Why is bremsstrahlung radiation greater than acceleration radiation?

Why is breaking radiation stronger than accelerating radiation? Why is it that when an electron comes accelerating toward a nucleus radiate weaker than when it goes decelerating away from the nucleus?

Is it because when it decelerates, at the same time it changes direction?

Or is it because it is decelerating much quicker than it accelerates because it also changes direction?

Where did you get this from? According to the Larmor formula (classical radiation formula) in non-relativistic approximation the irradiated power is
$$P=\frac{q^2 a^2}{6 \pi c^3}$$
(in Heaviside-Lorentz units), which does not depend on the direction of the acceleration of the charge.

vanhees71 said:

Where did you get this from? According to the Larmor formula (classical radiation formula) in non-relativistic approximation the irradiated power is
$$P=\frac{q^2 a^2}{6 \pi c^3}$$
(in Heaviside-Lorentz units), which does not depend on the direction of the acceleration of the charge.

Oh, okay I get it now thanks, but then my question should be, 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.

You didn't answer @vanhees71 question: Where did you get this from? We can't track down the misunderstanding without it.

Vanadium 50 said:

You didn't answer @vanhees71 question: Where did you get this from? We can't track down the misunderstanding without it.

From the book called Physics of Radiotherapy, Khan Page 71 :

Again, because of its small mass, an electron may interact with the electromagnetic field of a nucleus and be decelerated so rapidly that a part of its energy is lost as bremsstrahlung. The rate of energy loss as a result of bremsstrahlung increases with the increase in the energy of the electron and the atomic number of the medium.

Literally all sources tells that bremsstruhlung is a deceleration radiaition.

Wikipedia says the same https://en.wikipedia.org/wiki/Bremsstrahlung :

"In particle physics, from German bremsen 'to brake', and Strahlung 'radiation') is electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. The moving particle loses kinetic energy, which is converted into radiation (i.e., photons), thus satisfying the law of conservation of energy. The term is also used to refer to the process of producing the radiation. Bremsstrahlung has a continuous spectrum, which becomes more intense and whose peak intensity shifts toward higher frequencies as the change of the energy of the decelerated particles increases."

" Broadly speaking, bremsstrahlung or braking radiation is any radiation produced due to the deceleration (negative acceleration) of a charged particle, which includes synchrotron radiation (i.e., photon emission by a relativistic particle), cyclotron radiation (i.e. photon emission by a non-relativistic particle), and the emission of electrons and positrons during beta decay."

@cemtu It’s the magnitude of the acceleration that matters, and that’s very high during the braking, not so high at other times. So all the sources that you’re quoting are (understandably but confusingly) limiting their discussion to the braking case.

Wikipedia’s description is a bit clearer (emphasis mine):

Broadly speaking, bremsstrahlung or braking radiation is any radiation produced due to the acceleration (positive or negative) of a charged particle, which includes synchrotron radiation (i.e., photon emission by a relativistic particle), cyclotron radiation (i.e. photon emission by a non-relativistic particle), and the emission of electrons and positrons during beta decay. However, the term is frequently used in the more narrow sense of radiation from electrons (from whatever source) slowing in matter.

Nugatory said:

@cemtu It’s the magnitude of the acceleration that matters, and that’s very high during the braking, not so high at other times. So all the sources that you’re quoting are (understandably but confusingly) limiting their discussion to the braking case.

Wikipedia’s description is a bit clearer (emphasis mine):

Shouldn't it be high when accelerating too if it is high when breaking as the law of conservation of energy dictates?

cemtu said:

Shouldn't it be high when accelerating too if it is high when breaking as the law of conservation of energy dictates?

Yes, and that’s what that quote from Wikipedia is saying.
It is the magnitude of the acceleration, whether positive (“accelerating”) or negative (“decelerating”) that matters - that’s an ##a^2## in the power formula so the sign cancels out. Your quotes are all in the context of situations in which the positive acceleration is much smaller than the negative acceleration.

You are “broadly speaking” in the terms of that Wikipedia quote, your sources are using the term “in the more narrow sense”.

cemtu said:

Shouldn't it be high when accelerating too if it is high when breaking as the law of conservation of energy dictates?

Is there evidence of situations where this occurs? If not then it explains why the term is used still.

It's not the only situation in which braking produces more acceleration than a 'power source' can. On a suitable bicycle (where the design prevents going over the handlebars), anyone can brake at over half g but can pedalling (alone) produce anything like that?

sophiecentaur said:

Is there evidence of situations where this occurs? If not then it explains why the term is used still.

There should be.

sophiecentaur said:

It's not the only situation in which braking produces more acceleration than a 'power source' can. On a suitable bicycle (where the design prevents going over the handlebars), anyone can brake at over half g but can pedalling (alone) produce anything like that?

it is the same nucleus that causes the acceleration and the same nucleus that causes the deceleration.

cemtu said:

There should be.

it is the same nucleus that causes the acceleration and the same nucleus that causes the deceleration.
View attachment 330811

If the system radiates energy then the KE of the incident electron must decrease because it's the only source of energy for the bremsstrahlung . I don't know of a way that a nucleus can be the source of the energy but I could be wrong.

sophiecentaur said:

If the system radiates energy then the KE of the incident electron must decrease because it's the only source of energy for the bremsstrahlung . I don't know of a way that a nucleus can be the source of the energy but I could be wrong.

acceleration of electron caused by nucleus so it is the source of attractive force thus energy thus Bremsstruhlung.

cemtu said:

acceleration of electron caused by nucleus so it is the source of attractive force thus energy thus Bremsstruhlung.

Treating this a a simple bit of dynamics, the massive nucleus can't lose significant KE because it's going so slowly (more or less stationary). It's the very fast electron that has the energy available for producing em emission. This is like a bullet hitting a block of lead; the energy of the lead doesn't change significantly but the bullet loses all its KE when it penetrates the lead. OR, if it bounces off (elastic) then it will keep most of its KE because momentum is conserved and KE is all about v².

This link is nice and chatty and it seems ok to me. The notion of the nucleus 'having energy' makes no sense; it's stuck in the experimental equipment . The energy of the em will be the difference between original and final KE of the electron so the electron has to slow down.
EDIT: If you choose to change the frame of reference to be that of the electron then you still get the same answer but a sore head.

PS The 'cause' of the attraction is the interaction of charges on both particles. The potential energy is a mutual effect and the relative masses means that the electron is the one that gets the KE change.

For acceleration parallel to the velocity,
$$P_{\parallel}=\frac{2}{3}\frac{q^2}{m^2}\left(\frac{\bf dp}{dt}\right)^2$$.
For acceleration perpendicular to the velocity,
$$P_{\perp}=\frac{2}{3}\frac{q^2\gamma^2}{m^2}\left(\frac{\bf dp}{dt}\right)^2$$.
This power is a factor ##\gamma^2## larger than the power radiated for
##\bf a## parallel to ##\bf v##,

In covariant form the fully relativistic formula reads (in Heaviside Lorentz units)
$$P=-\frac{q^2}{6 \pi m^2 c^3} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} p_{\mu}}{\mathrm{d} \tau},$$
where ##\tau## is the proper time of the accelerated particle.