its a famous equation $$\Delta s^2=-(c\Delta t)^2+(\Delta x)^2$$ but why do we put the minus sign i heard its there because of that space is ruled by non-Euclidean geometry rules and if that was true what is the difference between a Euclidean and non-Euclidean geometry?


[I will work in natural units where $c=1$. I will, for the most part, consider $4$ coordinates in a coordinate frame. The coordinates of an event would be $x^\mu$ where $\mu$ runs from $0$ to $3$. $x^0$ is the time coordinate and the rest are the space coordinates. I will use Einstein summation convention throughout the treatment.]

The reasoning that the interval is $-\Delta t^2 + \Delta x^2$ because the spacetime is ruled by a non-Euclidean geometry is wrong. It is actually the other way around. Since the physical arguments dictate that the interval must have the signs that they have, we conclude that the spacetime actually has a non-Euclidean geometry.

The physical reasoning behind the signs can be presented something like this: What we want to construct by a quantity called the interval is a frame-invariant measure of the separation between events in the spacetime. How would we know whether a given quantity, expressed in terms of the coordinate measures $\Delta x^\mu$ is a frame invariant quantity or not? By expressing the quantity in terms of the coordinate measures of a different frame $\Delta x'^\nu$ and then expressing the primed quantities in the terms of the unprimed quantities (using the appropriate transformation law) and checking whether the expression for our quantity reduces to the expression for the same in terms of the unprimed coordinates.

In other words, if the quantity that we want to check for its frame-invariance is $I$ and $I=f(\Delta x^\mu)$ then for $I$ to be frame-invariant $f(\Delta x'^\nu)=f(P(\Delta x^\mu ))$ must hold; where the function $P$ is determined by the transformation law between the primed and unprimed coordinates.

Now, Special Relativity tells us that the coordinate transformation law between two inertial coordinate frame is the Lorentz transformation law.

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

where $\Lambda$ is a constant matrix for the given two inertial frames with the constraint: $$\Lambda^T\eta\Lambda=\eta$$ where $\eta$ is the following $4\times 4$ matrix:

\begin{bmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

Now, if I consider a quantity $\eta_{\mu\nu}\Delta x^\mu\Delta x^\nu$ then it can be shown that it is frame invariant in the following manner:

$$\eta_{\alpha\beta}\Delta x'^\alpha \Delta x'^\beta=\eta_{\alpha\beta}\Lambda^\alpha_\mu\Delta x^\mu \Lambda^\beta_\nu\Delta x^\nu=(\Lambda^\alpha_\mu\eta_{\alpha\beta}\Lambda^\beta_\nu)\Delta x^\mu \Delta x^\nu = \eta_{\mu\nu} \Delta x^\mu\Delta x^\nu$$

As you can easily see, $\eta_{\mu\nu}\Delta x^\mu\Delta x^\nu$ is simply $-\Delta t^2 + \Delta x_1^2 + \Delta x_2^2 + \Delta x_3^2$. Thus, $-\Delta t^2 + \Delta x_1^2 + \Delta x_2^2 + \Delta x_3^2$ is the invariant interval in Special Relativity.


Go and browse the related questions for the math. I'll still provide a handwaving illustration, hopefully it helps in understanding.

The difference between Euclidean and Minkowskian metric is precisely that minus. In general relativity, the metric gets more complicated, because the way how $\Delta t$ and $\Delta x$ are composed into the spacetime interval is where gravity comes in.

You will hear a lot of times, that the minus distinguishes the time coordinate from the spatial coordinates. How it does that, is the fact that in Minkowski metric, the space-time splits into two parts relative to a chosen origin: the time-like part (which can be related causally to your current point) and space-like (what cannot possibly influence the origin). This gives time its direction by restricting it from "rotating by 90 degrees" if rotations still worked the way they do in Euclidean space.

What would have been rotation, becomes boost - changing velocity of the reference frame, and gives you the hard limit of the speed of light.

This can be very well compared to the distinction between Elliptic differential equations: for instance, for solving the shape of soap on a wire frame, you need to solve it all at once: the boundary conditions influence everything and every point influences every other point, the solution needs to be self-consistent. Just because the equation has a plus: $\frac{d^2 f}{dx^2}+\frac{d^2 f}{dy^2}=0$.

If you take a wave equation, (which is a hyperbolic differential equation) $\frac{d^2 f}{dx^2}-\frac{d^2 f}{dt^2}=0$ you can propagate the solution from the previous time: you need an initial condition and don't need to know anything about the future. All it takes is a little minus.

Compare to a circle ($x^2+y^2=1$) and hyperbola ($x^2-y^2=1$), one of which you can rotate properly, but the other has two asymptotes (which correspond to the speed of light in the light-cone defined by the zero space-time interval). In other words: a hyperbola in space-time is what defines the set of equal time-space distance (compare to the circle which defines points of equal distance in euclidean space).


The question raised was "Why" the interval equation has a negative sign. The answer that this is so because of Minkowski space lets the "why" in the air: "Why" Minkowski space?

A more in-deep answer could be that this minus sign is intrinsically related to the possibility of movement:

That our Universe is 4-dimensional is obvious from Albert Einstein's mass-energy equivalence $ E = mc^2 $ and the relativistic energy invariant $ E^2/c^2 - p⃗^2 = m_0^2 c^2 $. Measuring distance in light-seconds instead of meters, the speed-of-light c becomes =1, and we obtain the simple formula:

$ E^2 = m^2 = m_0^2 + p⃗^2 = m_0^2 + p_1^2 + p_2^2 + p_3^2 $

From this formula it is obvious that the rest mass $m_0$ is the fourth component of the momentum (or movement) vector, and that the energy or mass is the total length (absolute value) of the momentum vector.

A fundamental (empirical) law of physics stats that in a closed physical system energy, i.e. the above sum of four squares, is maintained during movement and during physical processes.

There is a mathematical identity, stating that a sum of four squares can always be written as the product of two sums of each four squares. (This holds as well for sums of two squares, and for sums of eight squares, but for nothing else).

Applying this to our formula for $E^2$, we can write:

$ (m_0^2 + p_1^2 + p_2^2 + p_3^2) = (r_0^2 + r_1^2 + r_2^2 + r_3^2)(M_0^2 + P_1^2 + P_2^2 + P_3^2) $

wherein the components (proof by algebraic evaluation) are:

$ m_0 = (r_0M_0 - r_1P_1 - r_2P_2 - r_3P_3) $

$ p_1 = (r_0P_1 + r_1M_0 + r_2P_3 - r_3P_2) $

$ p_2 = (r_0P_2 - r_1P_3 + r_2M_0 + r_3P_1) $

$ p_3 = (r_0P_3 + r_1P_2 - r_2P_1 + r_3M_0) $

Now, let's assume that these sums of each four squares are metric products (scalar products) of vectors with themselves, noteworthy the vectors

$ P⃗ = (M_0, P_1, P_2, P_3) $

$ R⃗ = (r_0, r_1, r_2, r_3) $ and

$ p⃗ = (m_0, p_1, p_2, p_3) $

Then we can rewrite the formula for $E^2$ as

$ (p⃗)^2 = (R⃗ * P⃗)^2 $

wherein the multiplication $R⃗ * P⃗$ is defined as given above.

Let's now assume that $ R⃗ $ is a physical process operator which changes $ P⃗ $ into $ p⃗ $, and that $ R⃗ $ does not change the total energy $E$ of the system, i.e. $ (r_0^2 + r_1^2 + r_2^2 + r_3^2 ) = 1 $.

Given the bilinearity of the product $R⃗ * P⃗$, the vector $ P⃗ = P⃗1 + P⃗2 + ...$ can be a sum of vectors, representing a complex physical system. Also the vector $ p⃗ = p⃗1 + p⃗2 + ...$ can be a sum of vectors, representing another complex physical system.

The formula:

$ p⃗ = (p⃗1 + p⃗2 + ...) = R⃗ * (P⃗1 + P⃗2 + ...) = R⃗ * P ⃗ $

describes now a general movement of a physical system $ P⃗ = (P⃗1 + P⃗2 + ...) $ into a physical system $ p⃗ = (p⃗1 + p⃗2 + ...) $ under the influence of a physical process operator $ R⃗ $, wherein the total energy $E$ is conserved.

Note that the energy contained in physical system $ P⃗ $ can be transferred, shared, or cumulated in the systems $ p⃗1 , p⃗2 , ... $ under the effect of operator $ R⃗ $.

Note as well that the formula for

$ m_0 = (r_0M_0 - r_1P_1 - r_2P_2 - r_3P_3) $

shows the metric signature (+,-,-,-) of Minkowsky space.

Conservation of energy and movement is possible in 4 dimensional space because of the existence of the 4-squares identity, and movement necessarily follows a negative (hyperbolic) metrics, by the mathematical reasons outlined.

The corresponding metric differential equation in 4-space is

$ ( (1/c)^2 (∂^2/(∂t^2) - (∂^2/∂x_1^2) - (∂^2/∂x_2^2) - (∂^2/∂x_3^2) )A = μ_0 J $

with $ A = (φ/c,(A_1,A_2,A_3 )) $ the 4-potential composed of electrostatic and vector potential, and
$ J = (ρc,(J_1,J_2,J_3 )) $ the 4-current density composed of charge and currents.

This equation is the 4-dimensional form of Maxwells Equations.

Due to its negative metrics, the solutions of this differential equation are wave functions. In other words, movement always and necessarily takes the form of a wave.

There are also stationary solutions to this differential equation:

$ ((1/c^2) (∂^2/∂t^2) - (∂^2/∂x_1^2) - (∂^2/∂x_2^2) - (∂^2/∂x_3^2))Ψ = λΨ $

and they represent the massive particles and their possible compounds.

  • $\begingroup$ Why is the product of 2 four vectors makes another four vector introduced? $\endgroup$ – Kyle Kanos Mar 6 at 17:47
  • $\begingroup$ The conservation of energy claims that I can only change the system $ (m_0,(p_1,p_2,p_3)) $ into another system $ (M_0,(P_1,P_2,P_3)) $ if the latter has the same total energy $ E $. Given that $ E^2 = m_0^2 + p_1^2 + p_2^2 + p_3^2 $ is a four-square, which can always be written as a product of two four-squares according to Euler's formula, there exist indeed unitary double-rotation operators which transform $ (m_0,(p_1,p_2,p_3)) $ into $ (M_0,(P_1,P_2,P_3)) $. This is the reason for the existence of physical processes. $\endgroup$ – Edgar Mueller Mar 7 at 19:01

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