Newton's theory of gravity is covariant under Galilean transformations












1












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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 EOM 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?






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  • $begingroup$
    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
    28 mins ago
















1












$begingroup$


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 EOM 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.







$endgroup$












  • $begingroup$
    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
    28 mins ago














1












1








1


1



$begingroup$


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 EOM 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.







$endgroup$




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 EOM 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




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






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edited 1 hour ago









G. Smith

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  • $begingroup$
    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
    28 mins ago


















  • $begingroup$
    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
    28 mins ago
















$begingroup$
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
28 mins ago




$begingroup$
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
28 mins ago










1 Answer
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$begingroup$

Under Galilean transformations, the potential $phi$ is a 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.) So is 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.



(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.)



Therefore this equation 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.






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    $begingroup$

    Under Galilean transformations, the potential $phi$ is a 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.) So is 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.



    (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.)



    Therefore this equation 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.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Under Galilean transformations, the potential $phi$ is a 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.) So is 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.



      (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.)



      Therefore this equation 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.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Under Galilean transformations, the potential $phi$ is a 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.) So is 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.



        (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.)



        Therefore this equation 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.






        share|cite|improve this answer











        $endgroup$



        Under Galilean transformations, the potential $phi$ is a 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.) So is 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.



        (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.)



        Therefore this equation 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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 10 mins ago

























        answered 1 hour ago









        G. SmithG. Smith

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        6,6521023






















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