Newton's theory of gravity is covariant under Galilean transformations
$begingroup$
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
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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?
newtonian-mechanics newtonian-gravity inertial-frames galilean-relativity invariants
edited 2 hours ago
G. Smith
6,6621023
asked 3 hours ago
CosmologeeCosmologee
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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 $rho$.
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– Ben Crowell
40 mins ago
add a comment |
$begingroup$
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
edited 2 hours ago
G. Smith
6,6621023
asked 3 hours ago
CosmologeeCosmologee
63
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
edited 2 hours ago
G. Smith
6,6621023
asked 3 hours ago
CosmologeeCosmologee
63
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.
edited 2 hours ago
G. Smith
6,6621023
asked 3 hours ago
CosmologeeCosmologee
63
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Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
G. Smith
6,6621023
edited 2 hours ago
G. Smith
6,6621023
edited 2 hours ago
G. Smith
6,6621023
6,6621023
asked 3 hours ago
CosmologeeCosmologee
63
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Cosmologee is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 3 hours ago
CosmologeeCosmologee
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asked 3 hours ago
CosmologeeCosmologee
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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
40 mins ago
add a comment |
$begingroup$
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
40 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
40 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
40 mins ago
add a comment |
1 Answer
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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.
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.
answered 1 hour ago
G. SmithG. Smith
6,6621023
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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.
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.
answered 1 hour ago
G. SmithG. Smith
6,6621023
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add a comment |
$begingroup$
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.
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.
answered 1 hour ago
G. SmithG. Smith
6,6621023
$endgroup$
add a comment |
$begingroup$
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.
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.
answered 1 hour ago
G. SmithG. Smith
6,6621023
$endgroup$
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.
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.
answered 1 hour ago
G. SmithG. Smith
6,6621023
edited 3 mins ago
answered 1 hour ago
G. SmithG. Smith
6,6621023
answered 1 hour ago
G. SmithG. Smith
6,6621023
answered 1 hour ago
G. SmithG. Smith
6,6621023
6,6621023
add a comment |
add a comment |
Cosmologee is a new contributor. Be nice, and check out our Code of Conduct.
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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
40 mins ago