LOG#242. Hyperbolic magic.

Beta-gamma fusion! Live dimension!

Do you like magic? Mathemagic and hyperbolic magic today. Master of magic creates an “illusion”. In special relativity, you can simplify calculations using hyperbolic trigonometry!


(1)   \begin{equation*} E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\beta^2}} \end{equation*}

(2)   \begin{equation*} p=Mv=m\gamma v \end{equation*}

are common relativistc equation. Introduce now:

(3)   \begin{equation*} \tcboxmath{\beta=\dfrac{v}{c}=\tanh\varphi}\;\; 0\leq \beta<1,  -\infty<\varphi<\infty \end{equation*}

as the rapidity. Then:

(4)   \begin{equation*}  \gamma=\dfrac{E}{mc^2}=\dfrac{1}{\sqrt{1-\beta^2}}=\dfrac{1}{1-\tanh^2\varphi}}=\sqrt\dfrac{\cosh^2}{\cosh^2-\sinh^2}}=\cosh\varphi \end{equation*}


(5)   \begin{equation*} \tcboxmath{\gamma=\cosh\varphi}\;\;\; \gamma\geq 1, -\infty<\varphi<\infty \end{equation*}

Similarly, you get that

(6)   \begin{equation*} p=m\gamma v=mc\beta\gamma=mc\tanh\vaphi\cosh\varphi=mc\sinh\varphi \end{equation*}

and thus

(7)   \begin{equation*} \tcboxmath{p=mc\sinh\varphi}\;\;\; -\infty<\varphi<\infty \end{equation*}

Also, you can write

(8)   \begin{equation*} \tcboxmath{\tanh\varphi=\dfrac{pc}{E}}\;\; 0\leq \beta<1,  -\infty<\varphi<\infty \end{equation*}

(9)   \begin{equation*} \tcboxmath{\dfrac{vE}{mc^3}=\beta\gamma}\;\;\; 0\leq \beta<1,  -\infty<\varphi<\infty, \gamma\geq 1 \end{equation*}


(10)   \begin{equation*} \tcboxmath{\beta=\dfrac{pc}{E}} \;\; \;\; 0\leq \beta<1,  -\infty<\varphi<\infty \end{equation*}

The above equations can be inverted, and it yields

(11)   \begin{equation*} \tcboxmath{\beta=\dfrac{v}{c}=\tanh\varphi=\tanh\left(\sinh^{-1}\left(\dfrac{p}{mc}\right)\right)} \end{equation*}

(12)   \begin{equation*} \tcboxmath{\beta=\dfrac{v}{c}=\tanh\left(\cosh^{-1}\left(\gamma\right)\right)=\tanh\left(\cosh^{-1}\left(\dfrac{E}{mc^2}\right)\right)=\sqrt{1-\left(\dfrac{mc^2}{E}\right)^2}} \end{equation*}

(13)   \begin{equation*} \tcboxmath{\gamma=\dfrac{1}{\sqrt{1-\beta^2}}=\cosh\left(\tanh^{-1}\left(\beta\right)\right)} \end{equation*}

(14)   \begin{equation*} \tcboxmath{\gamma=\dfrac{1}{\sqrt{1-\beta^2}}=\cosh\left(\tanh^{-1}\left(\dfrac{pc}{E}\right)\right)} \end{equation*}

(15)   \begin{equation*} \tcboxmath{\varphi=\tanh^{-1}\left(\beta\right)=\tanh^{-1}\left(\dfrac{pc}{E}\right)} \end{equation*}

Hyperbolic functions also simply the Lorentz transformations into a more symmetric form! Consider the spacetime interval:

(16)   \begin{equation*} s^2=x^\mu x_\mu=x^2-(ct)^2=x^2+(ict)^2 \end{equation*}

and a rotation matrix

(17)   \begin{equation*} R(\theta)^T=R^{-1}=\begin{pmatrix}\cos \theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix} \end{equation*}

Now, make a rotation with imaginary angle \varphi=i\theta and apply it to the vector X=(x,ict):

(18)   \begin{equation*} \begin{pmatrix} x'\\ ict'\end{pmatrix} =\begin{pmatrix} \cos i\theta & \sin i\theta\\ -\sin i\theta & \cos i\theta\end{pmatrix}\begin{pmatrix} x\\ ict\end{pmatrix} \end{equation*}


(19)   \begin{equation*} \begin{pmatrix} x'\\ ict'\end{pmatrix} =\begin{pmatrix} \cosh\theta & i\sinh \theta\\ -i\sinh \theta & \cosh\theta\end{pmatrix}\begin{pmatrix} x\\ ict\end{pmatrix} \end{equation*}

(20)   \begin{equation*} \begin{pmatrix} x'\\ ict'\end{pmatrix}=\begin{pmatrix}\beta & i\beta\gamma\\ -i\beta\gamma & \beta\end{pmatrix}\begin{pmatrix} x\\ ict\end{pmatrix}=\begin{pmatrix} \beta x -\beta\gamma ct\\ i\left(-\beta\gamma x+\beta ct\right)\end{pmatrix} \end{equation*}

and thus

(21)   \begin{equation*} \begin{pmatrix}x'\\ ct'\end{pmatrix}=\begin{pmatrix} \beta & -\beta\gamma\\ -\beta\gamma & \beta\end{pmatrix}\begin{pmatrix}x\\ ct\end{pmatrix} \end{equation*}

That is the Lorentz transformation! A Lorentz transformation is just a rotation matrix of an imaginary angle with imaginary time! But you can give up imaginary numbers using hyperbolic functions! Indeed, L(\varphi)=L^T for Lorentz transformations, while R(\theta)=(R^{-1})^T for rotations.

Finally, something about particle spin and “rotations”, secretly related to Lorentz transformations of spinors. Spin zero particles are the same irrespectively how you see them, so if you turn them 0 degrees (radians), spin zero particles remain invariant. Vector spin one particles like A_\mu are the same if you turn them 360^\circ=2\pi \;rad. Tensor spin two particles like h_{\mu\nu} are the same if you rotate them about 180^\circ=\pi \; rad. Now, the weird stuff. Electrons and spin one-half fermions are the same only if you rotate them…720^\circ=4\pi\;rad!!! They see a largest world than the one we observe! The hypothetical gravitino field remains invariant only when you twist it about 240^\circ=4\pi/3\; rad. You can also iterate the argument for higher spin particles. Even you could consider the case with infinite (continous) spin.

Remark(I): in natural units with c=\hbar=k_B=1 you can prove that

    \[1kg=6.61\cdot 10^{35}GeV\]

    \[1K=8.617\cdot 10^{-14}GeV\]

    \[1m=8.07\cdot 10^{14}GeV^{-1}\]

Remark(II): in natural units with G_N=c=1 you also get

    \[1kg=7.42\cdot 10^{-28}m\]

    \[1kg=1.67 ZeV=8.46\cdot 10^{27}m^{-1}\]

Now, perhaps you have time for a little BIG RIP in de Sitter spacetime with phantom energy \omega<-1

(22)   \begin{equation*} \tcboxmath{T_{BRip}=-\dfrac{2}{3(1+\omega)H_0\sqrt{1-\Omega_{m,0}}}} \end{equation*}

Perhaps, now you face the proton decay crisis of your life due to pandemic, any time?

Challenges for you:

Challenge 1. Some recent reviews of proton decay in higher dimensional models derive the estimate

    \[\tau_{proton}\sim\left(\dfrac{M_P}{M_{proton}}\right)^D\dfrac{1}{M_{proton}}$$ For $D=4$, it yields about $\tau\sim 10^{52}s\sim 10^{45}yrs\]

However, Hawking derived a similar but not identical estimate

    \[\tau_{proton}\sim\left(\dfrac{M_P}{M_{proton}}\right)^8\dfrac{1}{M_{proton}}\sim 10^{120}yrs\]

using processes with virtual black holes and spacetime foam. I want to understand this formulae better, so I need to understand the origin of the powers and the absence (or presence if generalized GUT/TOE arises) of gauge couplings. In short:

1) What is the reason of the D-dependence in the first formula and the 8th power in the second formula?

2) Should the proton decay depend as well and in which conditions of gauge (or GUT,TOE) generalized couplings too?

Challenge 2. 

Derive the formulae



Finally, string theory…To crush you even more…Gravitational constant is just derived from the string coupling and the dilaton field in superstring theories. The recipe is

(23)   \begin{equation*} \langle e^\phi\rangle =e^{\phi_\infty} \end{equation*}

such as

(24)   \begin{equation*} g_s(d)=\langle e^\phi\rangle_0 \end{equation*}

Define the string tension \alpha'=L_s^2, and the string lenth L_s=\sqrt{\alpha'}. Then, in a 10d Universe

(25)   \begin{equation*} \tcboxmath{G_N(10d)=8\pi^6g_s^2\left(\alpha')^4=8\pi^6 g_s^2 L_s^8} \end{equation*}

Furthermore, with n compactified dimensions, you get

(26)   \begin{equation*} \tcboxmath{G_N(10d)=G_N(n)V_{10-n}} \end{equation*}


(27)   \begin{equation*} \tcboxmath{g_s^2(10d)=\dfrac{V_{10-d}}{(2\pi L_s)^{10-n}(g_s^{(n)})^2}} \end{equation*}

In summary, you can obtain

(28)   \begin{equation*} \tcboxmath{\dfrac{g_s^2(2\pi L_s)^{10-n}}{16\pi G_N(10d)}=\dfrac{g_s^2(n)}{16\pi G_N(n)}} \end{equation*}

Have I punched hard?

See you in another blog post dimension!

Liked it? Take a second to support amarashiki on Patreon!
Become a patron at Patreon!

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.