mocatgesex

2022-10-08

When a free particle move in space with a known momentum and energy then what is the physical process that gives mass to that free (relativistic) particle?

What is role does the Higgs field in that process, if any?

What is role does the Higgs field in that process, if any?

Brendan Bradley

Beginner2022-10-09Added 11 answers

It should be stated that the mass of any actual thing that you've encountered in life has almost nothing to do with the Higgs. Relativity says that energy and mass are equivalent. This means that if you clump massless particles together with $E$ potential energy, and then put those particles into a box, it will appear that that box has mass $E/{c}^{2}$.

Neat, but who cares? Well, it turns out that the strong interaction does exactly this, clumping quarks together in boxes we call protons and neutrons. Now, quarks DO get a mass from the Higgs mechanism, but this mass is much, much smaller (roughly $1/1000$) than their "relativistic mass." So, you would say that you and me, or a rock or a baseball has mass not due to the Higgs mechanism, but rather because all of these things are made mostly of bound quarks.

Neat, but who cares? Well, it turns out that the strong interaction does exactly this, clumping quarks together in boxes we call protons and neutrons. Now, quarks DO get a mass from the Higgs mechanism, but this mass is much, much smaller (roughly $1/1000$) than their "relativistic mass." So, you would say that you and me, or a rock or a baseball has mass not due to the Higgs mechanism, but rather because all of these things are made mostly of bound quarks.

ter3k4w8x

Beginner2022-10-10Added 4 answers

I'll focus on the Higgs aspect of this. There is, potentially, a separate issue to your question, in that you may be thinking about "relativistic mass", the energy an object gains as it speeds up. Modern physicists don't use this concept and I'll explain why underneath.

Ignore the all-too-common answers that tell you the Higgs field slows down particles, as if it were a sea of molasses. This suggests that the Higgs field imparts mass by exerting a drag effect on them, and is simply incorrect.

So how does it work? First, consider what is meant by "mass". Whilst you may be used to thinking of mass as a property that determines how much something weighs, or how hard it is to move (it is this latter aspect that the $\mathit{\text{molasses}}$ analogy appeals to), Einstein taught us another way to think of it (the way modern physicists use the word): as "rest energy".

Energy is a property of a particle that can come either from moving ("kinetic energy") or from its position in a force field ("potential energy"). Einstein's special relativity revealed that some particles also have another form; an intrinsic amount of energy that is always present, regarldess of where the particle is or how fast it is moving. It is this $\mathit{\text{intrinsic}}$ or $\mathit{\text{rest}}$ energy that we call mass. (Hence "relativistic mass" is simply the energy gained by speeding up; it is, in other words, just kinetic energy. The word and concept "mass" is best reserved for "rest mass").

We could, if we like, declare that certain particles are born with intrinsic rest energy and leave it there, but for technical reasons this is incompatible with what we know about particle physics, so we are forced to conclude that, intrinsically, all particles are massless, and this rest energy must be given to some of them by something called the Higgs field:

Consider an electric or magnetic field, which emanates from a charged object. Other particles which have an electric charge (or are magnetised) will gain potential energy when they are placed in those respective fields (the amount of energy will depend both on the charge of the particle and the strength of the field). The Higgs field is like an electric field, with one crucial difference: whereas an electric field diminishes in strength the further you get from its source, the Higgs field is constant (and non-zero) throughout space and time. Thus a particle with "Higgs charge" will have the same, constant amount of "Higgs potential energy" wherever it is. This potential energy is, in other words, intrinsic energy, or mass. Larger Higgs-charge leads to larger intrinsic energy = larger mass.

There is, of course, much more to be said on the subject, but that's the essence of how the Higgs field works.

Ignore the all-too-common answers that tell you the Higgs field slows down particles, as if it were a sea of molasses. This suggests that the Higgs field imparts mass by exerting a drag effect on them, and is simply incorrect.

So how does it work? First, consider what is meant by "mass". Whilst you may be used to thinking of mass as a property that determines how much something weighs, or how hard it is to move (it is this latter aspect that the $\mathit{\text{molasses}}$ analogy appeals to), Einstein taught us another way to think of it (the way modern physicists use the word): as "rest energy".

Energy is a property of a particle that can come either from moving ("kinetic energy") or from its position in a force field ("potential energy"). Einstein's special relativity revealed that some particles also have another form; an intrinsic amount of energy that is always present, regarldess of where the particle is or how fast it is moving. It is this $\mathit{\text{intrinsic}}$ or $\mathit{\text{rest}}$ energy that we call mass. (Hence "relativistic mass" is simply the energy gained by speeding up; it is, in other words, just kinetic energy. The word and concept "mass" is best reserved for "rest mass").

We could, if we like, declare that certain particles are born with intrinsic rest energy and leave it there, but for technical reasons this is incompatible with what we know about particle physics, so we are forced to conclude that, intrinsically, all particles are massless, and this rest energy must be given to some of them by something called the Higgs field:

Consider an electric or magnetic field, which emanates from a charged object. Other particles which have an electric charge (or are magnetised) will gain potential energy when they are placed in those respective fields (the amount of energy will depend both on the charge of the particle and the strength of the field). The Higgs field is like an electric field, with one crucial difference: whereas an electric field diminishes in strength the further you get from its source, the Higgs field is constant (and non-zero) throughout space and time. Thus a particle with "Higgs charge" will have the same, constant amount of "Higgs potential energy" wherever it is. This potential energy is, in other words, intrinsic energy, or mass. Larger Higgs-charge leads to larger intrinsic energy = larger mass.

There is, of course, much more to be said on the subject, but that's the essence of how the Higgs field works.

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Screen of the pinhole camera is invertedIn order to calculate the cross-section of an interaction process the following formula is often used for first approximations:

$\sigma =\frac{2\pi}{\hslash \phantom{\rule{thinmathspace}{0ex}}{v}_{i}}{\left|{M}_{fi}\right|}^{2}\varrho \left({E}_{f}\right)\phantom{\rule{thinmathspace}{0ex}}V$

${M}_{fi}=\u27e8{\psi}_{f}|{H}_{int}|{\psi}_{i}\u27e9$

${M}_{fi}=\u27e8{\psi}_{f}|{H}_{int}|{\psi}_{i}\u27e9$

$\varrho \left({E}_{f}\right)=\frac{\mathrm{d}n\left({E}_{f}\right)}{\mathrm{d}{E}_{f}}=\frac{4\pi {{p}_{f}}^{2}}{{\left(2\pi \hslash \right)}^{3}}\frac{V}{{v}_{f}}$

The derivation of this equation in the context of the non relativistic Schrödinger equation. Use this formula in the relativistic limit: ${v}_{i},{v}_{f}\to c\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{1em}{0ex}}{p}_{f}\approx {E}_{f}/c$

Very often books simply use this equation with matrix element derived from some relativistic theory, e.g. coupling factors and propagators from the Dirac equation or Electroweak interaction. How is this justified?Why does the azimuthal angle, $\varphi $, remain unchanged between reference frames in special relativity?

I think this comes from the aberration formula, showing dependence only on the polar angle, $\theta $.

The aberration formula is:

$\mathrm{tan}\theta =\frac{{u}_{\perp}}{{u}_{||}}=\frac{{u}^{\prime}\mathrm{sin}{\theta}^{\prime}}{\gamma ({u}^{\prime}\mathrm{cos}{\theta}^{\prime}+v)}$

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Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,

$\mathrm{\Psi}({x}_{1},{x}_{2})=-\mathrm{\Psi}({x}_{2},{x}_{1})$

If we assume that we can factorize the wave function in terms of single particle wave functions we can write

$\mathrm{\Psi}({x}_{1},{x}_{2})={\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})-{\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})$

which fulfills the anti-symmetry requirement. The plane wave single particle states are given by,

${\psi}_{\mathbf{k},{m}_{s}}(x)={u}_{\mathbf{k},{m}_{s}}(s)\varphi (\mathbf{k}\cdot \mathbf{r})$

So expect the total wavefunction to be

$\begin{array}{rl}\mathrm{\Psi}({x}_{1},{x}_{2})& ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\\ & ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\end{array}$

However

$u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})$

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How to calculate $ds$?