Newton's theory of gravity is covariant under Galilean transformations












2












$begingroup$


We know from classical mechanics that the gravitational field equation for the scalar potential takes the form $$nabla^2phi=4pi rho,$$ where $rho$ is mass density (which, can depend on time and space). Also, the associated equation of motion for point particle takes $$ddot{x}+nablaphi=0.$$ One of the basic requirement for a classical theory is that it should not depend on the inertial reference frame we are choosing. In particular, for a non-relativist theory such as the one described above, I would expect the theory to keep its form under Galilean transformations. I am, however, not sure how to do this rigorously with a general $phi$. Any ideas?










share|cite|improve this question









New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
    $endgroup$
    – Ben Crowell
    yesterday
















2












$begingroup$


We know from classical mechanics that the gravitational field equation for the scalar potential takes the form $$nabla^2phi=4pi rho,$$ where $rho$ is mass density (which, can depend on time and space). Also, the associated equation of motion for point particle takes $$ddot{x}+nablaphi=0.$$ One of the basic requirement for a classical theory is that it should not depend on the inertial reference frame we are choosing. In particular, for a non-relativist theory such as the one described above, I would expect the theory to keep its form under Galilean transformations. I am, however, not sure how to do this rigorously with a general $phi$. Any ideas?










share|cite|improve this question









New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
    $endgroup$
    – Ben Crowell
    yesterday














2












2








2


2



$begingroup$


We know from classical mechanics that the gravitational field equation for the scalar potential takes the form $$nabla^2phi=4pi rho,$$ where $rho$ is mass density (which, can depend on time and space). Also, the associated equation of motion for point particle takes $$ddot{x}+nablaphi=0.$$ One of the basic requirement for a classical theory is that it should not depend on the inertial reference frame we are choosing. In particular, for a non-relativist theory such as the one described above, I would expect the theory to keep its form under Galilean transformations. I am, however, not sure how to do this rigorously with a general $phi$. Any ideas?










share|cite|improve this question









New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




We know from classical mechanics that the gravitational field equation for the scalar potential takes the form $$nabla^2phi=4pi rho,$$ where $rho$ is mass density (which, can depend on time and space). Also, the associated equation of motion for point particle takes $$ddot{x}+nablaphi=0.$$ One of the basic requirement for a classical theory is that it should not depend on the inertial reference frame we are choosing. In particular, for a non-relativist theory such as the one described above, I would expect the theory to keep its form under Galilean transformations. I am, however, not sure how to do this rigorously with a general $phi$. Any ideas?







newtonian-mechanics newtonian-gravity inertial-frames galilean-relativity invariants






share|cite|improve this question









New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 19 hours ago









Will Ness

1033




1033






New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked yesterday









CosmologeeCosmologee

134




134




New contributor




Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
    $endgroup$
    – Ben Crowell
    yesterday


















  • $begingroup$
    The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
    $endgroup$
    – Ben Crowell
    yesterday
















$begingroup$
The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
$endgroup$
– Ben Crowell
yesterday




$begingroup$
The equation $nabla^2phi=4pirho$ is not a dynamical equation, it's more like a constraint. Cf. physics.stackexchange.com/a/20072/4552 . In your two equations, $rho$ only appears in one, so we can just take it as a definition of $rho$. Although $rho$ transforms trivially, even if it didn't, we wouldn't care; it wouldn't affect the truth-value of the equations. To make this a predictive theory, you need to couple your two equations somehow, probably by adding in an equation of continuity or something that relates motion of particles ($ddot{x}$) to changes in $rho$.
$endgroup$
– Ben Crowell
yesterday










1 Answer
1






active

oldest

votes


















8












$begingroup$

The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.



A translation looks like



$$x'=x-X\y'=y-Y\z'=z-Z$$



where $X$, $Y$, and $Z$ are constants.



A rotation looks like



$$x_i'=R_{ij}x_j$$



where $R$ is a constant rotation matrix.



A boost looks like



$$x'=x-V_xt\y'=y-V_yt\z'=z-V_zt$$



where $V_x$, $V_y$, and $V_z$ are constants.



Under any Galilean transformations, the potential $phi$ is assumed to be scalar satisying $phi’(mathbf{r}’, t)=phi(mathbf{r}, t)$. Here $mathbf{r}$ and $mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.



The same applies to the mass density $rho$.



The Laplacian operator can be shown to be a scalar with transformation $nabla’^2=nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $partial^2/partial x^2+partial^2/partial y^2+partial^2/partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.



Therefore your first equation



$$nabla^2phi=4pirho$$



has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently



$$nabla^2phi(mathbf{r},t)=4pirho(mathbf{r},t)$$



implies



$$nabla’^2phi’(mathbf{r’},t)=4pirho’(mathbf{r’},t),$$



which shows that it is form-invariant.



The second equation,



$$ddot{mathbf{r}}=-nablaphi,$$



is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.



So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.



Put another way, this equation implies



$$ddot{mathbf{r’}}=-nabla’phi’,$$



so it is form-invariant.



Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is a really clear answer! Thank you so much!
    $endgroup$
    – Cosmologee
    15 hours ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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/3.0/"u003ecc by-sa 3.0 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
});


}
});






Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f459113%2fnewtons-theory-of-gravity-is-covariant-under-galilean-transformations%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.



A translation looks like



$$x'=x-X\y'=y-Y\z'=z-Z$$



where $X$, $Y$, and $Z$ are constants.



A rotation looks like



$$x_i'=R_{ij}x_j$$



where $R$ is a constant rotation matrix.



A boost looks like



$$x'=x-V_xt\y'=y-V_yt\z'=z-V_zt$$



where $V_x$, $V_y$, and $V_z$ are constants.



Under any Galilean transformations, the potential $phi$ is assumed to be scalar satisying $phi’(mathbf{r}’, t)=phi(mathbf{r}, t)$. Here $mathbf{r}$ and $mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.



The same applies to the mass density $rho$.



The Laplacian operator can be shown to be a scalar with transformation $nabla’^2=nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $partial^2/partial x^2+partial^2/partial y^2+partial^2/partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.



Therefore your first equation



$$nabla^2phi=4pirho$$



has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently



$$nabla^2phi(mathbf{r},t)=4pirho(mathbf{r},t)$$



implies



$$nabla’^2phi’(mathbf{r’},t)=4pirho’(mathbf{r’},t),$$



which shows that it is form-invariant.



The second equation,



$$ddot{mathbf{r}}=-nablaphi,$$



is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.



So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.



Put another way, this equation implies



$$ddot{mathbf{r’}}=-nabla’phi’,$$



so it is form-invariant.



Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is a really clear answer! Thank you so much!
    $endgroup$
    – Cosmologee
    15 hours ago
















8












$begingroup$

The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.



A translation looks like



$$x'=x-X\y'=y-Y\z'=z-Z$$



where $X$, $Y$, and $Z$ are constants.



A rotation looks like



$$x_i'=R_{ij}x_j$$



where $R$ is a constant rotation matrix.



A boost looks like



$$x'=x-V_xt\y'=y-V_yt\z'=z-V_zt$$



where $V_x$, $V_y$, and $V_z$ are constants.



Under any Galilean transformations, the potential $phi$ is assumed to be scalar satisying $phi’(mathbf{r}’, t)=phi(mathbf{r}, t)$. Here $mathbf{r}$ and $mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.



The same applies to the mass density $rho$.



The Laplacian operator can be shown to be a scalar with transformation $nabla’^2=nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $partial^2/partial x^2+partial^2/partial y^2+partial^2/partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.



Therefore your first equation



$$nabla^2phi=4pirho$$



has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently



$$nabla^2phi(mathbf{r},t)=4pirho(mathbf{r},t)$$



implies



$$nabla’^2phi’(mathbf{r’},t)=4pirho’(mathbf{r’},t),$$



which shows that it is form-invariant.



The second equation,



$$ddot{mathbf{r}}=-nablaphi,$$



is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.



So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.



Put another way, this equation implies



$$ddot{mathbf{r’}}=-nabla’phi’,$$



so it is form-invariant.



Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That is a really clear answer! Thank you so much!
    $endgroup$
    – Cosmologee
    15 hours ago














8












8








8





$begingroup$

The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.



A translation looks like



$$x'=x-X\y'=y-Y\z'=z-Z$$



where $X$, $Y$, and $Z$ are constants.



A rotation looks like



$$x_i'=R_{ij}x_j$$



where $R$ is a constant rotation matrix.



A boost looks like



$$x'=x-V_xt\y'=y-V_yt\z'=z-V_zt$$



where $V_x$, $V_y$, and $V_z$ are constants.



Under any Galilean transformations, the potential $phi$ is assumed to be scalar satisying $phi’(mathbf{r}’, t)=phi(mathbf{r}, t)$. Here $mathbf{r}$ and $mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.



The same applies to the mass density $rho$.



The Laplacian operator can be shown to be a scalar with transformation $nabla’^2=nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $partial^2/partial x^2+partial^2/partial y^2+partial^2/partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.



Therefore your first equation



$$nabla^2phi=4pirho$$



has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently



$$nabla^2phi(mathbf{r},t)=4pirho(mathbf{r},t)$$



implies



$$nabla’^2phi’(mathbf{r’},t)=4pirho’(mathbf{r’},t),$$



which shows that it is form-invariant.



The second equation,



$$ddot{mathbf{r}}=-nablaphi,$$



is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.



So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.



Put another way, this equation implies



$$ddot{mathbf{r’}}=-nabla’phi’,$$



so it is form-invariant.



Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.






share|cite|improve this answer











$endgroup$



The Galilean group consists of three different types of coordinate transformations between two different inertial reference frames: translations, rotations, and boosts.



A translation looks like



$$x'=x-X\y'=y-Y\z'=z-Z$$



where $X$, $Y$, and $Z$ are constants.



A rotation looks like



$$x_i'=R_{ij}x_j$$



where $R$ is a constant rotation matrix.



A boost looks like



$$x'=x-V_xt\y'=y-V_yt\z'=z-V_zt$$



where $V_x$, $V_y$, and $V_z$ are constants.



Under any Galilean transformations, the potential $phi$ is assumed to be scalar satisying $phi’(mathbf{r}’, t)=phi(mathbf{r}, t)$. Here $mathbf{r}$ and $mathbf{r}’$ represent the same point in two different reference frames. The potential is just a single value at each point, and all observers agree on what that value is.



The same applies to the mass density $rho$.



The Laplacian operator can be shown to be a scalar with transformation $nabla’^2=nabla^2$. The easy argument is that it is the scalar product of the gradient vector operator with itself. For a more careful argument, work out what happens to $partial^2/partial x^2+partial^2/partial y^2+partial^2/partial z^2$ under translations, rotations, and Galilean boosts, using the transformation equations above.



Therefore your first equation



$$nabla^2phi=4pirho$$



has the covariant form scalar=scalar under translations, rotations, and boosts. Put differently



$$nabla^2phi(mathbf{r},t)=4pirho(mathbf{r},t)$$



implies



$$nabla’^2phi’(mathbf{r’},t)=4pirho’(mathbf{r’},t),$$



which shows that it is form-invariant.



The second equation,



$$ddot{mathbf{r}}=-nablaphi,$$



is covariant because both acceleration and the gradient operator are vectors under rotations and scalars under translations and boosts; and the potential is a scalar under all three.



So under rotations, this equation has the covariant form vector=vector, and under translations and boosts it has the covariant form scalar=scalar.



Put another way, this equation implies



$$ddot{mathbf{r’}}=-nabla’phi’,$$



so it is form-invariant.



Note: In the case of rotations, you get these same-form equations after “cancelling” the rotation matrix that the rotation introduces on both sides. Just multiply both sides by the inverse matrix to get rid of it and restore the original form.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









G. SmithG. Smith

6,7361123




6,7361123












  • $begingroup$
    That is a really clear answer! Thank you so much!
    $endgroup$
    – Cosmologee
    15 hours ago


















  • $begingroup$
    That is a really clear answer! Thank you so much!
    $endgroup$
    – Cosmologee
    15 hours ago
















$begingroup$
That is a really clear answer! Thank you so much!
$endgroup$
– Cosmologee
15 hours ago




$begingroup$
That is a really clear answer! Thank you so much!
$endgroup$
– Cosmologee
15 hours ago










Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.













Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.












Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.
















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f459113%2fnewtons-theory-of-gravity-is-covariant-under-galilean-transformations%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

"Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

Alcedinidae

RAC Tourist Trophy