LOG#031. Entropic Gravity (II).

gravity-articlelarge

We will generalize the entropic gravity approach to include higher dimensions in this post. The keypoint from this theory of entropic gravity, according to Erik Verlinde, is that gravity does not exist as “fundamental” force and it is a derived concept. Entropy is the fundamental object somehow. And it can be generalized to a a d-dimensional world as follows.

The entropic force is defined as:

    \[ \boxed{F=-\dfrac{\Delta U}{\Delta x}=-T\dfrac{\Delta S}{\Delta x}}\]

The entropic force is a force resulting from the tendency of a system to increase its entropy. Since \Delta S>0 the sign of the force (whether repulsive or attractive) is determined by how we take the definition of \Delta x as it is related to the system in question.

An arbitrary mass distribution M induces a holographic screen \Sigma  at some distance R that has encoded on it gravitational information. Today, we will consider the situation in d spatial dimensions.Using the holographic principle, the screen owns all physical information contained within its volume in bits on the screen whose number N is given by:

    \[ \boxed{N=\dfrac{A_\Sigma (R)}{l_p^{d-1}}}\]

This condition implies the quantization of the hyperspherical surface, where the hyperarea (from the hypersphere) is defined as:

    \[ \boxed{A_\Sigma=\dfrac{2\pi^{d/2}}{\Gamma \left(\frac{d}{2}\right)}R^{d-1}}\]

By the equipartition principle:

    \[\boxed{E=Mc^2=\dfrac{N}{2}k_BT}\]

Therefore,

    \[k_BT=\dfrac{2Mc^2l_p^{d-1}}{A_\Sigma}\]

The entropy shift due to some displacement is:

    \[ \Delta S=2\pi k_B \dfrac{\Delta x}{\bar{\lambda}}=2\pi k_B mc\dfrac{\Delta x}{\hbar}\]

Plugging the expression for the temperature and the entropy into the entropic force equation, we get:

    \[ F=-T\dfrac{\Delta S}{\Delta x}=-\dfrac{2Mc^2}{k_B}\dfrac{2\pi k_B mc}{\hbar}\dfrac{l_p^{d-1}}{A_\Sigma}=-\dfrac{4\pi Mmc^3l_p^{d-1}}{\hbar A_\Sigma}\]

and thus we finally get

    \[ F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}=-G_d\dfrac{Mm}{R^{d-1}}\]

i.e.,

    \[ \boxed{F=-G_d\dfrac{Mm}{R^{d-1}}}\leftrightarrow \boxed{F=-\dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}Mmc^3}{\hbar R^{d-1}}}\]

where we have defined the gravitational constant in d dimensions to be

    \[ \boxed{G_d\equiv \dfrac{2\pi^{1-\frac{d}{2}}\Gamma \left(\frac{d}{2}\right) l_p^{d-1}c^3}{\hbar }}\]

LOG#030. Entropic Gravity (I).

erikverlinde

In 2010, Erik Verlinde made himself famous once again. Erik Verlinde is a theoretical physicist who has made some contributions to String Theory. In particular, the so-called Verlinde formula. However, this time was not apparently a contribution related to string theory. He guessed a way to derive both the Newton’s second law and the Newton’s law of gravity. He received a prize time later, and some critical voices against his approach were raised.

I will review in this post his deductions.

A. Newton’s second law.

There are some hypothesis to begin with:

1st. Entropic force ansatz. Forces aren’t really fundamental, they are derived from some entropy functional. More precisely, forces are ”entropy fluxes”. Mathematically speaking:

    \[ F\Delta x=T\Delta S\]

or

    \[ \boxed{F=T\dfrac{\Delta S}{\Delta x}}\]

2nd. Acceleration has a temperature. Equivalently, this is the well known Unruh’s effect from QFT in curved spacetime. Any particle that is accelerated is equivalent to some thermal system. This parallels the Hawking’s effect in black hole physics as well. It stands mysterious for me yet, since indeed, the temperature is relative to some vacuum or rest system.

    \[ T=\dfrac{\hbar a}{2\pi k_B c}\]

3rd. Holographic principle.   A variant of the holographic principle is postulated to hold. The idea is that a particle separated certain distance from a “holographic screen” has an entropy shift:

    \[ \Delta S = 2\pi k_B \dfrac{mc}{\hbar} \Delta x\]

Then, plugging the holographic entropy and the Unruh’s temperature into the entropic force ansatz, we get easily

    \[ F= 2\pi k_B \dfrac{mc}{\hbar} \dfrac{\hbar a}{2\pi k_B c}\]

i.e. we get the Newton’s second law of Dynamics

    \[F=ma\]

B. Newton’s gravity.  In order to get the Newton’s law of gravitation, we have to modify a bit the auxiliary hypothesis but yet we conserve the core approach.

1st. The entropic force ansatz. Again,

    \[ \boxed{F=T\dfrac{\Delta S}{\Delta x}}\]

2nd. Holographic principle. Again,

    \[ \Delta S = 2\pi k_B \dfrac{mc}{\hbar} \Delta x\]

3rd. Equipartition principle of relativistic energy.   Temperature is obtained at the statistical level when you distribute N quanta in thermal equilibrium, and they equal the relativistic energy formula. Equivalently,

    \[ \dfrac{Nk_BT}{2}=Mc^2\]

4th. Microscopical degrees of freedom and minimal length ( or area quantization). The number of allowed microscopical quanta or microstates can not exceed and must match in the extreme case the ration of the area available and the square of Planck’s length ( or some other squared fundamental length). In other words, the number of bits can not overcome the area of a ball in Planck’s units. EQuivalently, the (hyper)area must be quantized (through a number N). Mathematically speaking,

    \[ N=\dfrac{A}{l_{p}^{2}}\]

Then, we plug the hypothesis 2 and 3 into 1, to have:

    \[ F=\dfrac{2Mc^2}{k_B N}2\pi k_B \dfrac{mc}{\hbar}\]

and now we use 4, in order to get

    \[ F=\dfrac{2Mc^2}{k_B A}l_{p}^{2} 2\pi k_B \dfrac{mc}{\hbar}\]

that is

    \[ F=4\pi Mm \dfrac{l_{p}^{2}c^3}{\hbar A}\]

And now, recalling that in 3 spatial dimensions, a ball has an area A=4\pi r^2 and that the Newton’s constant of gravity is indeed given as function of Planck’s length as G=\dfrac{l_{p}^{2}c^3}{\hbar}, we have what we wanted to derive

    \[ F=4\pi Mm \dfrac{G}{4\pi r^2}\]

i.e. the Newton’s gravitational law has been derived from the entropic force too

    \[ F=G\dfrac{Mm}{r^2}\]

It is done. Is it just a trick or something deeper is behing all this stuff? Nobody knows for sure…People think he is probably wrong, but there are a whole line of research opened from his works. It is quite remarkable his approach is quite general and he suggests that every fundamental force is “entropic” or “emergent”, i.e., also electromagnetic fields, or even Yang-Mills fields could be entropic according to this approach.

Is he right? Time will tell…Some doubts arise from the fact that he has only derived the “temporal” components of Einstein Field Equations for gravity (a.k.a. newtonian gravitation) but, indeed, he and other physicists have been able to derive the remaining components as well. Perhaps, the strongest critics comes from neutron interferometry results. However, theoretical ideas like this one, like extra dimensions, could be saved by some clever argument.

LOG#029. Interstellar trips in SR.

bussard

My final article dedicated to the memory of Neil Armstrong. The idea is to study quantitatively the relativistic rocket motion with numbers, after all we have deduced the important formulae, and we will explain what is happening in the two frames: S’-frame (in motion), S-frame (in rest on Earth). There are many “variations” of this problem, also called “Langevin’s paradox” or “the Langevin’s interstellar trip” problem by some authors. Here, we will follow the approach suggested in the book Gravitation, by Misner, Wheeler and Thorne, and we will study the interstellar trip (in the frame of special relativity) with the following conditions:

1st. Spacetime is locally minkovskian (i.e., spacetime is flat). The solution we will expose would not be valid in the case of an interstellar trip to a very far away quasar, or a very very long distance (about thousand millions of lightyears) where we should take into account the effect of the expansion of the Universe, i.e., that on large scales, spacetime is curved (in particular, accordingly to the current data, it is pseudoriemannian). Thus, we can use special relativity in order to calculate distances, velocities and accelerations from the purely kinematical sense. After all, it is logical, since General Relativity says that locally, in small enough regions of spacetime, spacetime is described by a minkovskian metric.

2nd. We select a 4 stage accelerated motion with our rocket. We will assume that our rocket is 100% efficient in the sense it uses photons as propellant particles. The four stages are: acceleration from rest to g (acceleration step 1), decceleration to rest with -g until we approach the destination (that would be the one-way trip), acceleration with -g (seen from the S-frame) and decceleration with +g ( with respect to the S-frame) to reach Earth again in rest at the end of the round-way trip. Schematically, we can draw a spacetime sketch of this 4 stage journey:

LangevinScheme

Please, note the symmetry of the procedure and the different travel steps.

3rd. We set g=9.8m/s², or, as we saw in one of our previous posts, so we use g=1.03lyr/yr² (in units where c=1).

4th. We can not refuel during the travel.

5th. In the case we return to Earth, we proceed to come back after we stop at the destination immediately. In this case, there is no refuel of the starship or rocket in any point of the trip.

6th. We neglect any external disturbance which can stop us or even destroy us, e.g., micrometeorites, cosmic radiations, comets, and any other body that could alter our route. Indeed, this kind of stuff has to be seriously considered in any realistic travel, but we want to solve an ideal problem consisting in an ideal interstellar trip according to the current knowledge.

The main quantities we have to compute are:

1. In the S’-frame of the rocket \tau, 2\tau,4\tau, in years. They are, respectively, the time we are accelerating with +g, the time we are decelerating with -g, and the total time accelerating supposing we return to earth immediately from our destination target.

2. Distance in which we are accelerating (x, in light years), seen from the S-frame (we will not discuss the problem in the S’-frame, since it involves lenght contraction and it is more subtle in the calculations). According to our previous studies, we have

    \[ \boxed{x=\dfrac{c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}\]

3. Maximum depth in space D=2x. This allows us to select our target or destination to set the remaining parameters of the trip. Due to the symmetry of our problem, we have

    \[ \boxed{D=2x=\dfrac{2c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}\]

4. Total length travelled by the spaceship/rocket (in the S’frame) in the roundtrip ( to the destination and back).

    \[ \boxed{L=4x=\dfrac{4c^2}{g}\left(\cosh \left[\dfrac{g\tau}{c}\right]-1\right)}\]

5. Maximum speed to an specific destination, after accelerating in a given proper time. It reads:

    \[ \boxed{V=c\tanh \left(\dfrac{g\tau}{c}\right)}\]

Indeed, the relativistic gamma factor is \gamma =\cosh \dfrac{g\tau}{c}

6. S-frame duration t(years) of the stage 1 (acceleration phase 1). It is

    \[ \boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}\]

7. One way duration of the trip (according to the S-frame):

    \[ \boxed{t'=2t=\dfrac{2c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}\]

8. Total duration of the trip in the S-frame ( round way trip):

    \[ \boxed{t''=4t=\dfrac{4c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}\]

9. Mass ratio of the final mass with the initial mass of the spaceship after 1 stage, 2 stages ( one way trip) and the 4 stages total round way trip:

    \[ \boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{g\tau}{c}\right)}\]

    \[ \boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{2g\tau}{c}\right)}\]

    \[ \boxed{R=\dfrac{M_f}{M_i}=\exp \left(-\dfrac{4g\tau}{c}\right)}\]

10. Fuel mass vs. payload mass ratio (FM/PM) after 1 stage, 2 stages (until the destination) and the total trip:

    \[ \boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{g\tau}{c}\right)-1}\]

    \[ \boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{2g\tau}{c}\right)-1}\]

    \[ \boxed{R=\dfrac{M}{m}=\exp \left(\dfrac{4g\tau}{c}\right)-1}\]

We can obtain the following data using these expressions varying the proper time:

Table1interstellar

Table2interstellar

Table3interstellar

Table4interstellarTable5interstellar

Here in the last two entries, data provided to be out of the limits of our calculator ( I used Libre Office to compute them). Now, we can make some final observations:

1st. We can travel virtually everywhere in the observable universe in our timelife using a photon rocket, a priori. However, it shows that the mass-ratio turns it to be theoretically impossible.

2nd. Remember that in the case we select very long distances, these calculations are not valid since we should use General Relativity to take into account the expansion of spacetime.

3rd. Compare \tau with t, 2\tau with 2t, 4\tau with 4t. For instance, for 4\tau=40 then we get 57700 years, and v=0.99999999773763c, FM/PM=30000 (multiplying por m-kilograms gives the fuel mass).

4th. In SR, the technological problems are associated to the way photon rockets have, the unfavorable mass ratios (of final mass with initial mass) and fuel mass/payload rations necessary in the voyage (supposing of course, current physics and that we can not refuel during the trip).

5th. You can choose some possible distance destinations (D=2x) and work out the different parameters of the travel with the above equations.

In this way, for instance, for some celebrated known space marks, arriving at complete stop ( you can make other assumptions and see how the answer changes varying g or varying the arrival velocity too), we can easily get:

Example 1: 4.3 ly Nearest star (Proxima Centauri) in t=2\tau=3.6 yrs (S’-frame).

Example 2: 27 ly Vega (Contact movie and book) star in t=2\tau=6.6 yrs (S’-frame).

Example 3: 30000ly, our galactic center, in t=2\tau=20 yrs (S’-frame).

Example 4: 2000000ly , the Andromeda galaxy, in t=2\tau=28 yrs(S’-frame).

Example 5: Generally, you can travel n ly anywhere (neglecting curved spacetime) in t=2\tau=1.94 \cosh^{-1}(n/1.94+1) years (S’-frame).


		

LOG#028. Rockets and relativity.

Bussard_Interstellar_Ramjet_Engine

The second post in this special thread of 3 devoted to Neil Armstrong memory has to do with rocketry.

Firstly, for completion, we are going to study the motion of a rocket in “vacuum” according to classical physics. Then, we will deduce the relatistic rocket equation and its main properties.

CLASSICAL NON-RELATIVISTIC ROCKETS

The fundamental law of Dynamics, following Sir Isaac Newton, reads:

    \[ \mathbf{F}=\dfrac{d\mathbf{p}}{dt}\]

Suppose a rocket with initial mass M_i and initial velocity u_i=0. It ejects mass of propellant “gas” with “gas speed” (particles of gas have a relative velocity or speed with respect to the rest observer when the rocket move at speed \mathbf{v}) equals to u_0 (note that the relative speed will be u_{rel} and the propellant mass is m_0. Generally, this speed is also called “exhaust velocity” by engineers. The motion of a variable mass or rocket is given by the so-called Metcherski’s equation:

    \[ \boxed{M\dfrac{d\mathbf{v}}{dt}=-\mathbf{u_0}\dfrac{dM}{dt}+\mathbf{F}}\]

where -\mathbf{u_0}=\mathbf{v_{gas}}-\mathbf{v}. The Metcherski’s equation can be derived as follows: the rocket changes its mass and velocity so M'=M+dM and V'=V+dV, so the change in momentum is equal to M'V'=(M+dM)(V+dV), plus an additional term v_{gas}dv_{gas} and -mV. Therefore, the total change in momentum:

    \[ dP=Fdt=(M+dM)(V+dV)+v_{gas}dv_{gas}-mV\]

Neglecting second order differentials, and setting the conservation of mass (we are in the non-relativistic case)

    \[ dM+dm_{gas}=0\]

we recover

    \[ MdV=v_{rel}dM+Fdt\]

that represents (with the care of sign in relative speed) the Metcherski equation we have written above.

Generally speaking, the “force” due to the change in “mass” is called thrust.  With no external force, from the remaining equation of the thrust and velocity, and it can be easily integrated

    \[ u_f=-u_0\int_{M_i}^{M_f}dM\]

and thus we get the Tsiolkowski’s rocket equation:

    \[ \boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{M_i}{M_f}}\]

Engineers use to speak about the so-called mass ratio R=\dfrac{M_f}{M_i}, although sometimes the reciprocal definition is also used for such a ratio so be aware, and in terms of this the Tsiokolski’s equation reads:

    \[ \boxed{\mathbf{u_f}=\mathbf{u_0}\ln \dfrac{1}{R}}\]

We can invert this equation as well, in order to get

    \[ \boxed{R=\dfrac{M_f}{M_i}=\exp\left(-\dfrac{u_f}{u_0}\right)}\]

Example: Calculate the fraction of mass of a one-stage rocket to reach the Earth’s orbit. Typical values for u_f=8km/s and u_0=4km/s show that the mass ratio is equal to R=0.14. Then, only the 14\% of the initial mass reaches the orbit, and the remaining mass is fuel.

Multistage rockets offer a good example of how engineer minds work. They have discovered that a multistage rocket is more effective than the one-stage rocket in terms of maximum attainable speed and mass ratios. The final n-stage launch system for rocketry states that the final velocity is the sum of the different gains in the velocity after the n-th stage, so we can obtain

    \[ \displaystyle{u_f=\sum_{i=1}^{n}u_i^f=u_1^f+\cdots+u_n^f}\]

After the n-th step, the change in velocity reads

    \[ u_i^f=c_i\ln \dfrac{1}{R_i}\]

where the i-th mass ratios are defined recursively as the final mass in the n-th step and the initial mass in that step, so we have

    \[ \displaystyle{u_f=\sum_i c_i\ln \dfrac{1}{R_i}}\]

and we define the total mass ratio:

    \[ \displaystyle{R_T=\prod_i R_i}\]

If the average effective rocket exhaust velocity is the same in every step/stage, e.g. c_i=c, we get

    \[ \displaystyle{u_f=c\ln \left( \prod_{i=1}^{n} R_i^{-1}\right)}\]

or

    \[ \displaystyle{u_f=c \ln \left[ \left(\dfrac{M_0}{M_f}\right)_1\left(\dfrac{M_0}{M_f}\right)_2\cdots \left(\dfrac{M_0}{M_f}\right)_n\right]=c\ln \left[\left(\dfrac{M_0}{M_f}\right)_T\right]}\]

The influence of the number of steps, for a given exhaust velocity, in the final attainable velocity can be observed in the next plots:

n-stages1

nstages2nstages4

 RELATIVISTIC ROCKETS

We proceed now to the relativistic generalization of the previous rocketry. An observer in the laboratory frame observes that total momentum is conserved, of course, and so:

    \[ M'du'=-u'_0dM'\]

where du' is the velocity increase in the rocket with a rest mass M’ in the instantaneous reference frame of the moving rocket S’. It is NOT equal to its velocity increase measured in the unprimed reference frame, du. Due to the addition theorem of velocities in SR, we have

    \[ u+du=\dfrac{u+du'}{1+\dfrac{udu'}{c^2}}\]

where u is the instanteneous velocity of the rocket with respect to the laboratory frame S. We can perform a Taylor expansion of the denominator in the last equation, in order to obtain:

    \[ u+du=(u+du')\left(1-\dfrac{udu'}{c^2}\right)\]

and then

    \[ u+du=u+du'\left(1-\dfrac{u^2}{c^2}\right)\]

and finally, we get

    \[ du'=\dfrac{du}{1-\dfrac{u^2}{c^2}}=\gamma^2_u du\]

Plugging this equation into the above equation for mass (momentum), and integrating

    \[ \displaystyle{\int_{0}^{u_f}\dfrac{du}{1-\dfrac{u^2}{c^2}}=-u'_0\int_{M'_0}^{M'_f}dM}\]

we deduce that the relativistic version of the Tsiolkovski’s rocket equation, the so-called relativistic rocket equation, can be written as:

    \[ \dfrac{c}{2}\ln \dfrac{1+\dfrac{u_f}{c}}{1-\dfrac{u_f}{c}}=u'_0\ln\dfrac{M'_i}{M'_f}\]

We can suppress the primes if we remember that every data is in the S’-frame (instantaneously), and rewrite the whole equation in the more familiar way:

    \[ \boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}{1+\left(\dfrac{M_f}{M_0}\right)^{\frac{2u_0}{c}}}=c\dfrac{1-R^{\frac{2u_0}{c}}}{1+R^{\frac{2u_0}{c}}}}\]

where the mass ratio is defined as before R=\dfrac{M_f}{M_i}. Now, comparing the above equation with the rapidity/maximum velocity in the uniformly accelerated motion:

    \[ u_f=c\tanh \left(\dfrac{g\tau}{c}\right)\]

we get that relativistic rocket equation can be also written in the next manner:

    \[ u_f=c\tanh \left[ -\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)\right]\]

or equivalently

    \[ u_f=c\tanh \left[ \dfrac{u_0}{c}\ln R\right]\]

since we have in this case

    \[ \dfrac{g\tau}{c}=-\dfrac{u_0}{c}\ln \left(\dfrac{1}{R}\right)=\dfrac{u_0}{c}\ln R\]

and thus

    \[ R^{\frac{u_0}{c}}=\left(\dfrac{M_f}{M_i}\right)^{\frac{u_0}{c}}=\exp \left(-\dfrac{g\tau}{c}\right)\]

If the propellant particles move at speed of light, e.g., they are “photons” or ultra-relativistic particles that move close to the speed of light we have the celebrated “photon rocket”. In that case, setting u_0=c, we would obtain that:

    \[ \boxed{u_f=c\dfrac{1-\left(\dfrac{M_f}{M_0}\right)^{2}}{1+\left(\dfrac{M_f}{M_0}\right)^{2}}=c\dfrac{1-R^{2}}{1+R^{2}}=c\tanh \ln R}\]

and where for the photon rocket (or the ultra-relativistic rocket) we have as well

    \[ \dfrac{g\tau}{c}=-\ln \left(\dfrac{1}{R}\right)=\ln R\]

Final remark: Instead of the mass ratio, sometimes is more useful to study the ratio fuel mass/payload. In that case, we set M_f=m and M_0=m+M, where M is the fuel mass and m is the payload. So, we would write

    \[ R=\dfrac{m}{m+M}\]

so then the ratio fuel mass/payload will be

    \[ \dfrac{M}{m}=R^{-1}-1=\exp \left(\dfrac{g\tau}{c}\right)-1\]

We are ready to study the interstellar trip with our current knowledge of Special Relativity and Rocketry. We will study the problem in the next and final post of this fascinating thread. Stay tuned!

LOG#027. Accelerated motion in SR.

FarewellArmstrong

Neil-Armstrong-jpg

Hi, everyone! This is the first article in a thread of 3 discussing accelerations in the background of special relativity (SR). They are dedicated to Neil Armstrong, first man on the Moon! Indeed,  accelerated motion in relativity has some interesting and sometimes counterintuitive results, in particular those concerning the interstellar journeys whenever their velocities are close to the speed of light(i.e. they “are approaching” c).

Special relativity is a theory considering the equivalence of every  inertial frame ( reference frames moving with constant relative velocity are said to be inertial frames) , as it should be clear from now, after my relativistic posts! So, in principle, there is nothing said about relativity of accelerations, since accelerations are not relative in special relativity ( they are not relative even in newtonian physics/galilean relativity). However, this fact does not mean that we can not study accelerated motion in SR. The own kinematical framework of SR allows us to solve that problem. Therefore, we are going to study uniform (a.k.a. constant) accelerating particles in SR in this post!

First question: What does “constant acceleration” mean in SR?   A constant acceleration in the S-frame would give to any particle/object a superluminal speed after a finite time in non-relativistic physics! So, of course, it can not be the case in SR. And it is not, since we studied how accelerations transform according to SR! They transform in a non trivial way! Moreover, a force growing beyond the limits would be required for a “massive” particle ( rest mass m\neq 0). Suppose this massive particle (e.g. a rocket, an astronaut, a vehicle,…) is at rest in the initial time t=t'=0, and it accelerates in the x-direction (to be simple with the analysis and the equations!). In addition, suppose there is an observer left behind on Earth(S-frame), so Earth is at rest with respect to the moving particle (S’-frame). The main answer of SR to our first question is that we can only have a constant acceleration in the so-called instantaneous rest frame of the particle.  We will call that acceleration “proper acceleration”, and we will denote it by the letter \alpha. In fact, in many practical problems, specially those studying rocket-ships, the acceleration is generally given the same magnitude as the gravitational acceleration on Earth (alpha=g\approx 9.8ms^{-2}\approx 10 ms^{-2}).

Second question: What are the observed acceleration in the different frames? If the instantaneous rest frame S’ is an inertial reference frame in some tiny time dt', at the initial moment, it has the same velocity as the particle (rocket,…) in the S-frame, but it is not accelerated, so the velocity in the S’-frame vanishes at that time:

    \[ \mathbf{u}'=(0,0,0)\]

Since the acceleration of the particle is, in the S’-frame, the proper acceleration, we get:

    \[ \mathbf{a}'=(a'_x,0,0)=(\alpha,0,0)=(g,0,0)=\mbox{constant}\]

Using the transformation rules for accelerations in SR we have studied, we get that the instantaneous acceleration in the S-frame is given by

    \[ \mathbf{a}=(a_x,0,0)=\left(\dfrac{g}{\gamma^3},0,0\right)\]

Since the relative velocity between S and S’ is always the same to the moving particle velocity in the S-frame, the following equation holds

    \[ v=u_x\]

We do know that

    \[ a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g\]

Due to time dilation

    \[ dt'=dt/\gamma\]

so in the S-frame, the particle moves with the velocity

    \[ du_x=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g dt\]

We can now integrate this equation

    \[ \int_0^{u_x}\dfrac{1}{(c^2-u_x^2)^{3/2}}du_x=\dfrac{g}{c^3}\int_0^t dt\]

The final result is:

    \[ \boxed{u_x=\dfrac{g t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}\]

We can check some limit cases from this relativistic result for uniformly accelerated motion in SR.

1st. Short time limit: gt<< c\longrightarrow u_x\approx gt=\alpha t. This is the celebrated nonrelativistic result, with initial speed equal to zero (we required that hypothesis in our discussion above).

2nd. Long time limit: t\rightarrow \infty. In this case, the number one inside the root is very tiny compared with the term depending on acceleration, so it can be neglected to get u_x\approx \dfrac{gt}{gt/c}=c. So, we see that you can not get a velocity higher than the speed of light with the SR framework at constant acceleration!

Furthermore, we can use the definition of relativistic velocity in order to integrate the associated differential equation, and to obtain the travelled distance as a function of t, i.e. x(t), as follows

    \[ u_x=\dfrac{dx}{dt}=\dfrac{gt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}\]

    \[ \int_0^x dx=\int_0^t\dfrac{gt dt}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}=\int_0^t\dfrac{ctdt}{\sqrt{\dfrac{c^2}{g^2}+t^2}}\]

We can perform the integral with the aid of the following known result ( see,e.g., a mathematical table or use a symbolic calculator or calculate the integral by yourself):

    \[ \int \dfrac{ctdt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+\mbox{constant}=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}+C\]

From this result, and the previous equation, we get the so-called relativistic path-time law for uniformly accelerated motion in SR:

    \[ x=c\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}-\dfrac{c^2}{g}\]

or equivalently

    \[ \boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}\]

For consistency, we observe that in the limit of short times, the terms in the big brackets approach 1+\frac{1}{2}\left(\frac{gt}{c}\right)^2, in order to get x\approx \frac{1}{2}gt^2, so we obtain the nonrelativistic path-time relationship x=\frac{1}{2}gt^2 with g=a_x. In the limit of long times, the terms inside the brackets can be approximated to gt/c, and then, the final result becomes x\approx ct. Note that the velocity is not equal to the speed of light, this result is a good approximation whenever the time is “big enough”, i.e., it only works for “long times” asymptotically!

And finally, we can write out the transformations of acceleration between the two frames in a explicit way:

    \[ a_x=\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{3/2}g\]

that is

    \[ \boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}\]

Check 1: For short times, a_x\approx g=\mbox{constant}, i.e., the non-relativistic result, as we expected!

Check 2: For long times, a_x\approx \dfrac{c^3} {g^2t^3}\rightarrow 0. As we could expect, the velocity increases in such a way that “saturates” its own increasing rate and the speed of light is not surpassed. The fact that the speed of light can not be surpassed or exceeded is the unifying “theme” through special relativity, and it rest in the “noncompact” nature of the Lorentz group due to the \gamma factor, since it would become infinity at v=c for massive particles.

It is inevitable: as time passes, a relativistic treatment is indispensable, as the next figures show

V-tSRuniformAcceleration

v-tPlotSRaccUniform

AccTimeSR

The next table is also remarkable (it can be easily built with the formulae we have seen till now with any available software):

SRpos-vel-acc

Let us review the 3 main formulae until this moment

    \[ \boxed{a_x=\dfrac{1}{\left[1+\left(\dfrac{gt}{c}\right)^2\right]^{3/2}}g}\]

    \[ \boxed{u_x=\dfrac{\alpha t}{\sqrt{1+\left(\dfrac{g t}{c}\right)^2}}}\]

    \[ \boxed{x=x(t)=\dfrac{c^2}{g}\left(\sqrt{1+\left(\dfrac{gt}{c}\right)^2}-1\right)}\]

We have calculated these results in the S-frame, it is also important and interesting to calculate the same stuff in the S’-frame of the moving particle. The proper time \tau=t' is defined as:

    \[ \boxed{d\tau=dt\sqrt{1-\left(\dfrac{u_x}{c}\right)^2}}\]

Therefore,

    \[ d\tau=dt\left[1-\dfrac{\left(\dfrac{gt}{c}\right)^2}{1+\left(\dfrac{gt}{c}\right)^2}\right]^{1/2}\]

We can perform the integral as before

    \[ \displaystyle{\int_0^\tau d\tau=\int_0^t\dfrac{dt}{\sqrt{1+\left(\dfrac{gt}{c}\right)^2}}}\]

and thus

    \[ \tau=\dfrac{c}{g}\int_0^\tau\dfrac{dt}{\sqrt{\left(\dfrac{c}{g}\right)^2+t^2}}=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{\left(\dfrac{gt}{c}\right)^2+1}\right)\bigg|_0^t\]

Finally, the proper time(time measured in the S’-frame) as a function of the elapsed time on Earth (S-frame) and the acceleration is given by the very important formula:

    \[ \boxed{\tau=\dfrac{c}{g}\ln \left(\dfrac{gt}{c}+\sqrt{1+\left(\dfrac{gt}{c}\right)^2}\right)}\]

And now, let us set z=gt/c, therefore we can write the above equation in the following way:

    \[ \dfrac{g\tau}{c}=\ln \left( z+\sqrt{1+z^2}\right)\]

Remember now, from our previous math survey, that

    \[ \sinh^{-1}z=\ln \left( z+\sqrt{1+z^2}\right)\]

, so we can invert the equation in order to obtain t as function of the proper time since:

    \[ \boxed{\tau=\dfrac{c}{g}\sinh^{-1}\left(\dfrac{gt}{c}\right)}\]

    \[ \boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}\]

Inserting this last equation in the relativistic equation path-time for the uniformly accelerated body in SR, we obtain:

    \[ x=x(\tau)=\dfrac{c^2}{g}\left(\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}-1\right)\]

i.e.,

    \[ \boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}\]

Similarly, we can calculate the velocity-proper time law. Previous equations yield

    \[ u_x=\dfrac{c\sinh\left(\dfrac{g\tau}{c}\right)}{\sqrt{1+\sinh^2\left(\dfrac{g\tau}{c}\right)}}=\dfrac{c\sinh \left(\dfrac{g\tau}{c}\right)}{\cosh \left(\dfrac{g\tau}{c}\right)}\]

and thus the velocity-proper time law becomes

    \[ \boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}\]

Remark: this last result is compatible with a rapidity factor \varphi= \left(\dfrac{g\tau}{c}\right).

Remark(II):

    \[ a_x=\dfrac{du_x}{dt}=\left(1-\dfrac{u_x^2}{c^2}\right)^{3/2}g=\left(1-\tanh^2\left(\dfrac{g\tau}{c}\right)\right)^{3/2}g=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g\]

.

From this, we can read the reason why we said before that constant acceleration is “meaningless” unless we mean or fix certain proper time in the S’-frame since whenever we select a proper time, and this last relationship gives us the “constant” acceleration observed from the S-frame after the transformation. Of course, from the S-frame, as this function shows, acceleration is not “constant”, it is only “instantaneously” constant. We have to take care in relativity with the meaning of the words. Mathematics is easy and clear and generally speaking it is more precise than “words”, common language is generally fuzzy unless we can explain what we are meaning!

As the final part of this log entry, let us summarize the time-proper time, velocity-proper time, acceleration-proper time-proper acceleration and distance- proper time laws for the S’-frame:

    \[ \boxed{t=\dfrac{c}{g}\sinh \left(\dfrac{g\tau}{c}\right)}\]

    \[ \boxed{u_x=c\tanh \left(\dfrac{g\tau}{c}\right)}\]

    \[ \boxed{a_x=\dfrac{1}{\cosh^3\left(\dfrac{g\tau}{c}\right)}g}\]

    \[ \boxed{x=x(\tau)=\dfrac{c^2}{g}\left[\cosh \left(\dfrac{g\tau}{c}\right)-1\right]}\]

My last paragraph in this post is related to express the acceleration g\approx 10ms^{-2} in a system of units where space is measured in lightyears (we take c=300000km/s) and time in years (we take 1yr=365 days). It will be useful in the next 2 posts:

    \[ g=10\dfrac{m}{s^2}\dfrac{1ly}{9.46\cdot 10^{15}m}\dfrac{9.95\cdot 10^{14}s^2}{1yr^2}=1.05\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}\]

Another election you can choose is

    \[ g=9.8\dfrac{m}{s^2}=1.03\dfrac{lyr}{yr^2}\approx 1\dfrac{lyr}{yr^2}\]

so there is no a big difference between these two cases with terrestrial-like gravity/acceleration.

LOG#026. Boosts, rapidity, HEP.

zzeeee

In euclidean two dimensional space, rotations are easy to understand in terms of matrices and trigonometric functions. A plane rotation is given by:

    \[ \boxed{\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}}\leftrightarrow \boxed{\mathbb{X}'=\mathbb{R}(\theta)\mathbb{X}}\]

where the rotation angle is \theta, and it is parametrized by 0\leq \theta \leq 2\pi.

Interestingly, in minkovskian two dimensional spacetime, the analogue does exist and it is written in terms of matrices and hyperbolic trigonometric functions. A “plane” rotation in spacetime is given by:

    \[ \boxed{\begin{pmatrix}ct'\\ x'\end{pmatrix}=\begin{pmatrix}\cosh \varphi & -\sinh \varphi \\ -\sinh \varphi & \cosh \varphi \end{pmatrix}\begin{pmatrix}ct\\ x\end{pmatrix}}\leftrightarrow \boxed{\mathbb{X}'=\mathbb{L}(\varphi)\mathbb{X}}\]

Here, \varphi = i\psi is the so-called hiperbolic rotation angle, pseudorotation, or more commonly, the rapidity of the Lorentz boost in 2d spacetime. It shows that rapidity are a very useful parameter for calculations in Special Relativity. Indeed, it is easy to check that

    \[ \mathbb{L}(\varphi_1+\varphi_2)=\mathbb{L}(\varphi_1)\mathbb{L}\mathbb(\varphi_2)\]

So, at least in the 2d spacetime case, rapidities are “additive” in the written sense.

Firstly, we are going to guess the relationship between rapidity and velocity in a single lorentzian spacetime boost. From the above equation we get:

    \[ ct'=ct\cosh \varphi -x\sinh \varphi \]

    \[ x'=-ct\sinh \varphi +x\cosh \varphi\]

Multiplying the first equation by \cosh \varphi and the second one by \sinh \varphi, we add the resulting equation to obtain:

    \[ ct'\cosh\varphi+x'\sinh \varphi =ct\cosh^2 \varphi -ct\sinh^2 \varphi =ct\]

that is

    \[ ct'\cosh\varphi+x'\sinh \varphi =ct\]

From this equation (or the boxed equations), we see that \varphi=0 corresponds to x'=x and t'=t. Setting x'=0, we deduce that

    \[ x'=0=-ct\sinh \varphi +x\cosh \varphi\]

and thus

    \[ ct\tanh \varphi =x\]

or

    \[ x=ct\tanh\varphi\]

.

Since t\neq 0, and the pseudorotation seems to have a “pseudovelocity” equals to V=x/t, the rapidity it is then defined through the equation:

    \[ \boxed{\tanh \varphi=\dfrac{V}{c}=\beta}\leftrightarrow\mbox{RAPIDITY}\leftrightarrow\boxed{\varphi=\tanh^{-1}\beta}\]

If we remember what we have learned in our previous mathematical survey, that is,

    \[ \tanh^{-1}z=\dfrac{1}{2}\ln \dfrac{1+z}{1-z}=\sqrt{\dfrac{1+z}{1-z}}\]

We set z=\beta in order to get the next alternative expression for the rapidity:

    \[ \varphi=\ln \sqrt{\dfrac{1+\beta}{1-\beta}}=\dfrac{1}{2}\ln \dfrac{1+\beta}{1-\beta}\leftrightarrow \exp \varphi=\sqrt{\dfrac{1+\beta}{1-\beta}}\]

In experimental particle physics, in general 3+1 spacetime, the rapidity definition is extended as follows. Writing, from the previous equations above,

    \[ \sinh \varphi=\dfrac{\beta}{\sqrt{1-\beta^2}}\]

    \[ \cosh \varphi=\dfrac{1}{\sqrt{1-\beta^2}}\]

and using these two last equations, we can also write momenergy components using rapidity in the same fashion. Suppose that for some particle(object), its  mass is m, its energy is E, and its (relativistic) momentum is \mathbf{P}. Then:

    \[ E=mc^2\cosh \varphi\]

    \[ \lvert \mathbf{P} \lvert =mc\sinh \varphi\]

From these equations, it is trivial to guess:

    \[ \varphi=\tanh^{-1}\dfrac{\lvert \mathbf{P} \lvert c}{E}=\dfrac{1}{2}\ln \dfrac{E+\lvert \mathbf{P} \lvert c}{E-\lvert \mathbf{P} \lvert c}\]

This is the completely general definition of rapidity used in High Energy Physics (HEP), with a further detail. In HEP, physicists used to select the direction of momentum in the same direction that the collision beam particles! Suppose we select some orientation, e.g.the z-axis. Then, \lvert \mathbf{P} \lvert =p_z and rapidity is defined in that beam direction as:

    \[ \boxed{\varphi_{hep}=\tanh^{-1}\dfrac{\lvert \mathbf{P}_{beam} \lvert c}{E}=\dfrac{1}{2}\ln \dfrac{E+p_z c}{E-p_z c}}\]

In 2d spacetime, rapidities add nonlinearly according to the celebrated relativistic addition rule:

    \[ \beta_{1+2}=\dfrac{\beta_1+\beta_2}{1+\frac{\beta_1\beta_2}{c^2}}\]

Indeed, Lorentz transformations do commute in 2d spacetime since we boost in a same direction x, we get:

    \[ L_1^xL_2^x-L_2^xL_1^x=0\]

with

    \[ L_1^x=\begin{pmatrix}\gamma_1 & -\gamma_1\beta_1\\ -\gamma_1\beta_1 &\gamma_1 \end{pmatrix}\]

    \[ L_2^x=\begin{pmatrix}\gamma_2 & -\gamma_2\beta_2\\ -\gamma_2\beta_2 &\gamma_2 \end{pmatrix}\]

This commutativity is lost when we go to higher dimensions. Indeed, in spacetime with more than one spatial direction that result is not true in general. If we build a Lorentz transformation with two boosts in different directions V_1=(v_1,0,0) and V_2=(0,v_2,0), the Lorentz matrices are ( remark for experts: we leave one direction in space untouched, so we get 3×3 matrices):

    \[ L_1^x=\begin{pmatrix}\gamma_1 & -\gamma_1\beta_1 &0\\ -\gamma_1\beta_1 &\gamma_1 &0\\ 0& 0& 1\end{pmatrix}\]

    \[ L_2^y=\begin{pmatrix}\gamma_2 & 0&-\gamma_2\beta_2\\ 0& 1& 0\\ -\gamma_2\beta_2 & 0&\gamma_2 \end{pmatrix}\]

and it is easily checked that

    \[ L_1^xL_2^y-L_2^yL_1^x\neq 0\]

Finally, there is other related quantity to rapidity that even experimentalists do prefer to rapidity. It is called: PSEUDORAPIDITY!

Pseudorapidity, often denoted by \eta describes the angle of a particle relative to the beam axis. Mathematically speaking is:

    \[ \boxed{\eta=-\ln \tan \dfrac{\theta}{2}}\leftrightarrow \mbox{PSEUDORAPIDITY}\leftrightarrow \boxed{\exp (\eta)=\dfrac{1}{\tan\dfrac{\theta}{2}}}\]

where \theta is the angle between the particle momentum \mathbf{P}  and the beam axis. The above relation can be inverted to provide:

    \[ \boxed{\theta=2\tan^{-1}(e^{-\eta})}\]

The pseudorapidity in terms of the momentum is given by:

    \[ \boxed{\eta=\dfrac{1}{2}\ln \dfrac{\vert \mathbf{P}\vert +P_L}{\vert \mathbf{P}\vert -P_L}}\]

Note that, unlike rapidity, pseudorapidity depends only on the polar angle of its trajectory, and not on the energy of the particle.

In hadron collider physics,  and other colliders as well, the rapidity (or pseudorapidity) is preferred over the polar angle because, loosely speaking, particle production is constant as a function of rapidity. One speaks of the “forward” direction in a collider experiment, which refers to regions of the detector that are close to the beam axis, at high pseudorapidity \eta.

The rapidity as a function of pseudorapidity is provided by the following formula:

    \[ \boxed{\varphi=\ln\dfrac{\sqrt{m^2+p_T^2\cosh^2\eta}+p_T\sinh \eta}{\sqrt{m^2+p_T^2}}}\]

where p_T is the momentum transverse to the direction of motion and m is the invariant mass of the particle.

Remark: The difference in the rapidity of two particles is independent of the Lorentz boosts along the beam axis.

Colliders measure physical momenta in terms of transverse momentum p_T instead of the momentum in the direction of the beam axis (longitudinal momentum) P_L=p_z, the polar angle in the transverse plane (genarally denoted by l \phi) and pseudorapidity \eta. To obtain cartesian momenta (p_x,p_y,p_z)  (with the z-axis defined as the beam axis), the following transformations are used:

    \[ p_x=P_T\cos\phi\]

    \[ p_y=P_T\sin\phi\]

    \[ p_z=P_T\sinh\eta\]

Thus, we get the also useful relationship

    \[ \vert P \vert=P_T\cosh\eta\]

This quantity is an observable in the collision of particles, and it can be measured as the main image of this post shows.

LOG#025. Minkovski diagrams.

600px-Minkowski_diagram_-_asymmetric.svg

“(…)The views of space and time which I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.(…)”
— Hermann Minkowski, in ‘Space And Time’, a translation of an address delivered at the 80th Assembly of German Natural Scientists and Physicians, at Cologne, 21 Sep 1908. In H.A. Lorentz, H. Weyl, H. Minkowski, et al., The Principle of Relativity: A Collection of Original Memoirs on the Special and General Theory of Relativity (1952), 74.

In Special Relativity, via Lorentz transformations, space and time are intrinsically united. Therefore, it makes sense to call the abstract set of space and time together spacetime, as we have seen before in our relativistic thread. Hermann Minkovski, a former professor of Albert Einstein guessed a quite simple and pictorical representation of “events” in “space-time”. That representation shares very similarities with usual cartesian geometry in the plane, but with some differences. Let me explain it better. Generally, the position of a point P in the real plane is expressed by Cartesian ( also called parallel or rectangular) coordinates P(x_0,y_0). But one could just as well use two oblique-angled axes, x' and y' instead, i.e., we can choose “a rotated frame”. In order to determine the new coordinates x'_0,y'_0 of any point P in such coordinate system, we draw parallels to the oblique axes through P, and then, we can calculate x'_0,y'_0 at the intersection points with the new frame. This result is completely general, and the units of length on the new axes don’t need to be the same as those we used in the old coordinates. See the following diagram:

MD1png

Lorentz transformations are what mathematicians call “affine transformations”, i.e., a one-to-one (a.k.a. bijective) mapping of a plane on itself that preserves parallelism and rectilinearity of straight lines! For a 2D Minkovski spacetime transformation, being general, we have

    \[ ct'=x'_0=L_{00}ct+L_{01}x+b_0\]

    \[ x'=L_{10}ct+L_{11}x+b_1\]

Setting that b_0=b_1=0 fixes a common origin of coordinates (i.e. an unchanged point of origin). It is very simple to calculate the direction and units of length of the new primed axes in the coordinate system of the unprimed (old) coordinates. We have to set

    \[ ct'=0, x'=1\]

into the inverse 2d Lorent transformations above, and we obtain the expected result ct=\gamma, x=\beta \gamma. It is a common practice to represent the time in the vertical axis, and the position in the horizontal axis. Curiously, it is the opposed practice in the more conventional plane motion with which we are more familiar,  and where we use to represent horizontally the time and vertically the position of some object. Of course, that is only a conventional election, we could draw the axis as we wanted to. We only have to establish the “rules”. And we will follow the common practices here. Furthermore, a Lorentz transformation does not cause the coordinate axes to rotate unidirectionally, since with the rotation of the coordinate frame around its origin, they will rotate counterdirectionally! Elementary trigonometry helps us now ( we reviewed it in a previous log), and the angle between the primed (new) and the unprimed (old) axes is showed to be:

    \[ \boxed{\tan \delta=\beta=\dfrac{v}{c}}\]

The units of length, L, on the new reference frame in spacetime can be calculated with a simple application of the Pythagoras theorem:

    \[ L=\sqrt{\gamma^2+\beta^2\gamma^2}\]

and thus

    \[ \boxed{L=\sqrt{\dfrac{1+\beta^2}{1-\beta^2}}}\]

This stuff can be reviewed in the following cool diagram:

MD2png

Example: the relative velocity between S and S’ is c/2. It provides \tan \delta=\beta=1/2 and L=\sqrt{5/3}\approx 1.29. Therefore, coming back to the new reference frame, the spacing of the “tick” marks (the scale) is stretched by a factor of 1.29. Suppose in addition that the units of length are given in light-years (ly). Then, time coordinates will be given too in light-years since we have ct, ct' in the temporal axes. Finally, an important remark to relativity beginners: the obliquity of the x’-axis in a Minkovsky spacetime diagram DOES NOT IMPLY that there is an angle between the x and the x’ axes IN SPACE. The angles between frames are between space-time axis in general, it is not necessary that the x-x’ axes form an angle a priori.

Some practical exercises will aid us to experiment the power of the Minkovski diagramas to solve commonly given examples in Special Relativity.

Exercise 1. Events in space-time.

Observe the following Minkovski diagram

MD4png

The event E is located 5 years in the future and 3 ly on the right of the origin (0,0), as it is given in the S frame. The S’ frame moves with velocity \beta=0.5 to the right. The question is: when and where does event E happen in the S’ frame? A simple calculation shows it:

    \[ \beta=0.5\rightarrow \gamma = \dfrac{2}{\sqrt{3}}\]

Therefore, the coordinates of the event E in the primed frame will be:

    \[ ct'=\gamma (ct-\beta x)=\dfrac{2}{\sqrt{3}}(5ly-0.5\cdot 3ly)\approx 4.04ly\]

    \[ x'=\gamma(x-\beta c t)=\dfrac{2}{\sqrt{3}}(3ly-0.5\cdot 5ly)\approx 0.59ly\]

In the primed reference frame S’, the event E happens in such a way that E’ is about 4.04 years in the future and about 0.58 ly on the right of the origin! Fascinating and elementary, from our limited knowledge at least. Consider a variation of this exercise, where the event E is located 2 years in the past and 2 ly on the left of the origin, as given in the S-frame. The S’-frame moves with velocity \beta=-0.6 to the left. Again, the question is: when and where does the event E happen in the S’-frame? Looking at the new Minkovski diagram:

MD5png

We can proceed to make a simple calculation as before we did:

    \[ \beta=-0.6\rightarrow \gamma =\dfrac{5}{4}\]

And thus, the new coordinates will be boosted by Lorentz transformations:

    \[ ct'=\gamma(ct-\beta x)=\dfrac{5}{4}(-3ly-(-0.6)(-2ly))=-4ly\]

    \[ x'=\gamma (x-\beta ct)=\dfrac{5}{4}(-2ly-(-0.6)(-2ly))=-4ly\]

In the primed reference frame S’, the event is located 4 years in the past and 4 light-years (ly) on the left of the common origin of both frames.

Exercise 2. Simultaneity in action.

We will suppose now that \beta=0.5. All the events that are located on a straight line parallel to the x-axis (dashed in the next Minkovski diagram) are simultaneous in S.

MD6

Therefore, every event located on a straight line parallel to the x’-axis (dotted in our Minkovski diagrama) are simultaneous in S’. It is evident and obvious that two arbitrary events, E_1, E_2, that are simultaneous in one inertial frame can NOT be simultaneous in any other inertial frame. This fact drives many of the “paradoxes” of Special Relativity. They are not really paradoxes in general, since usually we can solve the apparent contradictions with our language using a proper language and a right physical insight.

The relativity of simultaneity can also be understood with other Minkovski diagram for different events:

MD14

Exercise 3. The world lines in space-time.

Choose any inertial reference frame, e.g., the S-frame and place some object in space with spacetime coordinates E(x,ct)=(x_0,ct_0). This point of the spacetime will be denoted as E_0. In a posterior moment, the object moves towards any other arbitrary location, e.g., x=x_1 and ct=ct_1. We define the event E_1(x,ct)=(x_1,ct_1).

If we track the object through the spacetime, we obtain a sequence of events! This sequence of events is called “worldline”.

The worldline of any particle/object in nothing but a path-time diagram, i.e., the diagram length-time of the particle, with the path plotted to the right an the time plotted towards the top. Moreover, the worldline of any object AT REST in the S-frame is parallel to the ct-axis. The faster the object is in the S-frame, the flatter its worldline is. If the object moves at the speed of light, the worldline is a straight line with certain slope that can be only either +1 or -1, since it moves one unit in the x-direction (1 ly) in one unit of the ct-direction ( 1year times the speed of light). No material object can move faster than the speed of light in Special Relativity (i.e., under the hypothesis of Special Relativity, any body is restricted to move wih velocities lower than “c”. In addition, worldlines intersecting the x-axis at angles less than 45º are thus excluded ( unless we allow the notion of tachyons, to be discussed in the near future in this blog).

These notions can be understood with the following diagram, sketching the worldline:

MD7
Imagine a light flash propagating from the event l E(ct,x)=(2ly,2ly) in a spherical shell expanding in all directions of space. In the one dimensional representation of space, given by the following Minkovski spacetime diagram, the spherical shell is, at any instant of time, reduced to two points ( intersection of a line with the spherical shell). As time flows, the two points turn into two diverging world lines, we drawed them dashed. We can read from the Minkovski diagram that the light reaches the observer at t=4yr in the S-frame, and t'\approx 2.3 yrs in the S’-frame (\beta=0.5). The calculation for S is trivial, but for S’ some harder calculation is to be done.

MD8

Exercise 4. Time dilation via diagrams.

Again we take \beta=0.5. The event P'(ct',x')=(3ly,0ly) has oblique coordinate axis. In the cartesian frame at rest, S, we have ct\approx 3.5 ly. Explanation: a process extending from the origin to E’ and it takes 3 years in the S’-frame, while it will table about 3.5 years in the S-frame. The observer at rest in S concludes that the “clocks” of the S’-observer are slower than his own clocks.

The point Q(ct,x)=(3ly,0ly) in the cartesian S-frame has the time coordinate ct'\approx 3.5 ly in the S’-frame. Meaning: a process, extending from the origin to Q, it takes 3 years in the S-frame and it will take about 3.5 years in the S’-frame. The S’-observer claims that the clocks of the S-observer are slower that his own clocks.

The diagram is:

MD9

Remember: there is no contradiction in the fact that both observers “measure” that their own clocks are both slower than the other, since, at last, the events P’ and Q are not happening in the same point in spacetime!

Exercise 5. Length contraction via diagrams.

Once again, \beta=0.5, and the Minkovski diagram is now this one

MD10

We put a rod extended from the origin to the point P'(ct',x')=(0ly,3.5ly) and it will be initially at rest in the S’-frame. Its length in that frame equals 3.5 ly. The position of its left end is a function of time, i.e., its worldline IS the ct’-axis while the worldline of its right end is parallel to the ct’-axis through P’ (marked as a dotted line). At any instant t’ in teh S’-frame, the reod is placed on a paralllel line to the x’-axis.

By the other hand, in the S-frame the ros is extended from the worldline of the left end (ct’-axis) to the worldline of the right end (dotted line in the diagram). However, at any moment t in the S-frame, the rod is placed on a parallel line to the x-axis. At time t=0, the rod is extending from the origin to the point P(ct,x)=(0,x\approx 3). Thus, its length is approximately 3ly instead of 3.5 ly, and the rod is contracted.

In a similar way, a rod at rest in the S-frame, extending from the origin to the point Q(ct,x)=(0ly,3.5ly) will be contracted in the S’-frame to the distance from the origin to the point Q'(ct',x')=(0ly,x\approx 3ly).

Remark: the phenomenon of length contraction has to do with simultaneity as well, length measurements of moving objects are reasonable only if the positions of both ends can be measured simultaneously! Since S and S’ don’t agree on simultaneity, they cannot agree on the results of their length measurements.

Exercise 6. The past, the future and causality.
The next Minkovski diagram owns the coordinate axes of three different reference frames S, S’, and S”. The beta parameters are respectively \beta=0, \beta '=-0.6, \beta ''=0.6. There are two wordlines of light rays (flashes of light) passing through the origin ( with the dashed bisectors). The projections of the events P, Q and R on the different time axis are dotted and clearly distinguished from one to another:

MD11

We observe and distinguish:

1st. P is located in the future, corresponding to the S’-frame ct’>0, in the present according to the S-frame since ct=0, and in the past in the S”-frame because of the relationship ct”<0.

2nd. Q is located in the future in every frame, R is located in the past in every frame.

We can simplify the above diagram with a more simple Minkovski diagram

MD12

The 2 sectors on the left and right, in medium-gray colour, comprise zones where, depending on the frame, can be comprehended as past, present or future. Every worldline are at least as steep as the worldlines of light (dashed here), and everyworldline created from any event therein have to intersect the ct-axis above the origin. Some worldlines, created at P, have been added to the figure. No event in the medium-gray zones can affect the origin. This is a notion of causality. Furthermore, no event in the medium-gray zones can be reached or be influenced by a worldline from the origin. These points outside the lightcone are said to be out of the causal influence of the past and future.

Events contained in the darker zone are regarded as future by every frame. A world line from the origin can reach any event therein. In this way, we can call this zone “absolute future”. In the same fashion, events in the lower zone (the lighter gray) are past for every frame. They are absolute past and any event happening inside that zone can be connected to the origin by a worldline as well.

Remark: it is impossible, a priori, for a worldline to run from the upper to the lower zone, i.e., for events like Q to influence R is impossible (according to SR). By no means can be Q the cause of R. However, R CAN be the cause of Q. The sequence of cause and effect, causality, cannot be inverted by the special theory of relativity. It it happens, it would be an argument against the theory.

Exercise 7. Faster than the speed of light?

Consider the next gedanken experiment (thought or imaginary experiment). Let us suppose that information can be transmitted at a speed faster than light. Observe then the following diagram and once again set \beta=0.5

MD13

And next, imagine the next experiment:

1st. A transmission tower is at rest in the origin of the S-frame and a relay station is at rest in the S’-frame at x'=x'_0.

2nd. At the time t=t'=0, a signal will be transmitted from the common origin of the two frames to the relay station at speed 10c, as measured from the S-frame (square dots in the diagram).

3rd. It is being received by the relay station in the event A, and being re-emitted after a short period of time to the source at speed of -10c, according to the S’-frame (round dots in the diagram).

4th. The signal is being received in the event B at a time t<0, i.e., at the source, but BEFORE the signal was emitted there.

Conclusion: If the signal carries a destructive energy, it can destroy the transmitter. Cause and effect would occur in an inverted sequence!

LOG#024. Strange derivative.

drstrange492012

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I have been fascinated (perhaps I am in love too with it) by Mathematics since I was a child. As a teenager in High School, I was a very curious student ( I am curious indeed yet)  and I tried to understand some weird results I got in the classroom. This entry is devoted to some of those problems that made me wonder and think a lot out of class, at home. It took me some years and to learn complex variable function theory to understand one of the issues I could not understand before I learned complex variable. 3 years to understand a simple derivate! Yes, it is too much time. However, understanding stuff deeply carries time. Sometimes more, sometimes less.

This is the problem. A very simple problem indeed! The following “weird” (real) trigonometric-like function has a null derivative:

    \[ g(x)= \tan^{-1}((1+x)/(1-x))-\tan ^{-1} (x)\]

where the \tan^{-1} (x) is the arctangent function \arctan (x)( the inverse of the tangent function \tan (x)).

Proof:

    \[ g'(x) = \dfrac{dg}{dx}=\dfrac{\dfrac{(1-x)+(1+x)}{(1-x)^2}}{1+\left(\dfrac{(1+x)}{(1-x)}\right)^2}-\dfrac{1}{1+x^2}\]

    \[ g'(x) = \dfrac{\dfrac{2}{(1-x)^2}}{1+\left(\dfrac{1+x}{1-x}\right)^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{2}{\left(1-x\right)^2+\left(1+x\right)^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{2}{2+2x^2}-1/(1+x^2)\]

    \[ g'(x) = \dfrac{1}{1+x^2}-\dfrac{1}{1+x^2}\]

    \[ g'(x)=0 \]

q.e.d.

Since the derivative g'(x) is zero, the two functions must differ by a constant. We can guess that constant with usual real variable calculus. It’s quite simple:

    \[ g(x)=constant.$ => $ g(x=0)= constant = \tan^{-1}((1+0)/(1-0))-\tan ^{-1} (0)\]

    \[ g(0)= \tan^{-1}(1)-\tan ^{-1} (0)=\pi /4\]

So, the difference is \pi/ 4. However, this calculation does not explain why the two functions differ by a constant. The secret lies in the complex function origin of the arctangent. In complex variable function theory, it can be proved that

    \[ \tan^{-1}(x)=\dfrac{1}{2i} \ln ((1+ix)/(1-ix))\]

since

    \[ \tan(x)=y=\dfrac{\left[\dfrac{1}{2i} (\exp (ix)-\exp (-ix))\right]}{\left[\dfrac{1}{2} \left(\exp (ix)+\exp (-ix)\right)\right]}=\dfrac{1}{i}\dfrac{\exp (ix)-\exp (-ix)}{\exp (ix)+\exp (-ix)}\]

then iy=\left(\exp (2ix)-1)/(\exp (2ix)+1\right) and thus \exp (2ix)=(1+iy)/(1-iy) and therefore x=\dfrac{1}{2i} \ln \left((1+iy)/(1-iy)\right) Q.E.D.

Now, we can calculate the functions in terms of complex variables:

    \[ \tan^{-1}\left((1+x)/(1-x)\right)=\arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left(\dfrac{(1+i\left(\frac{1+x}{1-x}\right)}{(1-i\left(\frac{1+x}{1-x}\right)}\right)\]

We can make some algebra inside the logarithm function to get:

    \[ \arctan \left((1+x)/(1-x)\right)=\dfrac{1}{2i} \ln \left( \dfrac{(1-x)+i(1+x)}{(1-x)-i(1+x)} \right)= \dfrac{1}{2i} \ln \left( \dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)} \right)\]

By the other hand, we also have

    \[ \tan^{-1}(x)=\arctan (x)=\dfrac{1}{2i} \ln \left( \dfrac{1+ix}{1-ix} \right)\]

Thus,

    \[ g(x) = \dfrac{1}{2i} \left( \arctan ((1+x)/(1-x)) - \arctan (x) \right) = \dfrac{1}{2i} \left( \ln \left [ \dfrac{\dfrac{(1+ix)+i(1+ix)}{(1-ix)-i(1-ix)}}{ \dfrac{1+ix}{1-ix}} \right]\right)\]

i.e.,the terms depending on x cancel to get a pure complex number! The number is

    \[ g(x)=number=\dfrac{1}{2i} \ln \left( \dfrac{1+i}{1-i}\right)\]

We have to calculate the logarith of the complex number

    \[ z=\dfrac{1+i}{1-i}=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}=\dfrac{2i}{2}=i=exp(i\pi/2)\]

Then, \ln ( i ) = i\pi/2. Of course, that is we were expecting to get since in this case

    \[ g(x)=\dfrac{1}{2i} \dfrac{i\pi}{2}=\dfrac{\pi}{4}\]

as before! That is, we recover the phase difference we also got with real calculus. The origin of the cancellation was in the complex origin of the arctangent function! Beautiful mathematics! The complex world is fascinating but very enlightening even for real functions!

LOG#023. Math survey.

hypeuclElliptic

What is a triangle? It is a question of definition in Mathematics. Of course you could disagree, but it is true. Look the above three “triangles”. Euclidean geometry is based in the first one. The second “triangle” is commonly found in special relativity. Specially, hyperbolic functions. The third one is related to spherical/elliptical geometry.

Today’s summary: some basic concepts in arithmetics, complex numbers and functions. We are going to study and review the properties of some elementary and well known functions. We are doing this in order to prepare a better background for the upcoming posts, in which some special functions will appear. Maybe, this post can be useful for understanding some previous posts too.

First of all, let me remember you that elementary arithmetics is based on seven basic “operations”: addition, substraction, multiplication, division, powers, roots, exponentials and logarithms. You are familiar with the 4 first operations, likely you will also know about powers and roots, but exponentials and logarithms are the last kind of elementary operations taught in the school ( high school, in the case they are ever explained!).

Let me begin with addition/substraction of real numbers (it would be also valid for complex numbers z=a+bi or even more general “numbers”, “algebras”, “rings” or “fields”, with suitable extensions).

    \[ a+b=b+a\]

    \[ (a+b)+c=a+(b+c)\]

    \[ a+(-a)=0\]

    \[ a+0=a\]

Multiplication is a harder operation. We have to be careful with the axioms since there are many places in physics where multiplication is generaliz loosing some of the following properties:

    \[ kA=\underbrace{A+..+A}_\text{k-times}\]

    \[ AB=BA\]

    \[ (AB)C=A(BC)=ABC\]

    \[ 1A=A1=A\]

    \[ (A+B)C=AC+BC\]

    \[ A(B+C)=AB+AC\]

    \[ A^{-1}A=AA^{-1}=1\]

Indeed the last rule can be undestood as the “division” rule, provided A\neq 0 since in mathematics or physics there is no sense to “divide by zero”, as follows.

    \[ A^{-1}=\dfrac{1}{A}\]

Now, we are going to review powers and roots.

    \[ x^a=\underbrace{x\cdots x}_\text{a-times}\]

    \[ (x^a)^b=x^{ab}\]

    \[ x^{-a}=\dfrac{1}{x^a}\]

    \[ x^ax^b=x^{a+b}\]

    \[ \sqrt[n]{x}=x^{1/n}\]

Note that the identity x^a+y^a=(x+y)^a is not true in general. Moreover, if x\neq 0 then x^0=1 as well, as it can be easily deduced from the previous axioms. Now, the sixth operation is called exponentiation. It reads:

    \[ \exp (a+b)=\exp (a)\exp (b)\]

Sometimes you can read e^{a+b}=e^{a}e^{b}, where \displaystyle{e=\lim_{x\to\infty}\left(a+\dfrac{1}{n}\right)^n} is the so-called “e” number. The definitions is even more general, since the previous property is the key feature for any exponential. I mean that,

    \[ a^x=\underbrace{a\cdots a}_\text{x-times}\]

    \[ a^xb^x=(ab)^x\]

    \[ \left(\dfrac{a}{b}\right)^x=\dfrac{a^x}{b^x}\]

We also get that for any x\neq 0, then 0^x=0. Finally, the 7th operation. Likely, the most mysterious for the layman. However, it is very useful in many different places. Recall the definition of the logarithm in certain base “a”:

    \[ \log_{(a)} x=y \leftrightarrow x=a^y\]

Please, note that this definition has nothing to do with the “deformed” logarithm of my previous log-entry. Notations are subtle, but you must always be careful about what are you talking about!

Furthermore, there are more remarks:

1st. Sometimes you write \log_e=\ln x. Be careful, some books use other notations for the Napier’s logarithm/natural logarithm. Then, you can find out there \log_e=L or even \log_e=\log.

2nd. Whenever you are using a calculator, you can generally find \log_e=\ln and \log_{(10)}=\log. Please, note that in this case \log is not the natural logarithm, it is the decimal logarithm.

Logarithms (caution: logarithms of real numbers, since the logarithms of  complex numbers are a bit more subtle) have some other cool properties:

    \[ \log_{(a)}(xy)=\log_{(a)}x+\log_{(a)}y\]

    \[ \log_{(a)}\dfrac{x}{y}=\log_{(a)}x-\log_{(a)}y\]

    \[ \log_{(a)}x^y=y\log_{(a)}x\]

    \[ \log_{(a)}=\dfrac{\log_{(b)}}{\log_{(b)}a}\]

Common values of the logarithm are:

    \[ \ln 0^+=-\infty;\; \ln 1=0;\; \ln e=1;\; \ln e^x=e^{\ln x}=x\]

Indeed, logarithms are also famous due to a remarkable formula by Dirac to express any number in terms of 2’s as follows:

    \[ \displaystyle{N=-\log_2\log_2 \sqrt{\sqrt{\underbrace{\cdots}_\text{(N-1)-times}2}}=-\log_2\log_2\sqrt{\underbrace{\cdots}_\text{(N)-times}2}}\]

However, it is quite a joke, since it is even easier to write N=\log_a a^N, or even N=\log_{(1/a)}a^{-N}

Are we finished? NO! There are more interesting functions to review. In particular, the trigonometric functions are the most important functions you can find in the practical applications.

EUCLIDEAN TRIGONOMETRY

Triangles are cool! Let me draw the basic triangle in euclidean trigonometry.

400px-TrigonometryTriangle.svg

The trigonometric ratios/functions you can define from this figure are:

i)The function (sin), defined as the ratio of the side opposite the angle to the hypotenuse:

    \[ \sin A=\dfrac{\textrm{opposite}}{\textrm{hypotenuse}}=\dfrac{a}{\,c\,}\]

ii) The function (cos), defined as the ratio of the adjacent leg to the hypotenuse.

    \[ \cos A=\dfrac{\textrm{adjacent}}{\textrm{hypotenuse}}=\dfrac{b}{\,c\,}\]

iii) The function (tan), defined as the ratio of the opposite leg to the adjacent leg.

    \[ \tan A=\dfrac{\textrm{opposite}}{\textrm{adjacent}}=\dfrac{a}{\,b\,}=\dfrac{\sin A}{\cos A}\]

The hypotenuse is the side opposite to the 90 degree angle in a right triangle; it is the longest side of the triangle, and one of the two sides adjacent to angle ”A”. The ”’adjacent leg”’ is the other side that is adjacent to angle ”A”. The ”’opposite side”’ is the side that is opposite to angle ”A”. The terms ”’perpendicular”’ and ”’base”’ are sometimes used for the opposite and adjacent sides respectively. Many English speakers find it easy to remember what sides of the right triangle are equal to sine, cosine, or tangent, by memorizing the word SOH-CAH-TOA ( a mnemonics rule whose derivation and meaning is left to the reader).

The multiplicative inverse or reciprocals of these functions are named the cosecant (csc or cosec), secant(sec), and cotangent (cot), respectively:

    \[ \csc A=\dfrac{1}{\sin A}=\dfrac{c}{a} \]

    \[ \sec A=\dfrac{1}{\cos A}=\dfrac{c}{b}\]

    \[ \cot A=\dfrac{1}{\tan A}=\dfrac{\cos A}{\sin A}=\dfrac{b}{a}\]

The inverse trigonometric functions/inverse functions are called the arcsine, arccosine, and arctangent, respectively. These functions are what in common calculators are given by \sin^{-1},\cos^{-1},\tan^{-1}. Don’t confuse them with the multiplicative inverse trigonometric functions.

There are arithmetic relations between these functions, which are known as trigonometric identities.  The cosine, cotangent, and cosecant are so named because they are respectively the sine, tangent, and secant of the complementary angle abbreviated to “co-“. From the goniometric circle (a circle of radius equal to 1) you can read the Fundamental Theorem of (euclidean) Trigonometry:

    \[ \cos^2\theta+\sin^2\theta=1\]

Indeed, from the triangle above, you can find out that the pythagorean theorem implies

    \[ a^2+b^2=c^2\]

or equivalently

    \[ \dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}=\dfrac{c^2}{c^2}\]

So, the Fundamental Theorem of Trigonometry is just a dressed form of the pythagorean theorem!
The fundamental theorem of trigonometry can be rewritten too as follows:

    \[ \tan^2\theta+1=\sec^2\theta\]

    \[ \cot^2+1=\csc^2\theta\]

These equations can be easily derived geometrically from the goniometric circle:

goniometric

The trigonometric ratios are also related geometrically to this circle, and it can seen from the next picture:

338px-Circle-trig6.svg

Other trigonometric identities are:

    \[ \sin (x\pm y)=\sin x \cos y\pm \sin y\cos x\]

    \[ \cos (x\pm y)=\cos x \cos y -\sin x \sin y\]

    \[ \tan (x\pm y)=\dfrac{\tan x\pm \tan y}{1\mp \tan x \tan y}\]

    \[ \cot (x\pm y)=\dfrac{\cot x \cot y\mp 1}{\cot x\pm \cot y}\]

    \[ \sin 2x=2\sin x \cos x\]

    \[ \cos 2x=\cos^2 x-\sin^2 x\]

    \[ \tan 2x=\dfrac{2\tan x}{1-\tan^2 x}\]

    \[ \sin \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{2}}\]

    \[ \cos \dfrac{x}{2}=\sqrt{\dfrac{1+\cos x}{2}}\]

    \[ \tan \dfrac{x}{2}=\sqrt{\dfrac{1-\cos x}{1+\cos x}}\]

    \[ \sin x\sin y=\dfrac{\cos(x-y)-\cos(x+y)}{2}\]

    \[ \sin x\cos y=\dfrac{\sin(x+y)+\sin(x-y)}{2}\]

    \[ \cos x\cos y=\dfrac{\cos (x+y)+\cos(x-y)}{2}\]

The above trigonometric functions are also valid for complex numbers with care enough. Let us write a complex number as either a binomial expression z=a+bi or like a trigonometric expression z=re^{i\theta}. The famous Euler identity:

    \[ e^{i\theta}=\cos \theta + i\sin \theta\]

allows us to relate both two expressions for a complex number since

    \[ z=r(\cos \theta + i\sin \theta)\]

implies that a=r\cos\theta and b=r\sin\theta. The Euler formula is also useful to recover the identities for the sin and cos of a sum/difference, since e^{iA}e^{iB}=e^{i(A+B)}

The complex conjugate of a complex number is \bar{z}=a-bi, and the modulus is

    \[ z\bar{z}=\vert z\vert^2=r^2\]

with \theta =\arctan \dfrac{b}{a}, \;\; \vert z \vert=\sqrt{a^2+b^2}

Moreover, \overline{\left(z_1\pm z_2\right)}=\bar{z_1}\pm\bar{z_2}, \vert \bar{z}\vert=\vert z\vert, and if z_2\neq 0, then

    \[ \overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\bar{z_1}}{\bar{z_2}}\]

We also have the so-called Moivre’s formula

    \[ z^n=r^n(\cos n\theta+i\sin n\theta)\]

and for the complex roots of complex numbers with w^n=z the identity:

    \[ w=z^{1/n}=r^{1/n}\left(cos\left(\dfrac{\theta +2\pi k}{n}\right)+i\sin\left(\dfrac{\theta +2\pi k}{n}\right)\right)\forall k=0,1,\ldots,n-1\]

The complex logarithm (or the complex power) is a multivalued functions (be aware!):

    \[ \ln( re^{i\theta})=\ln r +i\theta +2\pi k,\forall k\in \mathbb{Z}\]

The introduction of complex numbers and complex values of trigonometric functions are fun. You can check that

    \[ \cos z=\dfrac{\exp (iz)+\exp (-iz)}{2}\]

and

    \[ \sin z=\dfrac{\exp (iz)-\exp (-iz)}{2i}\]

and

    \[ \tan z=\dfrac{\exp (iz)-\exp (-iz)}{i(\exp (iz)+\exp(-iz))}\]

thanks to the Euler identity.

In special relativity, the geometry is “hyperbolic”, i.e., it is non-euclidean. Let me review the so-called hyperbolic trigonometry. More precisely, we are going to review the hyperbolic functions related to special relativity now.

HYPERBOLIC TRIGONOMETRY

520px-Hyperbolic_functions-2.svg

We define the functions sinh, cosh and tanh ( sometimes written as sh, ch, th):

    \[ \sinh x=\dfrac{\exp (x) -\exp (-x)}{2}\]

    \[ \cosh x=\dfrac{\exp (x) +\exp (-x)}{2}\]

    \[ \tanh x=\dfrac{\exp (x) -\exp (-x)}{\exp (x)+\exp (-x)}\]

The fundamental theorem of hyperbolic trigonometry is

    \[ \cosh^2 x-\sinh^2 x=1\]

The hyperbolic triangles are objects like this:

250px-Hyperbolic_triangle.svg

The hyperbolic inverse functions are

    \[\sinh^{-1} x=\ln (x+\sqrt{x^2+1})\]

    \[ \cosh^{-1} x=\ln (x+\sqrt{x^2-1})\]

    \[ \tanh^{-1} x=\dfrac{1}{2}\ln\dfrac{1+x}{1-x}\]

Two specially useful formulae in Special Relativity (related to the gamma factor, the velocity and a parameter called rapidity) are:

    \[ \boxed{\sinh \tanh^{-1} x=\dfrac{x}{\sqrt{1-x^2}}}\]

    \[ \boxed{\cosh \tanh^{-1} x=\dfrac{1}{\sqrt{1-x^2}}}\]

In fact, we also have:

    \[ \exp(x)=\sinh x+\cosh x\]

    \[ \exp(-x)=-\sinh x+\cosh x\]

    \[ \sec\mbox{h}^2x+\tanh^2 x=1\]

    \[ \coth ^2 x-\csc\mbox{h}^2 x=1\]

There are even more identities to be known. The most remarkable and important are likely to be:

    \[ \sinh (x\pm y)=\sinh (x)\cosh (y)\pm\sinh (y)\cosh (x)\]

    \[ \cosh (x\pm y)=\cosh (x)\cosh (y)\pm\sinh (x)\sinh (y)\]

    \[ \tanh (x\pm y)=\dfrac{\tanh x\pm \tanh y}{1\pm \tanh x\tanh y}\]

    \[ \coth (x\pm y)=\dfrac{\coth x\coth y\pm 1}{\coth y\pm \coth x}\]

You can also relate euclidean trigonometric functions with hyperbolic trigonometric functions with the aid of complex numbers. For instance, we get

    \[ \sinh x=-i\sin ix\]

    \[ \cosh x=\cos ix\]

    \[ \tanh x= -i\tan ix\]

and so on. The hyperbolic models of geometry/trigonometry are also very known in arts. Escher’s drawings are very beautiful and famous:

escherhyperbolic

or the colorful variation of this theme

m.v.gagern_hyperbolization

I love Escher’s drawings. And I also love Mathematics, Physics, Physmatics, and Science. Equations are cool. And hyperbolic functions, and other functions we have reviewed here today, will arise naturally in the next posts.

LOG#022. Kaniadakis and relativity.

Hello, I am back! After some summer rest/study/introspection! And after an amazing July month with the Higgs discovery by ATLAS and CMS. After an amazing August month with the Curiosity rover, MSL(Mars Science Laboratory), arrival to Mars. After a hot summer in my home town…I have written lots of drafts these days…And I will be publishing all of them step to step.

We will discuss today one of interesting remark studied by Kaniadakis. He is known by his works on relatistivic physics, condensed matter physics, and specially by his work on some cool function related to non-extensive thermodynamics. Indeed, Kaniadakis himself has probed that his entropy is also related to the mathematics of special relativity. Ultimately, his remarks suggest:

1st. Dimensionless quantities are the true fundamental objects in any theory.

2nd. A relationship between information theory and relativity.

3rd. The important role of deformation parameters and deformed calculus in contemporary Physics, and more and more in the future maybe.

4nd. Entropy cound be more fundamental than thought before, in the sense that non-extensive generalizations of entropy play a more significant role in Physics.

5th. Non-extensive entropies are more fundamental than the conventional entropy.

The fundamental object we are going to find is stuff related to the following function:

    \[ exp_\kappa (x)=\left( \sqrt{1+\kappa^2x^2}\right)^{1/\kappa}\]

Let me first imagine two identical particles ( of equal mass) A and B, whose velocities, momenta and energies are, in certain frame S:

    \[ v_A, p_A=p(v_A), E_A=E(v_A)\]

    \[ v_B, p_B=p(v_B), E_B=E(v_B)\]

In the rest frame of particle B, S’, we have

    \[ p'_B=0\]

    \[ p'_A=p_A-p_B\]

If we define a dimensionless momentum paramenter

    \[ q=\dfrac{p}{p^\star}\]

    \[ \dfrac{p'_A}{p^\star}=\dfrac{p_A}{p^\star}-\dfrac{p_B}{p^\star}\]

we get after usual exponentiation

    \[\exp(q'_A)=\exp(q_A)\exp(-q_B)\]

Galilean relativity says that the laws of Mechanics are unchanged after the changes from rest to an uniform motion reference frame. Equivalentaly, galilean relativity in our context means the invariance under a change q'_A\leftrightarrow q_A, and it implies the invariance under a change q_B\rightarrow -q_B. In turn, plugging these inte the last previous equation, we get the know relationship

    \[ \exp (q)\exp (-q)=1\]

Wonderful, isn’t it? It is for me! Now, we will move to Special Relativity. In the S’ frame where B is at rest, we have:

    \[ v'_B=0, p'_B=0, E'_B=mc^2\]

and from the known relativistic transformations for energy and momentum

    \[ v'_A=\dfrac{v_A-v_B}{1-\dfrac{v_Av_B}{c^2}}\]

    \[ p'_A=\gamma (v_B)p_A-\dfrac{v_B\gamma (v_B)E_A}{c^2}\]

    \[ E'_A=\gamma (v_B)E_A-v_B\gamma (v_B)p_A\]

where of course we define

    \[ \gamma (v_B)=\dfrac{1}{\sqrt{1-\dfrac{v_B}{c^2}}}\]

    \[ p_B=m \gamma (v_B) v_B\]

    \[ E_B=m \gamma (v_B) c^2\]

After this introduction, we can parallel what we did for galilean relativity. We can write the last previous equations in the equivalent form, after some easy algebra, as follows

    \[ p'_A=p_A\dfrac{E_B}{mc^2}-E_A\dfrac{p_B}{mc^2}\]

    \[ E'_A=E_AE_B\dfrac{1}{mc^2}-\dfrac{p_Ap_B}{m}\]

Now, we can introduce dimensionless variables instead of the triple (v, p, E), defining instead the adimensional set (u, q, \epsilon):

    \[ \dfrac{v}{u}=\dfrac{p}{mq}=\sqrt{\dfrac{E}{m\epsilon}}=\vert \kappa \vert c=v_\star<c\]

Note that the so-called deformation parameter \kappa is indeed related (equal) to the beta parameter in relativity. Again, from the special relativity requirement \vert \kappa \vert c<c we obtain, as we expected, that -1< \kappa <+1. Classical physics, the galilean relativity we know from our everyday experience, is recovered in the limit c\rightarrow \infty, or equivalently, if \kappa \rightarrow 0. In the dimensionless variables, the transformation of energy and momentum we wrote above can be showed to be:

    \[ q'_A=\kappa^2q_A\epsilon_B-\kappa^2q_B\epsilon_A\]

    \[ \epsilon'_A=\kappa^2\epsilon_A\epsilon_B-q_Aq_B\]

In rest frame of some particle, we get of course the result E(0)=mc^2, or in the new variables \epsilon (0)=\dfrac{1}{\kappa^2}. The energy-momentum dispersion relationship from special relativity p^2c^2-E^2=-m^2c^2 becomes:

    \[ q^2-\kappa^2\epsilon^2=-\dfrac{1}{\kappa^2}\]

or

    \[ \kappa^4\epsilon^2-\kappa^2q^2=1\]

Moreover, we can rewrite the equation

    \[ q'_A=\kappa^2q_A\epsilon_B-\kappa^2q_B\epsilon_A\]

in terms of the dimensionless energy-momentum variable

    \[ \epsilon_\kappa (q)=\dfrac{\sqrt{1+\kappa^2q^2}}{\kappa^2}\]

and we get the analogue of the galilean addition rule for dimensionless velocities

    \[ q'_A =q_A\sqrt{1+\kappa^2q_B^2}-q_B\sqrt{1+\kappa^2q_A^2}\]

Note that the classical limit is recovered again sending \kappa\rightarrow 0. Now, we have to define some kind of deformed exponential function. Let us define:

    \[ \exp_\kappa (q) =\left(\sqrt{1+\kappa^2q^2}+\kappa q\right)^{1/\kappa}\]

Applying this function to the above last equation, we observe that

    \[ \exp_\kappa (q'_A)=\exp_\kappa (q_A) \exp_\kappa (-q_B)\]

Again, relativity means that observers in uniform motion with respect to each other should observe the same physical laws, and so, we should obtain invariant equations under the exchanges q'_A\leftrightarrow q_A and q_B\rightarrow -q_B. Pluggint these conditions into the last equation, it implies that the following condition holds (and it can easily be checked from the definition of the deformed exponential).

    \[ \exp_\kappa (q)\exp_\kappa (-q)=1\]

One interesting question is what is the inverse of this deformed exponential ( the name q-exponential or \kappa-exponential is often found in the literature). It has to be some kind of deformed logarithm. And it is! The deformed logarithm, inverse to the deformed exponential, is the following function:

    \[ \ln_\kappa (q)=\dfrac{q^{\kappa}-q^{-\kappa}}{2\kappa}\]

Indeed, this function is related to ( in units with the Boltzmann’s constant set to the unit k_B=1) the so-called Kaniadakis entropy!

    \[ S_{K}=-\dfrac{q^{\kappa}-q^{-\kappa}}{2\kappa}\]

Furthermore, the equation \exp_\kappa (q)\exp_\kappa (-q)=1 also implies that

    \[ \ln_\kappa \left(\dfrac{1}{q}\right)=-\ln_\kappa (q)\]

The gamma parameter of special relativity is also recasted as

    \[ \gamma =\dfrac{1}{\sqrt{1-\kappa^2}}\]

More generally, in fact, the deformed exponentials and logarithms develop a complete calculus based on:

    \[ \exp_\kappa (q_A)\exp_\kappa (q_B)=\exp (q_A\oplus q_B)\]

and the differential operators

    \[ \dfrac{d}{d_\kappa q}=\sqrt{1+\kappa^2q^2}\dfrac{d}{dq}\]

so that, e.g.,

    \[ \dfrac{d}{d_\kappa q}\exp_\kappa (q)=\exp_\kappa (q)\]

This Kanadiakis formalism is useful, for instance, in generalizations of Statistical Mechanics. It is becoming a powertool in High Energy Physics. At low energy, classical statistical mechanics gets a Steffan-Boltmann exponential factor distribution function:

    \[ f\propto \exp(-\beta E)=\exp (-\kappa E)\]

At high energies, in the relativistic domain, Kaniadakis approach provide that the distribution function departures from the classical value to a power law:

    \[ f\propto E^{-1/\kappa}\]

There are other approaches and entropies that could be interesting for additional deformations of special relativity. It is useful also in the foundations of Physics, in the Information Theory approach that sorrounds the subject in current times. And of course, it is full of incredibly beautiful mathematics!

We can start from deformed exponentials and logarithms in order to get the special theory of relativity (reversing the order in which I have introduced this topic here). Aren’t you surprised?