2
$\begingroup$

Since Newtonian gravity is also indistinguishable from acceleration, it should be possible to formulate it as a curvature in space, right? For example, if a body changes velocities purely under the influence of only gravity, we can define the new and the old velocities to be parallel transports of each other. So Newtonian gravity become a curvature in space. Do you know of any such formulation?

$\endgroup$
4
  • 3
    $\begingroup$ I think Newton-Cartan theory is what you're looking for. $\endgroup$ Commented Apr 1, 2022 at 2:27
  • $\begingroup$ @MaximalIdeal This makes me wonder...is the curvature formulation just for convenience? A body under gravity does change its velocity just like with any other force, but we simply define the new and old velocities to be equal. $\endgroup$
    – Ryder Rude
    Commented Apr 1, 2022 at 2:38
  • $\begingroup$ If you're talking about the necessity of spacetime curvature in GR, then I don't know. It plays a much deeper role in GR. $\endgroup$ Commented Apr 1, 2022 at 2:50
  • 1
    $\begingroup$ See this Lecture 9: Newtonian spacetime is curved! (International Winter School on Gravity and Light 2015): youtube.com/watch?v=IBlCu1zgD4Y $\endgroup$
    – asmaier
    Commented Feb 10, 2023 at 21:03

2 Answers 2

6
$\begingroup$

Yes, this is Newton-Cartan gravity. It is a little different from General Relativity. It has a pair of degenerate metrics on spacetime, and the connection is a little more difficult to define. But it shares many of the same basic features of GR. Gravity is locally an inertial force. Inertial objects travel on geodesics. The proper acceleration of an inertial object is 0. Tidal gravity is represented by curved spacetime. Etc.

$\endgroup$
2
  • $\begingroup$ "Gravity is locally an inertial force" Could you explain? $\endgroup$
    – Babu
    Commented Jul 6, 2022 at 5:17
  • $\begingroup$ This is also known as the equivalence principle. en.m.wikipedia.org/wiki/Equivalence_principle In a local free-falling frame there is no gravity. That is why astronauts float around the space station even though g is only about 10% weaker at that altitude. $\endgroup$
    – Dale
    Commented Jul 6, 2022 at 11:40
0
$\begingroup$

No. The curvature in space is zero, and is actually all curvature in time. This shows up as a curvature of worldlines, when time is displayed as a dimension. In turn, this curvature - in temporal language - is what spontaneous acceleration is and looks like. That is what you see as the free fall of gravity.

More explicitly, Newtonian gravity can be written as a curved chrono-geometry given by a line-element and constraint in the following way: $$dx^2 + dy^2 + dz^2 + 2 dt du - 2 V dt^2 = 0,$$ where $$V = -\sum_{0≤n<N} \frac{G m_n}{\sqrt{(x - x_n)^2 + (y - y_n)^2 + (z - z_n)^2}},$$ is the gravitational potential per unit mass for $N$ bodies with respective masses $(m_n: 0≤n<N)$ and positions $((x_n,y_n,z_n): 0≤n<N)$ that may, themselves, be functions of time.

In addition, the geometry imposes the condition that the proper time, denoted by $s$ and given by $s = t$ in non-relativistic physics, be an invariant with respect to changes in frames of reference, reorientations of the axes and spatial translations; and that the coordinate differential $ds = dt$ be invariant with respect to both these and time translations. A key distinguishing feature of non-relativistic physics is that the proper time $s$ coincides with coordinate time $t$.

You can think of this as the Steam-Punk version of General Relativity. In homage to this and to That Theme Song that was always used long ago by Carl Sagan in Cosmos, I did the following mash-up:

An Astronomer's Great Library, but with Alpha
https://www.youtube.com/watch?v=HKN-6djM8uw

As you play it alongside in a separate window (hint, hint) think of it as the venue where the developments about to be described are being laid out.

Orbital Motion - Geodesic Equations
One of two ways in which the equations of motion for a test body of, say, mass $m$ moving under the influence of the gravity given by this geometry can be derived is as the equations for the geometry's geodesics. One can directly write down the geodesic law that arises from the metric given by the line element.

As a footnote: in here and the following, I will adopt Einstein's summation convention, where repeated indices in each monomial are to be summed over.

The most convenient way to write out the geodesic equations, for a test body of mass $m$, for a given line element $g_{μν} dx^μ dx^ν$ where $g_{μν} = g_{νμ}$ are the components of the metric, is to first write out the components of its momentum $$p_μ = m g_{μν} \frac{dx^ν}{ds}.$$ The trajectory is taken with respect to the proper time $s$. One can then define a generalized potential, which I will denote as $W$, as: $$W\left(x, \frac{dx}{ds}\right) = \frac{m}{2} g_{μν}(x) \frac{dx^μ}{ds} \frac{dx^ν}{ds}.$$ The geodesic equations can then be written as: $$\frac{dp_μ}{ds} = \frac{∂W}{∂x^μ}.$$

Applying this to the line element just given, one finds: $$ 𝐩 ≡ \left(p_x, p_y, p_z\right) = m 𝐯, \hspace 1em 𝐯 ≡ \left(\frac{dx}{ds}, \frac{dy}{ds}, \frac{dz}{ds}\right), \\ H ≡ -p_t = m \left(2 V \frac{dt}{ds} - w\right), \hspace 1em μ ≡ p_u = m \frac{dt}{ds}, \hspace 1em w ≡ \frac{du}{ds},$$ subject to the conditions imposed by the geometry, itself: $$\frac{dt}{ds} = 1 ⇔ ds = dt, \hspace 1em |𝐯|^2 + 2 w - 2 V = 0. $$

Upon substitution into the equations for $H$, $μ$ and $w$, we find that: $$w = V - \frac{|𝐯|^2}{2}, \hspace 1em H = m \left(V + \frac{|𝐯|^2}{2}\right), \hspace 1em μ = m.$$ The coordinate $u$ is conjugate to the mass $m$, in the same way that $(x, y, z)$ is conjugate to the momentum $𝐩$ or $t$ to the negative $-H$ of the energy, which is denoted here as $H$.

For the function $W$, we have $$\begin{align} W\left(x, \frac{dx}{ds}\right) & = \frac{m}{2} \left(\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 + \left(\frac{dz}{ds}\right)^2 + 2 \frac{dt}{ds} \frac{du}{ds} - 2 V(x) \left(\frac{dt}{ds}\right)^2\right) \\ & = \frac{m}{2} \left(|𝐯|^2 + 2 w - 2 V(x)\right). \end{align}.$$ The respective gradients are: $$∇W = \left(\frac{∂W}{∂x}, \frac{∂W}{∂y}, \frac{∂W}{∂z}\right) = -m ∇V, \hspace 1em \frac{∂W}{∂t} = 0, \hspace 1em \frac{∂W}{∂u} = 0, $$ from which we derive the following as the geodesic equations, after setting $ds = dt$: $$\frac{d𝐩}{dt} = -m ∇V, \hspace 1em \frac{dH}{dt} = 0, \hspace 1em \frac{dm}{dt} = 0.$$

Orbital Motion - Constrained Action Principle
Alternatively, we can develop the equations of motion in terms of an action integral $$S = \int L ds,$$ with the a Lagrangian $L$ that directly embodies the two geometric invariants: $$L = \frac{m}{2}\left(\left(\frac{dx}{ds}\right)^2 + \left(\frac{dy}{ds}\right)^2 + \left(\frac{dz}{ds}\right)^2 + 2 \frac{dt}{ds} \frac{du}{ds} - 2 V \left(\frac{dt}{ds}\right)^2\right) + U \left(1 - \frac{dt}{ds}\right),$$ where $m$ and $U$ are Lagrange multipliers.

The equations of motion are then expressed in terms of the components $$p_μ = \frac{∂L}{∂(dx^μ/ds)},$$ of the canonical momentum $p$, by: $$\frac{dp_μ}{ds} = \frac{∂L}{∂x^μ}.$$ This actually also applies to the Lagrange multipliers, which are effectively treated as extra coordinates whose canonical-momentum components are all zero and so that the equations of motion yield the result that the constraints all be zero.

Using the same conventions as before: $$𝐩 = (p_x, p_y, p_z), \hspace 1em H = -p_t, \hspace 1em μ = p_u,$$ what we find is that: $$𝐩 = m 𝐯, \hspace 1em H = m\left(2V \frac{dt}{ds} - w\right) + U, \hspace 1em μ = m.$$

The constraints are, as before, $$\frac{dt}{ds} = 1 ⇔ ds = dt, \hspace 1em |𝐯|^2 + 2 w - 2 V = 0.$$ The equations of motion are, upon substitution of $ds$ by $dt$: $$\frac{d𝐩}{dt} = ∇L = -m ∇V, \hspace 1em \frac{dH}{dt} = 0, \hspace 1em \frac{dμ}{dt} = 0.$$

Solving for $w$, as before, we find $$w = V - \frac{|𝐯|^2}{2},$$ so that $$H = m\left(V + \frac{|𝐯|^2}{2}\right) + U.$$ The only difference, this time, is that we have an extra term $U$ for internal energy.

Deformation To Relativistic Form - Special Relativity
In the absence of gravity, $V = 0$, and the geometry reduces to one given by the following: $$ds = dt, \hspace 1em dx^2 + dy^2 + dz^2 + 2 dt du = 0.$$ When passing over to Special Relativity, $s$ and $t$ no longer coincide. Instead, the proper time is given in terms of the Minkowski metric and its associated line element by: $$-c^2 ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2,$$ where $c$ is light speed.

If you compare this to the non-relativistic form $$2 dt du ⇔ c^2 \left(ds^2 - dt^2\right) = (ds + dt) c^2 (ds - dt),$$ what you will find is that $du$ is non-relativistic limit of the time dilation $s - t$, multiplied by $c^2$. Unlike the time dilation $s - t$, itself, whose non-relativistic limit is zero, this has a finite, non-zero limit.

So, it is natural to carry $u$ over into relativistic form by just defining $u = c^2 (s - t)$. Substituting this into the expression for the proper time and the Minkowski line element, we find: $$ds = dt + \frac{1}{c^2} du, \hspace 1em dx^2 + dy^2 + dz^2 + 2 dt du + \frac{1}{c^2} du^2 = 0.$$

Deformation To Relativistic Form - General Relativity
Now, we turn gravity back on. This analysis will be restricted to only the case $N = 1$ of one body of mass $M = m_0$ located at $\left(x_0, y_0, z_0\right) = (0, 0, 0)$, with: $$V = -\frac{GM}{r}, \hspace 1em r = \sqrt{x^2 + y^2 + z^2}.$$

The limit to zero gravity corresponds to $V → 0$, while the non-relativistic limit corresponds to $α → 0$, which is expressed in terms of a "relativity parameter" by $α = (1/c)^2$ ... which brings us full circle to the theme song Alpha that you have (or had) playing in the background.

The most general line element we can form that has both the non-relativistic limit and zero gravity Special Relativistic limits has the form: $$dx^2 + dy^2 + dz^2 + 2 dt du + α du^2 - 2 V dt^2 + α V (⋯) = 0.$$

If we assume the line element is spherically symmetric about the origin $(x,y,z) = (0,0,0)$, then the missing terms may involve $dr$, $dt$ and $du$, where $r$ is given as above. Without further assumptions, that is as far as we can go.

Both the non-relativistic limit and the zero gravity limit have line elements whose metric $g$ has determinant $\det g = -1$. One possible assumption is that the extra term represent an actual deformation of spatial geometry - that is: a term expressing actual spatial curvature - while leaving everything else unaltered so that we may refine $α V (⋯)$ to $α V (⋯) dr^2$. Imposing the constraint that $\det g = -1$, we get: $$α V (⋯) dr^2 = -\frac{2 α V}{1 + 2 α V} dr^2,$$ resulting in the following geometry $$ ds = dt + α du, \\ dx^2 + dy^2 + dz^2 + 2 dt du + α du^2 - 2 V dt^2 - \frac{2 α V}{1 + 2 α V} dr^2 = 0. $$ Converting to spherical coordinates $$(x,y,z) = (r \cos φ \sin θ, r \sin φ \sin θ, r \cos θ),$$ and eliminating $du$ in favor of $ds$, the result is the following line element: $$-c^2 ds^2 = \frac{dr^2}{1 + 2 V/c^2} + r^2 \left(dθ^2 + \left(\sin θ dφ\right)^2\right) - c^2 (1 + 2 V/c^2) dt^2.$$ That's the Schwarzschild metric.

Were you to develop the equations of motion for a test body, the corresponding expressions for $𝐩 = \left(p_x, p_y, p_z\right)$, $H = -p_t$ and $μ = p_u$ would be more complicated, in both the Special Relativistic and General Relativistic cases. I'll leave it to the interested reader to try and work out what they and their corresponding equations of motion might be.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks for the detailed answer. Why did you set the metric to zero though? Isn't the metric a function from two vectors to numbers? $\endgroup$
    – Ryder Rude
    Commented Feb 6, 2023 at 10:26
  • 1
    $\begingroup$ It's a sneaky way of working backwards from the relativistic case, where the zero came about because I moved the ds^2 over to the same side as the line element for the Minkowski and Schwarzschild geometries. So, what this does - both for those geometries as well as the Newtonian versions of each - is embed the 4-dimensional geometries on light cones in the background 4+1 dimensional geometry. A metric can be presented as coefficients of a line element g_{μν} dx^μ dx^ν or equivalently as a linear functional on vectors g(X^μ ∂_μ, Y^ν ∂_ν) = g_{μν} X^μ Y^ν. $\endgroup$ Commented Feb 6, 2023 at 23:29
  • 1
    $\begingroup$ You might be overstating it a bit. It is true that the orbits are null geodesics in the 5D geometry and that the geodesics from a given point form a 4D manifold. But there is no foliation of "Newtonian" manifolds. In fact, your w is readily recognized as -L₄/m, where L₄ is the 4D-Lagrangian; and the 4D-action for the body is just S₄ = -mu! So, different curves (0,0,0,0,0)→(x,y,z,t,u) that land on a specific (x,y,z,t) may land on different u, where the geodesic lands on the largest u, and non-geodesics on smaller u. The difference of the u's is approximately c² their relativistic time-dilation. $\endgroup$
    – NinjaDarth
    Commented Feb 8, 2023 at 18:51
  • $\begingroup$ I think you're right. I didn't think through the matter of foliation. $\endgroup$ Commented Sep 15, 2023 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.