How Can Alternative Units Systems Enhance Your Understanding of Physics?

[marcus]

It can be useful to construct and try out alternative systems of units---mind-broadening and you can get a different perspective on physics.
And in several specialties they do this anyway because it is almost a precondition to being able to think. Whether QFT or Gen Rel they all have their own bastardized units and alternatives to straight metric that help them get a grip on things.

for instance in Gen Rel the main equation is the Einstein equation
G_{ab} = T_{ab}

basically curvature equals energy density, the concentration of matter curves space
When you see that equation you know that the person writing it is using a set of units in which the UNIT OF FORCE IS 500 TRILLION TRILLION TRILLION TONNES. (I am expressing it in the unofficial metric tonnes force because saying it in the official Newtons unit would be even worse.)

If you put everything back into the Einstein equation it says this

G_{ab} =\frac{8\pi G_{Newt}}{c^4} T_{ab}

And the central coefficient in the Einstein equation is the reciprocal of this huge force

F_{Einst} =\frac{c^4}{8\pi G_{Newt}}

and the relativist writing the equation adjusts his units to make that huge force unity, as well as making the speed of light unity, so he can streamline the equations and think conveniently.

The thing to remember is that just for dimensional reasons if you divide any energy density by a force you get a reciprocal area which is to say a CURVATURE. so to express the curving effect of a concentration of energy you have to divide by some force. the one Nature uses is this universal constant force F_Einst and that force is just built into how gravity works and how the universe is shaped.

So let's try taking this force seriously and constructing humanscale practical units that are in harmony with it, or easy power of ten relation to it-----and the same for other basic constants like hbar and c.

We can set up a system like that and try working some simple exercises to see how it goes. The exercises will be conventional vintage phyics and
the majority of them will be at the ATOMS AND MOLECULES scale, or larger, I guess. I will have some exercises about temperature, the speed of sound, electrostatic force between point charges, blackbody radiation,
so I guess the Atoms Molecules Solids forum is the right place for it.

 

[marcus]

so here's how the construction goes
we decide on power-of-ten sizes for three basic quantities
|F_Einst| = 10⁴³
|hbar| = 10^-32
|c| = 10⁹

Deciding that the constants will have those sizes (in our units) completely determines what our units of force, length, and time must be---and all the other units which can be derived from them.

What results is a length unit that is about one handbreadth
and unit of duration that is about 222 to the minute. I will call it a count because that is as fast as I can count (to twenty repeatedly, not all the way to 222!) And the unit force, which I call a mark of force just to have a name for it, and which must be 10^-43 of the great constant force, comes out to two ounces, or half a metric Newton.

F_Einst = 10⁴³ marks
hbar = 10^-32 mark hand count
c = 10⁹ hand per count

We will not ordinarily need to have metric equivalents because we will be able to work exercises in terms of these units, so only a rough idea of their size is really needed. But here at the beginning I will give multidigit equivalents, which afterwards we shall mostly ignore. BTW the derived energy unit, mark hand---pushing with mark force for unit distance---is about 1/100 of a calorie or 1/25 of metric joule. I will call it a jot.
The unit of mass is pound-size and will be called pound. The unit of volume, cubic hand, is pint-size.

hand 8.10263 centimeters (3 and 1/4 inch, obviously a handwidth)
count 0.270275 second
mark 0.4816 Newton
jot 0.039018 joule
pound 434.14 grams
sq. hand 65.65 sq. centimeters
cubic hand 532 cubic cm.
mark per sq.hand 73 pascal
hand per sq. count 1.11 meter per sq. second

for a bit more definition we set sizes for the elementary charge e and the Boltzmann constant k:

|k| = 10^-22
|e| = 10^-18

that makes the temperature step be half a Fahrenheit step
1000 degrees absolute equals 49 F on the conventional fahrenheit scale
the voltage unit is called a quartervolt (abbr. Q) and the microscopic energy unit eQ (eekyoo, electronquartervolt) is exactly 10^-18 of a jot, the macroscopic one.

the relation between photon energy and vacuum (angular) wavelength is given by
\hbar \times c = 10^{-5} \text{ eQ hand} = 10 \text{ eQ microhand}

As it happens a typical wavelength for green light is one microhand and the corresponding photon energy is 10 eQ

the relation between temperature and energy is given by

k = 10^{-4} \text{ eQ per degree}

It happens that a common estimate of the surface temp of the sun translates to just over 20,000 degrees. So kT for the solar surface is 2 eQ.

as an illustration, the average photon energy in thermal radiation from a surface at temp T is (in any system of units) 2.701 kT

So the average photon in sunlight has an energy of 5.4 eQ.

Now I hope very much that I have not frightened everybody off by this preliminary flood of information. Let's see what simple exercises we can construct, using these units.

 

[marcus]

Just had an example of calculating with different units
https://www.physicsforums.com/showthread.php?p=416892#post416892

It looks to me like Daniel (dextercioby) used the metric system and had a pretty complicated calculation and his answer is (IMHO) a little off.
https://www.physicsforums.com/showthread.php?p=416755&posted=1#post416755

We were calculating the number of photons per cubic meter at room temp

for me, on absolute scale, room temp is 1040

so I go:

\frac{1}{2.701}\times \frac{\pi^2}{15} \times 10400^3

that gives me 2.740E11 which is the number in a pintsize cubic hand volume.
Now I have to get back to metric, so I multiply by 1880 because there are that many cubic hands in a cubic meter.
that gives me 5.15 E14, which I think is right.

But dextercioby, as you see in that thread, has a rather complicated calculation and comes up with 6E14, which I think is mistaken. It is easy to make a mistake because he has messy metric constants like hbar, and k.

In our units this ratio of units k/(hbar c) = 10 exactly.
So I have multiplied the room temperature 1040 by 10 before cubing.
I have no more trouble with constants.
but he is using messy metric versions so he must look up the values and calculate some messy number for k/(hbar c). It will not be a simple 10!
Or he has to do some equivalent bother at some other point in the calculation.

We have room for a slight disagreement about room temp. The conventional metric figure is 293 kelvin. But I find 293 slightly too cold for comfort and prefer 294----this is why I used 1040.

To make everything comparable to dextercioby I should use 1037 degrees.

\frac{1}{2.701}\times \frac{\pi^2}{15} \times 10370^3

then I get 5.11E14 instead of 5.15E14. So it does not make much difference. But just to be careful.

------footnotes-------

the number 2.701 is interesting, it comes from the zeta function

2.701 = \frac{3\zeta(4)}{\zeta(3)}

in thermal glow of temp T the average photon energy is 2.701 kT.

the main formula i used, if you put back in the k, hbar, and c, would be:

\frac{1}{2.701}\times \frac{\pi^2}{15} \times( \frac{kT}{\hbar c})^3

but since k/(hbar c) has the value 10, one can simply multiply the temp by 10 and cube, which is what i did.

You might want to look at dextercioby calculation in the other thread.
It is very professional and comparatively elaborate. the difference is instructive about the system of units and the associated way of thinking.
https://www.physicsforums.com/showthread.php?p=416755&posted=1#post416755

 

[marcus]

Just as a review, here is how the construction goes.
there is this force in nature F_Eins which is basic to how gravity works and how space is shaped. And there's a basic energy-time quantity hbar, and a basic speed c, and a basic charge e. We set these things equal to some powers of ten. And also the Boltzmann k which relates the temperature scale to the energy scale.

F_Eins = 10⁴³ marks
hbar = 10^-32 mark hand count
c = 10⁹ hand per count
e = 10^-18 dram
k = 10^-22 mark hand per degree

the first equation defines the mark force unit. the next two define the hand and count (units of length and duration). the last two define the units of electric charge and temperature. The other types of units can be derived from these and are thus also determined.

The derived energy unit, mark hand---pushing with mark force for unit distance---is about 1/100 of a calorie or 1/25 of metric joule. I will call it a jot. The derived unit of mass is pound-size and will be called pound. The unit of volume, cubic hand, is pint-size. The unit of electric charge is about one sixth of a metric coulomb, and will be called a dram. The voltage unit deriving from this is about one quarter of a conventional volt and will be called a quartervolt, abbr. Q.

The temperature scale one gets from these assignments is an absolute scale with 1000 degrees being the same as 282.6 kelvin. It happens that 282.6 kelvin is close to 49 Fahrenheit. We take that temperature as a point of reference. the degree is about half a Fahrenheit step.

some temperature benchmarks:
1000 approx average Earth surface temp (serendipitous)
1040 room temperature
1100 approx. normal body temp
1320 boiling (at normal atmospheric pressure)
1600 moderate oven (350 for Fahrenheit cooks)
20000 approx. surface temp of sun

it is admittedly awkward not to have a relative scale like Celsius with a special zero at freezing. But this is an absolute scale and as such will have to do.

 

[marcus]

a little about speed and the speed of sound in cold air

the unit speed, 1 hand per count, is a billionth of the speed of light, so it is 2/3 mph (for people who think miles per hour) and 0.3 meters per second (for those who dont)

10 hand per count-----6.7 mph
100 hand per count---67 mph
1000 hand per count--670 mph, which is the speed of sound at typical aircraft cruising altitudes
10000 hand per count--earth orbit speed (30 km per second)

I want to make a point about the speed of sound and the weight of a proton.

In our units we have a mass unit called pound which is 434 grams (determined by the values given the constants earlier) and that means that protons are 2.6E26 to the pound. this is an important and useful number and it's lucky that it is easy to remember (because of a chance numerical "rhyme").

That means that because air molecules, on average, have atomic weight 29, the air molecule mass is
\frac{29}{2.6E26} \text{ pound}

The speed of sound in air is:
\sqrt{\frac{7}{5}\times \frac{kT}{ \text{mass of molecule} }

It is neat how it depends on very little besides the temperature of the air. And let's do a sample calculation for very cold air that has T = 800 degrees. That is actually a typical temperature 6 miles up, above the clouds and convection layer. One way to think of it is to say it is 200 degrees less than the reference temperature (1000 degrees = 49 Fahrenheit) and that those 200 degrees are each about half the size of F steps.

Or you can just multiply 800 by 0.2826 and get kelvin.

anyway, remember that in our units the value of k is E-22, so that for T = 800 we have kT = 8E-20, and let's calculate the speed

\sqrt{\frac{7}{5}\times \frac{8E-20 \times 2.6E26}{ 29} }= \sqrt{\frac{7}{5}\times \frac{8 \times 2.6\times 10^6}{ 29} }

and that turns out to be 1000 hands per count----a millionth of the speed of light

 

[marcus]

talking to dextercioby on the other thread, a kind of nice problem came to mind
I guess I thought of it but it certainly helps to have someone to converse with who takes casual dares and calculates stuff!

the problem is this
how massive should a black hole be for it to glow with green light---that is for the Bekenstein Hawking temperature T_BH
to be such that 2.701 k T_BH is the energy of a green photon.

that would make green photons average in the hawking glow of the hole.

well, in our system we have

hbar x c = 10 eQ microhand
and the ang. wavelength of green light is 1 microhand and the
photon energy of green light is 10 eQ (ten electronquartervolts)

and the eQ is exactly 10^-18 of the macroscopic energy unit (provisionally called "jot" because small, like 1/25 of a joule)

So I have to get the hole's mass M right so that the average photon energy is 10^-17

Here's what i calculated on the other thread:

IIRC the Bekenstein-Hawking temp formula (or perhaps it is solely due to Stephen Hawking?) is

kT = \frac{\hbar c^3}{8\pi G M}

now in the system I find most convenient the values of the constants are these powers of ten

|8 pi G| = 10^-7
|c| = 10⁹
|hbar| = 10^-32
|e| = 10^-18

kT = \frac{10^{-32} 10^{27}}{10^{-7} M} = \frac{100}{M}

the photon energy of green light is 10^-17 of the system's energy unit, so I must solve this for M

10^{-17} = 2.701 kT = 2.701 \frac{100}{M}

M = 2.701 \times 10^{17} \times 100

yes, so in terms of the system's mass unit it is 2.7E19,
or 27 continental trillion

and that will come to some 11 or 12 trillion kilograms

The main conclusion is that the mass of a green black hole is
2.701 x 10¹⁹ pounds

the reason for saying trillion is that in continental europe they say trillion for 10¹⁸

I am still amazed that the green black hole is so massive. I keep expecting someone to drop into this or the other thread and show me how I made some careless mistake. I am used to thinking of holes that are hot enough to glow visible as being very small.

I wonder how big this green black hole is?

 

[marcus]

the size of a green-glowing black hole

OK that can be another easy exercise. the aim of the thread is to test-drive the system by doing easy exercises.

We have a mass 2.701 E19 pounds
and I am going to calculate the Schwarzschild GIRTH of that mass.
the circumference, the 2 pi R of the thing.

It happens that in our system it is always (1/2) E-25 times the mass.

so we multiply 2.701 E19 X (1/2) E-25 = 1.35 E-6

wow!
that is 1.35 microhand
a microhand is the angular wavelength of green light!
the little mother is so small that it is down near the size of the wavelength.
How can it radiate green light at all?
Obviously there is a lot about this that I don't understand.
Well, maybe someone here can explain. I have just naively applied the famous BH temperature formula and got the mass that you need for the thing to be hot enough that the thermal radiation from it would be average green. It turns out that the mass was bigger than I imagined it would be and that the size is smaller than I expected and Hawking radiation turns out to be more of a puzzle than I had realized.

It is supposed to be regular old thermal radiation for the given temperature, but how can that be if the radiator is so small? Anybody want to help reconcile all this?

 

[marcus]

Part of a draft essay, maybe could use here:
You could say the premise of these 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.

\text{curvature} = \frac{1}{F_{Eins}}\text{energy density}

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:
G_{ab} = \frac{1}{F_{Eins}}T_{ab}

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.
F_{Eins} = \frac{c^4}{8\pi G_{Newt}}

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:
G_{ab} = \frac{8\pi G_{Newt}}{c^4}T_{ab}

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:
G_{ab} = T_{ab}

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

 

[Kea]

green cheesy smile

I wonder how big this green black hole is?

Marcus and Dex.

As I'm sure you know, using

kT = \frac{\hbar c^{3}}{8 \pi G M}

and substituting in Schwarzschild radius

r_{S} = \frac{2 G M}{c^{2}}

and then comparing to photon energy, one obtains

\frac{\lambda}{2 \pi r_{S}} = \frac{2}{2.701}

maybe modulo some factor of pi which is an easy mistake...so I'm
not surprised that the size of the hole matches the wavelength, no
matter what fancy units suitable for Martians that you may have
decided were nice...

Kea

 

[marcus]

Green cheesey smile

Marcus and Dex.
...
Kea

however these two smilies look like yellow butter balls, not green cheeses.

You could use

you have cleverly pointed out that ultimately units do not make any difference.

(but in various different specialties, physicists are always tinkering with units, to get cleaner equations of for whatever reason: maybe they tinker for no practical purpose, because they enjoy the activity, as parrots do)

Here is a problem for both of you, Kea and Dex:

John Baez in the spacecraft .

John Baez has been exploring the galaxy and on this occasion he discovers that he is in low circular orbit skimming over the surface of a small round planet which appears made of one uniform material.
After he has traveled one radian (1/2pi of the circumference) he looks at the ship's clock and sees that it has taken 31.5 minutes.

Question: What might the planet be made of?

Could it, for instance, be made of green cheese?

 

[marcus]

John Baez in the spacecraft

...
John Baez has been exploring the galaxy and on this occasion he discovers that he is in low circular orbit skimming over the surface of a small round planet which appears made of one uniform material.
After he has traveled one radian (1/2pi of the circumference) he looks at the ship's clock and sees that it has taken 31.5 minutes.

Question: What might the planet be made of?

Could it, for instance, be made of green cheese?

time units in our system are 222 to the minute, 31.5 minutes is 7000 counts.

In any system of units, metric included if you wish, if T is the minimal radian time of a low orbit then the reciprocal density of the planet is given by

\text{reciprocal density } = \frac{4\pi G}{3}T^2

In our case this is particularly simple to evaluate because 8 pi G is E-7, so we have the reciprocal density (in cubic hand per pound) given by

\text{reciprocal density } = \frac{10^{-7}}{6}T^2 = \frac{49}{60}

Baez sagely observes that the density of the planet is 60/49 or about 1.225
pounds per cubic hand, which is the density of water.

The planet is a great shimmering drop of water. As a Californian of course Baez has his swimming trunks handy, so after a moment's preparation he leaps from the spacecraft to enjoy a swim.

 

[Kea]

The planet is a great shimmering drop of water. As a Californian of course Baez has his swimming trunks handy, so after a moment's preparation he leaps from the spacecraft to enjoy a swim.

...which would be foolish, because I calculate that the density is around
1000 kg/m^3 (check please) which indicates water...except that correcting for the error in orbital radius would mean that the actual density was a little less...oops, no, more...maybe something like acetate...yuk (let us assume that John has checked the atmospheric conditions).

I assumed, as befits John Baez, that exploration of the galaxy requires a quantum gravitational spacecraft , which is of course not ever quite sure where it is going or where 'where' is ... and inadvertently John Baez has directed it to a planet which appears to made of (dense) green cheese with probability close to 1 on the large number of observations that John Baez's spacecraft has already made of the surface.

Use

\frac{G M}{r^{2}} = \frac{v^{2}}{r} [/itex]<br /> <br /> and the fact that the volume of a sphere is <br /> <br /> \frac{4}{3} \pi r^{3}<br /> <br /> to derive Marcus&#039; formula.<br /> <br /> <br /> <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/approve.gif" class="smilie" loading="lazy" alt=":approve:" title="Approve :approve:" data-shortname=":approve:" />

 

[marcus]

serious problem---count the beans of electricity

In some smalltown general stores they would put out a jar of beans and the customers got to guess-----then they would count and the person who guessed closest would get the prize.

Maybe they still do this in New Zealand but in California we have the state lottery so we don't need that old bean stuff.

now here we have two insulated balls with the charge distributed equally between them and we fix them 8 centimeters apart and you gauge the repulsion between them and decide that it is

THREE AND 1/2 NEWTONS

Anybody who knows metric units should kind of know what that feels like, it is about 4/5 of the weight of an oldfashioned conventional pound---which weighed about 4.4 Newtons. And a kilogram weighs 9.8 Newtons so this force is roughly a third of the weight of a kilogram. So everybody should have a real clear idea of the force between these two balls, shoving them apart.

Now i challenge you. If you think you understand the metric system and have some physics intuition, tell me HOW MANY extra ELECTRONS are on each ball? It is the same number extra on each one. It's two equal negative charges.

for people who want to try using the alternate units, the force is 7.3 mark (that is the equivalent to the 3.5 Newtons we said earlier)

BTW Kea thanks for the company! and the confirmation of the formula
I think Baez called one of his friends on the cell phone just before he jumped in for a swim. the friend has a sailboat and will pick him up and they can spend a few days fishing. fishing and sailing is about all they can do on this planet.

 

[marcus]

People who know metric units presumeably do not need any help because it is a straightforward problem.

For anyone who might wish to try out the alternate units, remember that the unit charge (dram) is the charge of 10¹⁸ electrons.

And I will tell you that the COULOMB CONSTANT that converts charge and separation into force is

\frac{1}{137} \times 10^{13}\text{ mark sq.hand per sq.dram}
where I am referring to a special number alpha = 1/137.036...
which we approximate 1/137.

Ordinarily, to find the force you multiply the two charges (in drams) and divide by the square of the distance (in hands) and multiply by this

\frac{1}{137} \times 10^{13}

and that will turn out to be the force (in marks)

But here it is slightly different because you know the force and must work back to find the two equal charges.

 

[Gamish]

I think we should make a standard time unit that fits into the metric system. But other than that, I hate tha American standard, I like the metric system. Sorry if this may be a bit irrelavent

Hitler was made of antimatter