# How can I interpret or mathematically formalize Maxwellian, Leibnizian, and Machian space-times?

I've been reading the book, World Enough and Space-Time, and I came across a rough list of classical space-times with varying structural significance.

Here is the same list, minus Machian Space-time, with good descriptions of what symmetries they have or world line structures they possess as inertial.

Machian comes straight after leibnizian with the only invariant being relative particle distances. Its structures including only absolute simultaneity and an structure on instantaneous spaces.

Symmetries (Machian):   For comparison, here are the symmetries for Neo-Newtonian space-time:  & its cousin Full Newtonian space-time:  . . . right from World Enough and Space-Time.

Aristotelian, Full-newtonian, and Neo-newtonian are very self explanatory and the #2 & #3 of these is closest to our everyday experiences as well as being introduced to us at an early age as a grounding for classical physics.

But how would a Maxwellian, Leibinzian, or Machian universe appear to any observer in it. Heck, it is already pretty mind-blowing trying to imagine acceleration as not absolute but relative. How would this work out. . . what would transformations in this space look like mathematically? Do these contain too little structure in their space-times to even be comparable to a galilean or full newtonian space-time? Are these too alien to us?

• I'd point out that most people misconstrue Aristotelian dynamics; his law of motion distinguishes between an internal force and external force, the internal law he calls a law of growth and the external force is what causes change; no external force, no change; this is essentially the first law of Newton - of course Newton is more precise but then again he came a long time after Aristotle. – Mozibur Ullah Dec 6 '17 at 8:48

## 1 Answer

I will describe the mathematical structure of the most common classical spacetimes, Galilean and Newtonian. This description may give you enough ammunition to expand it to the other classical spacetimes.

The Galilean spacetime is a tuple $(\mathbb{R}^4,t_{ab},h^{ab},\nabla)$ where $t_{ab}$ (temporal metric) and $h^{ab}$ (spatial metric) are tensor fields and $\nabla$ is the coordinate derivative operator specifying the geodesic trajectories.

A single metric does not work, because the speed of light is infinite, so time and space should be treated separately with the temporal metric:

$$t_{ab}=(\text{d}_a t)(\text{d}_b t)$$

and the spatial metric:

$$h^{ab}=\left(\dfrac{\partial}{\partial x}\right)^a\left(\dfrac{\partial}{\partial x}\right)^b+ \left(\dfrac{\partial}{\partial y}\right)^a\left(\dfrac{\partial}{\partial y}\right)^b+ \left(\dfrac{\partial}{\partial z}\right)^a\left(\dfrac{\partial}{\partial z}\right)^b$$

these essentially translate to

$$t'=t$$ $$\text{d}\mathbf{r}'^2=\text{d}\mathbf{r}^2$$

Finally, $\nabla$ on $\mathbb{R}^4$ is a unique flat derivative operator that for each coordinate $x^i$ satisfies:

$$\nabla_a\left(\dfrac{\partial}{\partial x^i}\right)^b=\mathbf{0}$$

While the space of Galilean 4-coordinates is not a Euclidean space, the space of Galilean velocities is a Euclidean space. Differentiating the Galilean transformation (for simplicity in two dimensions):

$$t'=t$$ $$x'=x-vt$$

we obtain $\text{d}t'=\text{d}t$ and therefore

$$\dfrac{\text{d}x'}{\text{d}t'}=\dfrac{\text{d}x}{\text{d}t}-v$$

If $v_R=\dfrac{\text{d}x}{\text{d}t}$ is the velocity of a body as observed from the frame $R$ and $v_{R'}=\dfrac{\text{d}x'}{\text{d}t'}$ is the velocity of the body as observed from the frame $R'$, then the result reveals the Euclidean symmetry

$$v_R=v_{R'}+v_{R'R}$$

In turn, the Newtonian spacetime is a tuple with an additional structure $(\mathbb{R}^4,t_{ab},h^{ab},\nabla,\lambda^a)$ where $\lambda^a$ is a field that adds the preferred frame of rest:

$$\lambda^a=\left(\dfrac{\partial}{\partial t}\right)^a$$

The Galilean choice is relativity, but the Galilean transformation is incompatible with the Maxwell equations. To address this problem, Newton adds the preferred rest frame, which can be viewed as the frame of the aether. This approach allows adding electromagnetism to mechanics by breaking the relativity principle for electromagnetism.

Because all classical spacetimes have separate time and space metrics, each spacetime can be described as a similar tuple with additional structures appropriate for each variant.

• You have an extra "l" in should at the begging and no "c" for "common". – The victorious truther Dec 6 '17 at 18:55
• Also, thank you for the extensive and informative comment. – The victorious truther Dec 6 '17 at 18:58
• Because linear acceleration is relative in maxwellian spacetime, is it curved or warped in some way? That seems like the only other route to go if acceleration is not to be relative. Course i'm not sure if this may mean spatial curvature on each instantaneous slice or curved planes of simultaneity. – The victorious truther Dec 8 '17 at 12:01
• @Thevictorioustruther No, it's not curved. It means no inertial forces in accelerating frames. You floor the gas pedal in your car, but your back doesn't push against the the seat. Or you suddenly break, but it doesn't throw you forward. Essentially all 3 Newton's laws don't hold. For example, you accelerate a ball to throw and it keeps flying with a constant acceleration instead of constant speed. Maxwell needed this, because his equations were incompatible with the Galilean spacetime, but the Maxwellian spacetime is incompatible with mechanics. Newton resolved this conflict (see my answer). – safesphere Dec 9 '17 at 7:27