arbitrarily have defined since he had seconds (though he couldn't have attributed any broader significance to it).

What can be deduced from this measurment? Well if you remember this from college physics there is a fiber stiffness coefficient k which gives the torque per radian of twist and

freq² = k/(mR²) = (E-44 Z)² = E-88 Z²

Now the crafty English eccentric stops the torsion pendulum and places massive balls near each knob to see how much the GRAVITY of the balls draws the knobs. and he discovers that this will produce a TINY TWIST of 1/500 radian!

this twist is maintained when the distance from knob to ball, at each end of the beam, is D = E33.

Now what, says Cavendish, is the experimental value of the Newtonian gravitational constant G that has been determined here?

marcus · Feb 23, 2005

the experimental value of G, from this data, turns out to be G = 1/25

(or if you like it better, 1/(8pi))

the way it comes out is call the deflection angle = theta and
[tex]G = \frac{k}{mR^2} \theta \frac{D^2R}{M}= (E\text{-}44)^2 \theta \frac{(E33)^2 2E33}{E10}= 2\theta E\text{-}88E89 = 20\theta[/tex]

so if the deflection angle is 1/500 radian, then the experimental value of G is 1/25.

marcus · Feb 24, 2005

On Sundays Toad operated the merrygoround in Golden Gate Park.
The painted animals went up and down. There was a bright-colored parrot you could ride on, and a unicorn with a blue saddle.

This is a big merrygoround, said Toad, the outside mounts are 10 paces from center. He turned on the calliope, which played Skater's Waltz and Stars and Stripes Forever. It wheezed and snorted and occasionally beat its drum and jangled bells.

Frog got on. He went to the outer circle of beasts, where he climbed onto a giant wooden chicken . Make it turn faster! said Frog. I am having a good time but if it turns faster I will have an even better time.

Cant, said Toad, it turns exactly E-43 of universal standard. It's a constant and won't change.

What was Frog's constant acceleration towards the center?

[tex]a = R \Omega^2 = E35 \times (E\text{-}43)^2 = E35 \times E\text{-}86[/tex]

[tex]a = E\text{-}51[/tex]

a gee is roughly E-50, so the acceleration towards center that Frog feels is about one tenth of gee.

BTW a pace here (E34 length units) is some 81 centimeters. so the radius of the carousel being ten steps or paces translates to E35. If I remember this is about right for the GG Park merrygoround. The rotationrate or angular freq is about right too. Remember that Kea's universal std freq. is 3.7E42 radians per sec and this one here is E-43 of that, so it is 0.37 radians per sec.

marcus · Feb 25, 2005

I want to calculate the magnetic field inside a superconducting coil
let's say 60 amps (E-22 natural) and 123 turns per centimeter (E-30 per nat. length)

mentioned earlier that a magn. field with strength E-53 measures very close to one tesla (coincidence is within one percent) so we can pretty much read off E-53 as tesla

the formula is B = 4pi x alpha x current x turns per unit length
B = (4pi/137) x E-22 x E-30 = 12.5/137 E-52 = 125/137 E-53 = 0.9 "tesla"

now we can go to this online calculator that does it in metric units
and check

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html

if you scroll down to the calculator and put in the metric equiv data
you do indeed get 0.9 tesla (enter 12340 turns per meter, 60 amps, relative perm k = 1 for vacuum, and in the formula right above the boxes where you enter the data click "magnetic field".)

this visit to georgiastate hyperfizix is mostly just to confirm the formula in natural units that, in nat. unit terms, magn. field strength inside coil is

[tex]|B| = 4\pi \alpha nI \approx \frac{4\pi}{137} \times \text{ turns per length} \times \text{ current} [/tex]

Using the force constant in equations

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