J*di Units: the F*rce is One

[Martin]

Different systems of units give a different perspective or handle on nature. You need experience with more than one system, and with converting.

Different specialties in physics and astronomy use their own units of convenience.

 

even the electron volt, which you see everywhere, is not a metric unit.

the metric unit of energy is the JOULE and the eevee is equal to

1.602176...x 10^-19 joule.

 

physicists are shamelessly uninhibited about using non-officially-sanctioned units. they will measure temperature and even mass in eevee.

(instead of in Kelvin and Kilograms). And of course astronomers have their Lightyears and their Megaparsecs. How many kilometers do you suppose...

 

So the real world is a dogs dinner unitwise and the one thing you can always use practice at is converting. Sorry

 

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

This thread is not about converting. We wont do conversions to metric most of the time. It is a chance to test-drive a set of practical size units (J*di) in which the MAIN NATURAL CONSTANTS ARE POWERS OF TEN.

the system aims to be more decimal than the metric system (which was established around 1790 when they didnt know from Planck's constant and hadnt realized how important the speed of light is.)

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

 

This thread was wiped in the newyears servertrouble Event, and I am going to try to reconstruct it.

 

I will define J*di units, and, in case you want to try them out, I will invent some easy basic physics exercises that use the units.

 

These exercises will involve a few major natural constants, like the speed of light and Planck's hbar, because those are the things which the system makes to be exact powers of ten---and I want to exhibit that feature.

[Ophiolite]

As someone raised on linear measures of rods poles and perches, and monetary units of florins, crowns, tanners and bobs, I think can handle that.

[Martin]

the premise of J*di units is that the coefficient in the 1915 Einstein equation is a real force. Or the reciprocal of a force.

 

the einstein equation is our main model of how gravity works and how matter curves spacetime. It says that if you divide the energy density by the force, you get the curvature. What makes space curved is the density of energy in a region and the way to find the resulting curvature is to take that energy density and divide by a certain universal constant force.

 

curvature is measured as a reciprocal area and it is a fact of life that if you divide any energy density (energy per unit volume) by a force you get one over some area.

 

[math]\text{curvature} = \frac{1}{F_{Eins}}\text{energy density}[/math]

 

the curvature is actually a tensor made up of a lot of curvatures in several directions and written G_ab and the energy density is also a tensor made up of lots of terms which are all equivalent to energy densities, and it is called

T_ab

So naturally the equation looks like this:

[math]G_{ab} = \frac{1}{F_{Eins}}T_{ab}[/math]

 

the thing to keep your eye on is the central coefficient because that is one over the Einstein force and it is at the heart of how gravity works and how matter and energy curves space and how the shape of the universe evolves.

 

this force F_Eins is about 50 trillion trillion trillion tonnes of force.

 

what that means is that what we think of as a lot of concentrated mass-energy only curves space a little bit (by our standards) because to get the curvature you are dividing by a force which is large (by our standards).

 

We can write F_Eins in terms of the newtonian gravity constant G_Newt and the speed of light.

[math]F_{Eins} = \frac{c^4}{8\pi G_{Newt}}[/math]

 

If you like using a calculator and know the speed of light and that in metric terms you can work it out and you will get some big number of metric newton force units, basically amounting to 50E36 tonnes force.

 

So now let's put that into the Einstein equation to see what it looks like in the textbooks:

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

 

One last shocker, the Relativists, the professionals who do General Relativity for a living, use their own private convenience system of

non-metric units in which the value of this force is ONE, so the coefficient in the Einstein equation completely disappears, and in their books and journals the equation what you often see is simply:

[math]G_{ab} = T_{ab}[/math]

 

Shall we do like them, and make that be our unit of force?

[Martin]

As someone raised on linear measures of rods poles and perches, and monetary units of florins, crowns, tanners and bobs, I think can handle that.

 

Good!

 

actually physics majors used to get several systems even in the first couple of undergrad years, I think there has been a trend towards uniformity in some senses for better in others for worse.

but here we are just doing a trial run, where we take a squint at the world thru alternate specs and then its back to normal.

 

Ophi, I think you glanced at the earlier version of this thread and so you know that it is not my intention to make this huge force constant have value One! So that we can get some practical humanscale units to do exercises with, I am just going to set the values of the main physical constants equal to POWERS OF TEN.

 

Our first three basic units, in the J*di system, are the mark of force, the handbreadth, and the count and they are determined by these three equations----since the natural constants are definite quantities we can define units in terms of them:

 

[math]F_{Eins} = 10^{43} \text{ marks}[/math]

[math]\hbar = 10^{-32} \text{ mark hand count}[/math]

[math]c = 10^{9} \text{ hand per count}[/math]

 

that makes the mark force unit come out to be half a newton (a couple of ounces)

it makes the handbreadth come out to be 8.1 centimeter (about 3 and 1/4 inch)

it makes the count unit of duration come out 222 to the minute.

 

We can give more precise metric equivalents for these things but we normally have no need for metric equivalents---wont be depending on them. And from these three basic units a lot of others derive:

the mass unit (pound-size, about 434 gram)

the energy unit (I call it "jot" since it's small, about 1/25 joule or 1/100 calorie)

the area unit (sq. hand, about 65.65 sq. cm.)

the volume unit (pint-size, about 532 cub. cm.)

the acceleration unit (about 1.11 meter per sq. second)

 

Besides what derives from just those three basic units, we can add two more equations and complete the job----by assigning power-of-ten values to the elementary charge and the Boltzmann temperature constant k.

So there is this list of 5 equations that defines all our units, for pretty much every type of physical quantity. Recall that we are calling the energy quantity (the mark hand----a unit force push for unit distance) by the name "jot".

 

[math]F_{Eins} = 10^{43} \text{ marks}[/math]

[math]\hbar = 10^{-32} \text{ jot count}[/math]

[math]c = 10^{9} \text{ hand per count}[/math]

[math]e = 10^{-18} \text{ dram}[/math]

[math]k = 10^{-22} \text{ jot per degree}[/math]

 

The dram unit charge turns out to be about 1/6 of a metric coulomb.

The voltage unit (jot per dram) deriving from it turns out to be about 1/4 of an ordinary volt and gets called a "quartervolt" abbr. Q.

the microscopic energy unit eQ (eekyoo, electron quartervolt) is exactly 10^-18 of the macroscopic energy unit "jot".

hbar and k can be expressed as powers of ten in terms of eQ.

 

[math]\hbar = 10^{-14} \text{ eQ count}[/math]

[math]\hbar \times c = 10 \text{ eQ microhand}[/math]

[math]k = 10^{-4} \text{ eQ per degree}[/math]

 

the relations between energies and wavelengths and frequencies and temperatures become very easy to follow, I will try to illustrate later.

 

One thing you really need to know that is not a simple power of ten is that the proton mass is 1/(2.6E26) of a pound. It is quite a useful fact. Our j*di pound mass unit (434 g) if it was made of protons, would consist of 2.6E26 of them.

 

Now these are kind of dry facts and it would probably help if I sketched some exercises using the units. It is really nice that the natural constants come out so often to be powers of ten---a big laborsaver--so the units can be fun to use.

 

Another thing it helps to know is that on an absolute scale, with J*di degrees, it turns out that 1000 degrees is the temperature conventionally known as 282.6 kelvin, which is right about Fahrenheit 49. So that 49F makes a good conceptual anchor to refer temperatures too-----you have room temp 1040 and body temp 1100 and a moderate (baking) oven 1600.

The temp scale takes a while to get used to because the temperatures always seem to be 1000 more than you think they should be, and because the J*di degree is only half a familiar Fahrenheit. Oh, by serendipity the estimated temperature of the sun's surface is right around 20,000.

So kT = 2 eQ (remember that k = 10^-4 eQ per degree) and we get a quick handle on the photon energies in sunlight and the visible colors.

Photon energies convert to (angular) wavelength by the earlier equation

[math]\hbar \times c = 10 \text{ eQ microhand}[/math]

 

so, for example, the wavelength for green is 1 microhand and the green photon has an energy of 10 eQ. A lot of serendipity showed up. now I really should shift gears. Ophi, comments would be really welcome pos or neg max nix. wd help to guide the exposition

[Martin]

here's an exercise for swansont

 

Swansont dreams that he is flying in a spacecraft in low orbit around a strange planet. His wrist feels a bit more encumbered than usual and he notices that he is wearing two watches---one measures minutes and the other measures time in counts (222 to the minute)

 

After a while he determines that it takes him 31.5 minutes to go (1/2pi) of the way around the planet. WHAT IS THE PLANET'S DENSITY?

[ed84c]

http://www.amazon.co.uk/exec/obidos/ASIN/0471125806/qid=1104767343/sr=1-3/ref=sr_1_8_3/202-7774367-3170234

 

No offence but Clifford Pickover does a way better explanation, but hes probably had more practice.

[Ophiolite]

But he charges £12.64 and he isn't one of us..

 

(I'm still digesting your exposition Martin.)

[Martin]

...

No offence but Clifford Pickover does a way better explanation' date=' but hes probably had more practice. [/quote']

 

Hi Ed,

yes an admirable character and a great explainer by all signs!

I have to go largely on other people's report 'cause I havent read his book

but I will copy the synopsis here just for definiteness:

 

Black Holes: A Traveler's Guide

 

"Black holes are the much publicized 'vacuum cleaners' of space that have such a strong gravitational pull that they suck in everything that comes too near to them, and from which nothing can escape. Even light rays are unable to pass through them which is why they appear black. Black holes are not solid objects, they are areas in space of extremely high gravity. Astrophysicists have recently theorized that black holes may actually serve as tunnels in space through which - if we could travel through them - we could enter other universes that exist parallel to our own. Clifford Pickover creates two fictional "Scientists of the Future" who travel to a black hole and perform a series of experiments designed to reveal all of the intriguing properties of black holes, such as experiments to see how close an object can get to a black hole without being sucked in, to show what would happen to an object that did get sucked in, and to see whether it would be possible to travel through a black hole. This introduction to the remarkable physics of black holes explains exactly what they are, at a level accessible to the popular science audience."

 

"CLIFFORD A. PICKOVER, PhD, (Yorktown Heights, New York) is a research scientist at the IBM T.J. Watson Research Center. He is the primary author of the "Brain Boggler" column in Discover magazine and author of the acclaimed Chaos in Wonderland, Mazes for the Mind, and Keys to Infinity. He is also an award-winning computer artist."

 

that's a whole other league, isn't it

 

At this point I am just trying to connect with a Freshman Physics fact that you may very well remember: If 2pi T is the low orbit period of a round planet, then multiply (4pi/3)G by T² and it tells you the DENSITY.

 

If you had freshman physics then you know that the units of G are

cubic meter per sq. second per kg

 

so if you multiply G by the square of a time, what you get is a quantity

cubic meters per kg

that is a reciprocal density

 

in this case flipping it over tells you the density of the planet.

[Martin]

...

 

(I'm still digesting your exposition Martin.)

 

Ophi, thanks for reading! Hope you are not disappointed but please to point out any flaws or shortcomings.

 

I am particularly concerned about the level of the exercises. I want them to be interesting but not at all hard.

 

I am going to assume that you (and other readers) know that associated with each temperature T there is a characteristic amount of energy kT

that shows up in all sorts of phenomena at that temp----like the kinetic energy of thermal motion, like photon energy in the thermal glow, at temp T.

 

Now this may be unfamiliar but it is so beautiful that if somebody doesnt know it then I can honestly say he SHOULD.

 

In the thermal glow at temp T, the AVERAGE photon energy is 2.701 kT

 

and this number 2.701 comes from the Riemann zeta function!!!

http://mathworld.wolfram.com/RiemannZetaFunction.html

scroll 3/5 of the way down the page to equation (53) and (54)

and you will see what zeta (4) and zeta(3) are. this number 2.701

is 3zeta(4)/zeta(3).

 

well, forget the zeta function, and just take the fact that in any system of units the average thermal photon has an energy of 2.701 kT.

 

HOW MANY PHOTONS PER CUBIC METER ARE IN THE ROOM WITH YOU RIGHT NOW?

 

Let's try that as a sample exercise----our first one except for the one where swansont is skimming over the surface of the strange planet

[Martin]

This thread doesnt have to be totally devoted to trying out the alternative units!

 

We can work things out in metric too, if people prefer.

 

Like the number of photons with you in the room, per unit volume.

 

There is a beautiful fact in physics that the ENERGY DENSITY in an empty space at temp T is equal to

 

[math]\frac{\pi^2}{15} \frac{k^4 T^4}{\hbar^3 c^3}[/math]

 

this would work in metric, if you knew the metric values of all the constants, and it would give joules per cubic meter.

 

It also works in our units, where the constants are the powers of ten I mentioned earlier.

 

Now to find the NUMBER OF PHOTONS (per unit volume) all we do is

divide the energy density by the average energy per photon at temp T,

which is equal to 2.701 kT.

 

Then the number of photons per cubic meter that is in the room with you is given by

 

[math]\frac{1}{2.701} \frac{\pi^2}{15} \frac{k^3 T^3}{\hbar^3 c^3}[/math]

 

We can work this out in metric using a comfortable T = 294 kelvin room temperature. Or we can work it out in "Jedi" using T = 1040 degrees, which is the same temperature.

 

doing it in "Jedi" will make it come out some number of photons per cubic hand. There are 1880 cubic hands in a cubic meter, so we can multiply by 1880 to change the answer into the number per cubic meter if we want.

 

the calculation is ridiculously easy in Jedi because the powers of ten cancel and we just get 10

 

[math]\frac{k}{\hbar c} = 10[/math]

 

You just multiply the temp (1040) by ten and cube, and the number per unit volume is:

 

[math]\frac{1}{2.701}\times \frac{\pi^2}{15} \times 10400^3[/math]

 

then if you want it to be in cubic meters you just multiply by 1880

 

WAIT! I was supposed to offer some exercises! I have gone and done one rather than letting people have a go first.

[ed84c]

I made these spreadsheets a while ago based on Cliffovers work see what you think the may help;

 

Black Holes Mass; http://sfn-download.biz.ly/Mass%20of%20a%20black%20hole%20SPU.xls

 

Black Holes Evaporating Time;

 

http://sfn-download.biz.ly/Evaporating%20Black%20Holes%20SPU.xls

[5614]

ed84c, links dont work:

 

"Forbidden

Remote Host: [iP address here]

 

You do not have permission to access http://sfn-download.biz.ly/Mass%20of%20a%20black%20hole%20SPU.xls

Data files must be stored on the same site they are linked from. "

[ed84c]

Here We are

 

http://sfn-download.biz.ly/Index.htm

 

Both are there, thanks for you patients

[Martin]

I made these spreadsheets a while ago based on Cliffovers work see what you think the may help;

 

Black Holes Mass; http://sfn-download.biz.ly/Mass%20of%20a%20black%20hole%20SPU.xls

...

 

thanks for the contributions Ed!

 

as far as relating the SIZE of a black hole to its MASS' date=' in J*di units, here is how it goes (in case anyone is interested)

 

for the usual kind of Schw. black hole you take the diameter

and multiply by 6.28 x 10[sup']25[/sup]

 

If you have been reading this thread you know our handbreadth unit and our pound unit

 

If the diameter of the black hole is 1 hand, then the mass is 6.28 x 10²⁵ pounds

 

diameter           mass
1 hand         6.28 x 10[sup]25[/sup] pounds
2 hands      12.56 x 10[sup]25[/sup] pounds
10 hands      62.8 x 10[sup]25[/sup] pounds
20 hands     125.6 x 10[sup]25[/sup] pounds

 

20 hands is right about 5 conventional feet

and 20,000 hands is about a mile

so if you just picture the size of black hole you have in mind

it is not hard to figure out its diameter in handbreadths (they are 8.1 cm)

and then it is easy to find the mass in pounds

 

I should put in a physics exercise about black holes.

[ed84c]

Did you see the spreadsheets that i had made Before making that post?, and if so how, i only fixed the links after you made the post.

[Martin]

While Ed is fixing the links to his spreadsheets I will just mention how to calculate the temperature of a black hole from its mass

 

in terms of our units (so I dont have to write them in) it is real simple:

 

[math] T = \frac{10^{24}}{M}[/math]

 

Mass         Temperature of BH with that mass

10[sup]24[/sup] pounds        1 degree
10[sup]23[/sup] pounds        10 degrees
10[sup]22[/sup] pounds        100 degrees
10[sup]21[/sup] pounds        1000 degrees
10[sup]20[/sup] pounds        10000 degrees

 

on our absolute temp scale, 1000 is about average surface temperature on earth, more exactly it is 282.6 kelvin or some 49 Fahrenheit.

 

and 10,000 is the temperature of the tungsten filament in an ordinary 100 watt lightbulb (approx. half the surface temperature of the sun)

 

so if you want a hole that glows like a lightbulb you know how many pounds it has to be----E20 pounds

and if you want one that is about yearreound average earth surface temperature then it has to be E21 pounds

 

what i have done is implement the Bekenstein Hawking BH temp formula

(circa 1976 IIRC) in Jedi units

 

Ed is getting us some links about BH evaporation and I guess you could say this is related. they evaporate because they are hot and radiate and the more they radiate the smaller they get and the smaller they get the hotter they get...

[ed84c]

Ive fixed the damn links

[Martin]

Did you see the spreadsheets that i had made Before making that post?, and if so how, i only fixed the links after you made the post.

 

No Ed, i havent tried your links yet. just going on general knowledge.

I have to skedaddle for the moment so cant look at the spreads right now but will later. thanks for contributing them! makes the thread more interesting.

[5614]

why are we know using J*di instead of Jedi ?

[ed84c]

I prosumed it was incase the other thread came back so there wouldnt be two of the same name.

[Martin]

why are we now using J*di instead of Jedi ?

 

Ed mentioned a good reason, but I was also wondering if it was fair

to George Lukas to adapt a name he dreamed up----not really a legal issue since our units aren't commercial but one of not wanting to seem too pushy.

 

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, however. I am going to make a table relating the DIAMETER of a hole (in handbreadths) to its TEMPERATURE (in our degrees). I guess the sizes of the holes will sometimes be in terms of

thousandths or millionths of a hand. because they have to be small in order to have any decent recognizable temperature

 

a hand is 8.1 centimeter

so a microhand would be 8.1 x 10^-8 meter = 81 nm.

 

Assuming it's in our units, so I dont have to write them, the formula for the diameter would be

 

[math]D = \frac{1}{62.8 T}[/math]

 

So if you want a hole with temperture of ONE DEGREE then the diameter of the hole is going to be 1/62.8 of a handbreadth. Real small hole already.

 

temperature       diameter
0.1 degree        1/6.28 hand
1 degree           1/62.8 hand
10 degrees         1/628 hand
100 degrees        1/6280 hand
1000 degrees      15.9 microhand

 

I have to say that holes have to be real real small before they get interestingly hot----or even by my standards warm.

 

maybe I should solve that exercise where swansont was in low orbit around a strange planet

[Martin]

here's an exercise for swansont

 

Swansont dreams that he is flying in a spacecraft in low orbit around a strange planet. His wrist feels a bit more encumbered than usual and he notices that he is wearing two watches---one measures minutes and the other measures time in counts (222 to the minute)

 

After a while he determines that it takes him 31.5 minutes to go (1/2pi) of the way around the planet. WHAT IS THE PLANET'S DENSITY?

 

 

One of the bad things about J*di units is that the normal density of water comes out to around 60/49 instead of one!!!!

Well that's the breaks, we have all these constants (c, hbar, the Force, k, the electron charge, and combinations of them....) be exact powers of ten and you cant have everything.

 

our cubic hand is pintsize but it is a BIG pint (532 cc) and our pound mass unit is poundsize but it is a SMALLISH pound (434 g) so our unit density is LESS than a gram per cc, and so, in our terms, the density of water is more than one pound per cubic hand!!!!

 

Well as it says in the hitchhiker's guide, dont panic. We can live with this.

Water density is around 1.225 pd per cubic hd.

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

 

It's a fact about round planets that the minimal time T_rad that it takes a satellite in low orbit to go a radian tells you the density. this works in any system of units

 

[math] \frac{4\pi}{3}G (T_{rad})^2 = \text{reciprocal density}[/math]

 

So if you like metric you can convert 31.5 minutes to some seconds, and square that, and multiply by G (if you know what the metric value of G is) and then by 4 pi/3. And that will give the reciprocal density of the planet, some number of cubic meters per kilogram.

 

In our case we measure time in counts---222 counts to the minute---and 31.5 minutes is the same as 7000 counts. There is some cancelation and the constants boil down to something pretty simple.

 

[math] \frac{10^{-7}}{6} 7000^2 = \frac{49}{60} \text{ cubic hand per pound}= \text{reciprocal density}[/math]

 

So swansont looks at his watch and sees that the radiantime is 7000 counts and he squares that and multiplies by (E-7)/6 and that's the recipr. density 49/60, so he knows that the density of the planet is 60/49 which is the density of WATER.

 

If he happens to have his bathing suit on he can jump out of the spacecraft and go swimming. the whole planet is this clear glistening quivering big drop of water. Maybe. Or maybe it is something else that has the same average density. Since he is very close to the surface and can look, we leave that for him to decide.

[Martin]

if you work in any system of units, with any absolute temp scale, you are going to get the same answer that there is a factor of TWENTY

between the sun surface temperature and the yearround all latitudes average surface temperature on earth

 

It just happens that in J*di units we have this Fahrenhalf degree and the absolute temp of the sun is 20,000

and the average earth surface temp is 1000

that is serendipity, that it worked out nice, so you see the factor of 20 right away

 

well this factor of 20 is basic and important to life etc so lets try to understand it and relate it to data, like the apparent size of the sun in the sky

 

this is Swiss Army Knife physics----how much can you figure out about the world with the barest minimum of data besides what you see with your own eyes.

 

THE ANGULAR SIZE OF THE SUN IS 1/107 of a radian-----I know, it is about half a degree of angle but i want it in radians of angle. the more accurate number is 107.5 and that is something that in principle one could measure. It is just how big an angle the sun makes in the sky----how big it looks.

 

So that means the distance to the sun is 215 TIMES THE SUN'S RADIUS

 

Now the square root of 215 is FOURTEEN AND 2/3, and I am going to tell you a StefanBoltzmann fact. This is a very nice fact. Because the sqrt 215 is 14 and 2/3

the equilibrium temperature of an unreflective flat onesided surface facing the sun will be the sun temperature divided by 14 and 2/3

 

I am going to explain that, it is because of the nice StefanBoltzmann fourthpower law. the flat surface has to get hot enough to radiate away just as much power as is coming in, and that is less than the sun surface is radiating by a factor of 215² and the fourth root of that is 14.66 so the temperature is less by that factor.

 

Maybe I will have to rewrite this to make it clearer.

 

In our units, just to be concrete, sun temp is 20,000 and the hottest a sidewalk (facing the sun with no filtering atmosphere) can get is nominally 1364 which is 20,000 divided by sqrt 215.

 

boiling is 1320 so this sidewalk gets hotter than boiling.

 

Now why doesnt the whole planet get this hot? well to a large extent it is

BECAUSE A BALL HAS 4 TIMES MORE SURFACE THAN ITS SILHOUETTE.

So it has more surface to radiate away energy than its cross-section for receiving sunlight energy. 4 pi R² ball surface versus pi R² cross-section silhouette---factor of 4

 

An average square meter of earth surface only has to get hot enough to radiate away

 

[math]\frac{1}{4}\times \frac{1}{215^2}[/math]

 

as much power as a square meter of sun surface is sending out.

 

the 1/4 is because earth is a ball and the 1/215² is because we are far away from sun.

 

So by the fourthpower radiation law we have to take the fourth root of that, which will tell us by what factor the equilibrium ball temperature is less.

 

Obviously it is going to be sqrt 430. except for some details that is where the factor of 20 comes from

 

to be concrete I will say it in J*di degrees, sun is 20,000 degrees

and ball equilibrium at earth distance from sun is 20,000 divided by

sqrt 430

which turns out to be around 964 degrees

 

NOW WE ARE IN BALLPARK because we know that the actual earth average is right close to 1000 degrees.

 

Why is it a little bit warmer here on the planet than the theoretical ball equilibrium?

 

Because of the greenhouse. there are two effects, albedo and greenhouse and they almost cancel but greenhouse barely wins and brings it up from 964 to 1000

albedo is the tops of the clouds reflect away some light and less light gets to the surface, so that would make it colder than ball equilibrium

but greenhouse blocks some radiation loss and makes it warmer and it balances out----at least for the time being---at what seems a very desirable temperature

[ed84c]

Heres an exciting graph i did of a black hole evaporating.

 

[ed84c]

Any of yuou computer wizz's know how i can make a programme just for the fun of watching a black hole's mass drop exponentially?