Can the equivalence principle be safely used in non-relativistic mechanics? Imagine an ideal pendulum in a train. While the train is in uniform motion, Newton's laws apply within the train, and we can easily write down the equations of motion for the pendulum. Now assume the train is being uniformly accelerated with acceleration $a$.
If we would like to directly approach this using Newton's laws, we would probably have to consider the force accelerating the train, the reaction force on the pivot of the pendulum, and so on. I think this would get quite cumbersome.
Naively invoking the equivalence principle as in general relativity, this would become easy again: let $\vec g$ be the gravitational acceleration, $\vec a$ the acceleration of the train, then in the train this would give us an equivalent gravitational field with uniform value $\vec g - \vec a$, and an equilibrium position in the direction of this vector, which corresponds to the angle $\theta_0 = \arctan\frac{-a}g$, and we conclude that the equation of motion for the angle $\theta - \theta_0$ is the same as that for an ordinary pendulum with gravitational acceleration $\sqrt{g^2 + a^2}$.
It is not at all clear (to me) that this reasoning is justified by Newton's laws.
My questions:

*

*Can the equivalence principle (possibly in some restricted form) be safely invoked in non-relativisting mechanics?

*Can its validity be derived from Newton's laws?

 A: If we write the acceleration of the pendulum as a combination of the uniform acceleration ($a_u$) and the acceleartion in the rest frame ($a_r$) of the train we get
$$\vec{a} = m \vec{a_u} + m \vec{a_r} = \vec{F_r} + \vec{F_g}$$
with $F_g$ the gravitational force acting on the pendulum and $F_r$ the reaction force of the string the pendulum is attached to.
What we now want to find is an expression for $a_r$, we write
$$m \vec{a_r} = \vec{F_r} - m(\vec{g} -\vec{a_u})$$
This is of course quite straightforward. If I understand your problem correct, you are wondering whether the reaction force $\vec{F_r}$ is now different then when the pendulum would swing in a uniform gravitational field with gravitational acceleration $\vec{g}-\vec{a_u}$.
One should remember how this reaction force is determined when solving the E.O.M. for a pendulum, it is uniquely determined by requiring that $\vec{a_r} \perp \vec{v}$ and thus that the pendulum swings on a circle. There is no difference in our case, we must still have $\vec{a_r}\perp\vec{v}$ and will find therefore exactly the same reaction force as we would for a pendulum in a $\vec{g}-\vec{a_u}$ gravitational field.
A: Yes, it can be. In classical mechanics the principle is nothing but the statement that inertial and gravitational masses are identical. As a consequence all inertial forces can be mathematically interpreted as gravitational forces using the standard mathematical machinery of Newtonian mechanics. Your solution is correct.
