J*di Units: the F*rce is One

[5614]

1) in that graph can i assume the time measurement is in years?

 

2) you could not program a program like that without a basic knowledge of programing, although im sure you could make an animation of it instead!

 

3) martin, your sig never displays any figures, im sure its meant to!

[Martin]

...

 

3) martin' date=' your sig never displays any figures, im sure its meant to![/quote']

 

what happens if you let the mouse rest lightly on one of the words in the sig-----putting the cursor on a line of the sig without pressing the mouse button?

 

when i do it words and numbers appear

[5614]

what happens if you let the mouse rest lightly on one of the words in the sig-----putting the cursor on a line of the sig without pressing the mouse button?

 

when i do it words and numbers appear

i have seen that before on other sites in other contexts, looking at the source code:

 <acronym title="269/day" style="color: #330033;">July average</acronym>

it is quite a neat little command that i didnt notice before.

[Martin]

5614 you have given me an idea

we are collecting easy basic exercises that use physics constants like Planck, Boltzmann, electron charge

it seems desirable that at least some be fun or humorous or at least casual

 

Why dont we hide the answer in a place that you have to mouse to see it?

 

that way there is no spoiling for people who want to try to work the exercise?

 

I was just now thinking of a couple of John Baez problems----physics exercises with a cameo role for the famous Usenet mathematician JohnBaez who was the moderator of sci.physics.research for many years and wrote some amusing mathematical physics-made-easy explanations.

I will try including the answers in that invisible way.

[Martin]

we have these J*di units where the Force is a power of ten and

hbar and c and Boltzmann k and the electron charge e are all powers of ten (in our units)

and those choices of powers of ten for the constants DETERMINE our units and they make the mass unit come out pound-size (in fact 434 grams)

 

And it happens that a pound of protons is 2.6E26 protons

that is an important number

 

HOW MANY HELIUM ATOMS DOES IT TAKE TO LIFT JOHN BAEZ?

 

John Baez is exploring the universe in a small spacecraft and is currently visiting a planet where the atmosphere is all O2.

 

Nothing is said about the gravity or the atmospheric pressure or what the temperature happens to be. Only that the air is oxygen molecules and that the people on this planet travel by helium balloon.

 

OK it happens that John Baez plus his balloon (empty) weigh 210 pounds.

How many atoms of helium must his friends put into the balloon in order to get him off the ground?

 

[acr= 19.5 x 10^26 helium atoms] How many atoms? [/acr]

[Martin]

BTW what volume does E26 molecules of pretty-good-if-not-ideal gas occupy at prevailing ambient temperature and pressure here on earth?

 

In J*di units the average temperature outdoors is 1000 degrees and

a common atmospheric pressure might be 1400 marks per sq. hand.

 

We have this law PV = N k T

and Boltzmann k = E-22 we know for damn sure

so PV = E26 x E-22 x 1000 = E7, in our units.

 

So put in for P, and find the volume in cubic hand.

 

[acr= 7140 cubic hands] Volume of E26 molecules?[/acr]

 

you can always divide the cubic hands by 1880 to get cubic meter, if you wish,

 

or do the whole thing in metric! At current ambient pressure (pascal) and temperature (kelvin) what volume (cubic meters) is occupied by 10²⁶ air molecules?

[Martin]

John Baez is visiting a popular vacation planet known for its Big Sun and its Mariachi bands.

 

their sun look bigger in the sky there than our sun looks in our sky

 

Our sun makes an angle of 1/107 radian in the sky (about half a degree).

 

this means the sun's radius makes an angle of 1/215 radian.

 

but this planet's sun makes an angle of 1/85 radian, so you notice that it is bigger-looking----also sunsets last longer.

 

the people are friendly as long as you have money, the women on the beach have big tits, and everybody knows how to make a good Margarita.

 

THEIR SUN IS HOTTER THAN THEIR PLANET BY WHAT FACTOR?

 

I think you know what i mean, the surface of our sun is 20 times hotter than our planet's global average surface temperature. Absolute (the only scale where such things make sense). Assume albedo and greenhouse effects cancel or are negligible. So how much hotter is their sun than their planet surface average?

 

[acr= 13 x sqrt(2) = 18] By this factor [/acr]

[Martin]

THE RECEDING HORIZON

 

Our intrepid voyager is visiting friends on a planet famed for its hiking boots, rock climbing gear, and camping goods of all sorts. Baez and a hiking companion climb from sealevel to a height of half a mile

and find they can see 50 miles out to sea (they spot some islands at a known distance)

 

they continue climbing to an altitude of one mile

and find they can now see 71 miles

 

they continue their climb to one and a half mile altitude

HOW FAR CAN THEY SEE?

and what is the diameter of the planet?

 

Dont worry how much a mile is, since everything is in miles in the problem.

If you insist, pretend it's really a kilometer.

 

[acr= they see 87 miles] how far?[/acr]

[acr=5000 miles] diameter?[/acr]

 

since the earth's diameter is around 8000 miles, assuming these are

ordinary miles, you can determine whether their planet is larger or smaller than earth

[Sayonara]

Why dont we hide the answer in a place that you have to mouse to see it?

 

that way there is no spoiling for people who want to try to work the exercise?

Use the

tags.

 

 

it is quite a neat little command that i didnt notice before.

The code you showed is actually an example of the acronym tag being misused.

[5614]

The code you showed is actually an example of the acronym tag being misused.

 

yes i know it is, however looking at the source code for this page it is clearly what is used!

 

martin, make sure people know the answer is there!!! put a message in your sig or something because only a few who have read this will know, and guests will not.

[Martin]

 

martin' date=' make sure people know the answer is there!!! put a message in your sig or something because only a few who have read this will know, and guests will not.[/quote']

 

 

5614 and Sayo, thanks for the suggestions!

 

I am not sure what to do about my sig, maybe let people who are curious figure it out for themselves what to do.

 

the acro function is perfect for what I have in mind. what I want is to have it say "how far?" and if you touch the question you get the answer

 

I want a feature that allows me to write questions at the end of a short discussion, like "what is the mass?" "what temperature?" and if you touch the question with the mouse you are told the answer. the acro seems perfect. (if you touch the acro it tells you what it means)

 

 

"spoiler" feature would be good for other things, like riddles, where you actually want to hide the answer because people actually try to guess the riddle. In the case of these physics problems I do not think anyone at SFN is going to try to work them, so I simply want to tell them the answer (but in a suitably delayed interactive way). One could use spoiler but it would be a bit of a kludge, and take more work on the reader's part

[Martin]

Baez ventures to a planet where dwells a Mad Inventor.

 

The Mad Inventor has constructed a room in which Alpha, the fine structure constant, can be different. There is a knob by the door.

 

Alpha is an important number in our universe which is usually equal to 1/137.036 and which for the sake of laziness we write 1/137. And if you turn the knob by the door you can make it 1/136 instead.

 

Some friendly advice: dont turn the knob too much or things will go haywire inside the room. there wont be the same elements on the periodic table and chemistry will be different and some things you didnt expect might be radioactive and so on. Just turn it a little bit.

 

the Inventor gives John Baez a lab-toy with two insulated balls, that can measure the force between. They place TEN TRILLION extra ELECTRONS

on each ball. And by ten trillion I mean 10¹³

 

The two balls are a handbreadth apart, center to center (i.e. 8.1 centimeters)

 

The smiling Boffin (Boffin is a UK word which may be used instead of Mad Inventor) turns the knob to 1/138.

 

WHAT IS THE FORCE BETWEEN THE BALLS?

 

If you are competent with metric units you should be able to solve purely in metric, without resorting on our J*di units. Because the fine structure constant is the same in any system, it is just a pure dimensionless number that tells the strength of the electromag interaction. And you are told the distance is 8.1 centimeters. And 10¹³ electrons is the same whatever system of units.

 

[acr= 1000/138 marks] the force in marks[/acr]

 

As you see one can write down the force in marks, the J*di unit, immediately. and then since a mark is half a newton, one can convert to newton and have the metric answer. but to work purely in metric would be rather involved so I will not bother. I will use the "spoiler" feature here (suggested by Sayo) to create a patch below where a more detailed explanation is hidden. you mouse the patch and the explanation shows up, in case anyone wants.

 

[hide] the unit charge dram = 10^18 electrons so each ball has a charge of 10^-5 drams

the coulomb constant is related to alpha by

coulomb = alpha x 10^13 marks sq. hand per sq. dram

so to get the force in marks one must just multiply

1/138 x 10^13 x 10^-5 x 10^-5 and divide by 1

and this gives 1000/138

which is 7.25 marks

 

if if you had to know it in metric it would be about 3.5 newton

 

one reason marks is convenient is that it is a power-of-ten fraction of the large Force which is the central coefficient in the Einstein equation so working with it is the next best thing to working with a natural unit [/hide]

[Sayonara]

"spoiler" feature would be good for other things, like riddles, where you actually want to hide the answer because people actually try to guess the riddle.

[hide]That's what it's for.[/hide]

[Martin]

The intrepid mathematician Baez is exploring the galaxy in a small spacecraft, and visiting friends along the way.

 

He encounters a star whose light is noticeably different from our sun's.

In the light from this star, a photon of average energy is green!

In this sense, the green photons are the ones typical in the light from this star.

 

This can be taken as an index of the star's surface temperature.

 

How hot is the star?

 

=======

Hint for metric users. If you want to solve this in metric units. Just recall what the energy E_green of a green photon is

and what the Boltzmann constant k is, in metric terms, and solve

this little equation for T

 

E_green = 2.701 kT

 

The number 2.701 is a mathematical constant which one would use regardless of which system of units one was working with. It relates the

characteristic energy kT, at temperature T, with the average photon energy. Starlight is assumed thermal here, as in fact it mostly is.

 

========

for a Jedi the answer is immediate, in terms of the "quartervolt" unit, a typical green photon's energy is 10 eQ

and Boltzmann k = 10^-4 eQ per degree (exactly)

 

Using the same equation as in the metric case one sees that the star's temp is

 

10⁵/2.701 degrees

 

[math]\frac{100,000}{2.701} = 37,000 \text{degrees}[/math]

 

By serendipity it happens that our sun's surface temp is 20,000 degrees, in Jedi. So it is easy to compare: this one is a bit less than twice as hot.

[ed84c]

Have you been able to see my spreadsheets, and if so are they ok/ good/ bad?

 

BTW martin where did u get ur sig stats from?

[Martin]

...

Ed I just tried your links and discovered to my chagrin that at least at the moment I have no working spreadsheet application. For now at least I cannot open the "XLS" format files that you've made.

 

You have inspired me' date=' however. I am going to make a table relating the DIAMETER of a hole (in handbreadths) to its TEMPERATURE (in our degrees). ...[/quote']

 

Have you been able to see my spreadsheets' date=' and if so are they ok/ good/ bad?

 

BTW martin where did u get ur sig stats from?[/quote']

 

Hi Ed, thanks for asking. No, my situation is unchanged from back in post #21 on this thread. For some reason my computer doesnt do spreads.

So I cannot comment. Tho I assume yours are quite OK because of the impression I get from what you have been posting.

 

You asked how i got the data on SFN "posts per day"

Around the beginning of the month I copy down the number of posts to date that it gives in the box on the Home page.

And then around the end of the month, and I subtract to get the number of posts for that month, and divide by the number of days.

It is not much work, you just have to remember to do it.

Would you like to take over the job?

I would be glad if someone reliable, who plans to be around indefinitely, would take it over. I could use my sig for something else.

[ed84c]

Well i would love to do it, however memory is not my strong point.

 

I was just hoping that there was a large amount of raw SFN statistics lying around somwhere. I would find them very interesting.

[Martin]

Well i would love to do it' date=' however memory is not my strong point.

 

I was just hoping that there was a large amount of raw SFN statistics lying around somwhere. I would find them very interesting.[/quote']

 

If you would love to do it, then you should!

 

Set up a copy of my stats, however you want to format it, in your sig and as soon as you have something like that I will see it and I will erase mine.

 

I am not bothering with December, BTW, because of holidays and because of that crash NewYears where so many posts were lost

 

I am starting to count again as of 3 january when the count was

118,228

 

If you care to know the time too it was around 10PM GMT but that doesnt matter very much.

 

I am serious, why dont you take it over if you like stats?

Accuracy, or remembering to do it regularly, is not so important as just having a rough gauge of whether or not the site is growing numerically.

[ed84c]

maybe i should do another statsitc, you know be original, but unless you particularly dont want the burden of it.

[Martin]

maybe i should do another statsitc, you know be original, but unless you particularly dont want the burden of it.

 

I would be really glad to trade off. It is not a big burden but i would like

to have no sig for a change or something completely different.

 

do what you think best.

 

if you pick up the monthly "posts per day" stat, then i will

be glad to shed it.

 

btw i dont feel obliged to do it for every month, just a few to

give an idea, but you would decide as you wished about this

[ed84c]

I think ill sleep on it

[Martin]

I think ill sleep on it

 

 

fine, whatever

 

I just went thru a messy calculation to find the evap time of a BH of mass M

I was using a site called halcyon

when I got the answer I wanted confirmation and went to Wiki

http://en.wikipedia.org/wiki/Hawking_radiation

WIKI HAD THE SAME FORMULA THAT I HAD DERIVED!

I am very relieved and proud to announce.

It should be interesting to see this in Jedi units, usually them make things cleaner or more streamlined

 

here is the formula

[math] \text{evaporation time} = \frac{5120 \pi G^2}{\hbar c^4} \text{mass}^3[/math]

 

I trust it because I got it from two independent directions

Well, Latex is not working so I will copy it in ordinary

 

evaporation time = (5120 pi G^2)/(hbar c^4) x mass^3

 

80 x (8piG)^2 divided by ( pi hbar c^4) times mass^3

 

(80/pi) E-14 E32 E-36 times mass^3

 

(80/pi) E-18 x mass^3

 

so our unit of mass is pound and our time is a count

if I plug in the mass in pounds here i will get the lifetime in counts

(they are 222 to the minute so if I then divide by 222 I get minutes)

So let us say the mass is E18 pounds

the lifetime will come out to

(80/pi) E36 counts

 

good so we have a handle on the BH evaporation time

[Martin]

jedi units are based on the Force F_Eins that is the central coefficient in the Einstein Equation of Gen Rel. the constant F that relates curvature tensor to energy density (stress energy tensor) see the first couple of posts in this thread about that.

this is the constant F that we make be a power of ten, in defining the units.

 

that is why lots of formulas, including the formula for the evaporation time of a black hole, simplify

 

here is the evap. time formula using F = F_Eins

[math] \text{evaporation time} = \frac{80}{\pi} \frac{c^4}{\hbar F^2} \text{mass}^3[/math]

 

the values of F, hbar, and c in our units are E43, E-32, and E9 so a lot of powers of ten cancel and we end up with a simple formula in Jedi units

 

[math] \text{evaporation time} = \frac{80}{\pi} 10^{-18} \text{mass}^3[/math]

 

If you dont use is the Force constant and write the formula using G instead then it is more complicated. We saw what it looks like before.

 

[math] \text{evaporation time} = \frac{5120 \pi G^2}{\hbar c^4} \text{mass}^3[/math]

 

It is the same formula just looks a little trimmer if you use F_Eins instead of G.

 

Here is the example I worked earlier:

so our unit of mass is pound and our time is a count

if I plug in the mass in pounds here i will get the lifetime in counts

So let us say the mass of the black hole is E18 pounds

the lifetime will come out to

(80/pi) E36 counts

[Martin]

Some gypsies were roaming this part of the galaxy and on their way thru the solar system they stole the moon

and replaced it by a black hole of the same mass

 

Dont worry, said the gypsies (when the people complained) you will still have tides and everything will be the same because the black hole we put in has the same mass----1.7E23 pounds---as the moon.

 

the people appointed John Baez the noted explorer to negotiate.

 

Give us back our moon, Baez told the gypsies, this black hole you gave us will eventually evaporate. It is not a fair exchange, you have cheated us.

 

the gypsies thought about it some. then they said "All right, we will give you one chance. If you can tell us how long the black hole will take to evaporate we will swap the moon in for the black hole and take our black hole away with us!"

 

John baez cubed 1.7E23 to get 4.9E69

 

then he multiplied that by (80/pi)E-18 to get 125E51

 

he knew gypsies like to use Jedi units so he said "125 x 10⁵¹ counts"

 

Very good, said one of the gypsies, but you are from Earth and measure time in minutes. Tell us the time in minutes so we can be sure you really understand.

 

john baez took 1250 x 10⁵⁰ counts and divided by 222 and said

"5.6 x 10⁵⁰ minutes"

 

All right said the gypsies and they swapped the familiar moon back into its orbit

[ydoaPs]

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