# LOG#072. The hG system.

Brazil is experimenting an increase of scientific production. Today, I am going to explain this brazilian paper http://arxiv.org/abs/0711.4276v2 concerning the number of fundamental constants.

The Okun cube of fundamental constant, firstly introduced by Gamov, Ivanenko and Landau, has raised some questions about what we should consider as fundamental “unit” we already had but now with more intensity. I mentioned the trialogue of fundamental constants between Veneziano, Duff and Okun himself more than a decade ago. Veneziano argued that 2 fundamental constants were well enought to fix everything. However, it is not the “accepted” and “more popular” approach in these days, but the brazilian paper about defends such a claim!

What do they claim? They basically argue that what we need is a convention for space and time measurements and nothing else. Specifically, they say that every physical observable $\mathcal{O}_i$ with $i=1,2,\ldots$ can be expressed as follows:

(1) $\boxed{\mathcal{O}_i=\Omega_i \sigma^{\alpha_i}\tau^{\beta_i}}$

and where $\alpha_i,\beta_i,\Omega_i$ are pure dimensionless numbers, while $\sigma, \tau$ denote “basic units” of space and time. We could argue that these two last “fundamental units” of “space and time” were “quanta” of “space” and “time”, the mythical “choraons” and “chronons” some speculative theories of Quantum Gravity seem to suggest, but it would be another different story not related to this post!

After introducing the above statement, they discuss 2 procedures to measure with clocks and rulers, what they call $G$-protocol and $h$-protocol. They begin assuming some quantity in the CGS system (note that the idea is completely general and they could use the MKSA or any other traditional system of units):

(2) $\boxed{\mathcal{D}_i= \Delta_i T^{\alpha_i}L^{\beta_i}M^{\gamma_i}}$

where $\Delta_i,\alpha_i,\beta_i,\gamma_i$ are dimensionless constants. And then, the 2 protocols are defined:

1st. G-protocol. Multiply the above equation (2) by $G^{\gamma_i}$ and identify $\mathcal{O}_i$ with $\mathcal{D}_i G^{\gamma_i}$ or $\mathcal{O}_i^{(G)}$. Rewriting all the physical quantities and laws in terms of this protocol in terms of $\mathcal{O}_i^{(G)}$ instead $\mathcal{D}_i$  we gain some bonuses:

i) The unit M from CGS “vanishes” or is “erased” from physical observables.

ii) G disappear from every physical law.

iii) Masses being measured in $cm^3/s^2$ imply that from Newton’s gravitational law $g=-G_Nm/d^2$ we deduce that

$\boxed{g=-m^{(G)}/d^2}$

where $m^{(G)}=mG$ are units with physical dimension $L^3T^{-2}$. $G_N$, the gravitational constant, is some kind of conversion factor between mass and “volume acceleration” $L^3/T^2$. This G-protocol applied to the Planck constant provides

$\boxed{h^G=hG}$

and it has dimensions of $L^5/T^3$.

2nd. h-protocol.  From equation (2), if we divide by $h^{\gamma_i}$ and we identigy $\mathcal{D}_i/h^{\gamma_i}$ with $\mathcal{O}_i^{(h)}$ we get the so-called h-protocol. The consequences are:

i) M units disappear from physical laws and quantities, as before.

ii) h is erased and vanishes from every equation, law and quantity.

iii) Masses are measured in units of $s/cm^2$, e.g., from the Compton equation we get in the h-protocol

$\Delta \lambda= \dfrac{1}{m^{(h)}c}\left(1-\cos\theta\right)$

and where $m^h=m/h$ are units of mass in the h-protocol with dimensions $T/L^2$. Therefore, h is the conversion factor between inverse areolar velocity $s/cm^2$ and mass $g$. In this protocol the inverse of the Compton length measures “inertia”, and indeed this fact fits with some recent proposals to determine a definition of kg independent from the old MKSA pattern (the famours iridium “thing”, which is know now not to have a 1 kg mass). Moreover, we also get that

$G^{(h)}=Gh$

and

$G^{(h)}=h^{(G)}$

The two protocols can be summarized in a nice table

They also derive the mysterious relations between charge and mass that we saw in the previous post about Pavšič units, i.e., they also derive

$e^{(G)}=2\cdot 10^{21}m_e^{(G)}$

and it is equivalent to $e=\kappa_0 m_e$. Somehow, and electron is more electrical/capacitive than gravitational/elastic!

Finally, in their conclusions, they remark that two constants, $(c, h^{(G)})$ instead three $(c,h,G_N)$ seems to be well enough for physical theories, and it squashes or squeezes the Gamov-Ivanenko-Landau-Okun (GILO) cube to a nice plane. I include the two final figure for completion, but I urge you to read their whole paper to get a global view before you look at them.

Are 2 fundamental constants enough? Are Veneziano (from a completely different viewpoint) and these brazilian physicists right? Time will tell, but I find interesting these thoughts!

See you soon in another wonderful post about Physmatics and system of units!

# LOG#040. Relativity: Examples(IV).

Example 1. Compton effect.

Let us define as  “a” a photon of frequency $\nu$. Then, it hits an electron “b” at rest, changing its frequency into $\nu'$, we denote “c” this new photon, and the electron then moves after the collision in certain direction with respect to the line of observation. We define that direction with $\theta$.

We use momenergy conservation:

$P^\mu_a+P^\mu_b=P^\mu_c+P^\mu_d$

We multiply this equation by $P_{\mu c}$ to deduce that

$P^\mu_a P_{\mu c}+P^\mu_{b}P_{\mu c}=P^\mu_c P_{\mu c}+P^\mu_d P_{\mu c}$

Using that the photon momenergy squared is zero, we obtain:

$P^\mu_a P_{\mu c}+P^\mu_bP_{\mu c}=P^\mu_dP_{\mu c}$

$P^\mu _a=\left(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0\right)$

$P^\mu_b=\left(mc,0,0,0\right)$

$P^\mu_c=\left(\dfrac{h\nu'}{c},\dfrac{h\nu'}{c}\cos\theta,\dfrac{h\nu'}{c}\sin\theta,0\right)$

Remembering the definitions $\dfrac{c}{\lambda}=\nu$ and $\dfrac{c}{\lambda'}=\nu'$ and inserting the values of the momenta into the respective equations, we get

$\dfrac{h^2}{c^2}\nu\nu'\left(1-\cos\theta\right)+mh\nu'=mh\nu$

$\dfrac{h^2}{\lambda\lambda'}\left(1-\cos\theta\right)+\dfrac{mhc}{\lambda'}=mhc\dfrac{\lambda'-\lambda}{\lambda\lambda'}$

$\boxed{\Delta \lambda\equiv \lambda'-\lambda=\dfrac{h}{mc}\left(1-\cos\theta\right)}$

or

$\boxed{\dfrac{1}{\nu'}-\dfrac{1}{\nu}=\dfrac{h}{mc^2}\left(1-\cos\theta\right)}$

$\boxed{\dfrac{1}{\omega'}-\dfrac{1}{\omega}=\dfrac{\hbar}{mc^2}\left(1-\cos\theta\right)}$

$\boxed{\dfrac{\omega'}{\omega}=\mbox{Energy transfer}=\left[1+\dfrac{\hbar \omega}{mc^2}\right]^{-1}}$

It is generally defined the so-callen electron Compon’s wavelength as:

$\lambda_C=\dfrac{h}{mc}$

$\bar{\lambda_C}=\dfrac{\hbar}{mc}\approx 2.42\cdot 10^{-12}m$

Remark: There are some current discussions and speculative ideas trying to use the Compton effect as a tool to define the kilogram in an invariant and precise way.

Example 2. Inverse Compton effect.

Imagine an electron moving “to the left” denoted by “a”, it hits a photon “b” chaging its frequency into another photon “c” and the electron changes its direction of motion, being the velocity $-u_b$ and the angle with respect to the direction of motion $\theta$.

$P^\mu_a=\left(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0\right)$

$P^\mu_b=\left(\gamma_b mc,-\gamma_b m u_b,0,0\right)$

$P^\mu_c=\left(\dfrac{h\nu'}{c},-\dfrac{h\nu'}{c},0,0\right)$

Using the same conservation of momenergy than above

$\dfrac{2EE'}{c^2}+\gamma_b mE'-\gamma_b m\dfrac{u_b}{c}E'=\gamma_b m E+\gamma_b \dfrac{mu_b E}{c}$

Supposing that $u_b\approx c$, and then $1-u_b/c\approx \dfrac{1}{2}\left(1+\dfrac{u_b}{c}\right)\left(1-\dfrac{u_b}{c}\right)=\dfrac{1}{2}\left(1-\dfrac{u_b^2}{c^2}\right)=\dfrac{1}{2}\dfrac{1}{\gamma_b^2}$

Thus,

$\dfrac{2EE'}{c^2}+\dfrac{mE'}{2\gamma_b}=2\gamma_b mE$

$\dfrac{E'}{E}=\dfrac{2\gamma_b m}{\dfrac{2E}{c^2}+\dfrac{m}{2\gamma_b}}=\dfrac{4\gamma_b^2}{1+\dfrac{4\gamma_b E}{mc^2}}$

This inverse Compton effect is important of importance in Astronomy. PHotons of the microwave background radiation (CMB), with a very low energy of the order of $E\approx 10^{-3}eV$, are struck by very energetic electrons (rest energy mc²=511 keV). For typical values of $\gamma_b >>10^8$, the second term in the denominator dominates, giving

$E'\approx \gamma_b\times 511keV$

Therefore, the inverse Compton effect can increase the energy of a photon in a spectacular way. If we don’t plut $u_b\approx c$ we would get from the equation:

$\dfrac{2EE'}{c^2}+\gamma_b mE'-\gamma_b m\dfrac{u_b}{c}E'=\gamma_b m E+\gamma m \dfrac{mu_b E}{c}$

$\gamma_b m E'\left(1-\dfrac{u_b}{c}+\dfrac{2E}{\gamma_b mc^2}\right)=\gamma_b m E\left(1+\dfrac{u_b}{c}\right)$

$\boxed{\dfrac{E'}{E}=\dfrac{1+\dfrac{u_b}{c}}{1-\dfrac{u_b}{c}+\dfrac{2E}{\gamma_b mc^2}}}$

If we suppose that the incident electron arrives with certain angle $\alpha_i$ and it is scattered an angle $\alpha_f$. Then, we would obtain the general inverse Compton formula:

$\boxed{\dfrac{E'_f}{E'_i}=\dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i\cos\alpha_f+\dfrac{E'_i}{\gamma_i mc^2}\left(1-\cos\theta\right)}}$

$\boxed{\dfrac{E'_\gamma}{E_\gamma}=\dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i\cos\alpha_f+\dfrac{E_\gamma}{E_e}\left(1-\cos\theta\right)}}$

In the case of $\alpha_f \approx 1/\gamma<<1$, i.e., $\cos\alpha_f\approx 1$, and then

$\dfrac{E'}{E}\approx \dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i}\approx \left(1-\beta_i\cos\alpha_i\right)2\gamma_i^2$

In conclusion, there is an energy transfer proportional to $\gamma_i^2$. There are some interesting “maximal boosts”, depending on the final energy (frequency). For instance, if $\gamma_i\approx 10^3-10^5$, then $E_f\approx \gamma_i^2\times 511 keV$ provides:

a) In the radio branch: $1GHz=10^9Hz$, a maximal boost $10^{15}Hz$. It corresponds to a wavelength about 300nm (in the UV band).

b) In the optical branch: $4\times 10^{14}Hz$, a maximal boost $10^{20}Hz\approx 1.6MeV$. It corresponds to photons in the Gamma ray band of the electromagnetic spectrum.

Example 3. Bremsstrahlung.

An electron (a) with rest mass $m_a$ arrives from the left with velocity $u_a$ and it hits a nucleus (b) at rest with mass $m_b$. After the collision, the cluster “c” moves with speed $u_c$, and a photon is emitted (d) to the left. That photon is considered “a radiation” due to the recoil of the nucleus.

The equations of momenergy are now:

$P\mu_aP_{a\mu}+2P^\mu_aP_{b\mu}+P^\mu_bP_{b\mu}=P^\mu_cP_{c\mu}+2P^\mu_cP_{d_\mu}+P^\mu_cP_{d\mu}$

$2P\mu_{a}P_{d\mu}+2P^\mu_bP_{d\mu}=2P\mu_cP_{d\mu}+2P^\mu_dP_{d\mu}$

$P^\mu_aP_{a\mu}+2P^mu_aP_{b\mu}+P^\mu_bP_{b\mu}-2P^\mu_aP_{d\mu}-2P^\mu_bP_{d\mu}=P^\mu_cP_{c\mu}-P^\mu_dP_{d\mu}$

$P^\mu_a=\left(\gamma_am_ac,\gamma_am_au_a,0,0\right)$

$P^\mu_b=\left(m_bc,0,0,0\right)$

$P^\mu_d=\left(\dfrac{E}{c},-\dfrac{E}{c},0,0\right)$

$(m_ac)^2+2\gamma_am_am_bc^2+(m_bc)^2-2\gamma_am_a(1+\beta_b)E-2m_bE=(m_a+m_b)^2c^2+0$

$2E\left(\gamma_am_a(1+\beta_a)+m_b\right)=2(\gamma_a-1)m_am_bc^2$

$\boxed{E=\dfrac{(\gamma_a-1)m_am_bc^2}{\gamma_a m_a(1+\beta_a)+m_b}}$

In clusters of galaxies, typical temperatures of $T\sim 10^7-10^8K$ provide a kinetic energy of proton and electron at clusters about $1.3-13keV$. Relativistic kinetic energy is $E_k=(\gamma_a-1)m_ac^2$ and it yields $\gamma_a\sim 1.0025-1.025$ for  hydrogen nuclei (i.e., protons $p^+$). If $\gamma_am_a(1+\beta_a)<<1$, then we have $E\approx (\gamma_a-1)m_ac^2=(\gamma_a-1)\times 511keV$. Then, the electron kinetic energy is almost completely turned into radiation (bremsstrahlung). In particular, bremsstrahlung is a X-ray radiation with $E\sim 1.3-13keV$.