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 1 hour ago
G. Smith
6,6521023
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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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
26 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 1 hour ago
G. Smith
6,6521023
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 1 hour ago
G. Smith
6,6521023
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.
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edited 1 hour ago
G. Smith
6,6521023
asked 3 hours ago
CosmologeeCosmologee
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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 1 hour ago
G. Smith
6,6521023
edited 1 hour ago
G. Smith
6,6521023
edited 1 hour ago
G. Smith
6,6521023
6,6521023
asked 3 hours ago
CosmologeeCosmologee
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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
26 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
26 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
26 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
26 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.
answered 1 hour ago
G. SmithG. Smith
6,6521023
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add a comment |
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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.
answered 1 hour ago
G. SmithG. Smith
6,6521023
$endgroup$
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.
answered 1 hour ago
G. SmithG. Smith
6,6521023
$endgroup$
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.
answered 1 hour ago
G. SmithG. Smith
6,6521023
$endgroup$
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.
answered 1 hour ago
G. SmithG. Smith
6,6521023
edited 7 mins ago
answered 1 hour ago
G. SmithG. Smith
6,6521023
answered 1 hour ago
G. SmithG. Smith
6,6521023
answered 1 hour ago
G. SmithG. Smith
6,6521023
6,6521023
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
26 mins ago