LOG#043. Tachyons and SR.

gsudarshanphoto

“(…)Suppose that someone studying the distribution of population on the Hindustan Peninsula cockshuredly believes that there are no poeple north of the Himalayas, because nobody ca pass throught the mountain ranges! That would be an absurd conclusion. The inhabitants of Central Asia have been born there; they are not obliged to be born in India and tehn cross the mountain ranges. The same can be said about superluminal particles(…)”  This is a quote by George Sudharhan.

I had the honour to meet George Sudarshan ( his full name is Ennackal Chandy George Sudarshan, http://en.wikipedia.org/wiki/Sudarshan,_E._C._George) some years ago, in Jaca (Huesca), during the Sudarshanfest celebrating his 75th birthday. I also met some really cool people like Susumo Okubo ( yes, the man behind the Gell-Mann-Okubo mass formula for hadrons-see http://en.wikipedia.org/wiki/Gell-Mann-Okubo_mass_formula! ). I really enjoy knowing (japanese) scientists when they are really gentle and generous. Mr. Okubo indeed gave me a copy of his wonderful book about non-associative stuff. By the other hand, Indian scientists are also fascinating because they use to be people very uncommon and some of them, like Ramanujan in Mathematics, own exceptional gifts and talents.

George Sudarshan has made contributions in many branches of Physics. He originally proposed as well the V-A nature of the electroweak interactions that recover the Fermi theory of weak interactions in the low energy limit. In addition to this, he has developed in the field of optical coherence the so-called Sudarhsan-Glauber representation in Quantum Optics, he invented the theory of the Quantum Zeno effect, he has worked in open quantum systems, the relationship spin-statistics in general QFT contexts, and he is one of the defenders of the existence of tachyons, via an interpretation of relativity including those faster than light particles created by Feinberg and studied by E. Recami, Feinberg, M.Pavsic, Gregory Benford (yes, the writter and physicist/astronomer) and other physicists. His works about relativity with tachyons are usually labelled under the name “metarelativity”. I would like to dedicate him this post.

According to wikipedia, a tachyon or tachyonic particle is a hypothetical particle that, a priori, always moves faster than light. The word comes from the Greek word: ταχύς or tachys, meaning “swift, quick, fast, rapid”, and was coined by Gerald Feinbergin a 1967 paper.Feinberg proposed that tachyonic particles could be quanta of a quantum field with negative squared mass. However, it was soon realized that excitations of such imaginary mass fields do not in fact propagate faster than light,but instead represent an instability known as tachyon condensation.  Nevertheless, they are still commonly known as “tachyons”, and have come to play an important role in modern physics, for instance, their role in string theory is still being studied. I will not discuss about advanced topics like tachyon condensation in this post, I will expect to do it in the near future some day.

Superluminality and velocity are subtle concepts in SR. I have to discuss more about (apparent) superluminality in SR, but the goal of this post is somewhat more modest. I am going to introduce you tachyonic particles and some of its curious features.

First point. From the addition theorem of velocities in special relativity:

    \[ V=\dfrac{v_1+v_2}{1+\dfrac{v_1 v_2}{c^2}}\]

we can see that, a priori, the ranges of the velocities is not restricted from that concrete formula. the real issue with faster than light particles in special relativity comes from the relativistic expression of energy:

    \[ E=Mc^2=m\gamma c^2=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}\]

This formula blows up at v=c for any finite value of the rest mass m! Indeed, we can say that particles slower than light, also called “tardyons”, have that energy-mass relation, while for photons ( or luxons), we do know that E=pc and they are massless particles. What happens if we forget our prejudices and we allow for v>c velocities saving the SR formula for mass-energy? Well, it is easy to realize that E becomes an imaginary number! Imaginary numbers are complex numbers without real part squaring a negative value. For instance, the solution to the equation x^2=-1 is the imaginary number unit x=i=\sqrt{-1}. Don’t forget that complex numbers are more general numbers verifying z=a+bi, with magnitude \vert z \vert ^2=a^2+b^2.

If we plug v>c in the SR formula for mass-energy, we get a negative number inside the square root. After some easy algebra, for v>c we obtain:

    \[ E=\dfrac{mc^2}{\sqrt{1-\dfrac{v^2}{c^2}}}=\dfrac{mc^2}{i\sqrt{\dfrac{v^2}{c^2}-1}}=\dfrac{\overline{m}c^2}{\sqrt{\dfrac{v^2}{c^2}-1}}\]

and where we have defined the imaginary mass quantity \overline{m}=\dfrac{m}{i}=-im. We could also have defined the “barred” gamma factor as

    \[ \boxed{\gamma \equiv -i\overline{\gamma}=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}}\leftrightarrow \boxed{\overline{\gamma}=\dfrac{1}{\sqrt{\dfrac{v^2}{c^2}-1}}}\]

so, for tachyons,

    \[ \boxed{E=\overline{m}\overline{\gamma}c^2=-im\overline{\gamma}c^2=\dfrac{\overline{m}c^2}{\sqrt{\dfrac{v^2}{c^2}-1}}}\]

Moreover, with this imaginary gamma/boost factor, we would define metarelativity or Lorentz transformations for tachyons:

    \[ x'=-i\overline{\gamma}(x-vt)\]

    \[ y'=y\]

    \[ z'=z\]

    \[ t'=-i\overline{\gamma}\left(t-\dfrac{v}{c^2}x\right)\]

and so on with any other transformation including imaginary boosts. Thus, we have extended the special relativity to the imaginary realm and we call this theory metarelativity since it includes superluminal transformations. For superluminal transformations, we also have the same invariants than those for usual relativity. For instance:

    \[ c^2t^{'2}-x^{'2}=c^2t^2-x^2=c^2\overline{t'}^2-\overline{x'}^2\]

The prize is that we have imaginary position and time coordinates ( or imaginary boosts, if you prefer that idea).

A curious property of the relationships \overline{m}=-im and \overline{\gamma}=i\gamma is that

    \[ \overline{m}\overline{\gamma}=m\gamma\]

This equation shows that a tachyonic imaginary mass and a imaginary gamma factor are equivalent, when they are multiplied, to the real valued common relativistic expression for the relativistic mass. So, we could handle tachyons with imaginary masses and imaginary gamma factors, at least in principle, in the same operational way we handle normal particles in SR, excepting for the sign inside the square root and the strange inertial properties of those tachyonic particles.

In addition to this fact, the energy for tachyonic particles some has interesting properties:

1st. E=\overline{m}\overline{\gamma}c^2 decreases as v increases! That is, tachyons are less energetic and so “more stable” when they have higher velocities. This behaviour is very different from the common inertial properties of normal matter. Unlike ordinary particles, the speed of a tachyon increases as its energy decreases.

2nd. A bradyon, also known as a tardyon or ittyon, is a particle that travels slower than light. The term “bradyon”, from Greek word βραδύς (bradys, “slow”), was coined to contrast with the name of the tachyon. Just as bradyons are forbidden to break the light-speed barrier, so too are tachyons forbidden from slowing down to below c, because infinite energy is required to reach the barrier from either above or below.

3rd. Einstein, Tolman and others noted that special relativity allowing for tachyons/superluminal transmissions would imply that they could be used to send signals backwards in time. This tool or use of tachyon pulses is called the tachyonic antitelephone device.

An electric charged tachyon would loose energy as Cherenkov radiation just as ordinary charged particles do when they exceed the local speed of light in a medium. A charged tachyon traveling in a vacuum therefore undergoes a constant proper time acceleration and, by necessity, its wordline forms a hyperbola in space-time.

However,  if we reduce the tachyon’s energy, then it increases its speed, so that the single hyperbola formed is of two oppositely charged tachyons with opposite momenta (same magnitude, opposite sign) which annihilate each other when they simultaneously reach infinite speed at the same place in space. (At infinite speed the two tachyons have no energy each and finite momentum of opposite direction, so no conservation laws are violated in their mutual annihilation. The time of annihilation is frame dependent.)

Even an electrically neutral tachyon would be expected to loose energy via gravitational Cherenkov radiation, at least in theory, because it has a gravitational mass, and therefore increase in speed as it travels, as described above. However, we have not detected gravitational Cherenkov radiation, as far as I know. If the tachyon interacts with any other particles, it can also radiate Cherenkov energy into those particles. Neutrinos interact with the other particles of the Standard Model and, recently,  Andrew Cohen and Sheldon Glashow used this to argue that the neutrino anomaly seen by the OPERA experiment could not be explained by making neutrinos propagate faster than light. Indeed, we do know that neutrino have a non-zero REAL mass from neutrino oscillation experiments.

Coming back to tachyons, we have seen that SR allows then if you allow for “imaginary energies”. A tachyon has the strange feature that its mass has the value

    \[ \mbox{TACHYON MASS}= \mbox{(SOMETHING)}\times i\]

so, while I can weight something like 70 kg, a tachyon clone of me would weight 70i kg. You can wonder what the imaginary mass means in terms of inertia with the SR equation above, but, of course, it is a weird result after all. And there are more “problems” and weird results for tachyons. For instance, tachyons, it they do gravitate according to Newton’s gravitation equation:

    \[ F_N=-G\dfrac{M_1 M_2}{d^2}\]

then they would experience “antigravitation”/antigravity. You can observe and “deduce” that the gravitational force between two tachyons with masses M_1=M_2=i separated by a distance of 1m. Then, the gravitational pull between those tachyons would be repulsive, since the sign of the gravitational Newton force would be positive instead of negative! Is it not amazing? Yes, you can wonder about the Dark Energy enigma, mysterious stuff out there, but there are quantum problems related with “superluminal” tachyons. I will discuss them in the future, I promise. So, it is not easy at all to associate a tachyonic field/mass origin to the Dark Energy. And of course, this hypothesis of antigravitating tachyons face problems when we think about what an imaginary gravitational force between a tachyon and a tardyon would mean. It shows that the mysteries of tachyons are yet not completely understood, and they are connected with the theory of scalar fields and the phenomenon, previously commented, of tachyon condensation. Let me know if you understand them better!

Morever, the transversal length contraction of a tachyon, and the time dilation of a tachyon in metarelativity are imaginary quantities as well. It is an easy exercise to derive the following relationships:

    \[ L_{\updownarrow}=iL_0\overline{\gamma}^{-1}\]

    \[ \tau=i\tau_0 \overline{\gamma}\]

A second post about tachyons and metarelativity is coming, but before that, you will have to wait for a while. I have other topics in my current agenda to be published, previously, to more tachyonic posts. I suppose I am not being beamed with tachyons from the future.

Let the tachyons be with you ;)!

LOG#042. Pulley with variable mass.

pulleyVariableMass

This interesting problem was recently found in certain public examination. I will solve it here since I found it fascinating and useful, since it is a problem with “variable mass”.

A pulley with negligible mass is given (see the figures above). Its radius is R. We tie two masses, m_1 and m_2 to the extremes of the pulley with the aid of a string. The string has a linear mass density \lambda and it has a total length equal to L. Calculate:

a) The forces in the extremes of the pulley over the masses and the acceleration when we release the system from the rest.

b) The velocity of the masses and the position of a the masses as a function of time.

Remark: We have the contraint q_2+\pi R+q_1=L. L is the total length of the string, R is the radius of the pulley, and q_2 and q_1 are the pieces of string unwrapped to the pulley for the masses m_2 and m_1, respectively. We can suppose that initially q_1=0 and q_2=l. Note that due to this constraint \dot{q_1}+\dot{q_2}=0, where the dot denotes derivative with respect to time, i.e., velocity of both masses, respecively too.

Solution.

a) Suppose m_1>m_2. Thus, the forces acting on the masses are:

\boxed{F_1=m_1 g+\lambda gx-T=\overline{m_1}a=(m_1+\lambda x)a}

\boxed{F_2=T-m_2 g-\lambda g (l-x)=\overline{m_2}a=(m_2+(L-x)\lambda) a}

Adding the two equations, we get

m_1g-m_2g+2\lambda g x-\lambda g l=(m_1+m_2+L\lambda)a

and then the acceleration will be

\boxed{a=\dfrac{m_1g-m_2g-\lambda gl-2\lambda gx}{m_1+m_2+\lambda L}=\dfrac{(m_1-m_2)g-l\lambda g+2g\lambda x}{m_1+m_2+\lambda L}}

b) Firstly, we have to derive the differential equations for the system with variable mass.
The equations for both masses are, respectively, using Newton’s laws:

(m_1+\lambda x)x''=m_1 g+\lambda gx-T

(m_2+(L-x)\lambda)x''=T-m_2g-\lambda g(L-x)

Adding these two equations we get the differential equation to be solved:

\boxed{a=x''=\dfrac{m_1g-m_2g-\lambda g L+2g\lambda x}{m_1+m_2+\lambda L}}

with x(0)=x'(0)=0

We can solve the differential equation

x''=\dfrac{(m_1-m_2)g-L\lambda g+2g\lambda x}{m_1+m_2+\lambda L}

with the initial conditions v(0)=x'(0)=0 and x(0)=0 easily, to obtain

\boxed{v(t)=x'(t)=\dfrac{((m_1-m_2)-\lambda L)\sqrt{\lambda g}}{\lambda \sqrt{2(m_1+m_2+\lambda L)}}\sinh \left(\sqrt{\dfrac{2\lambda g}{m_1+m_2+\lambda L}}t\right)}

\boxed{x(t)=\dfrac{(m_1-m_2-\lambda L)}{\lambda}\sinh^2 \left(\sqrt{\dfrac{\lambda g}{2(m_1+m_2+\lambda L)}}t\right)}

As the mass of the string depends on the linear density, the usual procedure to obtain the variation of the kinetic energy as the work done by the potential forces is something more complicated:

\Delta E_c=\Delta W (x)=\int_0^x F(x)dx

It can be shown that

\dfrac{1}{2}(m_1+m_2+\lambda L)v^2=(m_1-m_2+\lambda x-\lambda l)g x

v=\sqrt{\dfrac{2gx(m_1-m_2+\lambda x-\lambda l)}{m_1+m_2+\lambda L}}

Please, note that in this case, we have to be very careful with l and L. They are related with a contraint, but they are not the same thing.

LOG#041. Muons and relativity.

particleshower.svg

cosparticle

QUESTION: Is the time dilation real or is it an artifact of our current theories?

There are solid arguments why time dilation is not an apparent effect but a macroscopic measurable effect. Today, we are going to discuss the “reality” of time dilation with a well known result:

Muon detection experiments!

Muons are enigmatic elementary particles from the second generation of the Standard Model with the following properties:

1st. They are created in upper atmosphere at altitudes of about 9000 m, when cosmic rays hit the Earth and they are a common secondary product in the showers created by those mysterious yet cosmic rays.
2nd. The average life span is 2\times 10^{-6}s\approx 2ms.
3rd. Typical speed is 0.998c or very close to the speed of light.
So we would expect that they could only travel at most d=0.998c\times 2 \times 10^{-6}\approx 600m
However, surprisingly at first sight, they can be observed at ground level! SR provides a beautiful explanation of this fact. In the rest frame S of the Earth, the lifespan of a traveling muon experiences time dilation. Let us define

A) t= half-life of muon with respect to Earth.

B) t’=half-life of muon of the moving muon (in his rest frame S’ in motion with respect to Earth).

C) According to SR, the time dilation means that t=\gamma t', since the S’ frame is moving with respect to the ground, so its ticks are “longer” than those on Earth.

A typical dilation factor \gamma for the muon is about 15-100, although the value it is quite variable from the observed muons. For instance, if the muon has v=0.998c then \gamma \approx 15. Thus, in the Earth’s reference frame, a typical muon lives about 2×15=30ms, and it travels respect to Earth a distance

d'=0.998c\times 30ms\approx 9000m.

If the gamma factor is bigger, the distance d’ grows and so, we can detect muons on the ground, as we do observe indeed!

Remark:  In the traveling muon’s reference frame, it is at rest and the Earth is rushing up to meet it at 0.998c. The distance between it and the Earth thus is shorter than 9000m by length contraction. With respect to the muon, this distance is therefore 9000m/15 = 600m.

An alternative calculation, with approximate numbers:

Suppose muons decay into other particles with half-life of about 0.000001sec. Cosmic ray muons have speed now about v = 0.99995 c.
Without special relativity, muon would travel

d= 0.99995 \times 300000 km/s\times 0.00000156s=0.47 km only!

Few would reach earth’s surface in that case. It we use special relativity, then plugging the corresponding gamma for v=0.99995c, i.e.,  \gamma =100, then muons’ “tics” run slower and muons live 100 times longer. Then, the traveled distance becomes

d'=100\times 0.9995\times 300000000 m/s\times 0.000001s= 30000m

Conclusion: a lot of muons reach the earth’s surface. And we can detect them! For instance, with the detectors on colliders, the cosmic rays detectors, and some other simpler tools.

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:

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

l \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.

The momenergy reads

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 l \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.

LOG#039. Relativity: Examples(III).

Example 1. Absorption of a photon by an atom.

In this process, we have from momenergy conservation:

P^\mu_a P_{a\mu}+2P^\mu_a P_{b\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}

If the atom is the rest frame, before absorption we get

P^\mu_a =(m_a c,0,0,0)

P^\mu_b=(\dfrac{E_b}{c},p_{bx},0,0)=(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0)=(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0)

Description: an atom “a” at rest with mass m_a absorbs a photon “b” propagating in the x-direction turning itself into an excited atom “c”, moving in the x-axis (suffering a “recoil” after the photon hits it).

The atom after absorption has

P^\mu_c=(m_c c,0,0,0)

Therefore, since the photon verifies:

P^\mu_b P_{b\mu}=\left(\dfrac{h\nu}{c}\right)^2-\left(\dfrac{h\nu}{c}\right)^2=0

and it is true in every inertial frame. Thus,

l (m_a c)^2+2m_a c\dfrac{h\nu}{c}+0=(m_a c)^2

Then,

\boxed{m_c=\sqrt{m_a^2+2m_a\dfrac{h\nu}{c}}=m_a\sqrt{1+2\dfrac{h\nu}{m_a c^2}}}

Expanding the square root

m_c\approx m_a\left[ 1+\dfrac{h\nu}{m_a c^2}-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c}\right)^2+\mathcal{O}(h^3)\right]

In this case,

m_c\approx m_a+\dfrac{h\nu}{ c^2}-\dfrac{1}{2}\left(\dfrac{h\nu}{m_a c}\right)^2

Atom’s rest mass increases by an amount \dfrac{h\nu}{c^2} up to first order in the Planck’s constant, and it decreases up to second order in h a quanity -\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2 due to motion ( “recoil”). Therefore,

\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2=\dfrac{1}{2}m_a\left( \dfrac{u_a}{c}\right)^2

In this way, identifying terms: u_c=\dfrac{h\nu}{m_a c}

In the laboratory frame, the excited atom velocity is calculated by momenergy conservation. It is simple:

\dfrac{h\nu}{c}=\gamma_c m_c u_c

m_a c^2+h\nu=\gamma_c m_c c^2

\dfrac{h\nu}{u_c c}=m_a+\dfrac{h\nu}{c^2}

Then, we obtain that:

u_c=\dfrac{h\nu c}{m_a c^2+h\nu}=\dfrac{c}{\left(\dfrac{m_a c^2}{h\nu}\right)+1}\approx \dfrac{h\nu}{m_a}

where we have used in the last step m_ac^2>>h\nu.

Example 2. Emission of a photon by an atom.

An atom c at rest, with m_c the rest mass, emits a photon b with frequency \dfrac{E_b}{h}=\nu in the x-direction, turning itself into a non-excited atom “a”, with m_a. What is the energy shift \Delta E=E_c-E_a?

P^\mu_a P_{a\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}

where

P^\mu_a is P^\mu_a =(m_a,0,0,0) in the rest frame of “a”.

P^\mu_a in the rest from “c” reads P^\mu_a =\left(\gamma_a m_a c, -\dfrac{E_b}{c},0,0\right)

P^\mu_b in the rest frame of “c” is P^\mu_b =\left(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0\right)

P^\mu_a in the rest frame of “c” is P^\mu_c=(m_c c,0,0,0)

In this way, we have

(m_a c)^2+\gamma _a m_a E_b +\left( \dfrac{E_b}{c}\right)^2=\gamma _am_am_c c^2

\gamma_a m_a c^2+E_b=m_c c^2

m_a^2c^2+\dfrac{E_b}{c^2}\left(m_c c^2-E_b\right)^2=m_c^2c^2-m_c E_b

m_a^2c^2+m_cE_b=m^2_c c^2-m_c E_b

m_a^2c^2+m_cE_b=m_c^2c^2-m_cE_b

From this equations we deduce that

E_b=\dfrac{m_c c^2-m_a^2c^2}{2m_c}=\dfrac{E_c^2-E_a^2}{2E_c}=(E_c-E_a)\left(1-\dfrac{E_c-E_a}{2E_c}\right)

And from the definition E_c-E_a we get E_b=\Delta E\left(1-\dfrac{\Delta E}{2E_c}\right)

Note: the photon’s energy IS NOT equal to the difference of the atomic rest energies but it is less than that due to the emission process. This fact implies that the atom experieces “recoil”, and it gains kinetic energy at the expense of the photon. There is a good chance for the photon not to be absorbed by an atom of some kind. However, “resonance absorption” becomes problematic. The condition for recoilless resonant absorption to occur nonetheless, e.g., the reabsorption of gamma ray photons by nuclei of the some kind were investigated by Mössbauer. The so-called Mössbauer effect has been important not only to atomic physics but also to verify the theory of general relativity. Furthermore, it is used in materials reseach in present time as well. In 1958, Rudolf L. Mössbauer reported the 1st reoilless gamma emission. It provided him the Nobel Prize in 1961.

Example 3. Decay of two particles at rest.

The process we are going to study is the reaction C\rightarrow AB

The particle C decays into A and B. It is the inverse process of the completely inelastic collision we studied in a previous example.

From the conservation of the tetramomentum

(m_A c)^2+2\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)+(m_Bc)^2=(m_C c)^2

Choose the frame in which the following equation holds

P^\mu_AP_{A\mu}+P^\mu_BP_{A\mu}=P^\mu_{C}P_{A\mu}

Let u_A,u_B be the laboratory frame velocities in the rest frame of “C”. Then, we deduce

P^\mu_A=(\gamma_Am_Ac,\gamma_Am_Au_A,0,0)

P^\mu_B=(\gamma_Bm_Bc,\gamma_Bm_Bu_B,0,0)

P^\mu_C=(m_C c,0,0,0)

From these equations (m_Ac)^2+\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)=\gamma_Am_Am_C c^2

-(m_Ac)^2+(m_Bc)^2=(m_Cc)^2-2\gamma_Am_Am_Cc^2

2E_Am_C=(m_A^2+m_C^2-m_B^2)c^2

\boxed{E_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}}

\boxed{E_A^{kin}=T_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}-m_Ac^2=\dfrac{\left((m_C-m_A)^2-m_B^2\right)c^2}{2m_C}}

\boxed{E_B^{kin}=T_B=\dfrac{\left((m_C-m_B)^2-m_A^2)\right)c^2}{2m_C}}

\boxed{E_{kin}=T=T_A+T_B=(m_C-m_A-m_B)c^2}

Therefore, the total kinetic energy of the two particles A,B is equal to the mass defect in the decay of the particle.

Example 4. Pair production by a photon.

Suppose the reaction \gamma \rightarrow e^+e^-, in which a single photon (\gamma)decays into a positron-electron  pair.

That is, h\nu\rightarrow e^+e^-.

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

Squaring the momenergy in both sides:

P^\mu_a P_{\mu a}=P^\mu_c P_{\mu c}+2P^\mu_c P_{\mu d}+P^\mu_d P_{\mu d}

In the case of the photon: P^\mu_a P_{a\mu}=0

In the case of the electron and the positron: P^\mu_c P_{\mu c}=P^\mu_d P_{\mu d}=-(m_e c)^2=-(mc)^2

We calculate the components of momenergy in the center of mass frame:

P^\mu_c=\left( \dfrac{E_c}{c},p_{cx},p_{cy},p_{cz}\right)

P^\mu_d=\left( \dfrac{E_d}{c},-p_{dx},-p_{dy},-p_{dz}\right)

with E_c=E_d=mc^2. Therefore,

2 P^\mu_c P_{\mu d}=-2\left( \dfrac{E_c^2}{c^2}+p_x^2+p_y^2+p_z^2\right)

so

-2(mc)^2-2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0

2(mc)^2+2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0

This equation has no solutions for any positive solution of the photon energy! It’s logical. In vacuum, it requires the pressence of other particle. For instance \gamma \gamma \rightarrow e^+e^- is the typical process in a “photon collider”. Other alternative is that the photon were “virtual” (e.g., like QED reactions e^+e^-\rightarrow \gamma^\star\rightarrow e^+e^-). Suppose, alternatively, the reaction AB\rightarrow CDE. Solving this process is hard and tedious, but we can restrict our attention to the special case of three particles C,D,E staying together in a cluster C. In this way, the real process would be instead AB\rightarrow C. In the laboratory frame:

P^\mu_A+P^\mu_B=P^\mu_C

we get

P^\mu_A=\left( \dfrac{E_A}{c},\dfrac{E_A}{c},0,0\right)

P^\mu_B=\left( Mc,0,0,0\right)

and in the cluster reference frame

P^\mu_C=\left((M+2m)c,0,0,0\right)

Squaring the momenergy:

0+2E_AM+(Mc)^2=(M+2m)^2c^2

2ME_A=4Mmc^2+4m^2c^2

E_A=2mc^2+\dfrac{2m^2c^2}{M}

\boxed{E_A=2mc^2\left(1+\dfrac{m}{M}\right)}

In the absence of an extra mass M, i.e., when M\rightarrow 0, the energy E_A would be undefined, and it would become unphysical. The larger M is, the smaller is the additional energy requiere for pair production. If M is the electron mass, and M=m, the photon’s energy must be twice the size of the rest energy of the pair, four times the rest energy of the photon. It means that we would obtain

E=4mc^2=2m_{pair}c^2

and thus \gamma =4\rightarrow

\beta^2=1-\dfrac{1}{16}=\dfrac{15}{16}

\beta=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}\approx 0.97

v=\dfrac{\sqrt{15}}{4}c\approx 0.97c

In general, if m\neq M we would deduce:

\boxed{\beta=\sqrt{1-\dfrac{1}{4\left(1+\frac{m}{M}\right)^2}}}

Example 5. Pair annihilation of an  electron-positron couple.

Now, the reaction is the annihilation of a positron-electron pair into two photons, turning mass completely into (field) energy of light quanta. l e^+e^-\rightarrow \gamma \gamma implies the momenergy conservation

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

where “a” is the moving electron and “b” is a postron at rest. Squaring the identity, it yields

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

Then, we deduce

P^\mu_a=\left(\dfrac{E_a}{c},p_{ax},0,0\right)

P^\mu_b=\left( m_e c,0,0,0\right)

while we do know that

P^\mu_aP_{\mu a}=P^\mu_bP_{\mu b}=-(m_e c)^2 for the electron/positron and

P^\mu_cP_{\mu c}=P^\mu dP_{\mu d}=0 since they are photons. The left hand side is equal to -2(m_e c)^2+2E_am_e, and for the momenergy in the right hand side

P^\mu_c=\left(\dfrac{E_c}{c},p_{cx},0,0\right)

P^\mu_d=\left( \dfrac{E_d}{c},p_{dx},0,0\right)

Combining both sides, we deduce

(m_a c)^2+E_a m_e=\dfrac{E_cE_d}{c^2}-p_{cx}p_{dx}

The only solution to the right hand side to be not zero is when we select p_{cx}=\pm \dfrac{E_c}{c} and p_{dx}=\pm\dfrac{E_d}{c} and we plug values with DIFFERENT signs. In that case,

\boxed{(mc)^2+E_a m=\dfrac{2E_cE_d}{c^2}}

From previous examples:

P^\mu_aP_{\mu_b}+P^\mu_bP_{\mu b}=P^\mu_c P_{b\mu}+P^\mu_aP_{\mu b}

and we evaluate it in the laboratory frame to give

\boxed{E_a m+(mc)^2=E_c m+E_d m}

The last two boxed equations allow us to solve for E_d

\boxed{E_d=E_a-E_c+mc^2}

If we insert this equation into the first boxed equation of this example, we deduce that

(mc)^2+mE_a=\dfrac{2E_c}{c^2}\left( E_a-E_c+mc^2\right)

or

l \dfrac{1}{2}mc^2\left(mc^2+E_a\right)=-E_c^2+E_c(E_a+mc^2)

Solving for E_c this last equation

\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{1}{2}\left(E_a+mc^2\right)\pm\sqrt{\dfrac{1}{4}\left(E_a+mc^2\right)^2-\dfrac{1}{2}mc^2\left(mc^2+E_a\right)}}

\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{\left(E_a+mc^2\right)\pm \sqrt{\left(E_a+mc^2\right)\left(E_a-mc^2\right)}}{2}}

LOG#038. Relativity: Examples(II).

Example 1. Completely inelastic collision of 2 particles AB\rightarrow C.

We will calculate the mass and velocity after the collision, when a cluster is formed.

E_a+E_b=E_c

p_{ax}+p_{bx}=p_{cx}

p_{ay}+p_{by}=p_{cy}

p_{az}+p_{az}=p_{cz}

From the consevation of momentum in the collision, we get P^\mu_a+P^\mu_b=P^\mu_c

Then,

\left(\dfrac{E_a}{c},p_{ax},p_{ay},p_{az}\right)+\left(\dfrac{E_b}{c},p_{bx},p_{by},p_{bz}\right)=\left(\dfrac{E_c}{c},p_{ac},p_{ac},p_{ac}\right)

Squaring both sides:

\left(P_{\mu a}+P_{\mu b}\right)\left(P^{\mu a}+P^{\mu b}\right)=P^\mu_c P_{\mu c}

P^\mu_a P_{\mu a}+P^\mu_a P_{\mu b}+P^\mu_b P_{\mu a}+P^\mu_b P_{\mu b}=P^\mu_c P_{\mu c}

P^\mu_a P_{\mu a}+2P^\mu_a P_{\mu b}+P^\mu_b P_{\mu b}=P^\mu_c P_{\mu c}

In the rest frame of the particle labelled with “a”:

E_a=mc^2 \mathbf{p}_a=\mathbf{0}

P^\mu_a=(m_a c,0,0,0)

P^\mu_a P_{\mu a}=(-m_a^2 c^2)

P^\mu_b P_{\mu b}=(-m_b^2 c^2)

P^\mu_c P_{\mu c}=(-m_c^2 c^2)

Now, with E_a=\gamma _a m_a c^2 and E_b =\gamma _b m_b c^2

and

P^{\mu}_a=(\gamma _a m_a c, \gamma _a m_a u_a,0,0)

P^{\mu}_b=(\gamma _b m_b c, \gamma _b m_b u, 0, 0)

2P^\mu_a P_{\mu b}=2\gamma _a \gamma_ bm_a m_b (u_a u_b-c^2)

And thus,

-(m_a c)^2+2\gamma _a\gamma _b m_a m_b (u_a u_b-c^2)-(m_b c)^2=-(m_c c)^2

(m_a c)^2+2\gamma _a\gamma _b m_a m_b (c^2-u_a u_b)+(m_b c)^2=(m_c c)^2

From this equation, we get

\boxed{m_c=\sqrt{m_a^2+m_b^2+2m_am_b\gamma _a\gamma_b\left(1-\dfrac{u_au_b}{c^2}\right)}}

and

\gamma _a \gamma b \left(1-\dfrac{u_a u_b}{c^2}\right)=\dfrac{1-\dfrac{u_a u_b}{c^2}}{\sqrt{1-\dfrac{u_a^2}{c^2}}\sqrt{1-\dfrac{u_b^2}{c^2}}}

If u_a, u_b<<c, then m_c=m_a+m_b, as we should expect from classical physics ( Lavoisier’s law: the mass is conserved).

We observe that, in general, there is NO rest mass conservation law in relativistic physics ( as we have mentioned before). Momentum and energy can be transformed into rest mass and viceversa. From the next equations:

\gamma _am_a c^2+\gamma _b m_bc^2=\gamma _cm_c c^2

\gamma_a m_a u_a+\gamma_b m_bu_b=\gamma_c m_cu_c

We get

\boxed{u_c=\dfrac{\gamma _a m_a u_a +\gamma _bm_bu_b}{\gamma _am_a+\gamma_bm_b}}

is the center or mass velocity in the laboratory frame. If u_b=0 and \gamma_b=1, then

u_c=\dfrac{\gamma_a m_au_a}{\gamma_a m_a+m_b}=\dfrac{u_a}{1+\dfrac{m_b}{\gamma_a m_a}}=\dfrac{u_a}{1+\dfrac{m_b}{m_a}\sqrt{1-\dfrac{u_a^2}{c^2}}}

And if we write u_a\rightarrow 0, the formula provides

u_c=\dfrac{m_a}{m_a+m_b}u_a

And it is the classical center of mass equation for two particles, as we expected to be.

Example 2. Production of a proton-antiproton pair p-\overline{p} in the chemical highly energetic process given by pp\rightarrow p\overline{p} pp.

If we get an extremely energetic beam of protons. Then, we collide one proton A of the beam with another proton B, at rest. If the kinetic energy is enough, we can create a proton-antiproton pair! The question is, what is the minimum (total) energy, sometimes called threshold energy?What is the minimum kinetic energy, or threshold kinetic energy? And what is the velocity of the incident proton A in order to get the pair production? We are going to solve this really nice problem.

The proton B at rest has u_B=0 and then \gamma_B=1.

Note that we have m_A=m_B=m_{proton}=m_p=m and m_C=4m_p=4m, for the system after the collision when the pair is created.

Using the formulae we have studied

(mc)^2+2\gamma_A(mc)^2+(mc)^2=(4mc)^2

2\gamma_A m^2c^2=14m^2c^2

and thus

\boxed{E_A^{min}=7mc^2}

It implies \boxed{T_A^{min}=E_A^{min}-mc^2=6mc^2} for the kinetic threshold energy. The necessary velocity can be obtained in a simple way:

E_A=\dfrac{mc^2}{\sqrt{1-\dfrac{u_A^2}{c^2}}}=7mc^2\rightarrow u_A=c\sqrt{1-\dfrac{1}{7}^2}\approx 0.990c

And thus,

\left(\dfrac{c-u_a}{c}\approx 1\%\right)

Energetically, we find that the collision is more “favourable”, in general, when we get a storage ring with 2 protons and we crash them with V_A=-V_B. In that case, the energy threshold and the kinetic energy threshold, provide:

u_A=u u_B=-u \gamma_a=\gamma_b=\gamma_u m_a=m_b=m

m_C=4m_p=4m

(mc)^2+2\gamma^2(c^2+u^2)+(mc)^2=(4mc)^2

\gamma^2(c^2+u^2)=7c^2

c^2+u^2=7(c^2-u^2)

and then, the minimum velocity of each proton has to be: u=\sqrt{\dfrac{3}{4}}c=\dfrac{\sqrt{3}}{2}c\approx 0.866c<0.990c

Thus, the minimun threshold total energy of each proton is:

E=\dfrac{mc^2}{\sqrt{1-\frac{3}{4}}}=\dfrac{mc^2}{\sqrt{\frac{1}{4}}}=2mc^2

Therefore, the minimum kinetic energy, the kinetic threshold energy, is given by T=mc^2. For each proton PAIR, the total energy threshold is E_{pair}^{min}=4mc^2. This value differs from the relativistic value previously obtained by an amount 2mc^2. Now the kinetic threshold energy to obtain the pair equals T_{min}=4mc^2i.e., 1/3 (E=2mc^2)of the total energy in the previous example (6mc^2) is saved smashing 2 protons in opposite directions in order to create the pair proton-antiproton.

We can generalize the problem to the creation of n-pairs proton-antiproton. The equations yield:

E_A^{min}=\dfrac{(2n+2)^2(mc)^2-2(mc)^2}{2m}

and then

\boxed{E_A^{min,n-pairs}=\left(2(n+1)^2-1\right)mc^2=(2n^2+4n+1)mc^2}

The threshold kinetic energy will be in this case:

\boxed{T_A^{min,n-pairs}=\left(2n(n+2)\right)mc^2=(2n^2+4n)mc^2}

and the incident minimum velocity to hit the proton at rest is now:

u_A=c\sqrt{1-\left(\dfrac{mc^2}{E}\right)^2}=c\sqrt{1-\left(\dfrac{1}{2n^2+4n+1}\right)^2}

In the case of a very large number of pairs, n>>1 we get that

u_A\approx c\sqrt{1-\left(\dfrac{1}{2n^2}\right)^2}=c\sqrt{1-\left(\dfrac{1}{4n^4}\right)}\approx c

In the case of a storage ring,

\gamma_u (c^2+u^2)=(2n^2+4n+1)c^2

and from this equation, algebra provides

\boxed{u=\sqrt{\dfrac{2n^2+4n}{2n^2+4n+2}}c}

and

\boxed{E=\dfrac{mc^2}{\sqrt{1-\left(\dfrac{2n^2+4n}{2n^2+4n+2}\right)^2}}}

LOG#037. Relativity: Examples(I)

Problem 1. In the S-frame, 2 events are happening simultaneously at 3 lyrs of distance. In the S’-frame those events happen at 3.5 lyrs. Answer to the following questions: i) What is the relative speed between frames? ii) What is the temporal distance of events in the S’-frame?

Solution. i) x'=\gamma (x-\beta c t)

x'_2-x'_1=\gamma ((x_2-x_1)-\beta c (t_2-t_1))

And by simultaneity, t_2=t_1

Then \gamma=\dfrac{x'_2-x'_1}{x_2-x_1}=\dfrac{7}{6}

\beta=\sqrt{1-\gamma^{-2}}\approx 0.5

ii) ct'=\gamma (ct-\beta x)

c(t'_2-t'_1)=-\gamma \beta (x_2-x_1)

since we have simultaneity implies t_2-t_1=0. Then,

c\Delta t'\approx -1.8 lyrs

Problem 2. In S-frame 2 events occur at the same point separated by a temporal distance of 3yrs. In the S’-frame, D'=3.5yrs is their spatial separation. Answer the next questions: i) What is the relative velocity between the two frames? ii) What is the spatial separation of events in the S’-frame?

Solution. i) ct'=\gamma (ct-\beta x) with x_1=x_2

As the events occur in the same point x_2=x_1

c(t'_2-t'_1)=\gamma c (t_2-t_1)

\gamma=\dfrac{t'_2-t'_1}{t_2-t_1}=\dfrac{7}{6}

\beta=\sqrt{1-\gamma^{-1}}\approx 0.5

ii) x'=\gamma (x-\beta c t)

x_1=x_2 implies x'_2-x'_1=-\gamma \beta c (t_2-t_1)\approx -1.8 lyrs

Therefore, the second event happens 1.8 lyrs to the “left” of the first event. It’s logical: the S’-frame is moving with relative speed v\approx c/2 for 3.5 yrs.

Problem 3. Two events in the S-frame have the following coordinates in spacetime: P_1(x_0=ct_1,x_1=x_0), i.e., E_1(ct_1=x_0,x_1=x_0) and P_2(ct_2=0.5x_0, x_2=2x_0), i.e., E_2(ct_2=x_0/2,x_2=2x_0). The S’-frame moves with velocity v respect to the S-frame. a) What is the magnitude of v if we want that the events E_1,E_2 were simultaneous? b) At what tmes t’ do these events occur in the S’-frame?

Solution. a) ct'=\gamma (ct-\beta x)

t'_2-t'_1=0 and then 0=\gamma (c(t_2-t_1)-\beta (x_2-x_1))

\beta =\dfrac{c(t_2-t_1)}{(x_2-x_1)}=-\dfrac{0.5x_0}{x_0}=-0.5

b) t'=\gamma ( 1-\beta x/c)=\gamma ( 1-\beta x/c)

t'_1=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5x_0 /c)\right)\approx 1.7x_0/c

t'_2=\dfrac{1}{\sqrt{1-(-0.5)^2}} \left(x_0/c+0.5\cdot 2\cdot x_0/c)\right)\approx 1.7x_0/c

Problem 4. A spaceship is leaving Earth with \beta =0.8. When it is l x_0=6.66\cdot 10^{11}m away from our planet, Earth transmits a radio signal towards the spaceship. a) How long does the electromagnetic wave travel in the Earth-frame? b) How long does the electromagnetic wave travel in the space-ship frame?

For the spaceship, ct=x/\beta and for the signal ct=x+ct_0. From these equations, we get

\beta ct=x and ct=x+ct_0, and it yields \beta ct =ct-ct_0 and thus t=\dfrac{t_0}{1-\beta} for the intersection point. But, \beta=0.8=8/10=4/5 and 1-\beta=1/5. Putting this value in the intersection point, we deduce that the intersection point happens at l t_1=5t_0. Moreover,

t_1-t_0=4t_0=4\dfrac{x_0}{0.8c}\approx 11100 s=3.08h=3h 5min

b) We have to perform a Lorentz transformation from (ct_0,0) to (ct_1,x_1), with t'=t=0.

t_0=x_0/v=2775s and t_1=5t_0=13875s. Then x_1=vt_1=5vt_0=5x_0=5\cdot 6.66\cdot 10^{11}m=3.33\cdot 10^{12}m. And thus, we obtain that \gamma=5/3. The Lorentz transformation for the two events read

(t'_2-t'_1)=\gamma (t_1-t_0)=\gamma (t_1-t_0)-\beta/c(x_1-x_0)=3700 s\approx 1.03h=1h1m40s

Remarks: a) Note that t_1-t_0 and t'_2-t'_1 differ by 3 instead of 5/3. This is due to the fact we haven’t got a time interval elapsing at a certain location but we face with a time interval between two different and spatially separated events.

b)The use of the complete Lorentz transformation (boost) mixing space and time is inevitable.

Problem 5. Two charged particles A and B, with the same charge q, move parallel with \mathbf{v}=(v,0,0). They are separated by a distance d. What is the electric force between them?

E'=\left(0,\dfrac{1}{4\pi\epsilon_0}\dfrac{q}{d^2},0\right)

B'=\left(0,0,\gamma \dfrac{\gamma v q}{4\pi\epsilon_0d^2c^2}\right)

In the S-frame, we obtain the Lorentz force:

\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)=\left(0,\gamma k_C\dfrac{q^2}{d^2}-\gamma \beta^2\dfrac{q^2}{d^2},0\right)=\left(0,\dfrac{k_Cq^2}{\gamma d^2},0\right)

The same result can be obtained using the power-force (or forpower) tetravector performing an inverse Lorentz transformation.

Problem 6. Calculate the electric and magnetic field for a point particle passing some concrete point.

The electric field for the static charge is: E=k_C\dfrac{q}{x'^2+y'^2+z'^2}=k_C\dfrac{q}{r'^2}

with \mathbf{v}=(v,0,0) when the temporal origin coincides, i.e., at the time t'=t=0.  Suppose now two points that for the rest observer provide:

P=(0,a,0) and P'=(-vt',a,0). For the electric field we get:

E'=k_C\dfrac{q}{r'^3}(x',y',z') and E'(t')=k_C\dfrac{q}{(\sqrt{(x'^2+y'^2+z'^2})^3}(x',y',z')

E'_(t')=k_C\dfrac{q}{(v^2t'^2+a^2)^{3/2}}(-vt',a,0)

Then, E'_p\rightarrow E_p implies that l t'=\gamma t=\gamma (t-\dfrac{vx}{c^2})\vert_{x=0}

E'_p(t)=k_C\dfrac{q}{(\gamma^2v^2t^2+a^2)^{3/2}}(-\gamma v t,a,0)

B'=B'_p(t')=(0,0,0)=B'_p(t)

E_p(t)=(E'_{p_x}(t),\gamma E'_{p_y}(t),0)=k_C\dfrac{q}{\gamma^2 v^2t^2+a^2}(-\gamma v t,\gamma a,0)

B_p(t)=(B_{p_x},B_{p_y},B_{p_z})=(0,0,\gamma \dfrac{\gamma v E'_p(t)}{c^2})

B_p(t)=\dfrac{q}{\gamma^2v^2t^2+a^2}(0,0,\gamma \dfrac{v}{c^2}a)=(0,0,\dfrac{v}{c^2}E_{p_y}(t))

There are two special cases from the physical viewpoint in the observed electric fields:

a) When P is directly above the charge q. Then E_p(t=0)=(0,k_C\gamma \dfrac{q}{a^2},0)

b) When P is directly in front of ( or behind) q. Then, for a=0, E_p(t)=(-k_C\dfrac{vt}{\gamma^2(v^2t^2)^{3/2}},0,0)

Note that we have \dfrac{vt}{(v^2t^2)^{3/2}}\neq \dfrac{1}{v^2t^2} if t<0.

LOG#036. Action and relativity.

Least_action_principle.svg

The hamiltonian formalism and the hamiltonian H in special relativity has some issues with the definition. In the case of the free particle one possible definition, not completely covariant, is the relativistic energy

    \[ \boxed{E=\sqrt{m^2c^4+c^2p^2}=H}\]

There are two others interesting scalars in classical relativistic theories. They are the lagrangian L and the action functional S. The lagrangian is obtained through a Legendre transformation from the hamiltonian:

    \[ \boxed{L=pv-H}\]

From the hamiltonian, we get the velocity using the so-called hamiltonian equation:

    \[ \dot{\mathbf{q}}=\mathbf{v}=\dfrac{\partial H}{\partial \mathbf{p}}=c^2\dfrac{\mathbf{p}}{E}\]

Then,

    \[ L=\dfrac{E}{c^2}\mathbf{v}^2-E=E\left(\dfrac{v^2}{c^2}-1\right)=-\dfrac{E}{\gamma^2}=-\dfrac{m\gamma c^2}{\gamma^2}=-\dfrac{mc^2}{\gamma}\]

and finally

    \[ \boxed{L=-mc^2\sqrt{1-\dfrac{v^2}{c^2}}=-\dfrac{mc^2}{\gamma}=-mc\sqrt{-\dot{X}^2}}\]

The action functional is the time integral of the lagrangian:

    \[ \boxed{S=\int Ldt}\]

However, let me point out that the above hamiltonian in SR has some difficulties in gauge field theories. Indeed, it is quite easy to derive that a more careful and reasonable election for the hamiltonian in SR should be zero!

In the case of the free relativistic particle, we obtain

    \[ S=-mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt\]

Using the relation between time and proper time (the time dilation formula):

    \[ dt=\gamma d\tau\rightarrow \dfrac{dt}{\gamma}=d\tau\]

direct substitution provides

    \[ -mc^2\int \sqrt{1-\dfrac{v^2}{c^2}}dt=-mc^2\int d\tau\]

And defining the infinitesimal proper length in spacetime as ds=cd\tau, we get the simple and wonderful result:

    \[ \boxed{S=-mc\int ds}\]

Sometimes, the covariant lagrangian for the free particle is also obtained from the following argument. The proper length is defined as

    \[ ds^2=d\mathbf{x}^2-c^2dt^2\]

The invariant in spacetime is related with the proper time in this way:

    \[ ds^2=-c^2d\tau^2=d\mathbf{x}^2-c^2dt^2\]

Thus, dividing by dt^2

    \[ -c^2\dfrac{d\tau^2}{dt^2}=\mathbf{v}^2-c^2\]

and

    \[ d\tau^2=\gamma^{-2}dt^2=\dfrac{1}{\gamma^2}dt^2=\left(1-\dfrac{\mathbf{v}^2}{c^2}\right)dt^2\]

so

    \[ d\tau=\sqrt{1-\dfrac{\mathbf{v}^2}{c^2}}dt\]

    \[ cd\tau=ds=\sqrt{c^2-\mathbf{v}^2}dt=\sqrt{-\dot{X}^2}dt\]

that is

    \[ \boxed{ds=cd\tau=\sqrt{-\dot{X}^2}dt=\sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=\sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}\]

and the free coordinate action for the free particle would be:

    \[ \boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}dt=-mc^2\int \sqrt{-\dot{x}^\mu\dot{x}_\mu}dt=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}dt}\]

Note, that since the election of time “t” is “free”, we can choose t=\tau to obtain the generally covariant free action:

    \[ \boxed{S=-mc\int ds=-mc^2\int \sqrt{-\dot{X}^2}d\tau=-mc^2\int \sqrt{-\eta_{\mu\nu} \dot{x}^\mu\dot{x}^\nu}d\tau}\]

Remark: the (rest) mass is the “coupling” constant for the free particle proper length to guess the free lagrangian

    \[ \boxed{L=-mc^2\sqrt{-\dot{X}^2}}\]

Now, we can see from this covariant action that the relativistic hamiltonian should be a feynmanity! From the equations of motion,

    \[ P_\mu=\dfrac{\partial L}{\partial \dot{X}^\mu}=mc\dfrac{\dot{X}_\mu}{\sqrt{-\dot{X}^2}}\]

The covariant hamiltonian \mathcal{H}, different from H, can be build in the following way:

    \[ \mathcal{H}=P_\mu \dot{X}^\mu-L=mc\dfrac{\dot{X}_\mu \dot{X}^\mu}{\sqrt{-\dot{X}^2}}-mc\sqrt{-\dot{X}^2}=0\]

The meaning of this result is hidden in the the next identity ( Noether identity or “hamiltonian constraint” in some contexts):

    \[ \boxed{\mathcal{H}=P_\mu P^\mu+m^2c^2=0}\]

since

    \[ P_\mu P^\mu=m U_\mu mU^\mu=-m^2c^2\]

This strange fact that \mathcal{H}=0 in SR, a feynmanity as the hamiltonian, is related to the Noether identity E^\mu \dot{X}_\mu for the free relativistic lagrangian, indeed, a consequence of the hamiltonian constraint and the so-called reparametrization invariance \tau'=f (\tau). Note, in addition, that the free relativistic particle would also be invariant under diffeomorphisms x^{\mu'}= f^\mu (x)=f^\mu (x^\nu) if we were to make the metric space-time dependent, i.e., if we make the substitution \eta_{\mu\nu}\rightarrow g_{\mu\nu} (x). This last result is useful and important in general relativity, but we will not discuss it further in this moment. In summary, from the two possible hamiltonian in special relativity

    \[ H=E=\sqrt{\mathbf{p}^2c^2+(mc^2)^2}\]

    \[ \mathcal{H}=P_\mu P^\mu+m^2c^2=0\]

the natural and more elegant (due to covariance/invariance) is the second one. Moreover, the free particle lagrangian and action are:

    \[ \boxed{L=-mc^2\sqrt{-\dot{X}^2}}\]

    \[ \boxed{S=-mc^2\int d\tau=-mc\int ds=\int L dt}\]

Remark: The true covariant lagrangian dynamics in SR is a “constrained” dynamics, i.e., dynamics where we are undetermined. There are more variables that equations as a result of a large set of symmetries ( reparametrization invariance and, in the case of local metrics, we also find diffeomorphism invarince).

The dynamical equations of motion, for a first order lagrangian (e.g., the free particle we have studied here), read for the lagrangian formalism:

    \[ \boxed{\delta S=\delta \int L (q,\dot{q};t)dt =0\leftrightarrow\begin{cases}E(L)=0,\mbox{with E(L)=Euler operator}\\ E(L)=\dfrac{\partial L}{\partial q}-\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{q}}\right)=0\end{cases}}\]

By the other hand, for the hamiltonian formalism, dynamical equations are:

    \[ \boxed{\delta S=\delta \int H (q, p; t)dt =0\leftrightarrow\begin{cases}p=\dfrac{\partial L}{\partial \dot{q}},\;\mbox{with}\; \det\left(\dfrac{\partial ^2L}{\partial \dot{q}^i\partial \dot{q}^j}\right)\neq 0\\ \;\\ \dfrac{dq}{dt}=\dot{q}=\dfrac{\partial H}{\partial p}\\ \;\\\dfrac{dp}{dt}=\dot{p}=-\dfrac{\partial H}{\partial q} \\ \;\\ \dfrac{dH}{dt}=\dot{H}=\dfrac{\partial H}{\partial t}=-\dfrac{\partial L}{\partial t}\end{cases}}\]

LOG#035. Doppler effect and SR.

Blue-Shift-and-Red-Shift.

dopplerShift

The Doppler effect is a very important phenomenon both in classical wave motion and relativistic physics. For instance, nowadays it is used to the detect exoplanets and it has lots of applications in Astrophysics and Cosmology.

Firstly, we remember the main definitions we are going to need here today.

    \[ \omega =\dfrac{2\pi}{T}=2\pi\nu=2\pi f\]

Sometimes, we will be using the symbol f for the frequency \nu. We also have:

    \[ \mathbf{n}=\dfrac{\mathbf{k}}{\vert \mathbf{k}\vert}\]

    \[ k=\vert \mathbf{k}\vert=\dfrac{2\pi}{\lambda}=\dfrac{\omega}{c}\]

A plane (sometimes electromagnetic) wave is defined by the oscillation:

    \[ A(\mathbf{r},t)=A_0\exp\left(iKX\right)\]

where

    \[ KX=\mathbb{K}\cdot\mathbb{X}=\mbox{PHASE}=\mathbf{k}\cdot \mathbf{r}-\omega t\]

If K^2=0, then the wave number is lightlike (null or isotropic) and then (k^0)^2=\dfrac{\omega^2}{c^2}

Using the Lorentz transformations for the wave number spacetime vector, we get:

    \[ k'^0=\gamma \left(k^0-\beta k^1\right)\]

i.e., if the angle with the direction of motion is \alpha so k^1=\dfrac{\omega}{c}\cos\alpha

    \[ \dfrac{\omega '}{c}=\gamma \left(\dfrac{\omega}{c}-\beta\left(\dfrac{\omega}{c}\cos \alpha\right)\right)\]

we deduce that

    \[ \boxed{\omega=\omega_0\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}}\]

or for the normalized frequency shift

    \[ \boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{\sqrt{1-\beta^2}}{1-\beta\cos\alpha}-1}\]

This is the usual formula for the relativistic Doppler effect when we define \omega'=\omega_0, and thus, the angular frequency (also the frecuency itself, since there is only a factor 2 times the number pi of difference) changes with the motion of the source. When the velocity is “low”, i.e., \beta<<1, we obtain the classical Doppler shift formula:

    \[ \omega \approx (1+\beta\cos\alpha)\omega_0\]

We then calculate the normalized frequency shift from \Delta \omega=\omega-\omega_0

    \[ \boxed{D=\dfrac{\Delta \omega}{\omega_0}=\dfrac{V\cos\alpha}{c}}\]

The classical Doppler shift states that when the source approaches the receiver (\cos\alpha>0), then the frequency increases, and when the source moves away from the receiver (\cos\alpha<0), the frequency decreases. Interestingly, in the relativistic case, we also get a transversal Doppler shift which is absent in classical physics. That is, in the relativistic Doppler shift, for \alpha=\pi/2, we obtain

    \[ \boxed{\omega=\dfrac{\omega_0}{\gamma_V}=\omega_0\sqrt{1-\beta^2}}\]

and the difference in frequency would become

    \[ \boxed{D=\dfrac{\Delta \omega}{\omega_0}=\sqrt{1-\beta^2}-1}\]

There is an alternative deduction of these formulae. The time that an electromagnetic wave uses to run a distance equal to the wavelength, in a certain inertial frame S’ moving with relative speed V to another inertial frame S at rest, is equal to:

    \[ t=\dfrac{\lambda}{c-V}=\dfrac{c}{(c-V)f_s}=\dfrac{1}{(1-\beta_V)f_s}\]

where f_s is the frequency of the source. Due to the time dilation of special relativity

    \[ t=t_0\gamma\]

and thus

    \[ f_0=\dfrac{1}{t_0}=\gamma(1-\beta_V)f_s=\sqrt{\dfrac{1-\beta_V}{1+\beta_V}}f_s\]

so we get

    \[ \boxed{\dfrac{f_s}{f_o}=\dfrac{f_{source}}{f_{obs}}=\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}\]

The redshift (or Doppler displacement) is generally defined as:

    \[ \boxed{z=-D=\dfrac{\lambda_o-\lambda_s}{\lambda_s}=\dfrac{f_s-f_o}{f_o}=\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}-1}\]

When \beta\rightarrow 0, i.e., V\rightarrow 0,V<<c, we get the classical result

    \[ z\approx\beta=\dfrac{v}{c}\]

The generalization for a general non-parallel motion of the source/observer is given by

    \[ \boxed{f_0=\dfrac{f_s}{\gamma\left(1+\dfrac{v_s\cos\theta_o}{c}\right)}}\]

If we use the stellar aberration formula:

    \[ \boxed{\cos\theta_o=\dfrac{\cos\theta_s-\dfrac{v_s}{c}}{1-\dfrac{v_s}{c}\cos\theta_s}}\]

the last equation can be recasted in terms of \theta_s instead of \theta_o as follows:

    \[ \boxed{f_0=\gamma\left(1-\dfrac{v_s}{c}\cos\theta_s\right)}\]

and then

    \[ \boxed{z=\dfrac{f_s-f_0}{f_0}=-D=\dfrac{\Delta f}{f_0}=\dfrac{1}{\gamma\left(1-\dfrac{v_s\cos\theta_s}{c}\right)}-1}\]

Again, for a transversal motion, \theta_o=\pi/2, we get a transversal Doppler effect:

    \[ \boxed{f_0=\dfrac{f_s}{\gamma}=\sqrt{1-\beta^2}f_s} \leftrightarrow \boxed{z_T=-D=\dfrac{f_o-f_s}{f_s}=\sqrt{1-\beta_s^2}-1}\]

Remark: Remember that the Doppler shift formulae are only valid if the relative motion (of both source and observer/receiver) is slower than the speed of the (electromagnetic) wave, i.e., if v\leq c.

In the final part of this entry, we are going to derive the most general formula for Doppler effect, given an arbitrary motion of source and observer, in both classical and relativistic Physics. Recall that the Doppler shift in Classical Physics for an arbitrary observer is given by a nice equation:

    \[ \boxed{f'=f\left[\dfrac{v-v_o\cos\theta_o}{v-v_s\cos\theta_s}\right]}\]

Here, v is the velocity of the (electromagnetic) wave in certain medium, v_o is the velocity of the observer in certain direction forming an angle \theta_o with the “line of sight”, while v_s is the velocity of the source forming an angle \theta_s with the line of sight/observation. If we write

    \[ V_o=-v_o\cos\theta_o\]

    \[ V_s=-v_s\cos\theta_s\]

    \[ V_{so}=V_s-V_o\]

we can rewrite this last Doppler formula in the following way ( for v=c):

    \[ f'=\left(\dfrac{c+V_o}{c+V_s}\right)f\]

or with

    \[ f'=f_o\]

and

    \[ f=f_s\]

    \[ \boxed{f_o=f_s\left(\dfrac{c+V_s}{c+V_o}\right)=\left(1+\dfrac{V_s-V_o}{c+V_o}\right)f_s=\left(1+\dfrac{V_{so}}{c+V_o}\right)f_s}\]

and

    \[ \boxed{z=-D=\dfrac{f_0-f_s}{f_0}=\left(\dfrac{c+V_0}{c+V_s}\right)-1=\dfrac{V_o-V_s}{c+V_s}=\dfrac{-V_{so}}{c+V_s}}\]

The most general Doppler shift formula, in the relativistic case, reads:

    \[ \boxed{\dfrac{f_o}{f_s}=\dfrac{1-\dfrac{\vert\vert \mathbf{v}_o\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{co}}{1-\dfrac{\vert\vert \mathbf{v}_s\vert\vert}{\vert\vert \mathbf{c}\vert\vert}\cos\theta_{cs}}\sqrt{\dfrac{1-\dfrac{v_s^2}{c^2}}{1-\dfrac{v_o^2}{c^2}}}}\]

or equivalently

    \[ \boxed{\dfrac{f_o}{f_s}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}}\]

and where \mathbf{v}_s,\mathbf{v}_o are the velocities of the source and the observer at the time of emission and reception, respectively, \beta_s,\beta_o are the corresponding beta boost parameters, \mathbf{c} is the “light” or “wave” velocity vector and we have defined the angles \theta_{cs},\theta_{co} to be the angles formed at the time of emission and the time of reception/observation between the source velocity and the “wave” velocity, respectively, between the source and the wave and the observer and the wave. Two simple cases of this formula:

1st. Parallel motion with \mathbf{c}\parallel\mathbf{v}_s\rightarrow \theta_{cs}=0^\circ\rightarrow \cos\theta_{cs}=1\rightarrow f_o>f_s.

2nd. Antiparallel motion with \mathbf{c} going in the contrary sense than that of \mathbf{v}_s. Then, \theta_{cs}=180^\circ\rightarrow \cos\theta_{cs}=-1\rightarrow f_o<f_s.

The deduction of this general Doppler shift formula can be sketched in a simple fashion. For a signal using some propagating wave, we deduce:

    \[ \vert \mathbf{r}_o-\mathbf{r}_s\vert ^2=\vert\mathbf{C}\vert^2\left(t_o-t_s\right)^2\]

with

    \[ \mathbf{C}=\dfrac{\mathbf{r}_o-\mathbf{r}_s}{t_o-t_s}\]

Differentiating with respect to t_o carefully, it provides

    \[ \mathbf{C}\cdot\left[\mathbf{v}_o-\mathbf{v}_s\dfrac{dt_s}{dt_o}\right]=\vert \mathbf{C}\vert^2\left(1-\dfrac{dt_o}{dt_s}\right)\]

Solving for \dfrac{dt_o}{dt_s} we get

    \[ \dfrac{dt_o}{dt_s}=\dfrac{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_o }{\vert \mathbf{C}\vert^2-\mathbf{C}\cdot\mathbf{v}_s}=\dfrac{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_o\right)}{\mathbf{C}\cdot\left(\mathbf{C}-\mathbf{v}_s\right)}=\dfrac{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_o}{\vert\mathbf{C}\vert^2}}{1-\dfrac{\mathbf{C}\cdot\mathbf{v}_s}{\vert\mathbf{C}\vert^2}}\]

Using the known formula \vert \mathbf{r}\cdot\mathbf{s}\vert=\vert\mathbf{r}\vert\vert\mathbf{s}\vert\cos\theta_{rs}, we obtain in a simple way:

    \[ \dfrac{dt_o}{dt_s}=\dfrac{1-\dfrac{\vert\mathbf{v}_o\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_o}}{1-\dfrac{\vert\mathbf{v}_s\vert}{\vert\mathbf{C}\vert}\cos\theta_{\mathbf{c},\mathbf{v}_s}}=\dfrac{\vert\mathbf{C}-\mathbf{v}_o\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}}{\vert\mathbf{C}-\mathbf{v}_s\vert\cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}}\]

Finally, using the fact that \cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_s}=\cos\theta_{\mathbf{C},\mathbf{v}_s}, the similar result \cos\theta_{\mathbf{C},\mathbf{C}-\mathbf{v}_o}=\cos\theta_{\mathbf{C},\mathbf{v}_o}, and that the proper time induces an extra gamma factor due to time dilation

    \[ dt=d\tau\gamma\]

and we calculate for the frequency:

    \[ \dfrac{n_o}{n_s}=\dfrac{\nu_o}{\nu_s}=\dfrac{f_o}{f_s}=(\mbox{PREFACTOR})\dfrac{\gamma_s dt_s}{\gamma_odt_o}\]

where the PREFACTOR denotes the previously calculated ratio between differential times. Finally, elementary algebra let us derive the expression:

    \[ \dfrac{f_o}{f_s}=\dfrac{d\tau_s}{d\tau_o}=\dfrac{1-\vert\vert \beta_o \vert\vert \cos\theta_{co}}{1-\vert\vert \beta_s\vert\vert\cos\theta_{cs}}\sqrt{\dfrac{1-\beta_s^2}{1-\beta_o^2}}\]

Q.E.D.

An interesting remark about the Doppler effect in relativity: the Doppler effect allows us to “derive” the Planck’s relation for quanta of light. Suppose that a photon in the S’-frame has an energy E' and momentum (p'_x,0,0)=(-E',0,0) is being emitted along the negative x’-axis toward the origin of the S-frame. The inverse Lorentz transformation provides:

    \[ E=\gamma (E'+vp'_x)=\gamma (E'-\beta E')=\dfrac{1-\beta}{\sqrt{1-\beta^2}}E'\]

, i.e.,

    \[ E=\sqrt{\dfrac{1-\beta}{1+\beta}}E'\]

By the other hand, by the relativistic Doppler effect we have seen that the frequency f’ in the S’-frame is transformed into the frequency f in the S-frame if we use the following equation:

    \[ f=f'\sqrt{\dfrac{1-\beta}{1+\beta}}\]

If we divide the last two equations we get:

    \[ \dfrac{E}{f}=\dfrac{E'}{f'}=constant \equiv h\]

Then, if we write h=6.63\cdot 10^{-34}J\cdot s, and E=hf and E'=hf'.

AN ALTERNATIVE HEURISTIC DEDUCTION OF THE RELATIVISTIC DOPPLER EFFECT

In certain rest frame S, there is an observer receiving light beams/signals. The moving frame is the S’-frame and it is the emitter of light. The source of light approaches at velocity V, and it sends pulses with frequency f_0=\nu_0. What is the frequency that the observer at S observes? Due to time dilation, the observer at S observes a longer period

    \[ T=T_0\dfrac{1}{\sqrt{1-\beta_V^2}}\]

with

    \[ \beta_V=\dfrac{V}{c}\]

The distance between two consecutive light beams seen by the observer at S will be:

    \[ \lambda=cT-VT=(c-V)T=(c-V)\dfrac{T_0}{\sqrt{1-\beta_V^2}}\]

Therefore, the observer frequency in the S-frame is:

    \[ f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V)}=\nu_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}\]

i.e.

    \[ \boxed{f=f_0\sqrt{\dfrac{1+\beta_V}{1-\beta_V}}}\]

If the source approaches the observer, then f>f_0. If the source moves away from the observer, then f<f_0. In the case, the velocity of the source forms certain angle \alpha in the direction of observation, the same argument produces:

    \[ f=\nu=\dfrac{c}{\lambda}=\dfrac{c\sqrt{1-\beta_V^2}}{T_0(c-V\cos\alpha)}=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}\]

that is

    \[ \boxed{f=f_0\dfrac{\sqrt{1-\beta_V}}{1-\beta\cos\alpha}}\]

In the case of transversal Doppler effect, we get \alpha=90^\circ=\pi/2, and so:

    \[ \boxed{f=f_0\sqrt{1-\beta^2}}\]

Q.E.D.

Final remark (I): If \theta=\alpha=\dfrac{\pi}{2} or \theta=\alpha=\dfrac{3\pi}{2} AND v<<c, therefore we have that there is no Doppler effect at all.

Final remark (II): If v\approx c, there is no Doppler effect in certain observation directions. Those directions can be deduced from the above relativistic Doppler effect formula with the condition f=f_0 and solving for \theta. This gives the next angular direction in which Doppler effect can not be detected

    \[\boxed{\cos\theta=\dfrac{\sqrt{1-\beta^2}-1}{\beta}}\]

LOG#034. Stellar aberration.

aberration

In this entry, we are going to study a relativistic effect known as “stellar aberration”.

From the known Lorentz transformations of velocities (inverse case), we get:

    \[ v_x=\dfrac{v'_x+V}{1+\dfrac{v'_xV}{c^2}}\]

    \[ v_y=\dfrac{v'_y\sqrt{1-\beta^2}}{1+\dfrac{v'xV}{c^2}}\]

    \[ v_z=\dfrac{v'_z\sqrt{1-\beta^2}}{1+\dfrac{v'zV}{c^2}}\]

The classical result (galilean addition of velocities) is recovered in the limit of low velocities V\approx 0 or sending the light speed get the value “infinite” c\rightarrow \infty. Then,

    \[ v_x=v'_x+V\]

 

    \[v_y=v'_y\]

    \[v'_z=v_z\]

Let us define

    \[ \theta =\mbox{angle formed by x}\; \mbox{and}\; v_x\]

    \[ \theta' =\mbox{angle formed by x'}\; \mbox{and}\; v'_x\]

Thus, we get the component decomposition into the xy and x’y’ planes:

    \[ v_x=v\cos\theta\]

    \[v_y=v\sin\theta\]

    \[ v'_x=v'\cos\theta'\]

    \[ v'_y=v'\sin\theta'\]

From this equations, we get

    \[ \tan \theta=\dfrac{v'\sin\theta'\sqrt{1-\beta^2}}{v'\cos\theta'+V}\]

If v=v'=c

    \[ \boxed{\tan \theta=\dfrac{\sin\theta'\sqrt{1-\beta_V^2}}{\cos\theta'+\beta_V}}\]

and then

    \[ \boxed{\cos\theta=\dfrac{\beta_V+\cos\theta'}{1+\beta_V\cos\theta'}}\]

    \[ \boxed{\sin\theta=\dfrac{\sin\theta'\sqrt{1-\beta_V^2}}{1+\beta_V\cos\theta'}}\]

From the last equation, we get

    \[ \sin\theta'\sqrt{1-\beta_V^2}=\sin\theta\left(1+\beta_V\cos\theta'\right)\]

From this equation, if V<<c, i.e., if \beta_V<<1 and \theta'=\theta+\Delta\theta with \Delta \theta<<1, we obtain the result

    \[ \Delta \theta=\theta'-\theta=\beta_V\sin\theta\]

By these formalae, the angle of a light beam propagating in space depends on the velocity of the source respect to the observer. We can observe this relativistic effect every night (supposing a good approximation that Earth’s velocity is non-relativistic, as it shows). The physical interpretation of the above aberration formulae (for the stars we watch during a sky night) is as follows: due to the Earth’s motion, a star in the zenith is seen under an angle l \theta\neq \dfrac{\pi}{2}.

Other important consequence from the stellar aberration is when we track ultra-relativistic particles (\beta\approx 1). Then, \theta'\rightarrow \pi and then, the observer moves close to the source of light. In this case, almost every star (excepting those behind with \theta=\pi) are seen “in front of” the observer. If the source moves with almost the speed of light, then the light is “observed” as it were concentrated in a little cone with an aperture \Delta\theta\sim\sqrt{1-\beta_V^2}.