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Why only take into account deceleration radiation rather than the radiation caused by acceleration when going tangent towards the nucleus and acceleration caused by the change in direction when flyby happens?

In order for an electron to radiate Electromagnetic (EM) radiation, it needs to accelerate. Remember, acceleration means its velocity needs to change, which can be achieved either by changing the electron’s speed or its direction. Both types of change will produce an acceleration, and hence produce EM radiation.

Bremsstrahlung, electromagnetic radiation produced by a sudden slowing down or deflection of charged particles (especially electrons) passing through matter in the vicinity of the strong electric fields of atomic nuclei. Bremsstrahlung, for example, accounts for continuous X-ray spectra—i.e., that component of X-rays the energy of which covers a whole range from a maximum value downward through lower values.

Bremsstrahlung interactions account for ~100% of the X-ray beam.

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    $\begingroup$ Why do you think that deceleration radiation is stronger than acceleration radiation? The radiated power depends on the square of the acceleration, so is independent of the sign of the acceleration. "Bremsstrahlung" doesn't refer to radiation produced when a particle is decelerating away from a nucleus, but to all the radiation produced by a charged particle as it slows down in matter. $\endgroup$ Commented Aug 19, 2023 at 5:34
  • $\begingroup$ Can someone please get a medical physicist here? $\endgroup$ Commented Aug 19, 2023 at 15:17
  • $\begingroup$ @DavidBailey thanks, but I changed the question though. $\endgroup$ Commented Aug 19, 2023 at 19:28

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Are you asking why the x-rays in a clinical x-ray machine come from bremsstrahlung? It's because the acceleration when the electrons leave the cathode is relatively small, while the acceleration when they get close to a nucleus is large. Bremsstrahlung is just the name for the phenomenon of large acceleration near a nucleus in dense matter causing random velocity changes until the multiply scattered electron is stopped. The net effect is an approximately forward beam of x-rays.

There are x-ray machines that generate all of their output from acceleration induced by alternating magnetic fields - look up x-ray synchrotron. They're too big and complicated for use in clinical settings.

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  • $\begingroup$ Thank you, but bremsstrahlung is specifically used for a radiation production process that is generated by the deflection of electrons from the nucleus and their subsequent deceleration away from it. The source of radiation is deceleration. $\endgroup$ Commented Aug 19, 2023 at 19:27
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    $\begingroup$ @medicalphysics, Is your question why "deceleration" and not "acceleration"? I think in most contexts the word "acceleration" is preferred since it just means "change in velocity" and "deceleration" refers to a specific reference frame (the lab) and the dynamics of the electron beam in the target. $\endgroup$ Commented Aug 21, 2023 at 3:54
  • $\begingroup$ yes you understand my question correct. $\endgroup$ Commented Aug 21, 2023 at 8:06
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Bremsstrahlung does not just happen when an electron is deaccelerating away from a nucleus. The rigorous quantum electrodynamics (QED) lowest order calculation of the bremsstrahlung rate includes both of the following Feynman diagrams (borrowed from @anna-v's answer to "Quantum mechanics prediction for Bremsstrahlung?"):

Feynman diagram of bremsstrahlung of an electron

Radiation as the electron "approaches" and "leaves" the nucleus both contribute. This is not obvious from low energy bremsstrahlung angular distributions, but can easily be seen if we look at high energy ($E_e >> m_e$) scattering. For example, here is the calculated bremmstrahlung photon angle distribution for 4-momentum transfer squared $Q^2=7\,\mathrm{GeV^2}$, initial and final electron energies of $5.12$ and $1.42$ MeV, and an electron scattering angle of $57^\circ$.

Calculated bremmstrahlung angular distribtion

At high energies, bremsstrahlung is concentrated in the direction the electron is moving due to "relativistic searchlight effect", so the angle of the radiation tells us about the direction of the electron when the photon was emitted. The peak at $0^\circ$ is due to photons emitted by the electron accelerating towards the proton before it was deflected; the peak at $57^\circ$ is due to photons emitted after the electron was deflected and was decelerating.

The distribution is actually zero precisely at $0^\circ$ and $57^\circ$, but the dip is so narrow it can't be seen in this plot. Strictly speaking, we should also include radiation from the proton, but this is negligible. For more details, see "Radiative corrections for (e,e'p) reactions at GeV energies".

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  • $\begingroup$ I did not understand your graph can you explain it further, please? Why are there peaks at 0 and 60 degrees? Why there have to be peaks? And why are they relevant? Furthermore, why is this data so different from than X-Ray intensity VS Photon Energy graph? $\endgroup$ Commented Aug 23, 2023 at 5:38
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    $\begingroup$ @medical-physics There is no reason for this plot of intensity vs emission angle to be similar to an intensity vs photon energy graph. I have edited my answer to clarify that the two peaks demonstrate that bremsstrahlung is emitted both while the electron is accelerating towards the proton and decelerating away. The peaks are relevant because they visually demonstrate that the premise that we "only take into account deceleration radiation rather than the radiation caused by acceleration when going tangent towards the nucleus" is false. If that were true, there would not be a peak at 0°. $\endgroup$ Commented Aug 23, 2023 at 17:35

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