LOG#040. Relativity: Examples(IV).

Example 1. Compton effect.

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

We use momenergy conservation:

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

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

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

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

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

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

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

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

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

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

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

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

or

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

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

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

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

\lambda_C=\dfrac{h}{mc}

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

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

Example 2. Inverse Compton effect.

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

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 \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.

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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 aborption 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,

(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. 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

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

\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 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

$late \beta ct=x$ 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 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 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.