I was revisiting a couple of my earlier post, particularly one on the momentum of inertia of a photon when I got to thinking about 2 equations I derived in that post, the first equation was

$ E I = L^2$ .... (1)

Which says that energy multiplied by moment of inertia is the angular momentum squared. I also derived a second more specific version of this (originally equation 16), which shows

$ E I =\hbar^2 $ ... (2)

then it occurred to me that when a photon creates matter it has to obey the various conservation laws. A photon that converts to matter will form an electron/positron pair, rather than just a single electron. If it did form a single electron there would be conservation of charge violation and so on.

For two electrons (or an electron positron pair) though, we have

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

also the equivalent moment of inertia is

$ I = 2 I_e $ ... (4)

putting these into (2) we have

$ 4 m_e c^2 I_e = \hbar^2 $ ... (5)

so

$ m_e c^2 I_e = $ $\frac {\hbar^2} {4} $ ... (6)

but from (1) we have

$ E_e I_e = L_e^2 $ for an electron, where $E_e = m_e c^2$ giving

$L_e^2 =$ $\frac {\hbar^2} {4} $ ... (7) , which becomes

$L_e =$ $\frac {\hbar} {2}$ ...(8)

The intrinsic spin of an electron! That's nice eh? There is more to this though. If equation (2) is correct then does it imply all particles have spin. This makes me think we haven't got the Higgs particle sorted yet because it can't have a zero spin version. It's also a bit of a black eye for string theory which, I've been told, likes the idea of spinless particles.

This is not a problem if a particle can have zero moment of inertia, in which case $I=0$ and $L=0$. What would that mean?

After all

$I = m \lambda^2 $ ...(9) or

$I =$ $\frac {\hbar} {\omega}$ ...(10)

so $I$ can't be zero because if m is zero then we still have a moment of inertia because of the frequency of the massless particle given by

$E = \hbar \omega$

I'll figure what all that in the next post, hopefully!

wrote this while listening to this.

# physics rebel

A blog about physics things. Hopefully described in a way that makes sense to most people. You don't have to be Einstein to enjoy or even understand Physics. Whoever you are and what ever your eductional level, physics should be interesting and fun. I hope you enjoy it as much as I enjoy blogging about it.

## Thursday, 3 July 2014

## 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

q

$\alpha$ - 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.

$4 \pi \epsilon_0$ - 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 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.

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

q

_{p}- 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

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)

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.

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)

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.

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)

$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

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,

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.

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

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 SchrÃ¶dinger 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.

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)

----------------------------------------------------------------------------

----------------------------------------------------------------------------

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)

$r_e = 2.81794 10^{-15}$

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 SchrÃ¶dinger 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.

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.

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?

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.

$ 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.

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.

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