# Derivation of the Einstein Equivalence Principle

The derivation of the GLET Lagrangian follows from a few number of simple assumptions about the ether. But it is also interesting what one needs to derive only one of the main consequences of the GLET Lagrangian, namely the Einstein Equivalence Principle. Let's see what we need:

## Postulates

Postulate 1 (Newtonian background): We have a flat Euclidean space and absolute time.

Postulate 2 (Lagrange formalism): We have a field theory with Lagrange formalism.

Postulate 3 (Translational invariance): The theory has translational symmetry in space and time.

To describe the Newtonian background, we use a standard Cartesian coordinates - a time coordinate $$\mathfrak{t}$$ for absolute time and Euclidean coordinates $$\mathfrak{x}^i, i=1,2,3$$, for space. For simplicity of some formulas, we use also spacetime coordinates $$\mathfrak{x}^\mu, \mu=0,1,2,3$$, with $$\mathfrak{t}=\mathfrak{x}^0$$. We will also allow and consider arbitrary other coordinates in our Newtonian spacetime, they will be simply denoted by x. In such other coordinates, the preferred Cartesian coordinates will be expressed as functions $$\mathfrak{x}^i(x),\mathfrak{t}(x)$$, or $$\mathfrak{x}^\mu(x)$$.

## The short version

### A simple version of the Noether theorem

We use translational invariance to derive, in a quite simple, "covariant" version of the Noether theorem, the energy-momentum tensor of the theory simply as the Euler-Lagrange equation for the preferred coordinates: $0 = \partial_\nu t^{\mu\nu} = \frac{\delta}{\delta \mathfrak{x^\mu}} S(u^k(x),\mathfrak{x}^\mu(x)).$

This is straightforward, given that translational invariance $$S(u^k(x),\mathfrak{x}^\mu(x)) = S(u^k(x),\mathfrak{x}^\mu(x)+c^\mu)$$ means $$\frac{\partial}{\partial \mathfrak{x}^\mu} S = 0$$ and all other terms are full derivatives.

Then we use the stress-energy-momentum tensor $$t^{\mu\nu}(x)$$ to define the gravitational field, so that the energy-momentum conservation law becomes the harmonic condition, thus, $$t^{\mu\nu}(x) = g^{\mu\nu}(x) \sqrt{-g}$$. In general, more fields will be necessary to define the complete state of the theory. All fields beyond the gravitational field will be named "matter fields" and denoted $$u^k_{matter}(x)$$. So, what we have now, by the Noether theorem and by construction of the gravitational field $$g^{\mu\nu}(x)$$ and the matter fields $$u^k_{matter}(x)$$ is $\frac{\delta}{\delta \mathfrak{x^\mu}} S(g^{\mu\nu}, u^k_{matter},\mathfrak{x}^\mu) = \partial_\nu \left(g^{\mu\nu}\sqrt{-g}\right).$

### The "action equals reaction" symmetry of the Lagrange formalism

If some variable u leads to a nontrivial action on another variable v, it means that the equation of motion of v, the Euler-Lagrange equation $$\frac{\delta}{\delta v} S(u,v)$$ depends on u, and this "depends on u" means the variational derivative $\text{action of u on v } = \frac{\delta}{\delta u} (\text{ equ. for v }) = \frac{\delta}{\delta u} \frac{\delta}{\delta v} S(u,v).$ $\text{action of v on u } = \frac{\delta}{\delta u} (\text{ equ. for u }) = \frac{\delta}{\delta v} \frac{\delta}{\delta u} S(u,v).$ So, "action equals reaction" is nothing but the point that one can change the order of variational derivatives: $\frac{\delta}{\delta u} \frac{\delta}{\delta v} S(u,v) = \frac{\delta}{\delta v} \frac{\delta}{\delta u} S(u,v).$

### Applying "action equals reaction" symmetry to our theory

Let's now apply the action equals reaction symmetry to the action of the preferred coordinates on the matter fields.

We start with the definition of the matter fields as the degrees of freedom additional to the gravitational field, thus, additional to $$t^{\mu\nu}(x)$$ and therefore also independent of the $$g^{\mu\nu}(x)$$, so that by construction the Noether conservation law - the equation for the preferred coordinates - do not depend on the matter fields. Thus, $\frac{\delta}{\delta u^k_{matter}} \left(\frac{\delta}{\delta \mathfrak{x^\mu}} S(g^{\mu\nu}, u^k_{matter},\mathfrak{x}^\mu)\right) = \frac{\delta}{\delta u^k_{matter}} \left(\partial_\nu\left(g^{\mu\nu}\sqrt{-g}\right)\right) = 0.$

Now we can apply the "action equals reaction" symmetry and obtain $\frac{\delta}{\delta u^k_{matter}}\left( \frac{\delta}{\delta \mathfrak{x^\mu}} S(g^{\mu\nu},u^k_{matter},\mathfrak{x}^\mu)\right) = \frac{\delta}{\delta \mathfrak{x^\mu}} \left(\frac{\delta}{\delta u^k_{matter}} S(g^{\mu\nu},u^k_{matter},\mathfrak{x}^\mu)\right) = 0.$ But the term $$\frac{\delta}{\delta u^k_{matter}} S(g^{\mu\nu},u^k_{matter},\mathfrak{x}^\mu)$$ is nothing but the Euler-Lagrange equation for the matter field $$u^k_{matter}(x)$$. Thus, what we have derived is $\frac{\delta}{\delta \mathfrak{x^\mu}} (\text{ equ. for matter field } u^k_{matter}(x)) = 0.$ This is already the Einstein Equivalence Principle that the equations of the matter fields do not depend on the preferred coordinates $$\mathfrak{x}^\mu)$$.

## Derivation

In a field theory, the independent variables of the theory are fields, usually components of vector or tensor fields, like, for example, $$F^k(\mathfrak{x})$$. The Lagrange formalism is, then, defined by a scalar function L, the Langrange density, which depends on these fields and their partial derivatives in space and time $$F^k_{,\mu}(\mathfrak{x})=\frac{\partial}{\partial \mathfrak{x}^\mu} F^k(\mathfrak{x})$$, possibly also on higher order derivatives $$F^k_{,\mu\nu}(\mathfrak{x})$$, and so on. Thus, we have $L=L(F^k(\mathfrak{x}), F^k_{,\mu}(\mathfrak{x}), F^k_{,\mu\nu}(\mathfrak{x}),\ldots).$

where the dots denotes higher order partial derivatives.

### A weak covariant representation of the theory

Once our theory has preferred coordinates $$\mathfrak{x}^\mu$$, this means that the Lagrangian depends on these preferred coordinates.

Nonetheless, it is always possible to rewrite the Lagrange density so that it is valid in arbitrary coordinates. Developing general relativity, Einstein initially thought that this covariance of the formalism of general relativity is a distinguishing property of GR. Kretschmann has criticized this as wrong already in 1917, and proposed the thesis that every physical theory can be presented in a generally covariant form. And it appeared quite easy to find such covariant formulations of theories with preferred coordinates. For example, for SR there was found a presentation which uses a metric field $$\eta_{\mu\nu}(x)$$ with the equation that the Riemann tensor has to be zero as an additional equation. This seems typical, one usually has to introduce some additional fields in comparison with the theory given in preferred coordinates. Nonetheless, this is in general quite simple.

A particular variant for a construction of such a covariant version of a theory is to simply introduce, as additional field variables, the preferred coordinates $$\mathfrak{x}^\mu$$, which become, in general coordinates, functions $$\mathfrak{x}^\mu(x)$$. These preferred coordinates transform as four scalar functions. All usual expressions used in non-covariant Langrangians can be transformed into such a weak covariant form. For example, $$F^0(\mathfrak{x})$$ is equal to $$\mathfrak{t}_{,\mu}(x)F^\mu(x)$$. While $$\mathfrak{t}_{,\mu}(x)F^\mu(x)$$ depends, obviously, on absolute time, it looks formally, if we forget about the special role of $$\mathfrak{t}(x)$$ as preferred time, like a covariant expression.

So, without restriction of generality, we can assume that the Lagrange formalism is defined by such a covariant-looking Lagrange density: $L=L(F^\kappa(x), F^\kappa_{,\mu}(x), F^\kappa_{,\mu\nu}(x),\ldots,\mathfrak{x}^a(x),\mathfrak{x}^a_{,\mu}(x),\ldots).$

If one ignores the special geometric nature of the coordinates $$\mathfrak{x}^a(x)$$, and considers them as four scalar functions enumerated with some non-vector index a, this expression would be covariant.

### The Noether theorem gives an energy-momentum tensor

Once we have translational invariance, the theory does not depend on translations $$\mathfrak{x}^a(x) \to \mathfrak{x}^a(x) + c^a$$. Thus, we have a simplification $L=L(F^\kappa(x), F^\kappa_{,\mu}(x), F^\kappa_{,\mu\nu}(x),\ldots,\mathfrak{x}^a_{,\mu}(x),\ldots).$

so that the Lagrange density depends no longer on the coordinate functions $$\mathfrak{x}^a(x)$$ themself, but only on their partial derivatives.

Once we have a Lagrange density which explicitly depends on fields $$\mathfrak{x}^a(x)$$ and their derivatives, we can consider the corresponding Euler-Lagrange equations. And, once we have translational invariance, thus, no dependence on the $$\mathfrak{x}^a(x)$$ themself, only on their derivatives, these Euler/Lagrange equations immediately have the form of conservation laws: $\partial_\mu T^{\mu\nu} = 0.$

This can be easily seen, essentially much easier than in the usual formalism. All what one has to do is to use the general expression for the Euler-Lagrange equation for the preferred coordinates $$\mathfrak{x}^a$$ and then throw away what vanishes because of the translational invariance $$\mathfrak{x}^a(x) \to \mathfrak{x}^a(x) + c^a$$. For example, for the preferred time $$\mathfrak{t}$$ we obtain $\frac{\delta S}{\delta \mathfrak{t}} = \frac{\partial L}{\partial \mathfrak{t}} - \partial_{\mu} \frac{\partial L}{\partial \mathfrak{t}_{,\mu}} + \partial_{\mu}\partial_\nu \frac{\partial L}{\partial \mathfrak{t}_{,\mu\nu}} - \ldots .$

Now the first term, the only term which is not of the form of a partial derivative, vanishes because of translational invariance in time $$\mathfrak{t}(x) \to \mathfrak{t}(x) + t_0$$, and what remains has already the form necessary for a conservation laws $\frac{\delta S}{\delta \mathfrak{t}} = \partial_{\mu} \left(-\frac{\partial L}{\partial \mathfrak{t}_{,\mu}} + \partial_\nu \frac{\partial L}{\partial \mathfrak{t}_{,\mu\nu}} - \ldots\right) = \partial_\mu T^{\mu 0}.$

Similarly, the translations in space $$\mathfrak{x}^a(x) \to \mathfrak{x}^a(x) + c^a$$ give $\frac{\delta S}{\delta \mathfrak{x}^a} = \partial_{\mu} \left(-\frac{\partial L}{\partial \mathfrak{x}^a_{,\mu}} + \partial_\nu \frac{\partial L}{\partial \mathfrak{x}^a_{,\mu\nu}} - \ldots\right) = \partial_\mu T^{\mu a}$ with the energy-momentum tensor $T^{\mu a} = -\frac{\partial L}{\partial \mathfrak{x}^a_{,\mu}} + \partial_\nu \frac{\partial L}{\partial \mathfrak{x}^a_{,\mu\nu}} - \ldots .$

This is the particular case of the much more general Noether theorem for translational symmetries: Translation in time gives energy conservation, and translation in space gives momentum conservation.

### New variables: Energy-momentum tensor becomes the gravitational field

Once we have conservation laws from Noether's theorem, we have, with the energy-momentum tensor $$T^{\mu\nu}(x)$$, a quite natural set of variables, and it seems quite natural to use them as new independent variables. So, let's introduce now new field variables. Up to now, the field variables are in no way restricted by our axioms. Let's assume that we have more fields variables $$F^i(x)$$ than components $$T^{\mu\nu}(x)$$ of the energy-momentum tensor. Then we can use, as a new set of independent field variables, the energy-momentum tensor $$T^{\mu\nu}(x)$$, and, for the remaining degrees of freedom, some other, remaining fields $$\varphi^m(x)$$.

Now we can name (we are, last but not least, free to do this) the energy-momentum tensor $$T^{\mu\nu}(x)$$ "gravitational field", and all the other, remaining fields "matter fields".

Unfortunately, a change of the field variables does not leave the variational derivatives unchanged. Fortunately, the change of the field variables will not change the fact of translational symmetry as well as the weak variant of covariance. Thus, the Noether theorem remains working, and, therefore, the Euler-Lagrange equation for the preferred coordinates will be, again, a set of four conservation laws.

And, whatever a change of variables can do, it will not add new conservation laws. The old energy-momentum conservation law does not disappear with the change of variables. The new one is, also, one associated via Noether's theorem with translational invariance. Thus, these two sets of four conservation laws, in fact, define the same conservation law.

Unfortunately, even the same conservation law can have different expressions. One can add some appropriately chosen terms to give the conservation law a different form. We will ignore here this possibility (this is, last but not least, a popular website, not a scientific paper), and simply assume that the change of variables can be done in such a way that the conservation law remains unchanged. This would mean that we obtain a Lagrange formalism with the Lagrangian $L=L(T^{\mu\nu}(x), T^{\mu\nu}_{,\lambda}(x), T^{\mu\nu}_{,\kappa\lambda}(x),\ldots, \varphi^m(x), \varphi^m_{,\lambda}(x), \varphi^m_{,\kappa\lambda}(x),\ldots, \mathfrak{x}^a(x),\mathfrak{x}^a_{,\lambda}(x),\ldots),$ so that the conservation law gives the same formula $\frac{\delta S}{\delta \mathfrak{x}^a} = \partial_\mu T^{\mu a}$ as before.

### Action equals reaction symmetry

With these new variables, we can look what happens with the "action equals reaction" symmetry. The "action equals reaction" symmetry is a mathematical consequence of the Lagrange formalism — once there exists a Lagrange formalism, we also have an "action equals reaction" symmetry. One can understand the "action equals reaction" principle as a rule that variational derivatives can be exchanged: $\frac{\delta}{\delta u} \frac{\delta}{\delta v} S(u,v) = \frac{\delta}{\delta v}\frac{\delta}{\delta u} S(u,v)$

Here, $$\frac{\delta}{\delta v} S(u,v)$$ is the Euler-Langrange equation for v, and if this equation depends on u, this is expressed by $$\frac{\delta}{\delta u}$$ of the equation for v being nontrivial. So, the left hand side defines the "action" of u on v. Correspondingly, the right-handed side defines the action of v on u.

But now let's look at the particular case of the preferred coordinates $$\mathfrak{x}^a(x)$$, on the one hand, and the matter fields $$\varphi^m(x)$$ on the other hand. We have, by construction, $\frac{\delta S}{\delta \mathfrak{x}^a} = \partial_\mu T^{\mu a}(x).$ Because of our change of variables, the $$T^{\mu\nu}(x)$$ are independent variables describing the gravitational field, thus, the equation for the preferred coordinates do not depend on the matter fields at all. Thus, $\frac{\delta}{\delta \varphi^m} \frac{\delta S}{\delta \mathfrak{x}^a} = 0.$ That means, by the "action equals reaction" symmetry, that we have also $\frac{\delta}{\delta \mathfrak{x}^a} \frac{\delta S}{\delta \varphi^m} = 0.$ so that the Euler-Lagrange equation for the matter fields $$\frac{\delta S}{\delta \varphi^m}$$ does not depend on the preferred coordinates. That means, it has to be covariant.

That's all. This is already the Einstein Equivalence Principle. The equations for matter fields cannot depend on the preferred coordinates.