How is the speed of nucleons in the nucleus measured?What is an intuitive picture of the motion of nucleons?How do we know that the nucleus isn't a quark-gluon plasma?What do we know about the interactions between the protons and neutrons in a nucleus?binding energy of a nucleus is positive?Oscillation of nucleons mediated by polarizing electron charge on opposite ends of the nucleus?Are nucleons discrete within a nucleus?Properties of the nucleons and the nucleus from the interactions between quarks and gluonsRest mass of the nucleus and its constituent nucleonsIs the arrangement of nucleons within a nucleus identical between heavy elements of the same type?formation of atomic nuclei from nucleons
Advices to added homemade symbols
How to explain the strange behaviour about Function Nothing
Quote to show students don't have to fear making mistakes
Can/should you swim in zero G?
Was there an autocomplete utility in MS-DOS?
Generating Roman numerals with dice
As an interviewer, how to conduct interviews with candidates you already know will be rejected?
How can I find places to store/land a private airplane?
Why is my vegetable stock bitter, but the chicken stock not?
Why do we not always use the closed testing principle for multiple comparisons?
Does the US Armed Forces refuse to recruit anyone with an IQ less than 83?
Does test run is also count as a run in MS Flow?
Is right click on tables bad UX
What's the correct way to determine turn order in this situation?
Could the Queen overturn the UK Supreme Court ruling regarding prorogation of Parliament?
What is the difference between increasing volume and increasing gain?
The answer is the same (tricky puzzle!)
What does a textbook look like while you are writing it?
Did the Soviet army intentionally send troops (e.g. penal battalions) running over minefields?
What’s the BrE for “shotgun wedding”?
Was the whistleblower in the Ukraine scandal legally required to make his report?
Could Boris Johnson face criminal charges for illegally proroguing Parliament?
Booting Ubuntu from USB drive on MSI motherboard -- EVERYTHING fails
Airport Security - advanced check, 4th amendment breach
How is the speed of nucleons in the nucleus measured?
What is an intuitive picture of the motion of nucleons?How do we know that the nucleus isn't a quark-gluon plasma?What do we know about the interactions between the protons and neutrons in a nucleus?binding energy of a nucleus is positive?Oscillation of nucleons mediated by polarizing electron charge on opposite ends of the nucleus?Are nucleons discrete within a nucleus?Properties of the nucleons and the nucleus from the interactions between quarks and gluonsRest mass of the nucleus and its constituent nucleonsIs the arrangement of nucleons within a nucleus identical between heavy elements of the same type?formation of atomic nuclei from nucleons
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
$begingroup$
How do you measure or infer the speed of nucleons in the nucleus?
(For Aaron Stevens : link.)
experimental-physics nuclear-physics measurements neutrons protons
$endgroup$
add a comment
|
$begingroup$
How do you measure or infer the speed of nucleons in the nucleus?
(For Aaron Stevens : link.)
experimental-physics nuclear-physics measurements neutrons protons
$endgroup$
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago
add a comment
|
$begingroup$
How do you measure or infer the speed of nucleons in the nucleus?
(For Aaron Stevens : link.)
experimental-physics nuclear-physics measurements neutrons protons
$endgroup$
How do you measure or infer the speed of nucleons in the nucleus?
(For Aaron Stevens : link.)
experimental-physics nuclear-physics measurements neutrons protons
experimental-physics nuclear-physics measurements neutrons protons
edited 2 hours ago
dmckee♦
77.2k6 gold badges140 silver badges281 bronze badges
77.2k6 gold badges140 silver badges281 bronze badges
asked 8 hours ago
John OJohn O
2281 silver badge9 bronze badges
2281 silver badge9 bronze badges
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago
add a comment
|
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.
These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).
The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.
It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.
Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.
This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.
In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).
In the left-hand panel we plot the binding energy of the struck proton.
We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.
$endgroup$
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
add a comment
|
$begingroup$
There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.
$endgroup$
add a comment
|
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "151"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/4.0/"u003ecc by-sa 4.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f505045%2fhow-is-the-speed-of-nucleons-in-the-nucleus-measured%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.
These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).
The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.
It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.
Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.
This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.
In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).
In the left-hand panel we plot the binding energy of the struck proton.
We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.
$endgroup$
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
add a comment
|
$begingroup$
If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.
These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).
The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.
It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.
Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.
This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.
In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).
In the left-hand panel we plot the binding energy of the struck proton.
We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.
$endgroup$
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
add a comment
|
$begingroup$
If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.
These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).
The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.
It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.
Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.
This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.
In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).
In the left-hand panel we plot the binding energy of the struck proton.
We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.
$endgroup$
If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $mathrmMeV$ to a few $mathrmGeV$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.
These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).
The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.
It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.
Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.
This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 ,mathrmGeV$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 ,mathrmGeV^2$ on a carbon target.
In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $mathrmGeV/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 ,mathrmGeV/c^2$).
In the left-hand panel we plot the binding energy of the struck proton.
We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.
edited 5 hours ago
answered 7 hours ago
dmckee♦dmckee
77.2k6 gold badges140 silver badges281 bronze badges
77.2k6 gold badges140 silver badges281 bronze badges
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
add a comment
|
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Yeah. And thanks for the blue pencil work. I read what I meant instead of what I wrote for hours or days.
$endgroup$
– dmckee♦
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
Not a problem. Decades of coding has taught me to never underestimate the elusiveness of a determined typo. ;)
$endgroup$
– PM 2Ring
7 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
$begingroup$
This is a very nice answer, I learnt something today! +1
$endgroup$
– Gert
2 hours ago
add a comment
|
$begingroup$
There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.
$endgroup$
add a comment
|
$begingroup$
There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.
$endgroup$
add a comment
|
$begingroup$
There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.
$endgroup$
There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/capprox 0.1$ in this example.
answered 4 hours ago
Ben CrowellBen Crowell
61.2k6 gold badges179 silver badges343 bronze badges
61.2k6 gold badges179 silver badges343 bronze badges
add a comment
|
add a comment
|
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f505045%2fhow-is-the-speed-of-nucleons-in-the-nucleus-measured%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Do you have an example/source of this? More detail is needed.
$endgroup$
– Aaron Stevens
7 hours ago
$begingroup$
Closely related: physics.stackexchange.com/questions/35724/… where I posted some of the same figures.
$endgroup$
– dmckee♦
7 hours ago