Sunday 27 October 2013

Planck charge

I have recently done a couple of posts based on Planck units and this is another in the series. This one though is a little different. This is about charge.

When we thing about a Planck mass or Planck length there are a number of different things to compare them against. Take the Planck mass, we can take the ratio of the Planck mass to the mass of the electron, or the proton, or the neutron. Each of these will give a different value. Similarly if we take the ratio of the Planck length with the radius of an electron, or a proton or a neutron, again different values.

When we consider the charge however this is not the case, the charge on the electron is the same magnitude as that of the proton, only different in sign, positive or negative. So the ratio of Planck charge to the elementary charge will be the same, in fact it turns out that

$ \frac {e^2} {q_p^2} $ $ = \alpha  $    ....(1)

e - elementary charge
qp - Planck charge
$\alpha$ - fine structure constant

I did a post on the fine structure constant a while back, where I quote Richard Feynman describing it as a magic number, a true mystery. I like mysteries. So here it is, the ratio of the elementary charge and the Planck charge.

-----------------------------------------------------------------------------------------------
Aside : The number that Feynman describes is actually that given by

$ \frac {e} {q_p} $  $= \sqrt \alpha  = 0.08542455 $    ...(2)
-----------------------------------------------------------------------------------------------

What baffles me a little though is that the Planck charge does not appear to make sense. The Planck charge is 11.70623 times the charge of the electron, the elementary charge.

Apart from quarks that have a charge of 1/3, charges are integer multiples of the elementary charge. At least that is the way it appears at the moment. Quarks do not appear to exist in isolation so you do not get fractional charges.

In other words, it should not be possible to create a charge in isolation of 11.70623 e. So we can never actually create a particle with the charge equal to the Planck charge.

In which case the question remains, what exactly is the Planck charge?

The equation defining the Planck charge is 

$ q_p = $ $ \sqrt {4 \pi \epsilon_0 \hbar c}$ ... (3)

$\hbar$ - Reduced Planck constant - $1.0545717   10^{-34}  J.s$
$4 \pi \epsilon_0$ - inverse of Coulomb's constant - $1.11265   10^{-10} $
$\epsilon_0$ - permittivity of free space - $8.85418782   10^{-12}  m^{-3} kg^{-1} s^4 A^2$
c - speed of light - 299792458 $m s^{-1}$

which seems straight forward enough. Yes, it might look a little odd, but no more so than the other Planck units. Yet, unless it is some Fractional quantum Hall effect (FQHE) value, it is not real. In which case how can it give us the result in (2)? which is real and can be measured exceptionally accurately.

I can only think that it must be a valid value, does this imply it IS a FQHE value? or that there is another phenomena that has not been discovered yet that will allow a non-integer value of the elementary charge?

Let's say the value of the Planck charge could exist because of the FQHE. Does this give us any insights into why (1) is true?

The Fractional quantum Hall effect (FQHE)

the FQHE is a quantum mechanical version of the Hall effect and it is observed in 2 dimensional systems at really low temperatures in strong magnetic fields. This was discovered in the 1980s. I will post on this later. There is a value called the Hall conductance that is given by

$\sigma =$  $ \frac {I_{channel}} {V_{Hall}} = \nu \frac {e^2} {h}$ ....(4)

$\nu$ is called the "filling factor" and provides the fractional value (1/3, 2/5, 2/3 , 12/5 ....)

So is there a Planck version of this?

 $\sigma_p =$  $ \nu \frac {e^2} {h} = \nu_p \frac {q_p^2} {h}$ ...(5)

giving

$\nu$ $\frac {e^2} {h}$ $= \nu_p$  $\frac{q_p^2} {h}$  ...(6)

which becomes

$\alpha = $ $\frac {e^2} {q_p^2} = \frac {\nu_p} {\nu} $    ...(7)

Returning to the Planck charge, does the above imply that the Planck charge may only be valid in situations where FQHE is valid? to consider it anywhere else, while mathematically meaningful, would actually be physically invalid.

Alternatively, is there another theory of physics where fractional charges can exists in other situations besides the FQHE?

Using the Fractional quantum Hall effect (FQHE) it is possible to get fractional elementary charge. If we think that this may explain the problem of the Planck charge not having a integral charge, we can try the following

take $\alpha$ = 137.035999074 (44)

so $\sqrt {\alpha}$ as 11.7062376

The fraction 35/3 = 11.666666666

11.70623 / 11.666666 = 1.003391, which is pretty close. I do not think it is close enough though.

what about 199/17 = 11.70588235,

11.7062376 / 11.70588235 = 1.0000303, closer, can we do better? Yes. Try writing a simple number cruncher, you find hundreds of examples with closer agreement.

Using equation (7) as a starting point we could try this

$\frac {a}{b} e = \frac {c}{d} q_p$     ....(8)

a,b,c and d are integers. Try a=261, b=25, c=404, d= 453, you have integer values that give the number

11.70623762, squared this is 137.035999215

11.7062376/11.70623762 $\approx$ 1 or

137.035999074/137.035999215 = 0.99999999897

given the degree of error in $\alpha$ this value is pretty close to 1!

What does it mean though? Say the value of a,b, c and d are correct, so

$\frac {261}{25} e = \frac {404}{453} q_p$     ....(9)

or

$\frac {261}{404} e = \frac {25}{453} q_p$     ....(9)

why these value? why not others? These are as much of a mystery as 137.03599....

To see if I can get any further with this idea I shall be taking a look at the Hall Effect and the Quantum Hall Effect in more detail in a future post.

Thursday 24 October 2013

Observations on the fine structure constant - Part 1

A while back I did an introductory post on the fine structure constant. More recently I did a post on what was not the fine structure constant. In this post I am going to take another look at this delightful mystery. This is probably going to end up 2 or even 3 posts by the looks of things. Let's get on.

The fine structure is described on wikipedia, it has a number of physical interpretations, for example, the square of the ratio of the elementary charge to the Planck charge. Despite this interpretation and the others listed on the wiki page we still don't know what it is all about. You would think that 8 different descriptions we would be able to figure out what it is. This is not the case. We just can't figure it.

What is apparent to me though is that once we do understand this number we will have a far greater understanding of the universe we inhabit. The fact that it pops up so often in so many places. This thing is a big deal and anyone who managers to explain it will really have made a major break through in physics.

I have been looking at the Planck units in recent months and have found out some great stuff, but once again I have drawn a total blank when it comes to the fine structure constant. Above we mentioned that it can be represented by

$\alpha = $ $ \frac {e^2} {q_p^2}$    ... (1)

$\alpha$ - fine structure constant
e - elementary charge
$q_p$ - Planck charge

This is an amazing result, after all the Planck charge is given by

$q_p = \sqrt {4 \pi \epsilon_0 \hbar c}$    ...(2)

$hbar$ - Reduced Planck constant
$1 / 4 \pi \epsilon_0$ - Coulomb's constant
c - speed of light

Why do we get the result in (1), why isn't it 1? $\pi$ - 3.141592653? natural log - 2.718281828? the golden ratio - 1.61803398875? or even 42?

It isn't any of these, it's $7.297352569   10^{-3}  \approx 1 / 137.03599917$

137.03599917! - seriously what is that? Some strange solution to a Kepler triangle, some bizarre integral? Even the root looks no better

$\sqrt {7.297352569   10^{-3}} = 0.085424543 \approx 1 / 11.7062376$

Well what do we know?

1) The mass of an electron divided by the Planck mass does not seem to have any obvious relationship to the fine structure constant. This is also true for the classic radius of an electron and the Planck length.

2) It is dimensionless. So it is not energy or momentum, acceleration, velocity, charge or temperature, it is a number.

3) It is a ratio. In equation (1) above it is the ratio of two charge values. Whenever it is a ratio the two values have to be of the same type, eg charge, time, mass, momentum, energy etc so that the dimensions cancel.

That said, it could be the ratio of a potential energy and a kinetic energy, both are energy. It could also be the ratio of energy density and pressure, both of these have the same dimensions.

4) It isn't 1, that may sound obvious but think about it for a minute. The Planck charge and the elementary charge are remarkably close, they are pretty much within an order of magnitude.

5) You cannot actually get the Planck charge. The Planck charge is NOT an integer number of the elementary charge. Given our current understanding of charge, you cannot get isolated charges that are not integer values of the elementary charge. Therefore, it is not possible to isolate an amount of charge that is equal to the Planck charge.

May need to rethink this for Fractional Quantum Hall Effect.


Planck charge

Lets take a look at equation (2), the definition of the Planck charge. if contains 4 parts,

$\pi$ - 3.141592653 - if it contains $\pi$ then it may have something to do with circles and waves
$\epsilon_0    -     8.854178817    10^{-12} F.m^{-1}$
$\hbar    -    1.05457172     10^{-34}  Js$
$c - 299 792 458    m  s^{-1}$

It does NOT contain anything relating to Gravity and big G. If we divide by the Planck mass it becomes

$\frac {q_p} {m_p} = \sqrt {4 \pi \epsilon_0 G}$    ...(3)

The charge to mass ratio does NOT contain $\hbar$ or c. It does contain big G. $\epsilon_0$ is present in both equations.

The Planck charge ratio, just like the Coulomb to Gravity ratio

$\frac {F_c} {F_g} = \frac {e^2} {m^2}  \frac {1} {4 \pi \epsilon_0 G}  $   ...(4)

which is just

$\frac {F_c} {F_g} = \frac {e^2} {m^2}  \frac {m_p^2} {q_p^2}  = \frac {\alpha} {\alpha_G}$   ...(5)

where

$\alpha_G = $ $\frac {m^2} {m_p^2}$     ....(6) 

Alternatively if we replace the elementary charge with the Planck charge we get

$F_c = $  $ \frac {q_p^2} {4 \pi \epsilon_0 r^2} $    ... (7)

which becomes

$F_c = $ $\frac {4 \pi \epsilon_0 \hbar c} {4 \pi \epsilon_0 r^2} = \frac {\hbar c} {r^2}$ ...(8) 

In equation (8) we have a form of Coulomb's law that does not involve charge! What does that mean?

Ok, I think I'll leave this post here. Will continue shortly.

I wrote this while listening to this.





Monday 21 October 2013

Planck equations

In a number of recent posts I have been taking a look at Planck units and have found a number of equations that have real life equivalents. By this I mean that there are a number of equations that you can derive purely from Planck units that also have recognizable every day version, for example

$E_p = m_p c^2$    ...(1)

$E_P$ - energy
$m_p$ - Planck mass
c - speed of light

this is the well known

$E = m c^2$    ...(2)

yet (2) could have been derived from the Planck units without any knowledge of relativity and mass energy equivalence. This is not the only case, here are a few more

Planck equation General equation
$F_p = m_p g_p$ $ F = m a$
$V_p = T_p k_b / q_p$ $V = T k_b / e$
$T_p k_b = m_p c^2$  $T k_b = m c^2$
$E_p = \hbar \omega_p$ $E = \hbar \omega$
$Z_p = \hbar / q_p^2$ $R_k = \hbar / e^2$

The question must arise as to whether it is possible to derive other Planck equations that do not currently have an every day equivalent? If so, should these equations already exist but it is just a case that we have not found them yet? For example,


Planck equation General equation
$g_p l_p = c^2$ $ g \lambda = c^2$
$E_p c = \hbar g_p$ $E c = \hbar g$
$g_p l_p^2 = m_pG $ $g l_p^2 = m G$
$g_p = m_p c^3 / \hbar $  $g = m c^3 / \hbar$
$F_p = \hbar g_p^2 / c^3$ $F = \hbar g^2 / c^3$
$E_p I_p = \hbar^2$ $E I = \hbar^2$


There are a fair number of these and I have mentioned some of them in previous posts with possible interpretations of what they may be. No doubt I'll cover some of the others in future posts.

Great. Well yes, potentially it is, because we may have a bunch of answers looking for questions which will then tell us what the answer really means. A bit like "42"!

Let's take those listed here;

$ g \lambda = c^2$    ...(1)

does this imply that there is an acceleration associated with a wavelength of a photon just in the same way

$E = h \nu$    ...(2)

links energy with frequency, but if that is the case then does

$E = $ $ \frac {\hbar g} {c} $   ...(3)

imply that there is a relationship between energy and acceleration. The acceleration of what? Photons? I discuss this one in a previous post in more detail.

What about 

$g  =$  $ \frac {m G} {l_p^2}$   ...(3) 

Does this give us insight into Newton's law of gravitation? It throws up at least one surprise for me that I cover here.

Then 

$E I = \hbar^2$    ...(4)

where $I$ has the same units as moment of inertia. Is it the moment of inertia of a photon?  Again, I cover this elsewhere

I genuinely don't know if these are of value or if my interpretations have any validity, but I can't believe that they are not completely without worth. The equations above must mean something, it is really just a question of determining exactly what that is. I shall continue to post on these and give my view on what they may mean. 

Even though this approach yields some interesting results in the form of "new" equations, it does not give us any insight into why the fundamental constants are what they are. Take charge for example, we have

$ \alpha =$  $\frac {e^2} {q_p^2}$   ...(5)

$\alpha$ - fine structure constant
$e$ - charge on the electron
$q_p$ - Planck charge

$q_p = $ $\sqrt {4 \pi \epsilon_0 \hbar c}$    ...(6)

$\hbar$ - reduced Planck Constant
$ 1/ 4 \pi \epsilon_0$ - Coulomb constant

This shows us a relationship between the Planck charge and the charge on an electron, which turns out to be the fine structure constant, but it does not give us the reason why.

There are a number of equations that link mass, length and charge, so if an explanation for one of them can be found then the rest will fall into place. Though that is far easier to say than do.

The Planck units give us a number of Planck equations that may be giving us some new insights that we have not previously seen, but they do not appear to be giving out any clues as to why the electron has the charge, mass and radius it does have. Or is it just that I am missing something?

I shall continue to ponder.

I wrote this while listening to this.

Sunday 20 October 2013

Kepler, Newton and the fine structure constant

One of the things I love about physics is the little surprises it shows up. I did a post where I had "discovered" an interesting result that appears to have something to do with Kepler's 3rd law. This post follows on from one that extends a result about Newton's law of gravity.

In this case I consider the gravitational force when a positron/electron pair are created with a balancing Coulomb force. This may seem a little odd, since both force are attractive, but go with me on this one.

For Newton

$F_{Newton} = $ $ \frac {G m_e^2} {l_p^2}$   ...(1)

$F_{Newton}$ - Resulting force - $0.212013  N$
G- Gravitaional constant - $6.67384     10^{-11}  m^3   kg^{-1}   s{-2}$
$m_e$ - mass of the electron - $9.109382    10^{-31}  kg $, $2m_e$ - e/p pair
$l_p $ - Planck length - $1.616199    10^{-35}  m$
m - a value that was not stated


For Coulomb

$F_{Coulomb} =$ $ \frac{e^2} {4 \pi \epsilon_0 r^2}$     ...(2)

e - charge on the electron - $1.602176565    10^{-19}  C$
$\frac {1} {4 \pi \epsilon_0}$ - Coulombs constant - $8.98755178    10^{9}   kg  m^3   s^{-2}    c^{-2}$
r - separation of the electrons

What we are going to consider is the point where the two forces are actually equal in order to determine the value of r, the separation of the electrons.

$F_{Newton} = F_{Coulomb}$   ...(3)

so

$\frac { G m_e^2} {l_p^2} = \frac{e^2} {4 \pi \epsilon_0 r^2}$     ...(4)

rearrange to give r, so we have

$ r^2 = $ $ \frac {e^2 l_p^2} {4 \pi \epsilon_0 G m_e^2} $     ...(5)

----------------------------------------------------------------------------
Aside: it is possible to replace the $4 \pi \epsilon_0 G$ with the Planck equation

$4 \pi \epsilon_0 G = $ $\frac {q_p^2} {m_p^2}$     ...(A1)

resulting in

$ r^2 = $ $ \frac {e^2 l_p^2 m_p^2} {q_p^2 m_e^2} $     ...(A2)

or

$ r = $ $ \frac {e l_p m_p} {q_p m_e} $     ...(A3)

which gives

$ \frac {r m_e} {e} =  \frac {l_p m_p} {q_p} $     ...(A4)
----------------------------------------------------------------------------

putting in the numbers gives a value of 

$ r = 3.29874  10^{-14} m$

Now in a post on photons I derived a relationship (equation 27 in the post)

$ s = \lambda / 4 \pi$   ... (6)

In a later post I calculated a value for this for a photon with enough energy to create an electron/positron pair, it was

$s = 9.653976    10^{-14}   m  $    ...(7)

dividing r by this gives

$r / s = 0.341698$

divide by 4 and square gives and 

$(\frac {r} {4s})^2$  $ = 0.0072973$   ...(8)

the reciprocal of this is 137.035854, which is pretty close to the inverse fine structure constant. This is just

$(\frac {r \pi} {\lambda})^2$  $ = 0.0072973$   ...(9)

Also, if we take the classic electron radius

$r_e = 2.81794    10^{-15}$

divide this by r and square we get

$(\frac {r_e} {r})^2 $ $= 0.007300$    ...(10)

Let's continue this flight of fancy by saying that (8) and (10) are close enough to say they are equal, we ill revisit this later, so we end up with

$(\frac {r} {4s})^2 = (\frac {r_e} {r})^2$    ...(11)

rearranging this gives

$r^2 = r_e 4s$    ...(12)

which is also

$r^2 = r_e \lambda / \pi$    ...(12a)

if we then use the value for s given by 6 and

$r_e = 2 \pi \lambda \alpha$    ...(13)

we can show

$r^2 = 2 \alpha \lambda^2$    ...(14)

or

$r^2 =$  $ \frac {r_e^2} {2 \pi^2 \alpha}$   ... (15)

OK, so what does all this mean? if anything? We have defined a new value r that lies somewhere between the radius of an electron and the wavelength of a photon capable of creating an electron/positron pair.

Further to this let's take a look at the first part the gravitational force. This is the force that would occur between two electrons separated by a Planck length. Is that even valid to consider gravitation at the Planck length. Let us try something else, if we replace the $l_p$ with its definition

$l_p = $ $ \sqrt \frac {\hbar G} {c^3}$     ...(16)

then (1) becomes

$F_{Newton} = $ $ \frac {G m_e^2 c^3} {\hbar G}$

cancelling the G gives

$F_{Newton} = m_e a = $ $ \frac {m_e^2 c^3} {\hbar }$  ...(17)

a - acceleration, where

$ a = $  $\frac {m_e c^3} {\hbar}$    ...(18)

which does NOT involve the gravitational constant.  Does this help, I'm not sure.

If I am honest what appeals to me here is the fact that a new distance has been defined in equations (14) and (15) that lies in between the size of a photon (with enough energy to create an electron/positron pair) and the classic electron radius. It was derived by considering some balance point between the Coulomb force and, what appears to be, a gravitational force.

One problem though is if it is an electron/positron pair then both the Gravitational force and the Coulomb force are attractive, so how can there be a balance point? Even if we consider a pair of electrons in the vacuum of space, does this "balance point" actually tell us anything? Do we actually have pairs of electrons in a common orbit around each other at the distance calculated here? I find that hard to believe. Though I would think that it would be relatively easy to confirm by experiment.

One problem with orbiting electrons though is that Coulombs law is for static point like charges that are stationary relative to each other. If they are in orbit around each other are they stationary relative to each other?

If these electron pairs did exist then what would be their nature? In superconductivity there are "bound" Cooper pairs, would these electrons behave in a similar manner? The value of r is less than the wavelength of the individual electrons, so would they have to be bound in a "low orbit". In superconductivity, the coherence length, the distance between the electrons in a Cooper pair, is far far larger, typically 3 - 100 nm.

According to quantum mechanics electrons in a smaller orbit than the wavelength of each electron would have a total angular momentum of zero. This would the indicate that they may behave similarly to the ground state electron in a hydrogen atom and we could then take a look at the Schrödinger equation for that scenario. Will try this in a later post.

I'm going to finish this post here, though I can't help thinking I am going to revisit the idea in a later post once I have a better understanding of what has gone on.

I wrote this while listening to this.

Friday 18 October 2013

Energy and more energy

I recently did a post comparing forces. This is a similar exercise where I am comparing energies. So to start with the usual suspects, Einstein's classic

$ E_E = m c^2$    ...(1)

$E_E$ - Einsteins version of energy
m - mass
c - speed of light in vacuum

Next Planck's version

$ E_P = h \nu $    ...(2)

$E_P$ - Plancks version of energy
h - Planck constant
$\nu$ - a frequency

We are going to re-arrange this slightly using

$ \nu = c / r$     ...(3)

c - speed of light in vacuum
r -  a wavelength

so now Planck's version becomes

$E_P = $ $ \frac {hc} {r}$   ...(4)

Next we have the gravitational potential energy, this is typically represented by U and is negative, but today I have made it E and dropped the -ve sign, so

$E_G = $  $ \frac {G m_1 m_2} {r}$   ...(5)

$E_G$ - gravity version of energy
G - Newton's gravitational constant
$m_1   ,   m_2$ - mass of object 1 and object 2
r - distance between

let us set $m_1 = m_2 = m$ , so (5) becomes

$E = $ $ \frac {G m^2} {r}$   ...(6)

Finally we have the electric potential, given by

$E_C = $ $ \frac {e_1 e_2} {4 \pi \epsilon_0 r}$   ...(7)

$E_C$ - Coulombs version of energy
$e_1   ,  e_2$ - charges on two separate objects
$\epsilon_0$ - permittivity of free space
$\pi$ - 3.141592.....
r - distance between the charges.

Part 1: Gravity, Planck and Einstein

Let's re-arrange (1) in terms of mass and (4) in terms of r, so we have

$m = $ $\frac {E_E} {c^2}$    ...(8)

$\frac {1} {r} = \frac {E_P} {h c}$    ...(9)

lets assume that the distance between the masses in equation 6, is equal to the wavelength, r, given in (4) put these two results into (6)

$ E_G = $ $\frac {G E_E^2 E_P} {c^4 h c }$    ...(10)

Let us set all the energy values to be equal

$E_G = E_P = E_E = E$    ...(11)

so (10) becomes

$ E =$ $ \frac {G E^3} {c^5 h}  $   ... (11)

dividing by E and rearranging gives us

$ E^2 = $   $ \frac {c^5 h} {G}  $    ...(12)

which turns out to be the definition of the Planck Energy! OK, let's think about that for a minute. We set the masses equal, we set the energies equal and we said that the distance between the masses was equal to a wavelength. Everything balanced and worked out.

Part 2: Coulomb, Planck and Einstein

In the Coulomb version of energy there is no mass, so Einstein gets left out of this one. Let us do the same thing as we did for gravity, re-arrange Plancks version to give us (9) again and pop this into (7), in addition, set the charges to be equal, so

$E_C = $ $ \frac {e^2 E_P} {4 \pi \epsilon_0 h c}$    ...(13)

Let's set the energies to be equal as before this time giving

$E = $ $ \frac {e^2 E} {4 \pi \epsilon_0 h c}$    ...(14)

this time we can cancel E to give

$1 = $ $ \frac {e^2} {4 \pi \epsilon_0 h c}$   ...(15)

this however is incorrect, the correct value is

$\frac {\alpha} {2 \pi} = \frac {e^2} {4 \pi \epsilon_0 h c}$   ...(16)

$\alpha$ - is the fine structure constant. So where have we gone wrong?

Part 3: Planck Energy

In part (1), when we set the energies equal, what we had actually done, although it was not apparent, was to set them all equal to the Planck energy, so equations (1), (4) and (6) had become

$E = m_p c^2$    ...(17)

$E = $ $ \frac {hc} {2\pi l_p}$   ...(18)

$E = $ $ \frac {G m_p^2} {l_p}$   ...(19)

$m_p$ - Planck mass
$l_p$ - Planck length

These are just variations of the same equation

$ E = $   $\sqrt {\frac {c^5 \hbar} {G}}  $    ...(20)

Ok, so why didn't this work for part (2), isn't it likely that we are setting them both to the Planck energy again? Let's see, take the Planck version of equation (9)

$\frac {1} {l_p} = \frac {E_P} {h c}$   ...(21)

and let's have the Coulomb equation with r replaced with $l_p$ so

$E_C = $ $ \frac {e^2} {4 \pi \epsilon_0 l_p}$    ...(22)

but this is not $E_P$, the Coulomb equivalent of $E_P$ is

$E_{CP} = $ $ \frac {q_p^2} {4 \pi \epsilon_0 l_p}$    ...(23)

and $q_p \not= e$, so (22) and (23) are not equal. Does this mean we have the wrong choice of $q_p$? Should it be just the charge of an electron? This seems to be moving us away from the idea of Planck units though. If we are going to do that why not use the mass of the electron for Planck mass etc.

For Gravity, Einstein and Plancks relation we were able to find a common point where they all balanced which in itself is remarkable and I shall ponder this further. For Coulomb energy there was no common distance that worked. What does that tell us? I'm not sure at the moment, but if I figure it out I'll update this page. (This isn't strictly true, there is a common distance that works. I'll be covering it in a later post and will put in a link when I publish it.)

I wrote this while listening to this.

Thursday 17 October 2013

Forces and more forces

This is a short post on forces, Coulombs, Gravitational and Plancks. I will be doing more on this in later posts but this is just a little result that I stumbled upon today. I am certain others have been down this road, but it is a nice result so I thought I would share it with you.

I am not going to go into full detail on how I got this one, it is mostly irrelevant to this post and I will cover it elsewhere. So to start, Planck force is defined as

$F_p = $ $\frac {c^4} {G}$   ...(1)

c- speed of light
G - gravitational constant

Coulombs force

$F_c = $ $\frac {e^2} {4 \pi \epsilon_0 r^2}$    ...(2)

e - charge on the electron
$1 / 4\pi \epsilon_0$ - Coulombs constant
r - distance between the electrons

Gravitational force

$F_g = $ $ \frac {m_e^2 G} {r^2} $    ... (3)

$m_e$  - mass of the electron
r and G are the same as those above

Dividing (3) by (2) gives

$ \frac {F_g} {F_c} = \frac {m_e^2 G 4 \pi \epsilon_0} {e^2}$    ...(4)

the r term has cancelled here.

Now let's take a look at $r_e$, the classic electron radius, given by

$r_e = $ $\frac {e^2} {4 \pi \epsilon_0 m_e c^2}$   ...(5)

We are going to do two things here, the first is to multiple (2) by $r_e$, while also replacing $r$ with $r_e$ to give

$F_c r_e = $ $ \frac{e^2 r_e} { 4 \pi \epsilon_0 r_e^2} $

we can now cancel the $r_e$ on the right to give

$F_c r_e = $ $ \frac{e^2} { 4 \pi \epsilon_0 r_e} $  ...(6)

now we put (5) into (6) to give

$F_c r_e = $ $ \frac {e^2 4 \pi \epsilon_0 m_e c^2} { 4 \pi \epsilon_0 e^2} $

most of this cancels to leave

$F_c r_e = m_e c^2 = E $    ...(7)

Secondly put $r_e$ into (2) in place of r and sub in its value from (5) giving

$F_c = $ $ \frac {e^2 (4 \pi \epsilon_0)^2 m_e^2 c^4} { 4 \pi \epsilon_0 e^4} $

after some cancelling this becomes

$F_c = $ $ \frac { 4 \pi \epsilon_0 m_e^2 c^4} {  e^2} $ ...(8)

divide this bu $F_p$ to give

$\frac {F_c} {F_p} =  \frac { 4 \pi \epsilon_0 m_e^2 c^4 G} {   c ^4 e^2} $

cancelling leaves

$\frac {F_c} {F_p} =  \frac { 4 \pi \epsilon_0 m_e^2 G} { e^2} $    ...(9)

but this is just (4), so

$\frac {F_c} {F_p} =  \frac {F_g} {F_c} $    ...(10)

so when r = $r_e$ we have

$F_c^2 = F_p F_g $    ...(11)

Coulomb force squared is equal to the Planck Force multiplied by the gravitational force. Cool eh?

Wrote this while listening to this.

Wednesday 16 October 2013

This is NOT the fine structure constant

I came across something the other day, won't put a link to it. An alternative theory of everything type thing. You can't help stumble upon them if you are on the web looking at physics sites. Now this one had come up with an alternative idea where they did all these calculations to prove this and that.

The problem was that, although the figures appear to be right, the physics was just plain wrong. I won't give the example, but I will give you one of my own. Try this one on for size,

$ \alpha^{-1}$ $ = \frac {e^2 G^2} {m_e^2}$   ...(1)

$\alpha$  - to be calculated
$e$ - charge on the electron - $1.602176 10^{-19} C$
$G$ - Gravitational constant - $6.67384   10^{-11} m^3 kg^{-1} s^{-2}$
$m_e$ - mass of the electron - $9.109382  10^{-31} kg$

put in the numbers and what do you get?

$ \alpha^{-1} = 137.8876$    ...(2)

Hold on a minute that is real close to the fine structure constant, which is

$\alpha^{-1} = 137.035999$   ...(3)

In fact my value from (2) divided by (3) is 1.00621! I'm onto something here right. In fact if we just tweak the Gravitational constant a little, use

$G$ - Gravitational constant - $6.65573   10^{-11} m^3 kg^{-1} s^{-2}$

then it comes out exact. We all know that the gravitational constant is difficult to measure so who is to say I am not right with my value? We have all read that gravity is different if you are on the top of a mountain, or at the poles due to the shape of the earth right?

Not only that but I can go further, the real fine structure constant is just given by

$\alpha =$ $ \frac {e^2} {4 \pi \epsilon_0 \hbar c}$    ...(4)

using my value from (1) I can show

$ \frac {e^2} {4 \pi \epsilon_0 \hbar c} = \frac {e^2 G^2} {m_e^2}$   ...(5)

re-arrange this and you have

$m_e^2 = G^2 4 \pi \epsilon_0 \hbar c$   ...(6)

Oh my word, the mass of the electron is just the Gravitational constant multiplied by the other universal constants. This is a MAJOR result.

This is also completely WRONG!

Anyone checking this will notice I pulled a trick when I compared (4) to (1) and said they were equal, they are not. (1) is the inverse of (4), so (5) and (6) are mathematically wrong.

Now you may be thinking ok, but there is still something to (1) it is so close. Yes it is close, but it is not close enough. It is not good enough just to argue that the experimenters have it wrong and then dream up some daft excuse on why they have it wrong.

Acceleration due to gravity does vary across the earth, but the Gravitational constant does not. The Gravitational constant, as the name suggests, is constant!

Finally, even if they did have the value of the Gravitational constant wrong, and say they admitted it on the news tomorrow. "The actual Gravitational constant is wrong it should be ....", drum roll,

$G$ - Gravitational constant - $6.65573   10^{-11} m^3 kg^{-1} s^{-2}$

the value from (1) will equal the fine structure constant in numerical value, but it will still NOT be the fine structure constant!

Eh? how can it be numerically equal and not be the same? After all 2+2=4, 2+2 is the same as 4. Taught me that early on. So if equation (1) gives 137.035999 and the inverse of the fine structure constant is 137.035999 then they are the same. Have to be. So

$ \alpha^{-1}$ $ = \frac {e^2 G^2} {m_e^2}$   .

is true if $G$ - Gravitational constant - $6.65573   10^{-11} m^3 kg^{-1} s^{-2}$, right?

WRONG! and this is the point of this post. Although the numbers are the same, the equation does not balance when it comes to dimensions. In physics the dimensions have to balance. The fine structure constant has no units/dimensions, it is dimensionless. but the right hand side of the above has the units

$ Q^2 L^6 M^{-4} T^{-4} $  ... (7)

$Q$ - charge
$L$ - Length
$M$ - Mass
$T$ - time

which is far from dimensionless.  So even though the numbers are close numerically, they are not the same dimensionally. If you multiply a velocity by mass it is no longer a velocity, it is momentum.

So while equation (1) may look intriguing it is wrong. This result is just one of things.

Dimensions seems to be forgotten in many of the online Theories of everything.

The next time you come across an online Theory of everything check the dimensions, if they are wrong, then guess what? yep, it's wrong!

In order to make equation (1) unitless we would have to find something, lets call it K, with units of

$ Q^{-2} L^{-6} M^{4} T^{4} $  ... (8)

does such a thing exist? Actually it does

$K =$ $\frac {c Z_p} {G^3}$ $= 3.02352   10^{40}$ ...(9)

where $Z_p$ is the Planck Impedance, but this is not even close to 1 so while it would make the dimensions correct it would also change the value of $\alpha$ in (1) by a large amount!

Wrote this while listening to this.

Tuesday 15 October 2013

Planck, Schrödinger and positronium

In a previous post on natural units I derived a number of equations including

$m_\psi{'} = m_p \alpha^{5/2}$   ...(1)

$m_\psi{'}$ - alternative version of the Schrödinger mass
$m_p$ - Planck mass
$\alpha$ - fine structure constant

using Einsteins energy-mass formula

$ E = m c^2 $ ...(3)

we can derive a Schrödinger, Planck and electron version of energy such that

$ E_p = m_p c^2 $ ...(4)

$ E_\psi = m_\psi{'} c^2 $ ...(5)

$ E_e = m_e c^2 $ ...(6)

$m_e$ - mass of the electron
c - speed of light in a vacuum

Now, let us choose

$E_0 = m_0 c^2$   ...(7)

so that

$E_e E_\psi^2 = E_p^2 E_0$    ... (8)

this gives

$ m_e c^2 m_\psi{'}^2 c^4 = m_p^2 c^4 E_0$   ...(9)

now sub in the value from equation (1)

$ m_e c^2 m_p^2 \alpha^5 c^4 = m_p^2 c^4 E_0$    ...(10)

divide through by $m_p^2 c^4$ and use the Planck relationship

$E_0 = 2 \hbar \omega_0$   ...(11)
$\hbar$ - reduced Planck constant
$\omega_0 = 2 \pi \nu$
$\nu$ - frequency of the photon
factor of 2 is for 2 photons

to give

$2 \hbar  \omega_0 =  m_e c^2 \alpha^5$    ...(12)

re-arrange to give

$\frac {1} {\omega_0} = \frac {2 \hbar} {m_e c^2 \alpha^5} $  ... {13}

replace $\omega_0$ as follows

$t_0 =$  $ \frac {1} {\omega_0}$    ... (14)

$t_0$ has units of time, sub this into (13) and you end with

$t_0 =$  $ \frac {2\hbar} {m_e c^2 \alpha^5}$     ...(15)

If we put some numbers into (15) we get a value of

$t_0 = 1.244   10^{-10} s$   ...(16)

Here is the point of this post. So far we have used Einsteins equation and a  result we got from a previous post on natural units. That result, equation (1) shows  a relationship between two natural units of mass, one is an alternative Schrödinger mass, the other the Planck mass. We used these to derive equation (15).

Equation (15) also happens to be the mean life time of para-positronium (p-Ps)! Here is a link to a wiki page. How cool is that?

Will investigate this result in more detail in the next post. Note that there is a little ambiguity as to what the "2 photons" referred to in equation 11 actually are. There is also the value of $m_0$ mentioned in equation 7. These both need to explained in greater detail.

Wrote this while listening to this.

Stoney, Planck and Schrödinger

Recently I have been looking at the idea of universal or natural units. The two best known seem to be examples given by Stoney and Planck, but there are others, a good post can once again be found on the fabulous Wikipedia.

While doing this I also came across Schrödinger units. I will list them here along side Stoney and Planck


Quantity Stoney Planck Schrödinger
Length(L) $l_s=\sqrt \frac{Ge^2}{c^4 4\pi \epsilon_0}$ $l_p = \sqrt \frac{\hbar G} {c^3} $ $l_\psi = \sqrt \frac {\hbar^4 G (4\pi \epsilon_0 )^3} {e^6} $
Mass(M) $m_s=\sqrt \frac {e^2} {G 4 \pi \epsilon_0}$ $m_p=\sqrt \frac {\hbar c} {G} $ $m_\psi=\sqrt \frac {e^2} {G 4 \pi \epsilon_0}$
Time(T) $t_s = \sqrt \frac {Ge^2} {c^6 4 \pi \epsilon_0}$ $t_p = \sqrt \frac {\hbar G} {c^5}$ $t_\psi = \sqrt \frac {\hbar^6 G (4 \pi \epsilon_0)^5}{e^{10}}$
Electric Charge (Q) $q_s = e$ $q_p=\sqrt {4 \pi \epsilon_0 \hbar c}$ $q_\psi = e$
Temperature ($\theta$) $T_s = \sqrt \frac {c^4 e^2} {G 4 \pi \epsilon_0 k_B^2}$ $T_p = \sqrt \frac {\hbar c^5}{G k_b^2}$ $T_\psi = \sqrt \frac {e^{10}} {\hbar^4 (4 \pi \epsilon_0)^5 G k_b^2}$

First things first, I have not been able to confirm the Schrödinger units. The various sites out there look like they are quoting a single source. The danger of course is that if that source is wrong, then so is everyone else!

Let's proceed as if all is well and the values given above are correct. We can derive a number of results.

Length 

It can be shown that

$l_p^4 = l_s^3 l_\psi$     ...(1)

$\frac {l_s} {l_\psi}$ $ = \alpha^2$    ...(2)

$\frac {l_s^2} {l_p^2}$ $ =\alpha$    ...(3)

Mass

It can be shown that

$\frac {m_s^2} {m_p^2}$ $ =\alpha$  ...(4)

and

$m_s = m_\psi$    ...(5)

I think (5) introduces are a rather interesting problem, consider the following Planck result

$\frac {l_p m_p} {q_p^2}$  $= 10 ^{-7}  $   ...(6)

the Stoney version is

$\frac {l_s m_s} {q_s^2}$ $  = 10 ^{-7}  $   ...(7)

for an electron it is 

$\frac {r_e m_e} {e^2}$ $  = 10 ^{-7}  $   ...(8)

$r_e$ - classic electron radius, $m_e$ - mass of electron , $e$ - charge on an electron

where as for Schrödinger it is not the case 

$\frac {l_\psi m_\psi} {q_\psi^2}$ $  \not= 10 ^{-7}  $   ...(9)

The obvious place to look is at the Schrödinger mass term, use an alternative version where (9) now works

$\frac {l_\psi m_\psi^{'}} {q_\psi^2}$  $= 10 ^{-7}  $     ...(10)

Dividing (10) by (7) and using $q_s = q_\psi = e$

$ \frac {l_\psi m_\psi^{'}} {l_s m_s}$  $= 1$    ...(11)

using (2) and rearranging this gives

$ m_\psi{'} = m_s \alpha^2 = m_\psi \alpha^2 $   ...(12)

This would give the Schrödinger mass a value of 
$m_\psi{'} = 9.9   10^{-14}  kg$ 

which I think is a more appropriate value. I call this the Alternative Schrödinger mass.

Going forward and using (4) gives

$m_s^5 = m_p^4 m_\psi^{'}$    ...(13)

this can be substituted back into (10) along with the result obtained in  (1) gives

$\frac {l_\psi m_\psi^{'}} {e^2} = \frac {(\frac {m_s^5 l_p^4}{m_p^4 l_s^3})}{e^2} = \frac {l_s m_s} {e^2}$   ...(14)

taking the second half of the equation this just becomes

$\frac { m_s^5 l_p^4} {e^2 m_p^4 l_s^3} = \frac {m_s l_s}{e^2}$   ...(15)

which simplifies to

$m_s l_p = m_p l_s $    ...(16)

Using the Schrödinger mass derived in (12) we have 

$m_\psi{'} = $ $\sqrt \frac {e^{10}} {G (4 \pi \epsilon_0)^5 \hbar^4 c^4} $ $= m_p \alpha^{5/2}$   ...(17) 

This is probably the best place to wrap up this post, but before I do, I did notice something a little odd though when I was doing this, but I will cover it in another post.

Wrote this while listening to this.

Monday 14 October 2013

Something about symmetry.

The last couple of posts have been a little equation heavy. Most of the maths is not that difficult once you get passed the nomenclature and the subscripts and superscripts, I have done a post on this that just needs publishing, will get this out there soon. This post is a little lighter on equations, but there are still a few. Stick with it, if there is any truth in all this then it will be worth seeing.

Consider a high energy photon travelling through space. A photon cannot spontaneously change into mass because of the conservation of momentum. The momentum part of the photon is assimilated into the nearest mass object, a nucleus for example. It cannot be imparted as momentum into the newly created electron/positron pair because there is not enough energy (I will cover the maths of this in another post).

In a previous post I tried to explain an odd relationship involving Kepler's 3rd law of planetary motion. This came about from looking at the photon/mass interaction. One thing that came out of this post was Newton's law may actually extend all the way down to Planck lengths.

(I appreciate that there is currently experiments under way to test Newtons equation of gravity. For now though  I am going to proceed with the idea that it is still intact. It being subject to modification when we get into the large gravity arena when general relativity takes over.)

A photon needs something to hit in order to dissipate its momentum. Now you may be thinking the following;

Surely, if you fire to high energy lasers at each other then you could get two photons travelling in opposite directions to interact. This would then cancel the momentum while allowing mass to be created? Just like to billiard balls travelling in opposite directions. 

I am afraid not, in classical electrodynamics theory it is known that waves pass each other without interference. From Quantum Electro Dynamics (QED) it appears that photons cannot interact with each other because they do not carry charge.

NOTE: the previous paragraph will be explained in another post. The bit about QED is not strictly true, but will do for now.

I have highlighted a line there for two reasons. QED is the most accurate theory we have to date, so deserves some respect. Also it says that photons cannot interact because they do not carry charge. I take this to also mean that in order for mass to change to energy or vice versa according to

$E = m c^2$     ...(1)

we need charge. So while we always talk about the amount of energy we need to create mass, or the amount of energy that is liberated when mass is converted into energy, we actually need charge to be present? So, which is it? mass or charge or both? Of course, where there is charge there is mass.

In an earlier post I used a photon and mass to derive (1), this is very common. Imagine though, that this is just a consequence, what is really important is the amount of energy required to create charge, mass being a by product. Ages ago I did a post on conservation. Energy is conserved and so is charge and momentum, linear and angular.

What if the energy to create mass is actually the energy required to create charge?

why can't we have a neutral version of an electron much like a neutron? the nelectron! no that is not a typo. So a photon would just convert into a single nelectron, like so

$ \gamma \to e^0$

This never happens.

------------------------------------------------------------------
Aside: When it comes to photons creating electron/positron pairs, we argue that two particles have to be produced in order to balance the charge. Consider the neutrino/proton interaction.

$\bar{v_e} + p^+  \to n^0 + e^+  $

I have always assumed that no charges are created here. A positron is just liberated, taking away charge from the proton. Are we actually saying that we are creating two charges, an electron/positron pair? like this,

$\bar{v_e} + p^+  \to [p^+ + e^-] + e^+ $

The electron then binds to the proton to create a neutron and the positron is liberated?

$p^+ + e^- + e^+ \to n^0 + e^+  $

All this happening under the hood?

I'll do some research and then I will be doing a later post on about this.
-------------------------------------------------------------------

In positronium, the electron and the positron recombine and annihilate each other, converting into two or more photons, note that they NEVER annihilate into a single photon. Once again this is because of the conservation of momentum. Do we need mass/charge available in order for this to occur in addition to that provided by the electron and the positron? In other words, if the electron and the positron were out in the vacuum of inter galactic space (which is about the best vacuum you are going to find) would they still recombine and annihilate. My current understanding is that yes they would.

Let's try running it backwards and see what happens  (after all Feynman said that a positron is just an electron traveling backwards in time, more on this later). In this case we would see two photons come together, create an e/p pair that then recombine into a single photon.

Hold on a minute, there are two problems here. The first is that we have already stated that two photons cannot interact to form an e/p pair. Second at the end of the process e/p annihilation cannot create a single photon.

So this is not a symmetrical process. By looking at the results you would know if you were watching the film in reverse or not. I can't help thinking that there is something important here that I am missing.  

One thought did occur to me while writing this post, is it the case that whenever there is a energy to mass conversion there are at least 2 charges created, one positive and one negative? For a mass to energy conversion we destroy one positive and one negative charge? I like the idea, but can it be true?

A photon with enough energy being converted to mass will do so in the presence of charge. Is it the charge that creates +/- charge pair, or the photon itself? If my understanding of QED, which is very little, is correct then a photon in the electromagnetic field causes an excitation in the electron/positron field creating the charge pair. (Need to learn more about this I think.) So the electron/positron field creates the charge pair after being excited by a photon from the electromagnetic field? In which case we would always expect +/- pairs to be produced.

Does this mean we also have a muon/antimuon field and a tau/antitau field. I'm not buying that I'm afraid, the universe is much to elegant for such profligacy.

When an electron/positron pair annihilate they are in the presence of their own charge. Is this the reason they can annihilate? When the e/p do annihilate they cause excitations in the electromagnetic field in the form of photons. You might be tempted to argue that the neutron and anti neutron both have no charge and happily annihilate each other. But this is not strictly true because they are made up of quarks and anti quarks that do have charge (supposedly).

So to finish this post. The conversion of energy into mass and vise versa is dependent on conditions and other events happening in addition to the requirement given by equation 1. I am convinced that the solution will be far simpler than we think. Will let you know if I get anywhere.

Wrote this while listening to this.


Thursday 10 October 2013

Kepler part 2

In a previous post on Kepler I came across an odd link between a calculation I had been performing on photon - electron/positron production and Keplers 3rd Law,

$ \frac {4 \pi^2} {T^2} = \frac {G M} {R^3} $    ...(1)

G- gravitational constant
M - mass of the larger body
R - distance between the center of mass of the two bodies
T - period

So I thought I would take a look at this in more detail and lo and behold a nice solution drops out, Take a look at this, set

T = $1 / \nu$
$\nu$ - frequency of the photon
M - $2m_e$ - mass of 2 electrons
$R^3 = V = l_p^2 d$
$l_p$ - Planck length
$d =  \lambda / 2\pi$
$\lambda$ - wavelength of photon

So we now have

$ 4\pi^2 \nu^2 =$ $ \frac {G 2 m_e} {l_p^2 d} $   ...(2)

sub in the value for d and rearrange so

$ \frac {4\pi^2 \nu^2 \lambda} {2\pi} $  $ l_p^2 = G 2 m_e   $   ...(3)

using

$c = \nu \lambda$    ...(4)

c- speed of light

$ l_p^2 = $  $\frac {\hbar G} {c^3} $   ...(5)

$\hbar = h / 2\pi $ - reduced Planck constant

subbing these into (3) give

$ \frac { 2\pi \nu c \hbar G} {c^3} =$  $ G 2 m_e $    ...(6)

dividing through by G and multiplying each side by $c^2$ gives

$2\pi \nu \hbar = 2 m_e c^2 $    ...(7)

or as it is more usually written

$ E = h \nu = 2 m_e c^2 $     ...(8)

I think that is lovely. It is the use of

$R^3 = V = l_p^2 d$    ...(9)

that generates the solution.

So, how do we arrive at (9)? Consider this, Newton's equation for gravitational force is given by

$F_{grav} = G M_1 m_2 / R^2$   ...(10)

Then we have, energy is force x distance, so say we try this

$E = mc^2 = F_{net} d$    ...(11)

Then lets say that

$F_{grav} = F_{net}$    ...(12)

(does this imply $F_{net}$ is a centripetal force?) this becomes

$\frac {GM_1 m_2} {R^2} = \frac {mc^2} {d}$   ...(13)

lets say

$M_1 = 2m_e$
$R = l_p$
$m_2 = m$
$c = 2\pi d \nu$

so

$F_{grav} =$  $ \frac {G2m_e m} {l_p^2}$     ...(14)

cancelling the m on both sides (13) becomes

$\frac {G2m_e} {l_p^2} = \frac {4 \pi^2 \nu^2 d^2} {d}$     ....(15)

cancelling the d on the right and rearranging gives

$\frac {G 2 m_e} {l_p^2 d} $ $=4\pi^2 \nu^2 $     ....(16)

which is just equation (2) above. So what does that tell us? Let's take a look at (14)

$M_1 = 2m_e$

This value is because we have an electron positron pair, both with a mass of $m_e$. The distance, $l_p$, is the Planck length, given in (5) above.

What is this saying? at the moment of creation of the e/p pair there is a mixed state photon in orbit around the electron/positron pair. There is a force associated with this photon that is equivalent to a gravitational force given by (14) ?

There are a few things that concern me here. If we sub (5) into (14) then the G divides out. Is it still a gravitational force even though G is no longer in the equation? ( I think there may be an argument that can be used here to say that it is still a Gravitational force, but I will cover it in another post.) for now...

$F_{grav} =$  $ \frac {2m_e m c^3} {\hbar}$     ...(17)

which becomes

$F_{grav} =$  $ \frac {E_{ep} m c} {\hbar}$      ...(18)

$E_{ep} = 2 m_e c^2$   - creation energy of an electron/positron pair
also, Compton gave us

$  m \frac {\lambda} {2\pi} = \frac {\hbar}{c} $   ...(19)

subbing this into (17) and using the value of d from earlier

$d = 2 \pi \lambda$

gives

$F_{grav} =$  $ \frac {E_{ep}} {d} $ ...(20)

which is the same form as  (11).

Another question is whether the inverse square law survives at these distances. There is some evidence that it may not be true for sub-millimeter distances and yet here it is at the Planck length! I'm sure Erik Verlinde will be shaking his head at this point. On the plus side, even though it is an incredibly short distance, the masses involved are considerably smaller than the Planck mass and so should obey Newtonian gravity rather than having to worry about general relativity effects. Wouldn't it be magnificent if Newtonian gravity extended all the way down to the Planck length!

What is the meaning of m in (14)? It cancels out so makes life easy, but what is it?

In the Kepler version of (2), $l_p = R$, $d = R$, giving $R^3$. In the version above

$d = \lambda / 2\pi$ - the wavelength of the photon, the equations and the numbers work with this value rather than $l_p$. While this sort of makes sense I am still a little worried.

So while we may have got to the bottom of why Kepler's third law seems to hold for sub atomic particles, there are still some outstanding questions to be resolved.

There is no time dependency discussed here . Once the e/p pair are created they go off in there separate directions and the photon is no more. The analysis described here can only be valid for a certain period of time. What is this? Is it related to the idea of the photon creation/destruction time mentioned in previous posts? This happens to be

$ \delta t = 1 / 2 \pi \nu  = 6.44044   10^{-23} s$
$\nu$ - frequency of the initial photon

Lastly,  I am arguing that Newton's gravitational law holds all the way down to Planck's length for an electron/positron pair. These are charged particles!

What about the Coulomb force?

If Coulomb's law holds, at the distances we are talking about it should be colossal and because electrons and positrons are oppositely charged it should be attractive. Definitely need to explain this away. Could it be that in the initial phase of electron/positron creation the charge has not separated enough to be +/-, or does the weak interaction take care of the Coulomb force, who knows? not me!

Wrote this while listening to this.

Kepler, Acceleration and some numerology

NOTE: since posting this I have found the explanation for the results shown here. You can find it in this post.

This is a very short post just to highlight a little oddity I noticed in my blog on acceleration. The title of the post was the conservation of acceleration, click on the link for details. This is just a bit of fun, but it is also very odd indeed.

In the post I derive a number of values including the time it takes to create/destroy a photon, $\delta t$. In the example given the photon has just enough energy to create an electron/positron pair.

$\delta t = 1 / \omega = 6.4404433   10^{-22} s  $
$ \omega = 2 \pi \nu $
$ \nu$ - frequency of photon

a further calculation is performed where a volume is derived and found to be

$ V = l_p^2 d =  5.0434295   10^{-83} $    ...(1)
$ d= \lambda / 2 \pi $
$ \lambda$ - wavelength of the photon
$l_p$ - Planck length

the last was someting relating to moment of inertia of a photon given by

$I = \frac {\hbar c} {g} $

I - moment of inertia - $6.7919   10^{-56}   kg  m^{-2}$
$\hbar$ - reduced Plancks Constant
c - speed of light in a vacuum
$g$ - photon creation acceleration - $4.654845  10^{29} ms^{-2}$

Now the mass of electron and gravitational constant are given by

$m_e - 9.1093829  10^{-31}  kg$
$G - 6.67384   10^{-11}     m^3 kg^{-1} s^{-2} $

Kepler's 3rd law of planetary motion is

$ \frac {4 \pi^2} {T^2} = \frac {G M} {R^3} $    ...(2)

G- gravitational constant
M - mass of the larger body
R - distance between the center of mass of the two bodies
T - period

rearrange this to give

$ R^3 4\pi^2 = G M T^2 $    ...(3)

now set
T = $2\pi \delta t$   - time period for frequency $\nu$
M = $2m_e$    - mass of e/p pair

$ R^3 = 2 G m_e \delta t^2$
$ R^3  = 5.043433   10^{-83}  $

but this is just the value of V from earlier, in other words

$R^3  = V $  ... (4)

$R = 3.694668    10^{-28}m$

taking this value of R and putting it into

$I_m = \mu R^2$  ...(4)

where $\mu$ is some reduced mass, we get a moment of inertia of

$I_m = 1.36506   10^{-55} \mu  $

dividing by the value of $I$ mentioned earlier gives

$I_m/I = 2.00983 \mu$    ...(5)

if we set (5) = 1 then

$2.00983 \mu = 1$

$\mu = 0.4975546 = \frac {m_0 m_1} {m_0+m_1} $ ...(6)

$m_1 \approx \frac {m_0} {2.00983m_0 -1}  $ ....(7)

So let's get this straight, we carried out a calculation involving the creation of an electron positron pair. As part of this calculation we determined a volume related to the Planck constant and the wavelength of the photon, equation (1).

We then found those values actually work in Kepler's 3rd Law! How cool is that?

In addition, an earlier idea about moment of inertia of a photon has comparable values to that obtained using a traditional moment of inertia calculation, though I have to say that this is a very weak idea, need to take more of a look at it.

Seeing as this is just a bit of fun and an odd coincidence I am going to abandon my usual caution and make some guesses as to what I think is going on.

NOTE: These ideas are just guesses, this is NOT real physics, this is a bit of fun! After all, we have not ven considered the Coulomb force!

What is the meaning of R in (4) ? In Kepler's 3rd law it is the distance between the center of mass of the two bodies. Is this the distance between the e/p pair? I don't think so.

I think that e/p pair represent the larger body, M in equation (3). So R could be the distance between the e/p pair and the center of mass of the photon that creates the e/p pair. Will ponder this more and then update this post.


Wednesday 9 October 2013

Conservation of acceleration

In the previous post on the conservation of force and acceleration I looked at the idea that there may be an acceleration that links energy, mass and momentum. In this post I am going to have a go at putting in some numbers to the equations and see how things come out.

The equations of interest from the previous post are

$Ec = \hbar g $   ...(1)

E - energy
c - speed of light
$\hbar$ - reduced Planck constant
$g$ - acceleration

$ F = \frac {\hbar} {c^3} g^2 $   ....(2)

F - force

$ m = \frac {\hbar} {c^3} g $    ...  (3)

m - mass

$ g = c \omega $     ...(4)

$\omega$ = $ 2\pi \nu$
$\nu$ - frequency

$ g \hbar = p c^2 $    ...(5)

p - momentum.

$I =\frac {\hbar c} {g} $   ...(6)

To open we are going to use the standard equation for calculating the minimum energy required to create a positron/electron pair

$ \hbar  \omega = 2 m_e c^2 $    ...(7)

$m_e$ - mass of the electron
the factor of two is because we are creating a positron/electron pair. Replace $\omega$ using (4) to give

$ \frac {\hbar} {c} g = 2 m_e c^2 $    ...(8)

Using standard values
$m_e - 9.109382   10^{-31}$ kg
$c - 299792458$ m/s
$\hbar  - 1.054571  10^{-34}$  J.s
giving

$g = 4.654845  10^{29} ms^{-2} $     ...(9)

Putting this value into equations (2) , (5) and (6) gives

$F = 0.848055 N$

$p = 5.461848  10^{-22}$ kg m/s

$I = 6.7919  10^{-56} kg   m^2 $

also, the photon creation/destruction time and distance is given by

$ \delta t = 1 / \omega = 6.44  10^{-22} s $
$ s = 9.653976   10^{-14} m$

So let's take a look, the only new value is the $g$, the rest can all be calculated from existing theories. $g$ is massive. Consider a neutron star with its massive gravity, this may typically have a value of $10^{12} ms^{-2}$ but this is tiny compared to the value quoted in (9). The question though is does it actually mean anything, also, can we find out anything new from this value?

The value in (9) is the minimum acceleration required to create mass.

The next section relates to something I first discussed back when I was discussing physics units and dimensions. In that post I pointed out that Energy density and pressure have the same units.

In Planck units they are identically the same

$ \rho_p^E = \frac {E_p} {l_p^3} = \frac {c^7} {\hbar G^2} = \frac {g_p^2} {G} $ ...(10)

$\rho_p^E$ - Planck energy density
$E_p$ - Planck Energy
$l_p$ - Planck length
$g_p$ - Planck acceleration
G - gravitational constant
$\hbar$ - reduced Planck constant
c - speed of light in a vacuum

$ p_p = \frac {F_p} {l_p^2} = \frac {c^7} {\hbar G^2} = \frac {g_p^2} {G} $ ...(11)

$p_p$ - Planck Pressure
$F_p$ - Planck force

$ \rho_p^E = p_p$    ... (12)

further Planck Intensity or irradiance is given by

$ I_p = \rho_p^E c = p_p c = \frac {g_p^2 c} {G} $ ....(13)

the general form of irradiance is given by

$ p = \frac {I} {c}$ ...(14)

can we then say

$ \rho^E = \frac {g^2} {G} $ ...(15)

$ p = \frac {g^2} {G} $ ...(16)

if this is valid then putting in the value of $g$ into (15) and (16) gives

$ p = 3.24664   10^{69} = F / A $  ... (17)

A - area.

Using the value of F from earlier gives an area of

$ A = 2.612   10^{-70}$ , giving a length of $l = 1.61619  .10^{-35}  = l_p$ the Planck length.

Does this imply that the general form of (11) is

$ p = \frac {F} {l_p^2} =  \frac {g^2} {G} $ ...(18)

using the definition for $l_p$

$ l_p^2 = \frac {\hbar G } {c^3} $    ...(19)

putting this into (18) and rearranging results in (2) from earlier. Applying the same result from (17) to (15) gives

$\rho^E = 3.24664   10^{69}  = E / V $   ... (20)

E - energy - $1.63742  10^{-13}$ J, this is the energy required to create e/p pair.
V - volume - $5.0434295   10^{-83}    m^3$

If we take

$ V = l_p^2 d $   ...(21)

where d is a length, then

$d = 1.930699   10^{-13}  = 2 s $

where s is the creation/destruction distance calculated earlier. The wavelength of the photon that creates the e/p pair is given by

$ \lambda = c / \nu = 1.213156   10^{-12}   = 2 \pi d $

A further result from a previous post was this

$ g l = c^2 $     ... (22)

using the value of d calculated here for $l$ and using $g$ from above the value of c is found to be

$ c = 299784999 m s^{-1} $ which is 0.999975 of c.

So, what have we learned from this post. While the physical meaning of $g$ is still not proven the equations involving this variable are found to be true, at least for the example given here. The value of $g$ is found to be incredibly high and would be so even for relatively low energy photons.

Equations for energy density and pressure have been derived (guessed!) and values calculated for a photon converting to e/p pair. The energy density and pressure have been found to be very large and have a dependency on the gravitational constant. This I find quite remarkable, a microscopic result linked to the gravitational constant.

If we assume that the electron is the smallest particle mass that can be created, the the values for acceleration, energy density and so on can be considered the smallest values that allow the creation of matter  (e/p pair) to take place.

There is of course one small problem here, momentum. The amount of energy discussed here is just enough to create the e/p pair, but these would have no momentum. If they did it would require more energy because

$ E = 2\sqrt {(pc)^2 + (m_ec^2)^2}  $

So what happened to the momentum, after all it was calculated to be

$p = 5.461848  10^{-22}$ kg m/s

Traditional thinking is that this is imparted into the local mass, a local nucleus for example. This is also the reason why a photon cannot spontaneously convert itself into an e/p pair. It needs mass. Does this imply that mass is required to accelerate/decelerate energy, yes, I think it does. It is this that will be covered in a later post (I will add a hyperlink when I publish it).


Tuesday 8 October 2013

Conservation of Force (Acceleration)

Had a bit of a surprise earlier.  It dawned on me today that the entire universe may actually be governed by one major principle. That of force, well actually acceleration, so as usual I do the Google thing and low and behold there is nothing new under the sun.

Hermann Ludwig Ferdinand von Helmholtz: On The Conservation Of Force, 1863!

'According to Helmholtz, the primary curator of this principle, the "law of conservation of force", as he called, had been enunciated prior to him by Isaac Newton, Daniel Bernoulli, Benjamin Thomson, and Humphry Davy. Likewise, in the 1670s the theory of vis viva or “living force” of German mathematician Gottfried Leibniz was prominant. In 1837, German pharmacist Karl Mohr gave one of the earliest statements of the conservation of force...'

The law of conservation of force, according to Helmholtz, states:

“The quantity of force which can be brought into action in the whole of nature is unchangeable, and can neither be increased nor diminished.”

If anyone has read my previous posts on the nature of photons and the moment of inertia of a photon you would have seen that I have been doing a little playing around with the Planck Units. Out of this came the equation

$ E c = \hbar g $     ...(1)

E - energy
c - speed of light
$\hbar$ - Reduced Planck constant
g - acceleration

From this equation I suggested that maybe energy has an associated, intrinsic, acceleration given by equation (1). I went further to suggest that  photons are created in a finite time given by

$ \delta t = 1 / \omega $    ...(2)

$\omega$ - frequency of the photon

and that the creation process actually takes place over a distance of

$  s = \lambda / 4 \pi $    ...(3)

$\lambda$ - wavelength of the photon created.

There are a number of issues with the idea, but lets say for now that it may have some merit. Continuing on the same idea it is possible to derive the following from Planck Units

$ F_p = \frac {\hbar} {c^3} g_p^2 $    ...(4)

where
$g_p$ - Planck acceleration, defined below
$F_p$ - Planck Force, defined below

$ F_p = \frac {c^4} {G}   ,   g_p = \sqrt \frac {c^7} {G\hbar} $   ...(5)

In a post I will publish later it is possible to show that there are real world equivalents to many of the equations you can derive using the Planck units. This is were we get a little dodgy because if we apply this idea then from (4) we have

$ F = \frac {\hbar} {c^3} g^2 $    ...(6)

Where $g$ is the acceleration described in (1). In fact, using (1) and the standard Planck relationship

$ E = \hbar \omega$   ... (7)

it is possible to derive

$ E = F \frac {\lambda} {2\pi} $    ...(8)

The maths is not that difficult, but if you are struggling I am happy to post it in detail. Further if (6) is correct then using Newton's version

$F = m a$    ...(9)

with a = $g$ and

$m = \frac \hbar {c^3} g$     ...(10)

using this value of m in

$ E = mc^2 $    ...(11)

just gives us

$ E = \frac \hbar {c^3} g c^2 = \frac {\hbar} {c} g$     ...(12)

which just boils down to (1). From the previous post another equation was derived

$g = c    \omega $    ...(13)

$\omega$ - $2 \pi \nu$
$\nu$ - frequency

It is this last equation, along with (1) and (10) that this post is really about. Is it possible that

equation (1) really does represent a relationship between energy and acceleration?
equation (10) indicates that mass also as a direct relationship with the very same acceleration?
equation (13) a photon can also exist that is directly dependent on acceleration?

If there is any truth to any of this then is the implication that energy and mass are related not only by (11), but also (1), (10) and (13), namely, acceleration. Is acceleration really that significant?

Further, are the conservation of energy and the conservation of momentum laws actually a single law - the conservation of acceleration?

Hold on one minute. Conservation of momentum, we haven't even mentioned that yet, how did that sneak in? Take (12) and apply it to momentum

$ E = p c = \frac {\hbar} {c} g $   ... (14)

p - is momentum of a photon, so

$ g \hbar = p c^2 $     ...(15)

again a direct relationship between acceleration and momentum. Subbing this back into (6) gives

$ F = p \frac {g} {c} $    ... (16)

but using (13) this just gives

$ F = p   \omega $    ...(17)

using (2)

$ F \delta t = p $    ...(18)

which is the more traditional representation of momentum and force.

The final part of this post on acceleration concerns electric charge, from Planck units

$ V_p = m_p c^2 / q_p  = \frac {\hbar g_p} {q_p c} $     ...(19)

$q_p$ - Planck charge
$V_p$ - Planck voltage
$m_p$ - Planck mass

and the equivalent

$ V = m_e c^2 / e  $     ...(20)

$m_e$ - mass of the electron
c- speed of light
e - charge on the electron

This is just the unit conversion of electron mass into electron volts. Using (10) and rearranging slightly this gives

$ e V = \frac {\hbar} {c} g  $    ...(21)

This time implying a relationship between charge and acceleration. Equation (21) is a bit of an after thought and needs more investigation.

IMPORTANT NOTE. There is nothing here to prove that the two laws are actually one, and the assumption that energy may under go acceleration is unproven and without corroborating evidence as they say in the movies. This is a house of cards built on the assumption used to give (6). There are a number of instances where this is assumption is valid, (19) and (20) are certainly true and is one example.

I can't help thinking that there is something not quite right in all of this. After all it seems to imply that energy, momentum, photons, mass and charge are actually related to some intrinsic acceleration. Further that this acceleration is conserved and seems to show that there is no need for two laws, namely the conservation of energy and conservation of momentum.

Though Helmholtz was keen on the idea so it does have some historical merit.

I'm going to try and apply this idea to the creation of an electron / positron pair from a photon and see what numbers I get, will publish this in a future post.

Amendment: Got to pondering this one and there is more to this than meets the eye. will report soon on this one.

Monday 7 October 2013

No Cigar

This is one of those things that just seems so close that it must mean something. Take a look at this.

$ m_p k_e = 195.61406 = 10^8 / 511210 $     ...(1)

$ m_p = \sqrt \frac {\hbar c} {G}   $     ...(2)
$ k_e = \frac {1} {4 \pi \epsilon_0} $    ...(3)

G - Gravitational constant - 6.67384 .10-11 m3 kg-1 s-2
$ \hbar $ - reduced Planck constant - 1.0545717.10-34 kg m2 s-1 C-2
c - speed of light - 299792458 m s-1
$ \epsilon_0 $ - permittivity of free space - 8.9875517.109 kg m3 s-2 C-2

now granted this is going to start looking like some numerology exercise, but bare with me. It's the 511210 that caught my eye and this is why

$ \frac {m_e c^2} {e} = V_e =  510998.9   eV $    ...(4)

$ 5110210 / 510998.0 = 1.000415 $, that is pretty close to unity. So close in fact that it would only involve a small change in the Gravitational constant for it to work. Try this, say we replace the 511210 with equation (4) to give

$ m_p k_e \approx  \frac {e 10^8} {m_e c^2} = \frac {10^8} {V_e} $    ...(5)

sub in (2) and (3)

$  \sqrt \frac {\hbar c} {G} \frac {1} {4 \pi \epsilon_0} =  \frac {e 10^8} {m_e c^2} $    ...(6)

using (1), (2) & (3) we can show that

$ G = \frac {m_e^2 \hbar c^5} {e^2 (4 \pi \epsilon_0)^2  10^{16}} $    ...(7)

giving a value

G = 6.66837 .10-11 m3 kg-1 s-2

That is less than 1% difference. So I thought I would give it an hour to see if there was anything to this. After all if I understood where the 108 comes from maybe we would have something. So lets do some more rearranging to get charge to mass ratio

$ \frac {e} {m_e} = \frac {m_p c^2} {4 \pi \epsilon_0  10^8} =   \frac {E_p} {4 \pi \epsilon_0  10^8}  $   ....(8)

$E_p$ - Planck energy. This can easily be rearranged using

$ E_e = m_e c^2 $    ...(9)

and

$ c \mu_0 = 1 / c \epsilon_0 $    ....(10)

$ \mu_0$ - permeability of free space - $4\pi  10^{-7}$

to give

$ E_p E_e = e  10^{15}  $   ...(11)

so what is the $10^{15}  $? Let's take a look at the dimensions. Energy is Joules, e - charge is in Coulombs, so if we use

$ d = 10^{15}  $

$ E_p E_e = e  d  $   ...(12)

d has the units $ J^2 / C = V^2 C $, so let us try

$ d = V_e V_p q_p $    ... (13)

$V_p$ - Planck voltage - $1.04295   10^{27} V$
$q_p$ - Planck Charge - $1.875545   10^{-18} C$
$V_e$ - this is electron energy in electronVolts

so (12) becomes

 $ E_p E_e = V_p q_p  e V_e $   ...(14)

but

$ E_p = V_p q_p  $  and  $ E_e =  e V_e $ by definition!

using (9) and dividing the above we get

$ \frac {q_p V_p} {m_p} = \frac {e V_e} {m_e} = c^2 $   ...(15)

using (15) we get

$ m_p = {q_p V_p} \frac{m_e}{e V_e}$    ...(16)

multiplying by $k_e$ and rearranging gives

$ m_p k_e V_e = \frac {m_e} {e} q_p V_p k_e $    ....(17)

so substituting from (5) we should have

$ 10^8 \approx \frac {m_e} {e} q_p V_p k_e $    ....(18)

doing the calculation

$ \frac {m_e} {e} q_p V_p k_e = 99858416 \approx 10^8$    ....(19)

So that is what our $10^8$ from (5) actually is! Had I been a little smarter and looked at (15) first then I could have saved myself a couple of hours. Maybe next time.

We can go a little further with this, but I think that this is probably a good place to finish.

more like this

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