Septenary Units (split from Different Planck Units)

[Royston]

I don't want make seconds from kilogram, but I want make kilograms from second.

 

It would be irrelevant which way round you're deriving one unit with the other i.e seconds into kilograms and vice versa. They would be equivalent...but as already stated, a second isn't a natural unit.

 

EDIT: I'm going out with some friends tonight, but I'll start the thread as mentioned earlier over the weekend.

[TurricaN]

I made working formula in form:

((1 second)*(c²/h))-¹

Both second and (c²/h) are to -¹ power.

 

Thus I transform it to:

(1 second)-¹*(c²/h)-¹

 

Thus how to transform it further to make bare (1 second) on left side of multiplication (or division if needed) sign, and appropiate transformation on right side of multiplication (or division if needed) sign?

[daniel_haxby]

This debate is interesting me in both TurricaN's predicament and Martin's knowledge of Planck units (measurement?), so here is my attempt to put a different angle on things. I may be incorrect, so feel free to post corrections; that's why I'm posting. On the other hand, I may be writing exactly what TurricaN thinks he wants, in which case, it may illustrate Martin's point.

 

Start with [math]E=mc^2[/math] and [math]E=vh[/math], where v is frequency of a photon, and m is that photon's mass, c is light in a vacuum, and h is Planck constant (not h-bar yet).

 

So, equate to get [math]vh=mc^2[/math].

 

So, from what I believe Martin was saying, for this to make sense in the universal case, we need to use Planck units, or rather consider the thought experiments from which they are derived. We have figures for converting the Planck units to SI, so by using these in the correct manner we should get some conversion into SI.

 

For v, we mean [math]v=v_0 /t_{\textrm{unit}}[/math], where [math]v_0[/math] is a quantity, and [math]t_{\textrm{unit}}[/math] is the unit we are measuring in, which is Planck time. For universality, we pick 1 Planck time. Now we convert Planck time to seconds by multiplying t(unit) by the conversion factor to SI (seconds), which is [math]t_p[/math].

 

Now we have [math]h v_0 / t_p = mc^2[/math]

 

For m to make sense we must consider the photon whose frequency we were measuring, and its mass is [math]m=m_0 m_{\textrm{unit}}[/math] where m_0 is a quantity, and m(unit) is 1 Planck mass. To get kg, we use conversion factor m_p to say [math]m=m_0 m_p[/math]

 

Now, [math]h v_0 / t_p = m_0 m_p c^2[/math]

 

But [math]t_p = \bar{h} / m_p c^2[/math],

 

So [math]h v_0 / t_p = 2 \pi \bar{h} v_0 / t_p = 2 \pi \bar{h} v_0 c^2 m_p / \bar{h} = m_0 m_p c^2[/math]

 

Which means [math]v_0 2 \pi = m_0[/math]

 

which I presume means that a photon of mass m Planck units takes its own mass divided by 2pi units of Planck time to cross its own wavelength.

 

Comments or corrections to working or interpretation of result welcomed.

 

To TurricaN, I hope that final equation demonstrates what Martin was saying, that you can only equate time with mass in a meaningful way in the thought experiment in which Planck units are derived (defined?). You can't convert time into mass, but you can relate the frequency of a photon with its mass.

[daniel_haxby]

When I started writing my reply (#53) we were on post #46!

 

And I think I may have attempted to derive Compton time, as Jaques mentioned in #46!

 

This is me learning physics.

[TurricaN]

Sadly, Daniel Haxby's tips doesn't fit my needs. As you see, I made working formula in form:

((1 second)*(c²/h))-¹

Both second and (c²/h) are to -¹ power.

 

Thus I transformed it to:

(1 second)-¹*(c²/h)-¹

 

But how to transform it further to make bare (1 second) on left side of multiplication (or division if needed) sign, and appropiate transformation on right side of multiplication (or division if needed) sign?

[TurricaN]

I was only able to make working conversion formula from plain seconds to inverse kilograms (kg-¹), such as:

 

1 second*c²/h,

 

but how to transform it to formula that converts directly from plain seconds into plain kilograms?

[Martin]

I was only able to make working conversion formula from plain seconds to inverse kilograms (kg-¹), such as:

 

1 second*c²/h,

 

but how to transform it to formula that converts directly from plain seconds into plain kilograms?

 

I have already explained the situation to you several times. You keep asking the same question as if you either

1. are not listening.

2. cannot learn.

3. or do not understand simple algebra.

 

Please listen carefully this time. If you show me that you have not learned, I will ask for this thread to be locked and you will have to go away.

 

So if you want to be allowed to stay around and continue discussing, please listen carefully and allow your ideas to be changed by learning from what people tell you. In the next post there is another try at explaining.

[Martin]

...

 

Thus I transformed it to:

(1 second)-¹*(c²/h)-¹

 

But how to transform it further to make bare (1 second) on left side ...

 

I see that you mean by a conversion formula always involves multiplying a unit on the left by a COMBINATION OF FUNDAMENTAL CONSTANTS. Here for example

 

(1 second)-¹*(c²/h)-¹ = some number of kilograms, or some fraction of a kilogram

 

Actually it is a tiny tiny mass, a very small fraction of a kilogram. Let us call this quantity the "TurricaN mass."

 

Equation One

(1 second)-¹*(c²/h)-¹ = TurricaN mass

 

Now what you are asking for is some combination of fundamental constants XYZ which converts the plain second to the same mass. Like this:

 

Equation Two

(1 second) *XYZ = TurricaN mass

 

As I will prove to you with simple algebra, THE ONLY WAY THIS CAN HAPPEN IS IF YOU COULD FIND A COMBINATION OF FUNDAMENTAL CONSTANTS EQUAL TO ONE SECOND

 

So therefore you could redefine your program to have a new, but equivalent, goal: To find a way to write one second as a combination of fundamental physical constants. (To show that the second is a unit built into nature.)

 

I will prove this to you:

If you combine Equation One and Equation Two, you will get

 

Equation Three

(1 second) *XYZ = (1 second)-¹*(c²/h)-¹

 

Multiplying by (1 second) on both sides,

and dividing by XYZ on both sides,

you will get

 

Equation Four

(1 second)² = (c²/h)-¹ (XYZ)^-1

 

So now you would have (1 second)² written as a formula containing nothing but fundamental constants.

 

You could then take the square root of both sides, and you would have

written the second purely as an algebraic combination of universal natural physical constants.

 

Equation Five

(1 second) = sqrt( (c²/h)-¹ (XYZ)^-1)

 

But this cannot be done.

 

If it could be done, then the second would be a natural unit of time, built into the universe. If it was possible to do, then Max Planck would have already done it in 1899 when he was searching for a natural unit of time. What he then found was a unit of time, which we call the planck unit of time, that CAN be written using nothing but fundamental universal constants. THIS unit, not the second, is the unit of time which is woven into the universe.

 

1 natural time unit = [math]\sqrt{\frac{\hbar G}{c^5}}[/math]

[Martin]

If all you wanted to do was to take the bare second, multiply by some combination of fundamental constants, and get some mass, that would be easy.

 

We have already discussed that. The formula or combination of constants that works is

 

c³/G

 

If you take one second, and multiply by that, you will get some large number of kilograms

 

Equation Six

 

(1 second)* (c³/G) = large number of kilograms

 

This is the unique combination of universal natural constants that will do this (unique except for factors of order one like pi, one could for example use 2 pi G instead of G, but that is a trivial difference.)

 

You have rejected this conversion formula already because you say it is not "consistent" with the other one, which I called Equation Two.

 

In other words, you don't like Equation Six because the large mass is not equal to the TurricaN mass of Equation Two.

 

This contradiction happens because the second is not a natural unit. In effect the Universe "does not want you" to use the second as a unit of time.

 

Please show that you have learned something from this discussion or I will be forced to ask that this thread be closed.

 

thanks for your interest in these matters.

[TurricaN]

I now understand your formulas, but I now want to look at their dimension:

 

For example, Google Scholar accepts this:

(h/c^2)/(h*G/c^5)^0.5)=

http://scholar.google.com/scholar?hl=en&lr=&q=%28h%2Fc%5E2%29%2F%28h*G%2Fc%5E5%29%5E0.5%29%3D&btnG=Search

 

but rejects this:

((h/c^2)/(h*G/c^5)^0.5)^0.5=

http://scholar.google.com/scholar?q=%28%28h%2Fc%5E2%29%2F%28h*G%2Fc%5E5%29%5E0.5%29%5E0.5%3D&hl=en&lr=&btnG=Search

 

How to force Google to perform calculation?

 

for XYZ I substituted ((h*g)/c^5)^0.5

 

I want to check dimension of this, but Google Scholar treats my input as search, instead of calculating it. Do you know alternative for Google calculator?

[Martin]

When I use your first link this is what I get

(h / (c^2)) / (((h * G) / (c^5))^0.5) = 54.5604236 micrograms

 

Let me see if that is dimensionally correct.

 

The denominator of your fraction is a well-known amount of time----the natural unit of time that is built into the universe.

 

(except that you are always using h instead of hbar but that only changes things by factors like 2pi and the squareroot of 2pi)

 

And in the numerator of your fraction you have h-----its dimensions are energy x time, so if you divide it by time you get an energy.

 

Yes and then if you divide an energy by c^2 you should get a mass.

 

OK Google calculator is correct. Your formula should give a mass and it should be larger (by a factor of sqrt 2pi) than what physicists normally consider the natural unit of mass because you are using h instead of hbar.

 

If you will divide your 54.56... micrograms by sqrt(2 pi) then you will probably get the NIST value for the Planck mass, if Google is accurate.

 

I hope Google calculator let you divide by sqrt(2 pi), or equivalently by (2*pi)^0.5

and then you will see the NIST value for the natural mass unit

which is around 21 or 22 micrograms

 

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

 

Your next calculation looks like you are trying to take the square root of the first thing you did.

 

This is THE SQUARE ROOT OF A MASS. Among physicists the idea of the square root of a mass does not make sense. Therefore Google was right when it refused to calculate it for you.

 

One can take the square root of an area---this is a length and that makes sense. One cannot take the square root of a mass because there is no physical unit which is the square root of mass.

[Martin]

I just went to ordinary google and put in

 

sqrt(hbar*G/c^5)=

 

it gave me the natural unit of time

 

http://www.google.com/search?hl=en&q=sqrt%28hbar*G%2Fc%5E5%29%3D&btnG=Google+Search

 

sqrt((hbar * G) / (c^5)) = 5.3907205 × 10-44 seconds

 

 

I am very glad that Google has hbar. In the physics literature it is generally preferred to Planck's original h.

[TurricaN]

Someone from Physics Forums wrote me something like:

[c] = L T^-1

[h] = M L² T^-1

 

[c²/h] = L² T^-2 M^-1 L^-2 T¹ = M^-1 T¹ = T/M.

 

Therefore, in SI units:

 

1 second = c²/h * 7.372 x 10^-51 kg

1 kg = h/c² * 1.356 x 10⁵⁰ second

 

I deduced that these factors:

 

7.372 x 10^-51

1.356 x 10⁵⁰

 

are derived from:

h/c²

and

c²/h

 

but unit dimension doesn't match, because of introducing of dimensionless unit made by multiplication by inversion. Thus how finally these two expressions:

 

1 second = c²/h * 7.372 x 10^-51 kg

1 kg = h/c² * 1.356 x 10⁵⁰ second

 

are formulated using combination of constants only?

[Martin]

Hi TurricaN,

 

I would like to ask you, when you are working with Google, putting in combinations of constants, would you please use

 

hbar

 

instead of h.

 

It is generally preferred by physicists who actually have to use it, as the better form of the Planck constant.

 

As it is usually written it looks like the lowercase letter h

together with a Christian cross

 

 

[math]\displaystyle \hbar[/math]

 

but the resemblance is accidental and you should not let it distract you.

Just use hbar because it is more of a real physicist thing than the earlier version h.

[Martin]

typing hbar into google will be recognized as [math]\hbar[/math]

 

it would be nice if you would at least try using

 

c²/hbar

 

and

 

hbar/c²

[TurricaN]

I finally avoided this ugly contradiction, by choosing another conversion factor for temperature, thus I got:

 

*for conversion from meter to '''second to one power''' - c defining how many meters are in one second

*for conversion from kilogram to '''second to one power''' - c³/G defining how many kilograms are in one second

*for conversion from second to '''second to one power''' - no conversion factor, because 1s=1s¹

*for conversion from ampere to '''second to zero power''' - no conversion factor, because it is dimensionless number

*for conversion from kelvin to '''second to minus one power''' - c⁵/(G*k) defining how many kelvins are in one second (not in second to minus one power)

*for conversion from mole to '''second to zero power''' - no conversion factor, because it is dimensionless number

*for conversion from candela to '''second to zero power''' - no conversion factor, because it is dimensionless number

 

I cross-checked these conversion factors by converting from obtained from second these values of both meter and kilogram back to kelvin, both using direct conversions from second to kelvin, and too from meter to kelvin multiplying by (c⁴/(G*k)) [K/m] and from kilogram to kelvin multiplying by (c²/k) [K/kg], and finally obtained the same results as in direct conversion from second to kelvin. That means that this set of conversion formulas is fully cross-consistent.

 

But how transform this conversion factor c⁵/(G*k) to change its dimension from '''kelvins per second to one power''' (K/s¹) into '''kelvins per second to minus one power''' (K/s^-1) ?

[Martin]

Congratulations on achieving the goal, to your own satisfaction.

I believe the original plan was to reform the Roman Catholic calendar or some other traditional time-keeping institution. (sorry the ultimate goal was never entirely clear to me). So you are only part-way there and still have plenty more work to do.

 

I wish you the best of luck in your further endeavors!

[TurricaN]

Yes, I want to purge Catholic units from babylonian influences, retaining only God's septenary system and nothing else. (Even sixth day is complemented with seventh, and twelve Apostles are complemented with thirteenth Mary-Queen of the Apostles and fourteenth Jesus-King of the Apostles) For example look at this: http://the-light.com/cal/veseptimal.html

 

But you forgot answer me question how to transform this conversion factor c⁵/(G*k) to change its dimension from '''kelvins per second to one power''' (K/s¹) into '''kelvins per second to minus one power''' (K/s^-1) ? This is needed because in LUFE Matrix (drawn here:http://en.wikipedia.org/wiki/Talk:Geometrized_unit_system) kelvin is expressed as '''dimension to minus one power'''. I want to derive hertz-to-kelvin ratio from second-to kelvin ratio, because it is consistent with the rest. How to do this?

[TurricaN]

Because you refused to answer me, I made some research and I'm answering myself to present solution of this problem to public: LUFE Matrix treats temperature as dimension to minus one power instead of dimension to one power, because it has contradiction at kelvin position between "°K" and "per °K" between these tables: http://www.brooksdesign-cg.com/Code/Html/Lm/LMunitSI.htm http://www.brooksdesign-cg.com/Code/Html/Lm/LMGsota.htm Additionally, because c⁵/(G*k) [K/s] is consistent with other conversion factors, while h/k [K/Hz] is not consistent with other conversion factors, LUFE Matrix is in error at Kelvin position, while Wikipedia set of conversion factors placed here: http://en.wikipedia.org/wiki/Geometrized_unit_system and in related Wikipedia article discussion is valid. That fact finally denies possibility of changing dimension of abovementioned ratio from K/s¹ into K/s^-1.

[Martin]

Yes, I want to purge Catholic units from babylonian influences, retaining only God's septenary system and nothing else. (Even sixth day is complemented with seventh, and twelve Apostles are complemented with thirteenth Mary-Queen of the Apostles and fourteenth Jesus-King of the Apostles) For example look at this: http://the-light.com/cal/veseptimal.html

...

 

You are wonderful. I wish all Catholics had as much common sense and reasonableness as you. The world would be a better place. As I said before, good luck with your project of time and calendar reform!

[foodchain]

Really, I am not posting in this thread because I want to keep it around, I have a valid reason in my own opinion.

 

Time, many questions about it, that’s great, not really what we are talking about but it applies. What is time based on, well going from the dictionary its a continuum or some would say a dimension in which allows for such to be put into names and numbers for physical judgment of whatever absolute reality time might actually be, I for one don’t think we have it yet but hey, I am not a physicist.

 

So in relation to the debate question, what’s the point I ask? If the world hundreds of years ago made a week nine days long and factored the rest into our solar systems environment such as orbits would it be some great cause for alarm? I mean why do we have a leap year? If the day was only 23 tigers long instead of 24 hours, and each tiger was a different unit of measurement such as instead of second you had turtle, would it make some profound difference to anybody as long as it worked in regards to the natural world? I mean we call winter because its a particular event if you will, that occurs in a timely fashions right? The day gets shorter though, and in some places during the year the sun never sets or comes up for the matter!

 

So really I am not against change, but I would really like to think about the reason requesting the change and does it hold any real rational or positive reason for such. I mean the watchmaker might be blind, so such is why we need systems of time to be developed, but that does not mean we have to be blind, and that’s why we have physicists I think. You can cut a unit of time down to whatever smallest possible fashion, and then into whatever longest unit of such, and everything in-between, but in reality the time it takes for the sun to rise and set is just that, the name or system around it is based on our ability to understand the reason we need to word time to exist in the first place, not about how time should be in relation to us, its kind of like arguing with gravity and jumping from a height to mock its rule really in my opinion.

 

Well anyways, thanks for the time to read this!

[TurricaN]

My reason of making all units as septenary is following: I want to keep both God's Commandment of keeping seventh day holy and consistency in and between all units at once, without ever slight contradiction or inconsistency. I finished my Excel table and discovered, that Biblical non-time units roughly matches septenary multiples and submultiples of daynight expressed in metres, kilograms seconds and kelvins that I derived from one 86400 second period using abovementioned energy equivalent ratios. Units such as ampere, mole and candela are dimensionless, thus I discarded them. I even use for this septenary system of units following set of digits such as 0,1,2,3,4,5,6,10 as is natively in base seven system. This septenary system has only one unit called daynight and its septenary multiples and submultiples, that abovementioned Excel table expresses too in metres, kilograms seconds and kelvins. I even propose to use draconitic-like year that has exactly 343 days. In these daynights and its septenary multiples and submultiples is possible to measure not only time, but too length, mass, and temperature, as is in geometrized unit system. This septenary system is much simpler than SI, because in this system exists only one unit that fits for measuring of all scientific quantities.

[Jacques]

year that has exactly 343 days

Do you plan to change earth orbits???

[TurricaN]

No. But there exists such thing as draconitic year, described here: http://en.wikipedia.org/wiki/Year#Draconic_year , that is very similar to 343 day year, because it has 346 days, but best solution to keep septenary consistence would be abandoning synchronization with period of going Earth around the Sun completely.

[Jacques]

but best solution to keep septenary consistence would be abandoning synchronization with period of going Earth around the Sun completely.

Why ? I don't see how it can help...

I don't understand the purpose of your work. Why convert everything into seconds... 100 seconds of mass + 3 second of volt = 103 seconds ...

Completely useless.