34

One can quantize linearized spacetime perturbations in General Relativity and compute the effect of photons scattering elastically by exchanging virtual gravitons. This theory isn’t consistent at Planck-scale photon energies but is believed to be fine at the energies of photons we observe... even very high-energy gamma rays. All the energy coming in has to ...


13

Yes. It is possible to treat linearized gravity like a QFT, write the graviton propagator and study the scattering of two photon thought a graviton exchange. The amplitude, given the momentum of the two photons $k_{1}$ and $k_{2}$, in the limit where the graviton propagator momentum $q \rightarrow 0$ behaves proportionally to $$\sim \frac{(k_{1} \cdot p_{1})...


9

It is "easy" to show that the cross-section of the 2 reactions: $$ (1)~~~\nu_{\mu} + d \to \mu^- + u~~~~~~(2)~~~\bar{\nu}_{\mu} + u \to \mu^+ + d$$ are different. The naive calculation gives a factor $\sigma_1/\sigma_2 = 3$ as shown below. The fact that the figure in the particle data group gives actually almost a factor 2 needs to take into account the ...


9

We need to be precise: the total energy of the system is always greater in the lab frame than the center of momentum frame (unless the lab frame is the same as the center of momentum frame). If you have a collection of particles and you want to know the total energy you can treat them all as just one object. In the center of momentum frame the total object ...


8

Welcome to the world of nuclear physics, where the answer is "It's a little more complicated than that." Density of the solid You can rule this out: cross sections are tabulated per target atom. Size of the nucleus, i.e., strictly increasing with (N+Z). This is a good guess, but you miss an important feature of thermal neutron physics: ...


7

Peskin and Schroeder assume the laboratory frame, as is evident from the top of page 106: The difference $|v_A-v_B|$ is the relative velocity of the beams as viewed from the laboratory frame. As a result, as you (and they) have also pointed out, the cross section is not Lorentz invariant. They furthermore explain that it is only invariant with respect ...


7

By definition, Thomson scattering is the elastic scattering of light by a free charged particle. Atoms cannot be described as such, but the electrons in an atom may approximate to free electrons if their binding energy is much lower than the photon energy. This might be true for X-ray wavelengths, although if the photon energy gets too high then elastic ...


6

The Figure 2 of this paper http://arxiv.org/pdf/hep-ph/0611148v1.pdf doesn't show a factor-of-ten difference at all! Extract the ratio properly on the log scale and you will see it is less than four, just slightly greater than 1/2 of the height corresponding to the decade. Note that 1/2 of the weight corresponds to the factor of $\sqrt{10}\sim 3.16$. ...


6

How "wide" is a photon, The photon is a point particle in the standard model of particle physics. It has no extent to be described with "wide". Interactions of photons with charges or magnets can have a "width" in the sense of "measurable range" if any, of its electromagnetic fields? Is there any physical length measurement of these two orthogonal fields,...


6

Muons are easy to measure. They're extremely long-lived relative to most other particles, and they're pretty massive compared to electrons, which means they have substantial penetrating power that is essentially unmatched by other charged particles. There's a reason that muon detection systems are put on the very outer edges of detectors - muons can punch ...


5

It is simply a matter of notation. The $p_1$ (and hence $E_1$ and $E_2$) in $$\int d\Pi_2=\int d\Omega\frac{p_1^2}{16\pi^2E_1E_2}(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$$ is no longer an integration variable; it has the fixed value that satisfies the delta function $\delta(E_{cm}-E_1-E_2)$ in the previous integral. The factor $(\frac{p_1}{E_1}+\frac{p_1}{...


5

I am not sure to understand your question because the vlasov equation is only valid for a collisionless plasma ... The interactions between particles is done through the long range mean electromagnetic field. If you want to include the collision operator you need to work with the Landau's equation or Fokker-Planck one. For an uncharged gas, the Boltzmann's ...


5

You typically measure the probability of the event. For example, you shoot X-rays at a known blob of material and measure the fraction of Xryas that were deflected / absorbed. You can then imagine each atom in the material to have a certain cross section, and imagine whether you could "shoot past" all those atoms for a given average cross section. Make ...


5

It is not possible to get the cross section from the decay width. The reason is that when calculating the decay width, one squares the amplitude and then sums over the momenta of B and C. While the amplitude squared part is the same for cross section, summing over final momenta means you cannot use it for specific momenta of B and C in the cross section.


5

That the total cross section is infinite just means that every charged particle that passes by the (bare) nucleus is scattered to some extent. This is a consequence of the Coulomb potential being long range. Classically, it is sufficient for the potential to be nonzero for all radii in order to have an infinite total cross section. Quantum mechanically, ...


5

I haven't work out the math but looks like expression you get is correct, you just need to use the following identity $$ f(x)\delta^n(x)=f(0)\delta^n(x) $$ Therefore $$ (E-H)G=e^{ik|r-r'|}\delta(r-r')=\delta(r-r') $$


5

[The figure shown in the OP question above ...] is experimental data for the ratio $R = $ [...] as a function of the centre of mass energy $\sqrt{s}$ The so called cross section ratio $$R[~\sqrt{s}~] = \frac{\sigma^{(0)}[~e^+~e^- \rightarrow \text{hadrons}, \sqrt{s}~]}{\sigma^{(0)}[~e^+~e^- \rightarrow \mu^+~\mu^-, \sqrt{s}~]}$$ ?? Actually, surprisingly, ...


5

There will be as many formulae as models, because it depends what one supposes dark matter is composed of. If WIMP In particle physics and astrophysics, Weakly Interacting Massive Particles, or WIMPs, are among the last hypothetical particle physics candidates for dark matter. The term “WIMP” is given to a dark matter particle that was produced by ...


5

The paper does not say that this is the number of detected events, it says that the "number of events scales as". Whether it is exact depends on how you specify the flux and the cross-section. If the cross-section is in units of $m^2$ per "target" (nucleus) in the detector and the flux is numbers of neutrinos per square metre per unit time, then the ...


5

In your expression, you have a term of the form $$I = \epsilon^{\alpha \mu \beta \nu} S_{\mu\nu}$$ where $S_{\mu\nu} = - g_{\mu\nu} + k_\mu k_\nu / m_W^2$ is a symmetric tensor. The contraction of an antisymmetric tensor with a symmetric one is always zero, because $$I = \epsilon^{\alpha \nu \beta \mu} S_{\nu\mu} = - \epsilon^{\alpha \mu \beta \nu} S_{\mu\nu}...


5

By properly redefining the gauge field $$ eA_\mu \rightarrow A_\mu, $$ one can reduce the coupling constant to the YM term only, $$ \sim \frac{1}{4e^2} F^{\mu\nu}F_{\mu\nu}. $$ Therefore, the absolute sign of $e$ does not matter, since only $e^2$ matters. (Of course, if you are concerned with the sign in front of $\pm \frac{1}{4e^2}$, it surely matters. ) ...


5

To first order, muon pair production: $$ e^+ + e^- \rightarrow \mu^+ + \mu^- $$ only proceeds via the $s$-channel. That is, the electron position pair annihilate into a virtual photon or Z-boson, which then decays to the final state. Meanwhile: $$ e^+ + e^- \rightarrow e^+ + e^- $$ has both an $s$ and $t$ channel amplitude, making comparison with $q\bar q$ ...


4

Part of your confusion here probably comes from your notation. Usually we reserve the index $\mu$ for spacetime. The generators of $SU(N)$ are more commonly labelled with Latin indices $t^a$. See for example here. We can split the amplitude into two parts, according to whether they concern color or kinematics. You are just interested in the color part. Each ...


4

I would say the approach here is based on scale. I have no references on this, I am exposing my educated guess. I think there is a relation between the energy transfer and the relevant scale receiving this energy. That is how I would explain the difference between pushing a wood piece, with a rather slow movement, while a skilled martial artist is able to ...


4

I consider the scattering process $A+B \to 1 + 2$. The differential cross-section is always given by \begin{equation} \begin{split}\label{eq1} d\sigma &= \frac{1}{(2E_A)(2E_B)|v_A - v_B|} \frac{d^3p_1}{(2\pi)^3} \frac{1}{2E_1} \frac{d^3p_2}{(2\pi)^3} \frac{1}{2E_2} \left| {\cal M} \right|^2(2\pi)^4 \delta^4( p_A + p_B - p_1 - p_2) \end{split} \end{...


4

The small number of "conceptually independent types of processes and calculations" is exactly a symptom of the theory's being fundamental! Even in classical physics, all calculations could have been mathematically reduced to the calculation of the final state that evolves from an initial state (or a state that is stationary etc.). In quantum mechanics, this ...


4

One can calculate a lot more in quantum field theory if one goes beyond asymptotic computations into thermal field theory. I recommend that you look at the book ''Nonequilibrium Quantum Field Theory'' by Calzetta and Hu.


4

It should be noted that the differential cross section at $\theta = 0$ is not really defined - what is the difference between a particle being scattered by angle zero and not being scattered at all? However, it's completely legitimate to ask about the behavior of $\frac{d\sigma}{d\Omega}$ as $\theta$ approaches close to zero. For Rutherford scattering, the ...


4

The probability density over the Lorentz invarant phase space (LIPS) is proportinal to the matrix element squared and a Dirac function for energy momentum conservation, $$ p(\text{LIPS}) \propto|\mathcal M|^2 \delta(p_i -p_f) $$ The right hand side appears in the formula for a cross section (NB I don't include the Dirac function in the $d\text {LIPS} $) $$ \...


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