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.

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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.
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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.

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.

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).

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.

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.

Moment of inertia of a photon

In a previous post I raised the question of whether energy can be accelerated. I also mentioned that a number of other interesting results come out of this analysis and idea. In this post I cover another of these. This one relates to inertia and starts out using Planck units. Consider the following

$ I_p = m_p l_p^2  $      .... (1)

Where m_p is Planck mass and l_p is Planck length. In this discussion I_p is Planck moment of inertia. It can be shown that this gives

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

t_p - Planck time.

Equation (2) can also be represented by

$ I_p \omega_p = \hbar  $    ... (3)

Planck energy is defined as

$ E_p = m_p c^2 =$  $ \frac {\hbar} {t_p}   $     ...(4)

using (2) to eliminate Planck time and rearranging this gives

$ E_p I_p = \hbar^2    $     ....(5)

Also, by dividing (4) by (2)

$ \frac {E_p} {I_p} = \frac {1} {t_p^2} = \omega_p^2   $     ...(6)

Planck Momentum is defined by

$ p_p =$  $ \frac {\hbar} {l_p} = \frac {I_p \omega_p} {l_p} $ ....(7)

Angular Momentum is defined as

$ L = r  x  p  $     ....(8)

p - momentum, r -position vector?. Replacing r with Planck Length and p with Planck momentum gives

$ L_p = l_p   p_p  $ ....(9)

giving

$ L_p = l_p I_p $ $\frac {\omega_p} {l_p} =$ $ I_p \omega_p $  ...(10)

but from (3) this gives

$ L_p = \hbar   $     ...(11)

Ok. Let's stop there for a minute. For starters, I am a little uncomfortable with (9), we have replaced a position vector with a scalar length. Not really sure we can do that. Yet the result (11) is just the lowest level of the angular momentum of electrons around the atom given by Bohr Theory.

$ L = mvr = n\hbar  $ ... (12)

This is just a first-order approximation of the hydrogen atom using quantum mechanics. If that is the case then combining (10) and (6) gives

$ E_p = L_p \omega_p    $   ... (13)

can we argue that there is a more general term given by

$ E = L \omega  $  ... (14)

using a similar argument for (3) and (5) becomes

$ I \omega = \hbar $    ...(15)

and

$ E I = \hbar^2   $    ...(16)

respectively.

In the post on accelerating energy it was shown that

$ E c = \hbar g$    ...(17)

substituting this into (16) gives

$ I =$   $ \frac {\hbar c} {g} $    ...(18)

$g$ - acceleration

This can also be represented by

$ I = $  $\frac {\hbar^2} {m c^2}$    ...(19)

What does this mean? Earlier I mentioned that (11) is the angular momentum of an electron in the Bohr model. Could it be more than that? could it also be true for photons? Photons have angular momentum. Fine. Does (15) imply that a photon as a moment of inertia, I? Does (16) imply a relationship between moment of inertia of a photon and energy?

There are a number of assumptions made here. The first when we derive (11) and another when we go from Planck specific terms to general terms in (14), (15) & (16). I am not really comfortable with this and feel as though I have pulled a little bit of "smoke and mirrors".

That said, the result is rather interesting, photons have a moment of inertia, and a rather nice relationship between this and a photons energy. There is also the relationship between inertia and acceleration that is explored more in this post on acceleration.

Will ponder this one some more and see if I can justify my assumptions.

In the next post I'll show some more of the results that come out of this type of analysis.