# Gauge invariance, Lorentz factors and Elasticity of Spacetime

In this piece of writing, I will try to analyze further the main features of my main equation. I will talk about gauge invariance and the importance of imaginary time as well, and the link to the Lorentz factor. Finally, I present a specific form for the general equation, which interprets potential energy as elasticity of space-time.

The main equation is

$\mathbf{H}=\frac{1}{2}\rho \left( \mathbf{A}\otimes \mathbf{A}\right) +\nu \left( \nabla^*\mathbf{A}+\nabla^* \mathbf{A}^T\right) +\mu \left(\nabla^*\mathbf{A}+\nabla^* \mathbf{A}^T\right)^2 + p\mathbf{I}$

What sort of Gauge invariance do we have?

As the total energy of the system is conserved, the divergence of the tensor $\mathbf{H}$ should disappear, that is

$\nabla^* \cdot \mathbf{H}=0$

This in turn means that when one considers the actual dynamic equation, we can have any such total energy tensor, whose divergence vanishes. This is the gauge freedom in the model. By taking the divergence of the main equation, one ends up with fairly complicated nonlinear system of partial differential equations. But the main point is to understand that the total energy tensor must be divergence free. This will allow us to choose conveniently that

$\mathbf{H}=0$

which is probably the most trivial choice, but I tend to believe that nature works in the most simple ways as is possible. So then our main equation reads

$\frac{1}{2}\rho \left( \mathbf{A}\otimes \mathbf{A}\right) +\nu \left( \nabla^*\mathbf{A}+\nabla^* \mathbf{A}^T\right) +\mu \left(\nabla^*\mathbf{A}+\nabla^* \mathbf{A}^T\right)^2 + p\mathbf{I}=0$

This actually means that the generalized Lagrangian is zero as well. As the Lagrangian of a system is the difference of the kinetic and potential energies, we actually have in this gauge that kinetic energy equals the negative of potential energy (the sign is not important). This is in turn interesting as it is somehow related to systems where time averages and ergodicity are considered. Think of a simple pendulum without friction. The pendulum divides its  total energy between kinetic and potential energies, but the time average converges to the gauge where kinetic and potential energies are equal (halfway in terms of y-cooordinate and halfway in terms of x-coordinate).

The role of imaginary time

It is worth mentioning that the imaginary time is crucial in the model. I already showed how the inner product of the space-time event  vector  corresponds to the correct relativistic measure. Now we should also consider proper velocities of the space-time event vector. Remember that the space-time event vector is defined as

$\vec{r}=\begin{bmatrix} -ict \\ x \\ y \\ z \\ \end{bmatrix}$

Now let us define the proper velocity as

$\vec{V}=\frac{\partial \vec{r}}{\partial (-ict)}$

this is then

$\vec{V}=\begin{bmatrix} 1 \\ \frac{i}{c}\frac{\partial x}{\partial t} \\ \frac{i}{c}\frac{\partial y}{\partial t} \\ \frac{i}{c}\frac{\partial z}{\partial t} \\ \end{bmatrix}$

Then we can consider the length of this proper velocity, which is a velocity in space-time:

$\sqrt{\vec{V}\cdot \vec{V}}=\sqrt{1-\frac{\vec{v}\cdot \vec{v}}{c^2}}$

Now this shows that with the use of imaginary time, we can infer naturally the Lorentz factor from special relativity; so that

$\sqrt{\vec{V}\cdot \vec{V}}=\frac{1}{\gamma}$

It is important to realize that this space-time velocity is a “proper” one, as the partial differentiation is taken with respect to $-ict$ instead of $t$. Notice that I have used capital letters for space-time and ordinary letter for space-velocity.

What about the Green-Lagrange Strain tensor?

The last thing is this posting I wanna talk about is the role of the stress tensor. Now we have used for the time being the very general form, where the stress tensor is quadratic in the symmetric tensor. That might be indeed the most general case, but for intuitive reasons I want to discuss the more specific stress tensor of the form

$\mathbf{T}=\nu \left( \nabla^*\mathbf{A}+\nabla^*\mathbf{A}^T +\nabla^*\mathbf{A}\nabla^*\mathbf{A}^T \right)$

In the field of continuum mechanics, in ordinary space without the time-component this tensor is called the Green-Lagrange strain tensor, which provides the strain of elastic bodies. Think of a tennis ball which is squeezed, the amount of squeeze or strain is potential energy stored in the tennis ball. So my analogy is that maybe there is a similar elasticity of space-time (it can be indeed the same as in General relativity, but I don’t know that, so I just proceed with my intuition ).

Now if we indeed proceed with this assumption, the tensor equation reads

$\frac{1}{2}\rho \left ( \mathbf{A}\otimes \mathbf{A}\right ) +\nu \left( \nabla^*\mathbf{A}+\nabla^*\mathbf{A}^T +\nabla^*\mathbf{A}\nabla^*\mathbf{A}^T \right) +p\mathbf{I}=0$

The equation can be read as :

“The total kinetic energy of the system plus the potential energy stored in the elasticity of the space-time must add up to zero”