Geodesic equation: Bridging the gap between inertia and gravity

Question

The geodesic equation can be expressed as: $$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau ^2} + \Gamma _{\alpha \beta} ^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} \frac{\mathrm{d}x^\beta}{\mathrm{d}\tau} = 0$$ where $\tau$ denotes the proper time and $\Gamma _{\alpha \beta} ^\mu$ are the Christoffel symbols.

From my understanding:

The expression $$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau ^2} + \Gamma _{\alpha \beta} ^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} \frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$ is a tensor. However, neither $$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau ^2}$$ which signifies inertia, nor $$\Gamma _{\alpha \beta} ^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} \frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$ representing gravity or curved spacetime, are tensors when considered individually.

This suggests that inertia and gravity might be interconnected as a unified physical property.

Could I inquire about the physical explanation behind the relationship between inertia and gravity?

In the geodesic equation, mass does not appear explicitly; rather, it is the geometry of space-time, represented by Christoffel symbols, that plays a central role. This absence of mass in the equation likely stems from the equivalence of gravitational and inertial masses. Consequently, mass seems to cancel out, particularly when interpreting Christoffel symbols as representing gravitational force. Thus, there is not a direct equation relating force and change in momentum, which could be seen as an extension of Newton's Second Law of Motion: $$f^\nu = \frac{\mathrm{d}P^\nu}{\mathrm{d}\tau} = \gamma \frac{\mathrm{d}P^\nu}{\mathrm{d}t}$$ where $f^\nu$ represents force and $P^\nu$ denotes momentum.

However, in a flat Euclidean space, one may assume Christoffel symbols to be zero, indicating the absence of gravitational force. Nonetheless, other forces, such as electrical forces, may still act on the particle.

Furthermore, the classical derivation of kinetic energy in Newtonian physics, and as I have demonstrated, in Einstein's special relativity, arises from the presence of interaction force acting along an interaction path.

In light of this, it may be worthwhile to explore questions concerning momentum and energy transfer in interactions, particularly if they are related to inertia.

Let me quote from "Special and General Relativity (Great Works that Shape our World)," specifically from "The Meaning of Relativity" by Albert Einstein, which comprises Four Lectures Delivered at Princeton University in May 1921, particularly from "Lecture IV," titled "The General Theory of Relativity (continued)."

$$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}s^2} + \Gamma _{\alpha \beta}^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}s} \frac{\mathrm{d}x^\beta}{\mathrm{d}s} = 0$$

Equation expresses the influence of inertia and gravitation upon the material particle. The unity of inertia and gravitation is formally expressed by the fact that the whole left-hand side of the equation has the character of a tensor (with respect to any transformation of co-ordinaes), but the two terms taken separately do not have tensor character, so that, in analogy with Newton's equations, the first term would be regarded as the expression for inertia, and the second as the expression for the gravitational force.

In Einstein's geodesic equation, he employs space-time interval denoted by $s$ rather than proper time denoted by $\tau$. Although using proper time introduces a change in the equation, as indicated by

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

the equivalence should hold if we substitute $\tau$ for $s$.

Answer 1

The term $$\frac{\mathrm{d}^2x^{\mu}}{\mathrm{d}\tau^2}$$ does not signify inertia and $$\Gamma _{\alpha \beta} ^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} \frac{\mathrm{d}x^\beta}{\mathrm{d}\tau}$$ does not signify gravity or curved spacetime.

In flat spacetime, and even in Riemannian geometry (where $\tau$ will represent curve length rather than proper time), the geodesic equation will still be $$\frac{\mathrm{d}^2x^{\mu}}{\mathrm{d}\tau^2} + \Gamma _{\alpha \beta} ^\mu \frac{\mathrm{d}x^\alpha}{\mathrm{d}\tau} \frac{\mathrm{d}x^\beta}{\mathrm{d}\tau} = 0.$$ Notice I said this is true in flat spacetime, so this is true even in special relativity.

The thing you're missing is that these equations are highly coordinate-dependent (which is the meaning behind them not being tensors). Your interpretation in terms of inertia and gravity only holds in very specific coordinate frames, such as the Cartesian coordinates of Minkowski spacetime. Relativity, however, should work for any coordinate frame. For an explicit example, if you were to use spherical coordinates in flat spacetime, your interpretation would break down, because the Christoffell symbols are necessary to keep a straight motion in curvilinear coordinates.

Within general relativity, gravity is inertia. Things fall because they are moving inertially, and inertial motion is now motion through geodesics through spacetime. It doesn't make sense in splitting gravity and inertia because they are both the same thing.

Also, notice that

in a flat Euclidean space, one may assume Christoffel symbols to be zero

is false. This only holds in Cartesian coordinates. In spherical or cylindrical, or any other of the infinitely many coordinate systems you can use on Euclidean space, this is false. Nevertheless, there is no gravity (unless you want to call the Coriolis force, for example, "gravity").

Furthermore, you can define forces in general relativity as well. If we denote the body's four-velocity through $u^\mu$ and the four-momentum by $p^\mu$, the "four-force" $F^{\mu}$ would be given by $$F^{\mu} = u^\nu \nabla_\nu p^\mu,$$ i.e., the time-derivative of four-momentum. There are other notions of forces in Relativity, and I think Griffiths' book Introduction to Electrodynamics discusses them in Chap. 12.

Answer 2

Characteristically, the geodesics equation represents the acceleration of the tangent vector on a geodesic tangent to the coordinate line $dx^\sigma$ $$ u^\mu\left(x(\tau + d\tau)\right)) \ = \ u^\mu\left((x(\tau)\right)) \ + \ \left( {\Gamma(x(\tau))^\mu}_\nu \ u^\nu\right)_\sigma\ dx^\sigma$$ with $dx^\sigma= u^\sigma \ d\tau.$

Here the Christoffel Gamma term in brackets is the linear map of the basis of cotangent frame to nearby points on all geodesics passing in all directions
$$\left( dx(\tau)^\mu \right) \mapsto \left( dx(\tau+d\tau)^\mu \right) = {\Lambda(x)^{\mu}}_\nu dx^\nu$$

As a linear map of the tangent frame onto itself at the same point, in a locally euclidean space, $\Lambda$ is a vector (for each direction) of the matrices, elements of thhe Lie algebra of the rotation group, fixed at that point.

This mix of a vector quantity by tranport with values in Lie groups or algebras, rotating an orthonormal frame at a point, makes it so difficult, to understand its non-covariant indexing.

Any such 'forces of inertia', that accelate mass point in free fall, are independent of the mass and are quadratic in the velocities.

In Einsteins general theory of relativity, squares of the derivative of time wrt to proper time are included in the equations of motion; that is, the relativistic energy will contribute to inertia.

But the central theorem due to Gauss states, that one can always introduce a local inertial system of coordinates, such that all $\Gamma$ terms vanish and the curvature $R$ is hidden in the derivatives, even if the movement is locally free with constant velocities.

In this view, the Christoffel $Gamma$ terms are 'gauge potentials', only their curl derivatives matter intrinsically, making the difference between curvilinear coordinate systems in flat space and motion in manifolds with intrinsic curvature.

Answer 3

neither $$\frac{\mathrm{d}^2 x^\mu}{\mathrm{d}\tau ^2}$$ which signifies inertia

I disagree that $\mathrm{d}^2 x^\mu/{\mathrm{d}\tau ^2}$ signifies inertia. Usually inertia is understood to be the entirety of Newton’s first law. That is, that a free object will move in a straight line at constant speed.

So in Newtonian physics in a rectilinear coordinate system that would be written $$\frac{\mathrm{d}^2 x^i}{\mathrm{d}t ^2}=0$$ for a free object. Where $i$ indexes the spatial coordinate.

What about Newtonian physics in other coordinate systems? Newtonian physics can be done in other coordinate systems, such as polar coordinates. Then the Newtonian law of inertia would be $$\frac{\mathrm{d}^2 x^i}{\mathrm{d}t ^2} + \Gamma _{jk} ^i \frac{\mathrm{d}x^j}{\mathrm{d}t} \frac{\mathrm{d}x^k}{\mathrm{d}t} = 0$$

This generalizes quite nicely to the standard geodesic equation. So I would say that the whole geodesic equation represents inertia, not just the $\mathrm{d}^2 x^\mu/{\mathrm{d}\tau ^2}$ term.

The relationship between gravity and inertia is that the usual Newtonian gravitational field is part of the Christoffel symbol terms. Some of those Christoffel terms are like the terms that arise in polar coordinates, but some are like the terms that arise in a rotating reference frame (e.g. the centrifugal force) also called inertial forces. The Newtonian gravitational field looks like one of the latter.

This similarity is not just mathematical, it is also experimental. Just like inertial forces Newtonian gravity cannot be detected with an accelerometer. Newtonian gravity locally has all of the properties of an inertial force. The only distinguishing feature are the tidal effects of gravity, which is what is modeled by curvature. Where tidal gravity is insignificant, gravity is mathematically and experimentally indistinguishable from an inertial force.

Let me quote from "Special and General Relativity (Great Works that Shape our World)," specifically from "The Meaning of Relativity" by Albert Einstein

Note that the seminal author of a theory has the first word on a theory, not the last word. This quote was from only 6 years after the theory was first developed. Considerable refinement has occurred in the subsequent century of study. In particular, it is now understood that the Christoffel symbol terms do not represent just gravity.