Newtonian Physics¶

Introduction: Why Tensors¶

This section gives a brief introduction, and in the next sections we derive everything in detail. The Newton law is:

$m {\d^2 {\bf x}\over\d t^2} = {\bf F}$

and using a potential for ${\bf F}$ , we get:

${\d^2 {\bf x}\over\d t^2} = -\nabla \phi {\d^2 x^i\over\d t^2} = -\partial^i \phi$

the last two equations are two different equivalent ways to write a tensor equation in 3D, which means that this equation has the exact same form (is valid) in any (spatial) coordinate system (rotated, translated, in cartesian coordinates, spherical coordinates, ...). Each coordinate system has a different metric, but we can always locally transform into $g_{ij}=\diag(1, 1, 1)$ .

However, if our coordinate transformation depends on time (e.g. a rotating disk), then the above tensor equation changes (e.g. for the rotating disk, we get the Coriolis acceleration term), that’s because time is treated as a parameter, not as a coordinate.

To fix this, we need to work in 4D and treat time as a coordinate, so we introduce $x^0 = ct$ where $c$ is any constant speed (it can be any speed, doesn’t have to be the speed of light). Then in 4D, the above equations are not tensor equations anymore, because the operator ${\d\over \d t} = c \partial_0$ is not a tensor. The 4D tensor formulation happens to be the geodesic equation:

${\d x^\beta\over\d\lambda}\nabla_\beta {\d x^\alpha\over\d\lambda} = 0 R_{00} = 4\pi G\rho R_{ij} = 0$

Which (given that we know how to calculate the Ricci tensor in our coordinates) is valid in any coordinates, not only rotated, translated, cartesian, spherical, ..., but also with arbitrary time dependence, e.g. a rotating disk, accelerating disk, ...

After suitable local coordinate transformation, we can only get two possible metrics (that connect the time and spatial coordinates): $\diag(-1, 1, 1, 1)$ and $\diag(1, 1, 1, 1)$ . Inertial systems have no fictitious forces, so the metrics is one of the two above (possibly with $c\to\infty$ ). Transformation between inertial systems is such a coordinate transformation that leaves the metric intact, e.g.:

$g' = \Lambda^T g \Lambda$

There is no coordinate transformation that turns the metric $\diag(-1, 1, 1, 1)$ into $\diag(1, 1, 1, 1)$ , so we need to choose either one to describe one inertial system and then all other inertial systems will automatically have a metric with the same signature.

The Newton law is valid for small speeds compared to the speed of light, so when we want to extend the theory for all speeds, we only have 4 options: O(3, 1) with either $c\to\infty$ or $c$ finite and O(4) with either $c\to\infty$ or $c$ finite. If $c$ is finite, it has to be large enough, so that we still recover the Newton law for small speeds with the given experimental precision. All 4 cases give the correct Newton law, but give different predictions for large speeds. All we need to do to decide which one is correct is to perform such large speeds (relativistic) experiments. It turns out that all such relativistic experiments are in agreement with the O(3, 1) case where $c$ is the (finite) speed of light and with disagreement with the 3 other cases. For small speeds however (i.e. Newtonean physics), all 4 cases will work, as long as $c$ is chosen large enough.

Given a tensor equation, we can easily determine, if it transforms correctly under the Galilean ( $c\to\infty$ ) or Lorentz transformations ( $c$ is finite). All we have to do is to perform the limit $c\to\infty$ . For example the Newton second law is recovered if we do the $c\to\infty$ limit, but Maxwell equations are only recovered if we choose $c$ to be exactly the speed of light in the Maxwell equations.

The reason why we write equations as tensor equations in 4D is that we can then use any coordinates (including any time dependence), i.e. any observer, and the equations still have the exact same form. So specifying the metrics is enough to define the coordinates (observer) and since the equations has only one form, that is all we need. If we write equations only as tensors in 3D, we not only need to specify the (3D) metrics, but also how the observer accelerates with respect to some (usually inertial) frame where the equations (let’s say Newton law) is defined and we then need to transform all the time derivatives correctly. By using tensors in 4D, all those transformations are taken care of by the standard tensor machinery and all we need to care about is exactly one observer, defined by its metric tensor.

By choosing the correct metrics and $c$ (i.e. $\diag(-1, 1, 1, 1)$ and $c$ the speed of light), all equations are then automatically Lorentz invariant. If we choose $c\to\infty$ (and any metric), we automatically get all equations Galilean invariant.

High School Formulation¶

The usual (high school) formulation is the second Newton’s law:

$m {\d^2 {\bf x}\over\d t^2} = {\bf F}$

for some particle of the mass $m$ and position ${\bf x}$ . To determine the force $\bf F$ , we have at hand the Newton’s law of gravitation:

$|{\bf F}| = G {m_1 m_2\over r^2}$

that expresses the magnitude $|\bf F|$ of the force between two particles with masses $m_1$ and $m_2$ and we also know that the direction of the force is directly towards the other particle. We need to take into account all particles in the system, determine the direction and magnitude of the force due to each of them and sum it up.

College Formulation¶

Unfortunately, it is quite messy to keep track of the direction of the forces and all the masses involved, it quickly becomes cumbersome for more than 2 particles. For this reason, the better approach is to calculate the force (field) from the mass density function $\rho$ :

$\nabla\cdot{\bf F} = -4\pi Gm\,\rho(t, x, y, z)$

To see that both formulations are equivalent, integrate both sides inside some sphere:

$\int\nabla\cdot{\bf F}\,\d x\d y\d z = -4\pi Gm_2\int\rho\,\d x\d y\d z$

apply the Gauss theorem to the left hand side:

$\int\nabla\cdot{\bf F}\,\d x\d y\d z = \int{\bf F}\cdot{\bf n}\,\d S= 4\pi r^2\,{\bf F}\cdot{\bf n}$

where ${\bf n}={{\bf r}\over |{\bf r}|}$ and the right hand side is equal to $-4\pi G m_1m_2$ and we get:

${\bf F}\cdot{\bf n} = -G{m_1m_2\over r^2}$

now we multiply both sides with ${\bf n}$ , use the fact that $({\bf F}\cdot{\bf n}){\bf n} ={\bf F}$ (because ${\bf F}$ is spherically symmetric), and we get the traditional Newton’s law of gravitation:

${\bf F} = -G{m_1m_2\over r^2}{\bf n}$

It is useful to deal with a scalar field instead of a vector field (and also not to have the mass $m$ of the test particle in our equations explicitly), so we define a gravitational potential by:

${\bf F} = -m\nabla\phi(t, x, y, z)$

then the law of gravitation is

(1) $\nabla^2\phi = 4\pi G\rho$

and the second law is:

$m{\d^2 {\bf x}\over\d t^2} = -m\nabla\phi(t, x, y, z)$

Note about units:

$[r] = [{\bf x}] = \rm m$

$[m] = \rm kg$

$[\rho] = \rm kg\,m^{-3}$

$[F] = \rm kg\,m\,s^{-2}$

$[G] = \rm kg^{-1}\,m^3\,s^{-2}$

$[\phi] = \rm m^2\,s^{-2}$

Differential Geometry Formulation¶

There are still problems with this formulation, because it is not immediatelly clear how to write those laws in other frames, for example rotating, or accelerating – one needs to employ nontrivial assumptions about the systems, space, relativity principle and it is often a source confusion. Fortunately there is a way out — differential geometry. By reformulating the above laws in the language of the differential geometry, everything will suddenly be very explicit and clear. As an added bonus, because the special and general relativity uses the same language, the real differences between all these three theories will become clear.

We write $x, y, z$ and $t$ as components of one 4-vector

$x^\mu = \mat{t\cr x\cr y\cr z\cr}$

Now we need to connect the Newtonian equations to geometry. To do that, we reformulate the Newton’s second law:

${\d^2 x^i\over\d t^2} + \delta^{ij}\partial_j\phi =0$

by choosing a parameter $\lambda$ such, that ${\d^2 \lambda\over\d t^2}=0$ , so in general

$\lambda = at+b$

and

${\d^2\over\d t^2} = a^2{\d^2\over\d \lambda^2}$

so

${\d^2 x^i\over\d\lambda^2} + {1\over a^2}\delta^{ij}\partial_j\phi =0$

and using the relation ${\d \lambda\over\d a}=a$ we get

${\d^2 x^i\over\d\lambda^2} + \delta^{ij}\partial_j\phi \left({\d t\over\d\lambda}\right)^2 =0$

So using $x^0$ instead of $t$ , we endup with the following equations:

${\d^2x^0\over\d\lambda^2}=0$

${\d^2 x^i\over\d\lambda^2} + \delta^{ij}\partial_j\phi \left({\d x^0\over\d\lambda}\right)^2 =0$

But this is exactly the geodesic equation for the following Christoffel symbols:

(2) $\Gamma^i_{00} = \delta^{ij}\partial_j\phi$

and all other components are zero.

In order to formulate the gravitation law, we now need to express $\nabla^2\phi$ in terms of geometric quantities like $\Gamma^\alpha_{\beta\gamma}$ or $R^\alpha{}_{\beta\gamma\delta}$ . We get the only nonzero components of the Riemann tensor:

$R^j{}_{0k0} = -R^j{}_{00k} = \delta^{ji}\partial_i\partial_k\phi$

we calculate the $R_{\alpha\beta}$ by contracting:

$R_{00} = R^\mu{}_{0\mu0} = R^i{}_{0i0} = \delta^{ij}\partial_i\partial_j\phi$

$R_{ij} = 0$

and we see that the Newton gravitation law is

$R_{00} = 4\pi G\rho$

$R_{ij} = 0$

Thus we have reformulated the Newton’s laws in a frame invariant way — the matter curves the geometry using the equations:

$R_{00} = 4\pi G\rho$

$R_{ij} = 0$

from which one can (for example) calculate the Christoffel symbols and other things. The particles then move on the geodesics:

${\d^2 x^\alpha\over\d\lambda^2} + \Gamma^\alpha_{\beta\gamma} {\d x^\beta\over\d\lambda}{\d x^\gamma\over\d\lambda} = 0$

Both equations now have the same form in all coordinate systems (inertial or not) and it is clear how to transform them — only the Christoffel symbols (and Ricci tensor) change and we have a formula for their transformation.

Metrics¶

There is a slight problem with the metrics — it can be proven that there is no metrics, that generates the Christoffel symbols above. However, it turns out that if we introduce an invariant speed $c$ in the metrics, then calculate the Christoffel symbols (thus they depend on $c$ ) and then do the limit $c\to\infty$ , we can get the Christoffel symbols above.

In fact, it turns out that there are many such metrics that generate the right Christoffel symbols. Below we list several similar metrics and the corresponding Christoffel symbols (in the limit $c\to\infty$ ), so that we can get a better feeling what metrics work and what don’t and why:

$g_{\mu\nu} = \mat{-c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=-\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{-c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=-\partial_y\phi$

$\Gamma^3_{00}=-\partial_z\phi$

$g_{\mu\nu} = \mat{-c^2-2\phi & 0 & 0 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr}$

$\Gamma^1_{00}=-\partial_x\phi$

$\Gamma^2_{00}=-\partial_y\phi$

$\Gamma^3_{00}=-\partial_z\phi$

$g_{\mu\nu} = \mat{-c^2+45-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{-c^2-2\phi & 0 & 0 & 0\cr 0 & 1-{2\phi\over c^2} & 0 & 0\cr 0 & 0 & 1-{2\phi\over c^2} & 0\cr 0 & 0 & 0 & 1-{2\phi\over c^2}\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{-c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 0 & 0\cr 0 & c^2 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & {2\phi\over c^2}\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & c^2\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=-\infty$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 0 & 0\cr 0 & 1 & 0 & 5\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi-5\partial_z\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

$g_{\mu\nu} = \mat{c^2-2\phi & 0 & 5 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

If we do the limit $c\to\infty$ in the metrics itself, all the working metrics degenerate to:

$g_{\mu\nu} = \mat{\pm\infty & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

(possibly with nonzero but finite elements $g_{0i}=g_{i0}\neq0$ ). So it seems like any metrics whose limit is $\diag(\pm\infty, 1, 1, 1)$ , generates the correct Christoffel symbols:

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

but this would have to be investigated further.

Let’s take the metrics $\diag(-c^2-2\phi, 1-{2\phi\over c^2}, 1-{2\phi\over c^2}, 1-{2\phi\over c^2})$ and calculate the Christoffel symbols (without the limit $c\to\infty$ ):

$\Gamma^0_{\mu\nu}=\begin{pmatrix}- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}}\\- \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(- 2 \phi\left(t,x,y,z\right) - {c}^{2}\right)} & 0 & 0\\- \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & 0 & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(- 2 \phi\left(t,x,y,z\right) - {c}^{2}\right)} & 0\\- \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{- 2 \phi\left(t,x,y,z\right) - {c}^{2}} & 0 & 0 & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(- 2 \phi\left(t,x,y,z\right) - {c}^{2}\right)}\end{pmatrix} \Gamma^1_{\mu\nu}=\begin{pmatrix}\frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}} & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0 & 0\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\\0 & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0\\0 & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0 & \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\end{pmatrix} \Gamma^2_{\mu\nu}=\begin{pmatrix}\frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}} & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0\\0 & \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\\0 & 0 & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\end{pmatrix} \Gamma^3_{\mu\nu}=\begin{pmatrix}\frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}} & 0 & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\\0 & \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & 0 & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\\0 & 0 & \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{{c}^{2} \left(1 - 2 \frac{\phi\left(t,x,y,z\right)}{{c}^{2}}\right)}\end{pmatrix}$

By taking the limit $c\to\infty$ , the only nonzero Christoffel symbols are:

$\Gamma^1_{00}=\partial_x\phi$

$\Gamma^2_{00}=\partial_y\phi$

$\Gamma^3_{00}=\partial_z\phi$

or written compactly:

$\Gamma^i_{00}=\delta^{ij}\partial_j\phi$

So the geodesics equation

${\d^2 x^\alpha\over\d\lambda^2} + \Gamma^\alpha_{\beta\gamma} {\d x^\beta\over\d\lambda}{\d x^\gamma\over\d\lambda} = 0$

becomes

${\d^2 x^0\over\d\lambda^2}=0$

${\d^2 x^i\over\d\lambda^2} + \delta^{ij}\partial_j\phi \left({\d x^0\over\d\lambda}\right)^2 = 0$

From the first equation we get $x^0 = a\lambda+b$ , we substitute to the second equation:

${1\over a^2}{\d^2 x^i\over\d\lambda^2} + \delta^{ij}\partial_j\phi = 0$

or

${\d^2 x^i\over\d (x^0)^2} + \delta^{ij}\partial_j\phi = 0$

${\d^2 x^i\over\d t^2}=-\delta^{ij}\partial_j\phi$

So the Newton’s second law is the equation of geodesics.

Obsolete section¶

This section is obsolete, ideas from it should be polished (sometimes corrected) and put to other sections.

The problem is, that in general, Christoffel symbols have 40 components and metrics only 10 and in our case, we cannot find such a metrics, that generates the Christoffel symbols above. In other words, the spacetime that describes the Newtonian theory is affine, but not a metric space. The metrics is singular, and we have one metrics $\diag(-1, 0, 0, 0)$ that describes the time coordinate and another metrics $\diag(0, 1, 1, 1)$ that describes the spatial coordinates. We know the affine connection coefficients $\Gamma^\alpha_{\beta\gamma}$ , so that is enough to calculate geodesics and to differentiate vectors and do everything we need.

However, for me it is still not satisfactory, because I really want to have a metrics tensor, so that I can easily derive things in exactly the same way as in general relativity. To do that, we will have to work in the regime $c$ is finite and only at the end do the limit $c\to\infty$ .

We start with Einstein equations:

$R_{\alpha\beta}-\half Rg_{\alpha\beta}={8\pi G\over c^4}T_{\alpha\beta}$

or

$R_{\alpha\beta}={8\pi G\over c^4}(T_{\alpha\beta}-\half Tg_{\alpha\beta})$

$R^\alpha{}_\beta={8\pi G\over c^4}(T^\alpha{}_\beta-\half T)$

The energy-momentum tensor is

$T^{\alpha\beta} = \rho U^\alpha U^\beta$

in our approximation $U^i \sim0$ and $U^0 \sim c$ , so the only nonzero component is:

$T^{00} = \rho c^2$

$T = \rho c^2$

and

$R^i{}_j={8\pi G\over c^4}(-\half \rho c^2)=-{4\pi G\over c^2}\rho$

$R^0{}_0={8\pi G\over c^4}(\half \rho c^2)={4\pi G\over c^2}\rho$

We need to find such a metric tensor, that

$R^0{}_0={1\over c^2}\nabla^2\phi$

then we get (1).

There are several ways to choose the metrics tensor. We start We can always find a coordinate transformation, that converts the metrics to a diagonal form with only $1$ , $0$ and $-1$ on the diagonal. If we want nondegenerate metrics, we do not accept $0$ (but as it turns out, the metrics for the Newtonian mechanics is degenerated). Also, it is equivalent if we add a minus to all diagonal elements, e.g. $\diag(1, 1, 1, 1)$ and $\diag(-1, -1, -1, -1)$ are equivalent, so we are left with these options only: signature 4:

$g_{\mu\nu}=\diag(1, 1, 1, 1)$

signature 2:

$g_{\mu\nu}=\diag(-1, 1, 1, 1)$

$g_{\mu\nu}=\diag(1, -1, 1, 1)$

$g_{\mu\nu}=\diag(1, 1, -1, 1)$

$g_{\mu\nu}=\diag(1, 1, 1, -1)$

signature 0:

$g_{\mu\nu}=\diag(-1, -1, 1, 1)$

$g_{\mu\nu}=\diag(-1, 1, -1, 1)$

$g_{\mu\nu}=\diag(-1, 1, 1, -1)$

No other possibility exists (up to adding a minus to all elements). We can also quite easily find coordinate transformations that swap coordinates, i.e. we can always find a transformation so that we first have only $-1$ and then only $1$ on the diagonal, so we are left with: signature 4:

$g_{\mu\nu}=\diag(1, 1, 1, 1)$

signature 2:

$g_{\mu\nu}=\diag(-1, 1, 1, 1)$

signature 0:

$g_{\mu\nu}=\diag(-1, -1, 1, 1)$

One possible physical interpretation of the signature 0 metrics is that we have 2 time coordinates and 2 spatial coordinates. In any case, this metrics doesn’t describe our space (neither Newtonian nor general relativity), because we really need the spatial coordinates to have the metrics either $\diag(1, 1, 1)$ or $\diag(-1, -1, -1)$ .

So we are left with either (this case will probably not work, but I want to have an explicit reason why it doesn’t work):

$g_{\mu\nu} = \mat{1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

or (this is the usual special relativity)

$g_{\mu\nu} = \mat{-1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

It turns out, that one option to turn on gravitation is to add the term $-{2\phi\over c^2}\one$ to the metric tensor, in the first case:

$g_{\mu\nu} = \mat{1-{2\phi\over c^2} & 0 & 0 & 0\cr 0 & 1-{2\phi\over c^2} & 0 & 0\cr 0 & 0 & 1-{2\phi\over c^2} & 0\cr 0 & 0 & 0 & 1-{2\phi\over c^2}\cr}$

and second case:

$g_{\mu\nu} = \mat{-1-{2\phi\over c^2} & 0 & 0 & 0\cr 0 & 1-{2\phi\over c^2} & 0 & 0\cr 0 & 0 & 1-{2\phi\over c^2} & 0\cr 0 & 0 & 0 & 1-{2\phi\over c^2}\cr}$

The second law is derived from the equation of geodesic:

${\d^2 x^\alpha\over\d\lambda^2} + \Gamma^\alpha_{\beta\gamma} {\d x^\beta\over\d\lambda}{\d x^\gamma\over\d\lambda} = 0$

in an equivalent form

${\d U^\alpha\over\d\tau} + \Gamma^\alpha_{\beta\gamma}U^\beta U^\gamma = 0$

The only nonzero Christoffel symbols in the first case are (in the expressions for the Christoffel symbols below, we set $c=1$ ):

$\Gamma^0_{\mu\nu}= \begin{pmatrix}- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\- \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & 0\\- \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0\\- \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & 0 & \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\end{pmatrix} \Gamma^1_{\mu\nu}= \begin{pmatrix}\frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & 0\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\0 & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0\\0 & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\end{pmatrix} \Gamma^2_{\mu\nu}= \begin{pmatrix}\frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0\\0 & \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\0 & 0 & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\end{pmatrix} \Gamma^3_{\mu\nu}= \begin{pmatrix}\frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\0 & \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & 0 & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\0 & 0 & \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\\- \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 - 2 \phi\left(t,x,y,z\right)}\end{pmatrix}$

and in the second case, only $\Gamma^0_{\mu\nu}$ is different:

$\Gamma^0_{\mu\nu}= \begin{pmatrix}\frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & \frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)}\\\frac{\frac{\partial}{\partial x} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & 0 & 0\\\frac{\frac{\partial}{\partial y} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & 0\\\frac{\frac{\partial}{\partial z} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)} & 0 & 0 & - \frac{\frac{\partial}{\partial t} \phi\left(t,x,y,z\right)}{1 + 2 \phi\left(t,x,y,z\right)}\end{pmatrix}$

Now we assume that $\partial_\mu\phi \sim \phi \ll c^2$ , so all $\Gamma^\alpha_{\beta \gamma}$ are of the same order. Also $|U^i| \ll |U^0|$ and $U^0 = c$ , so the only nonnegligible term is

${\d U^\alpha\over\d\tau} + \Gamma^\alpha_{00}(U^0)^2 = 0$

Substituting for the Christoffel symbol we get

${\d U^i\over\d\tau} =-{\delta^{ij}\partial_j{\phi\over c^2}\over1-{2\phi\over c^2}} \, c^2 =-\delta^{ij}(\partial_j\phi)\ \left(1+O\left({\phi\over c^2}\right)\right) =-\delta^{ij}\partial_j\phi + O\left(\left({\phi\over c^2}\right)^2\right)$

and multiplying both sides with $m$ :

$m{\d U^i\over\d\tau} =-m\partial_j\phi\ \delta^{ij}$

which is the second Newton’s law. For the zeroth component we get (first case metric)

$m{\d U^0\over\d\tau} =m{\d\phi\over\d\tau}$

second case:

$m{\d U^0\over\d\tau} =-m{\d\phi\over\d\tau}$

Where $mU^0 = p^0$ is the energy of the particle (with respect to this frame only), this means the energy is conserved unless the gravitational field depends on time.

To summarize: the Christoffel symbols (2) that we get from the Newtonian theory contain $c$ , which up to this point can be any speed, for example we can set $c=1\rm\,ms^{-1}$ . However, in order to have some metrics tensor that generates those Christoffel symbols, the only way to do that is by the metrics

$\diag(-1, 1, 1, 1)-{2\phi\over c^2}\one$

then calculating the Christoffel symbols. If we neglect the terms of the order $O\left(\left(\phi\over c^2\right)^2\right)$ and higher, we get the Newtonian Christoffel symbols (2) that we want. It’s clear that in order to neglect the terms, we must have $|\phi| \ll c^2$ , so we must choose $c$ large enough for this to work. To put it plainly, unless $c$ is large, there is no metrics in our Newtonian spacetime. However for $c$ large, everything is fine.

Intertial frames¶

What is an inertial frame? Inertial frame is such a frame that doesn’t have any fictitious forces. What is a fictitious force? If we take covariant time derivative of any vector, then fictitious forces are all the terms with nonzero Christoffel symbols. In other words, nonzero Christoffel symbols mean that by (partially) differentiating with respect to time, we need to add additional terms in order to get a proper vector again – and those terms are called fictitious forces if we are differentiating the velocity vector.

Inertial frame is a frame without fictitious forces, i.e. with all Christoffel symbols zero in the whole frame. This is equivalent to all components of the Riemann tensor being zero:

$R^\alpha{}_{\beta\gamma\delta} = 0$

In general, if $R^\alpha{}_{\beta\gamma\delta} \neq 0$ in the whole universe, then no such frame exists, but one can always achieve that locally, because one can always find a coordinate transformation so that the Christoffel symbols are zero locally (e.g. at one point), but unless $R^\alpha{}_{\beta\gamma\delta} = 0$ , the Christoffel symbols will not be zero in the whole frame. So the (local) inertial frame is such a frame that has zero Christoffel symbols (locally).

What is the metrics of the inertial frame? It is such a metrics, that $\Gamma^\alpha{}_{\beta\gamma} = 0$ . The derivatives $\partial_\mu\Gamma^\alpha{}_{\beta\gamma}$ however doesn’t have to be zero. We know that taking any of the metrics listed above with $\phi=const$ we get all the Christoffel symbols zero. So for example these two metrics (one with a plus sign, the other with a minus sign) have all the Christoffel symbols zero:

$g_{\mu\nu} = \mat{\pm c^2 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

Such a metrics corresponds to an inertial frame then.

What are the (coordinate) transformations, that transform from one intertial frame to another? Those are all transformations that start with an inertial frame metrics (an example of such a metrics is given above), transform it using the transformation matrix and the resulting metrics is also inertial. In particular, let $x^\mu$ be inertial, thus $g_{\mu\nu}$ is an inertial metrics, then transform to $x'^\mu$ and $g'$ :

$g'_{\alpha\beta} = {\partial x^\mu\over\partial x'^\alpha} {\partial x^\nu\over\partial x'^\beta} g_{\mu\nu} = \left({\partial x\over\partial x'}\right)^T g \left({\partial x\over\partial x'}\right)$

if we denote the transformation matrix by $\Lambda$ :

$\Lambda^\mu{}_\alpha= {\partial x^\mu\over\partial x'^\alpha}$

then the transformation law is:

$g' = \Lambda^T g \Lambda$

Now let’s assume that $g'=g$ , i.e. both inertial systems are given by the same matrix and let’s assume this particular form:

$g'_{\mu\nu}=g_{\mu\nu} = \mat{\pm c^2 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

(e.g. this covers almost all possible Newtonian metrics tensors).

Lorentz Group¶

The Lorentz group is O(3,1), e.g. all matrices satisfying:

(3) $g = \Lambda^T g \Lambda$

with $g=\diag(-c^2, 1, 1, 1)$ . Taking the determinant of (3) we get $(\det\Lambda)^2=1$ or $\det\Lambda=\pm1$ . Writing the 00 component of (3) we get

$-c^2 = -c^2(A^0{}_0)^2+(A^0{}_1)^2+(A^0{}_2)^2+(A^0{}_3)^2$

or

$(A^0{}_0)^2 = 1 + {1\over c^2}\left((A^0{}_1)^2+(A^0{}_2)^2+(A^0{}_3)^2\right)$

Thus we can see that either $A^0{}_0\ge1$ (the transformation preserves the direction of time, orthochronous) or $A^0{}_0\le-1$ (not orthochronous). Thus we can see that the O(3, 1) group consists of 4 continuous parts, that are not connected.

First case: elements with $\det\Lambda=1$ and $A^0{}_0\ge1$ . Transformations with $\det\Lambda=1$ form a subgroup and are called SO(3, 1), if they also have $A^0{}_0\ge1$ (orthochronous), then they also form a subgroup and are called the proper Lorentz transformations and denoted by ${\rm SO}^+(3, 1)$ . They consists of Lorentz boosts, example in the $x$ -direction:

$\Lambda^\mu{}_\nu= \mat{ {1\over\sqrt{1-{v^2\over c^2}}}& -{{v\over c^2}\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr -{v\over\sqrt{1-{v^2\over c^2}}} & {1\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

which in the limit $c\to\infty$ gives

$\Lambda^\mu{}_\nu= \mat{ 1 & 0 & 0 & 0\cr -v & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

and spatial rotations:

$R_1(\phi)= \mat{ 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & \cos\phi & \sin\phi\cr 0 & 0 & -\sin\phi & \cos\phi\cr}$

$R_2(\phi)= \mat{ 1 & 0 & 0 & 0\cr 0 & \cos\phi & 0 & \sin\phi\cr 0 & 0 & 1 & 0\cr 0 & -\sin\phi & 0 & \cos\phi\cr}$

$R_3(\phi)= \mat{ 1 & 0 & 0 & 0\cr 0 & \cos\phi & \sin\phi & 0\cr 0 & -\sin\phi & \cos\phi & 0\cr 0 & 0 & 0 & 1\cr}$

(More rigorous derivation will be given in a moment.) It can be shown (see below), that all other elements (improper Lorentz transformations) of the O(3, 1) group can be written as products of an element from ${\rm SO}^+(3, 1)$ and an element of the discrete group:

$\{\one,\ P,\ T,\ PT\}$

where $P$ is space inversion (also called space reflection or parity transformation):

$P= \mat{ 1 & 0 & 0 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr}$

and $T$ is time reversal (also called time inversion):

$T= \mat{ -1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

Second case: elements with $\det\Lambda=1$ and $A^0{}_0\le-1$ . An example of such an element is $PT$ . In general, any product from ${\rm SO}^+(3, 1)$ and $PT$ belongs here.

Third case: elements with $\det\Lambda=-1$ and $A^0{}_0\ge1$ . An example of such an element is $P$ . In general, any product from ${\rm SO}^+(3, 1)$ and $P$ belongs here.

Fourth case: elements with $\det\Lambda=-1$ and $A^0{}_0\le-1$ . An example of such an element is $T$ . In general, any product from ${\rm SO}^+(3, 1)$ and $T$ belongs here.

Example: where does the reflection around a single spatial axis $(t, x, y, z)\to(t, -x, y, z)$ belong to? It is the third case, because the determinant is $\det\Lambda=-1$ and the 00 element is 1. Written in the matrix form:

$\Lambda = \mat{ 1 & 0 & 0 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} = \mat{ 1 & 0 & 0 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr} \mat{ 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr} = = \mat{ 1 & 0 & 0 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & -1 & 0\cr 0 & 0 & 0 & -1\cr} \mat{ 1 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr 0 & 0 & \cos\pi & \sin\pi\cr 0 & 0 & -\sin\pi & \cos\pi\cr} =PR_1(\pi)$

So it is constructed using the $R_1$ element from ${\rm SO}^+(3, 1)$ and P from the discrete group above.

We can now show why the decomposition ${\rm O}(3,1)={\rm SO}^+(3, 1)\times\{\one,\ P,\ T,\ PT\}$ works. Note that $PT=-\one$ . First we show that ${\rm SO}(3,1)={\rm SO}^+(3, 1)\times\{\one, -\one\}$ . This follows from the fact, that all matrices with $\Lambda^0{}_0\le-1$ can be written using $-\one$ and a matrix with $\Lambda^0{}_0\ge1$ . All matrices with $\det \Lambda=-1$ can be constructed from a matrix with $\det\Lambda=1$ (i.e. SO(3, 1)) and a diagonal matrix with odd number of -1, below we list all of them together with their construction using time reversal, parity and spatial rotations:

$\begin{align*} \diag(-1, 0, 0, 0) &= T\\ \diag(0, -1, 0, 0) &= PR_1(\pi)\\ \diag(0, 0, -1, 0) &= PR_2(\pi)\\ \diag(0, 0, 0, -1) &= PR_3(\pi)\\ \diag(0, -1, -1, -1) &= P\\ \diag(-1, 0, -1, -1) &= TR_1(\pi)\\ \diag(-1, -1, 0, -1) &= TR_2(\pi)\\ \diag(-1, -1, -1, 0) &= TR_3(\pi)\\ \end{align*}$

But $R_i(\pi)$ belongs to ${\rm SO}^+(3, 1)$ , so we just need two extra elements, $T$ and $P$ to construct all matrices with $\det\Lambda=-1$ using matrices from SO(3, 1). So to recapitulate, if we start with ${\rm SO}^+(3, 1)$ we need to add the element $PT=-\one$ to construct SO(3, 1) and then we need to add $P$ and $T$ to construct O(3, 1). Because all other combinations like $PPT=T$ reduce to just one of $\{\one, P, T, -\one\}$ , we are done.

The elements from ${\rm SO}^+(3, 1)$ are proper Lorentz transformations, all other elements are improper. Now we’d like to construct the proper Lorentz transformation matrix $A$ explicitly. As said above, all improper transformations are just proper transformations multiplied by either $P$ , $T$ or $PT$ , so it is sufficient to construct $A$ .

We can always write $A=e^L$ , then:

$\det A = \det e^L = e^{{\rm Tr}\,L} = 1$

so $\Tr L = 0$ and $L$ is a real, traceless matrix. Rewriting (3):

$g = A^T g A$

$A^{-1} = g^{-1} A^T g$

$e^{-L} = g^{-1} e^{L^T}g = e^{g^{-1}L^Tg}$

$-L = g^{-1}L^Tg$

$-gL = (gL)^T$

The matrix $gL$ is thus antisymmetric and the general form of $L$ is then:

$L= \mat{ 0 & {L_{01}\over c^2} & {L_{02}\over c^2} & {L_{03}\over c^2}\cr L_{01} & 0 & L_{12} & L_{13}\cr L_{02} & -L_{12} & 0 & L_{23}\cr L_{03} & -L_{13} & -L_{23} & 0\cr}$

One can check, that $gL$ is indeed antisymmetric. However, for a better parametrization, it’s better to work with a metric $\diag(-1, 1, 1, 1)$ , which can be achieved by putting $c$ into $(ct, x, y, z)$ , or equivalently, to work with $x^\mu=(t, x, y, z)$ and multiply this by a matrix $C=\diag(c, 1, 1, 1)$ to get $(ct, x, y, z)$ . To get a symmetric $\tilde L$ , we just have to do $Cx' = \tilde LCx$ , so to get an unsymmetric $L$ from the symmetric one, we need to do $C^{-1} \tilde L C$ , so we get:

$L = C^{-1} \mat{ 0 & \zeta_1 & \zeta_2 & \zeta_3\cr \zeta_1 & 0 & -\varphi_3 & \varphi_2\cr \zeta_2 & \varphi_3 & 0 & -\varphi_1\cr \zeta_3 & -\varphi_2 & \varphi_1 & 0\cr} C = -i\boldsymbol\varphi\cdot{\bf L}-i\boldsymbol\zeta\cdot C^{-1}{\bf M}C$

We have parametrized all the proper Lorentz transformations with just 6 parameters $\zeta_1$ , $\zeta_2$ , $\zeta_3$ , $\varphi_1$ , $\varphi_2$ and $\varphi_3$ . The matrices ${\bf L}$ and ${\bf M}$ are defined as:

$L_1=-i\mat{ 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & -1 & 0\cr}$

$L_2=-i\mat{ 0 & 0 & 0 & 0\cr 0 & 0 & 0 & -1\cr 0 & 0 & 0 & 0\cr 0 & 1 & 0 & 0\cr}$

$L_3=-i\mat{ 0 & 0 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & -1 & 0 & 0\cr 0 & 0 & 0 & 0\cr}$

$M_1=i\mat{ 0 & 1 & 0 & 0\cr 1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}$

$M_2=i\mat{ 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 0\cr 1 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr}$

$M_3=i\mat{ 0 & 0 & 0 & 1\cr 0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr 1 & 0 & 0 & 0\cr}$

Straightforward calculation shows:

$[L_i, L_j] = i\epsilon_{ijk}L_k$

$[L_i, M_j] = i\epsilon_{ijk}M_k$

$[M_i, M_j] = -i\epsilon_{ijk}L_k$

The first relation corresponds to the commutation relations for angular momentum, second relation shows that $M$ transforms as a vector under rotations and the final relation shows that boosts do not in general commute.

We get:

$A = e^{-i\boldsymbol\varphi\cdot{\bf L}-i\boldsymbol\zeta\cdot C^{-1}{\bf M}C} = C^{-1}\,e^{-i\boldsymbol\varphi\cdot{\bf L}-i\boldsymbol\zeta\cdot{\bf M}}\,C$

As a special case, the rotation around the $z$ -axis is given by $\boldsymbol\varphi=(0, 0, \varphi)$ and $\boldsymbol\zeta=0$ :

$A= e^{-i\varphi L_3} = \one-L_3^2+iL_3\sin\varphi+L_3^2\cos\varphi= \mat{ 1 & 0 & 0 & 0\cr 0 & \cos\varphi & \sin\varphi & 0\cr 0 & -\sin\varphi & \cos\varphi & 0\cr 0 & 0 & 0 & 1\cr}$

The boost in the $x$ -direction is $\boldsymbol\varphi=0$ and $\boldsymbol\zeta=(\zeta, 0, 0)$ , e.g.:

$A= C^{-1}e^{-i\zeta M_1}C = C^{-1}\left( \one-M_1^2+iM_1\sinh\zeta+ M_1^2\cosh\zeta\right)C=$

$= C^{-1} \mat{ \cosh\zeta & -\sinh\zeta & 0 & 0\cr -\sinh\zeta & \cosh\zeta & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} C = \mat{ \cosh\zeta & -{1\over c}\sinh\zeta & 0 & 0\cr -c\sinh\zeta & \cosh\zeta & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

from the construction, $-\infty<\zeta<\infty$ , so we may do the substitution $\zeta={v\over c}\,{\rm atanh}\left({v\over c}\right)$ , where $-c<v<c$ . The inverse transformation is:

$\cosh\zeta={1\over\sqrt{1-{v^2\over c^2}}}$

$\sinh\zeta={{v\over c}\over\sqrt{1-{v^2\over c^2}}}$

and we get the boost given above:

$A= \mat{ \cosh\zeta & -{1\over c}\sinh\zeta & 0 & 0\cr -c\sinh\zeta & \cosh\zeta & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} = \mat{ {1\over\sqrt{1-{v^2\over c^2}}}& -{{v\over c^2}\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr -{v\over\sqrt{1-{v^2\over c^2}}} & {1\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

Adding two boosts together:

$A(u)A(v) = \mat{ {1\over\sqrt{1-{u^2\over c^2}}}& -{{u\over c^2}\over\sqrt{1-{u^2\over c^2}}} & 0 & 0\cr -{u\over\sqrt{1-{u^2\over c^2}}} & {1\over\sqrt{1-{u^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} \mat{ {1\over\sqrt{1-{v^2\over c^2}}}& -{{v\over c^2}\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr -{v\over\sqrt{1-{v^2\over c^2}}} & {1\over\sqrt{1-{v^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =$

$= \mat{ {1\over\sqrt{1-{w^2\over c^2}}}& -{{w\over c^2}\over\sqrt{1-{w^2\over c^2}}} & 0 & 0\cr -{w\over\sqrt{1-{w^2\over c^2}}} & {1\over\sqrt{1-{w^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

with

$w = {u+v\over1+{uv\over c^2}}$

O(4) Group¶

The group of rotations in 4 dimensions is O(4), e.g. all matrices satisfying:

(4) $g = \Lambda^T g \Lambda$

with $g=\diag(c^2, 1, 1, 1)$ . Taking the determinant of (4) we get $(\det\Lambda)^2=1$ or $\det\Lambda=\pm1$ . Writing the 00 component of (4) we get

$c^2 = c^2(A^0{}_0)^2+(A^0{}_1)^2+(A^0{}_2)^2+(A^0{}_3)^2$

or

$(A^0{}_0)^2 = 1 - {1\over c^2}\left((A^0{}_1)^2+(A^0{}_2)^2+(A^0{}_3)^2\right)$

Thus we always have $-1\le A^0{}_0\le1$ . That is different to the O(3, 1) group: the O(4) group consists of only 2 continuous parts, that are not connected. (The SO(4) part contains the element $-\one$ though, but one can get to it continuously, so the group is doubly connected.)

Everything proceeds much like for the O(3, 1) group, so $gL$ is antisymmetric, but this time $g=\diag(c^2, 1, 1, 1)$ , so we get:

$L= \mat{ 0 & -{L_{01}\over c^2} & -{L_{02}\over c^2} & -{L_{03}\over c^2}\cr L_{01} & 0 & L_{12} & L_{13}\cr L_{02} & -L_{12} & 0 & L_{23}\cr L_{03} & -L_{13} & -L_{23} & 0\cr}$

and so we also have 6 generators, but this time all of them are rotations:

$A = C^{-1}\,e^{-i\varphi_a L_a}\,C$

with $a=1, 2, 3, 4, 5, 6$ . The spatial rotations are the same as for O(3, 1) and the remaining 3 rotations are $(t,x)$ , $(t,y)$ and $(t,z)$ plane rotations. So for example the $(t,x)$ rotation is:

$A= C^{-1} \mat{ \cos\varphi_4 & \sin\varphi_4 & 0 & 0\cr -\sin\varphi_4 & \cos\varphi_4 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} C = \mat{ \cos\varphi_4 & {1\over c}\sin\varphi_4 & 0 & 0\cr -c\sin\varphi_4 & \cos\varphi_4 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

Now we can do this identification:

$\sin\phi_4 = {{v\over c}\over\sqrt{1+({v\over c})^2}}$

$\cos\phi_4 = {1\over\sqrt{1+({v\over c})^2}}$

so we get the Galilean transformation in the limit $c\to\infty$ :

$A= \mat{ {1\over\sqrt{1+({v\over c})^2}} & {{v\over c^2}\over\sqrt{1+({v\over c})^2}} & 0 & 0\cr -{v\over\sqrt{1+({v\over c})^2}} & {1\over\sqrt{1+({v\over c})^2}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} \to \mat{ 1 & 0 & 0 & 0\cr -v & 1 & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

Adding two boosts together:

$A(u)A(v) = \mat{ {1\over\sqrt{1+{u^2\over c^2}}}& {{u\over c^2}\over\sqrt{1+{u^2\over c^2}}} & 0 & 0\cr -{u\over\sqrt{1+{u^2\over c^2}}} & {1\over\sqrt{1+{u^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} \mat{ {1\over\sqrt{1+{v^2\over c^2}}}& {{v\over c^2}\over\sqrt{1+{v^2\over c^2}}} & 0 & 0\cr -{v\over\sqrt{1+{v^2\over c^2}}} & {1\over\sqrt{1+{v^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr} =$

$= \mat{ {1\over\sqrt{1+{w^2\over c^2}}}& {{w\over c^2}\over\sqrt{1+{w^2\over c^2}}} & 0 & 0\cr -{w\over\sqrt{1+{w^2\over c^2}}} & {1\over\sqrt{1+{w^2\over c^2}}} & 0 & 0\cr 0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr}$

with

$w = {u+v\over1-{uv\over c^2}}$

However, there is one peculiar thing here that didn’t exist in the O(3, 1) case: by adding two velocities less than $c$ , for example $u=v=c/2$ , we get:

$w = {c\over 1-{1\over 4}}={4c\over3}>c$

(as opposed to $w = {c\over 1+{1\over 4}}={4c\over5}<c$ in the O(3, 1) case). So one can get over $c$ easily. By adding $u=v={4c\over3}$ together:

$w = {{8c\over 3}\over 1-{16\over 9}}=-{24c\over7}<0$

(as opposed to $w = {{8c\over 3}\over 1+{16\over 9}}={24c\over25}>0$ in the O(3, 1) case). So we can also get to negative speeds easily. One also needs to be careful with identifying $\cos\phi_4 = {1\over\sqrt{1+({v\over c})^2}}$ , because for $\varphi_4>\pi/2$ we should probably set $\cos\varphi_4 = -{1\over\sqrt{1+({v\over c})^2}}$ . All of this follows directly from the structure of SO(4), because one can get from $\Lambda^0{}_0>0$ to $\Lambda^0{}_0<0$ continuously (this corresponds to increasing $\varphi_4$ over $\pi/2$ ). In fact, by adding two speeds $u=v>c(\sqrt 2 - 1)$ , one always gets $w>c$ . But if $c(\sqrt 2 - 1)\doteq0.414c$ is larger than any speed that we are concerned about, we are fine.

Proper Time¶

Proper time $\tau$ is a time elapsed by (physical) clocks along some (4D) trajectory. Coordinate time $t$ is just some time coordinate assigned to each point in the space and usually one can find some real clocks, that would measure such a time (many times they are in the infinity). To find a formula for a proper time (in terms of the coordinate time), we introduce a local inertial frame at each point of the trajectory – in this frame, the clocks do not move, e.g. $x$ , $y$ , $z$ is constant (zero) and there is no gravity (this follows from the definition of the local inertial frame), so the metric is just a Minkowski metric.

For any metrics, $\d s^2$ is invariant:

$\d s^2 = g_{\mu\nu} \d x^\mu\d x^\nu$

so coming to the local inertial frame, we have $x$ , $y$ , $z$ constant and we get:

$\d s^2 = g_{00} \d\tau^2$

so:

$\d\tau=\sqrt{\d s^2\over g_{00}}$

since we are still in the local inertial frame (e.g. no gravity), we have $g_{00}=-c^2$ (depending on which metrics we take it could also be $+c^2$ ), so:

$\d\tau=\sqrt{-{\d s^2\over c^2}}$

This formula was derived in the local inertial frame, but the right hand side is the same in any inertial frame, because $\d s^2$ is invariant and $c$ too. So in any frame we have:

$\d\tau=\sqrt{-{\d s^2\over c^2}} =\sqrt{-{g_{\mu\nu} \d x^\mu\d x^\nu\over c^2}}$

We’ll explain how to calculate the proper time on the 1971 Hafele and Keating experiment. They transported cesium-beam atomic clocks around the Earth on scheduled commercial flights (once flying eastward, once westward) and compared their reading on return to that of a standard clock at rest on the Earth’s surface.

We’ll calculate it with all the metrics discussed above, to see the difference.

Weak Field Metric¶

Let’s start with the metrics:

$\d s^2=-\left(1+{2\phi\over c^2}\right)c^2 \d t^2 +\left(1-{2\phi\over c^2}\right)(\d x^2 +\d y^2 +\d z^2)$

Then:

$\tau_{AB} =\int_A^B\d\tau =\int_A^B\sqrt{-{\d s^2\over c^2}} =\int_A^B\sqrt{\left(1+{2\phi\over c^2}\right)\d t^2 -{1\over c^2}\left(1-{2\phi\over c^2}\right)(\d x^2 +\d y^2 +\d z^2)}=$

$=\int_A^B\d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)\left( \left(\d x\over\d t\right)^2 + \left(\d y\over\d t\right)^2 + \left(\d z\over\d t\right)^2\right)}=$

$=\int_A^B\d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)|{\bf V}|^2}$

where

$|{\bf V}|^2= \left(\d x\over\d t\right)^2 + \left(\d y\over\d t\right)^2 + \left(\d z\over\d t\right)^2$

is the nonrelativistic velocity. Then we expand the square root into power series and only keep terms with low powers of $c$ :

$\tau_{AB} =\int_A^B\d t\sqrt{\left(1+{2\phi\over c^2}\right) -{1\over c^2}\left(1-{2\phi\over c^2}\right)|{\bf V}|^2} =\int_A^B\d t\left(1+{\phi\over c^2}-{1\over 2c^2}|{\bf V}|^2\right)$

so

$\tau_{AB} =\int_A^B\d t\left(1-{1\over c^2}\left({1\over2}|{\bf V}|^2-\phi\right)\right)$

Now let $V_g=V_g(t)$ be the speed of the plane relative to the (rotating) Earth (positive for the eastbound flights, negative for the westbound ones), $V_\oplus={2\pi R_\oplus\over24}\,{\rm1\over h}$ the surface speed of the Earth, then the proper time for the clocks on the surface is:

$\tau_\oplus =\int_A^B\d t\left(1-{1\over c^2}\left({1\over2}V_\oplus^2-\phi_\oplus\right) \right)$

and for the clocks in the plane

$\tau =\int_A^B\d t\left(1-{1\over c^2}\left({1\over2}(V_g+V_\oplus)^2-\phi\right) \right)$

then the difference between the proper times is:

$\tau-\tau_\oplus=\Delta\tau ={1\over c^2}\int_A^B\d t\left(-{1\over2}(V_g+V_\oplus)^2+\phi +{1\over2}V_\oplus^2-\phi_\oplus\right) ={1\over c^2}\int_A^B\d t\left( \phi-\phi_\oplus-{1\over2}V_g(V_g+2V_\oplus) \right)$

but $\phi-\phi_\oplus=g h$ , where $h=h(t)$ is the altitude of the plane, so the final formula is:

$\Delta\tau ={1\over c^2}\int_A^B\d t\left( gh-{1\over2}V_g(V_g+2V_\oplus) \right)$

Let’s evaluate it for typical altitudes and speeds of commercial aircrafts:

$R_\oplus=6 378.1{\rm\,km}=6.3781\cdot10^6{\rm\,m}$

$V_\oplus={2\pi R_\oplus\over24}\,{\rm1\over h} ={2\pi R_\oplus\over24\cdot3600}\,{\rm1\over s} ={2\pi\,6.3781\cdot10^6\over24\cdot3600}{\rm m\over s}=463.83\rm\,{m\over s}$

$V_g = 870\,{\rm km\over h}=241.67\rm\,{m\over s}$

$h = 12{\rm\,km}=12000\rm\,m$

$t = {2\pi R_\oplus\over V_g} = {2\pi\,6.3781\cdot10^6\over 241.67}{\rm\,s} =165824.41{\rm\,s}\approx 46{\rm\,h}$

$c = 3\cdot10^8\rm\,{m\over s}$

For eastbound flights we get:

$\Delta\tau ={t\over c^2} \left( gh-{1\over2}V_g(V_g+2V_\oplus) \right) =-4.344\cdot10^{-8}{\rm\,s}=-43.44{\rm\,ns}$

and for westbound flights we get:

$\Delta\tau ={t\over c^2} \left( gh-{1\over2}V_g(V_g-2V_\oplus) \right) =3.6964\cdot10^{-7}{\rm\,s}=369.63{\rm\,ns}$

By neglecting gravity, one would get: eastbound flights:

$\Delta\tau ={t\over c^2} \left(-{1\over2}V_g(V_g+2V_\oplus) \right) =-260.34{\rm\,ns}$

and for westbound flights:

$\Delta\tau ={t\over c^2} \left(-{1\over2}V_g(V_g-2V_\oplus) \right) =152.73{\rm\,ns}$

By just taking the clocks to the altitude $12\rm\,km$ and staying there for 46 hours (without moving with respect to the inertial frame, e.g. far galaxies), one gets:

$\Delta\tau ={ght\over c^2}=216.90{\rm\,ns}$

Rotating Disk Metric¶

The rotating disk metrics is (taking weak field gravitation into account):

$\d s^2=-\left(1+{2\phi\over c^2}-{\omega^2\over c^2}(x^2+y^2)\right)c^2 \d t^2 +(\d x^2 +\d y^2 +\d z^2)-2\omega y\,\d x\d t + 2\omega x\,\d y\d t$

Then:

$\tau_{AB} =\int_A^B\d\tau =\int_A^B\sqrt{-{\d s^2\over c^2}}=$

$=\int_A^B\sqrt{\left(1+{2\phi\over c^2}-{\omega^2\over c^2}(x^2+y^2)\right) \d t^2 -{1\over c^2}(\d x^2 +\d y^2 +\d z^2) +{2\omega y\over c^2}\,\d x\d t - {2\omega x\over c^2}\,\d y\d t }=$

$=\int_A^B\d t\sqrt{\left(1+{2\phi\over c^2}-{\omega^2\over c^2}(x^2+y^2)\right) -{1\over c^2}|{\bf V}|^2 +{2\omega y\over c^2}\,{\d x\over\d t} - {2\omega x\over c^2}\,{\d y\over\d t}}$

where

$|{\bf V}|^2= \left(\d x\over\d t\right)^2 + \left(\d y\over\d t\right)^2 + \left(\d z\over\d t\right)^2$

is the nonrelativistic velocity. Then we expand the square root into power series and only keep terms with low powers of $c$ :

$\tau_{AB} =\int_A^B\d t\left(1+{\phi\over c^2}-{1\over 2c^2}|{\bf V}|^2 +{\omega y\over c^2}\,{\d x\over\d t} - {\omega x\over c^2}\,{\d y\over\d t} \right)$

so

$\tau_{AB} =\int_A^B\d t\left(1-{1\over c^2}\left({1\over2}|{\bf V}|^2-\phi -{\omega y}\,{\d x\over\d t} + {\omega x}\,{\d y\over\d t} \right)\right)$

Now as before let $V_g=V_g(t)$ be the speed of the plane (relative to the rotating Earth, e.g. relative to our frame), $V_\oplus={2\pi R_\oplus\over24}\,{\rm1\over h}$ the surface speed of the Earth, so $\omega R_\oplus=V_\oplus$ . For the clocks on the surface, we have:

$x = R_\oplus$

$y = 0$

$z = 0$

so

${\d x\over\d t}={\d y\over\d t}={\d z\over\d t}=0$

$|{\bf V}|^2=0$

then the proper time for the clocks on the surface is:

$\tau_\oplus =\int_A^B\d t\left(1-{1\over c^2}\left(-\phi_\oplus\right) \right)$

and for the clocks in the plane we have:

$x = (R_\oplus+h)\cos\Omega t$

$y = (R_\oplus+h)\sin\Omega t$

$z = 0$

where $\Omega$ is defined by $\Omega (R_\oplus+h)=V_g$ , so

${\d x\over\d t}=-(R_\oplus+h)\Omega\sin\Omega t$

${\d y\over\d t}=(R_\oplus+h)\Omega\cos\Omega t$

${\d z\over\d t}=0$

$|{\bf V}|^2=\Omega^2(R_\oplus+h)^2$

$\omega y {\d x\over\d t}=-\omega\Omega(R_\oplus+h)^2\sin^2\Omega t$

$\omega x {\d y\over\d t}=\omega\Omega(R_\oplus+h)^2\cos^2\Omega t$

and

$\tau =\int_A^B\d t\left(1-{1\over c^2}\left({1\over2} \Omega^2(R_\oplus+h)^2 - \phi +\omega\Omega(R_\oplus+h)^2\right) \right)$

then the difference between the proper times is:

$\tau-\tau_\oplus=\Delta\tau ={1\over c^2}\int_A^B\d t\left(-{1\over2}\Omega^2(R_\oplus+h)^2 -\omega\Omega(R_\oplus+h)^2 +\phi-\phi_\oplus\right) =$

$={1\over c^2}\int_A^B\d t\left( -{1\over2}V_g^2 -V_\oplus V_g \left(1+{h\over R_\oplus}\right) +\phi-\phi_\oplus \right) =$

$={1\over c^2}\int_A^B\d t\left( \phi-\phi_\oplus -{1\over2}V_g\left(V_g+2V_\oplus\left(1+{h\over R_\oplus}\right)\right) \right)$

but $\phi-\phi_\oplus=g h$ , where $h=h(t)$ is the altitude of the plane and we approximate

$\left(1+{h\over R_\oplus}\right)\approx 1 \,,$

so the final formula is the same as before:

$\Delta\tau ={1\over c^2}\int_A^B\d t\left( gh-{1\over2}V_g(V_g+2V_\oplus) \right)$

Note: for the values above, the bracket $\left(1+{h\over R_\oplus}\right)^2\doteq1.00377$ , so it’s effect on the final difference of the proper times is negligible (e.g. less than $1 \rm\,ns$ ). The difference is caused by a slightly vague definition of the speed of the plane, e.g. the ground speed is a bit different to the speed relative to the rotating Earth (this depends on how much the atmosphere rotates with the Earth).

Concluding Remarks¶

The coordinate time $t$ in both cases above is totally different. One can find some physical clocks in both cases that measure (e.g. whose proper time is) the particular coordinate time, but the beauty of the differential geometry approach is that we don’t have to care about this. $t$ is just a coordinate, that we use to calculate something physical, like a proper time along some trajectory, which is a frame invariant quantity. In both cases above, we got a different formulas for the proper time of the surface clocks (and the clocks in the plane) in terms of the coordinate time (because the coordinate time is different in both cases), however the difference of the proper times is the same in both cases:

$\Delta\tau ={1\over c^2}\int_A^B\d t\left( gh-{1\over2}V_g(V_g+2V_\oplus) \right)$

There is still a slight difference though – the $t$ here used to evaluate the integral is different in both cases. To do it correctly, one should take the total time as measured by any of the clocks and then use the right formula for the proper time of the particular clock to convert to the particular coordinate time. However, the difference is small, of the order of nanoseconds, so it’s negligible compared to the total flying time of 46 hours.

FAQ¶

How does one incorporate the fact, that there are only two possible transformations, into all of this? For more info, see: http://arxiv.org/abs/0710.3398. Answer: in that article there are actually three possible transformations, $K<0$ corresponds to O(4), $K>0$ to O(3, 1) and $K=0$ to either of them in the limit $c\to\infty$ .

What is the real difference between the Newtonian physics and special relativity? E.g. how do we derive the Minkowski metrics, how do we know we need to set $c=const$ and how do we incorporate gravity in it? Answer: there are only three possible groups of transformations: O(4), O(3, 1) and a limit of either for $c\to\infty$ . All three provide inequivalent predictions for high speeds, so we just choose the right one by experiment. It happens to be the O(3, 1). As to gravity, that can be incorporated in either of them.

Questions Without Answers (Yet)¶

How can one reformulate the article http://arxiv.org/abs/0710.3398 into the language of the O(4) and O(3, 1) groups above? Basically each assumption and equation must have some counterpart in what we have said above. I’d like to identify those explicitely.

What are all the possible metrics, that generate the Newtonian Christoffel symbols? (Several such are given above, but I want to know all of them) Probable answer: all metrics, whose inverse reduces to $g^{\mu\nu}=\diag(0, 1, 1, 1)$ in the limit $c\to\infty$ . I would like to have an explicit proof of this though.

What is the role of the different metrics, that generate the same Christoffel symbols in the limit ( $c\to\infty$ )? Can one inertial frame be given with one and another frame with a different form of the metrics (e.g. one with $g_{00}=c^2$ and the other one with $g_{00}=-c^2$ ?) Possible answer: there is no transformation to convert a metrics with signature +4 to signature +2, so one has to choose one and then all other inertial frames have the same one.

What are all the allowed transformations between intertial frames? If we assume that the inertial frames are given with one given metrics (see the previous question), then the answer is: representation of the O(3, 1) group if $g_{00}=-c^2$ or O(4) group if $g_{00}=c^2$ . But if one frame is $g_{00}=-c^2$ and we transform to another frame with $g_{00}=c^2$ , then it is not clear what happens. Possible answer: one has to choose some signature and stick to it, see also the previous question.

What is the real difference between Newtonian physics and general relativity? Given our formulation of Newtonian physics using the differential geometry, I want to know what the physical differences are between all the three theories are.

Newtonian Physics¶

Introduction: Why Tensors¶

High School Formulation¶

College Formulation¶

Differential Geometry Formulation¶

Metrics¶

Obsolete section¶

Intertial frames¶

Lorentz Group¶

O(4) Group¶