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I am reading Steven Weinberg's book Gravitation and Cosmology. On pages 67-68, he begins his discussion of the Principle of Equivalence of Gravitation and Indertia by saying the following:

The Principle of Equivalence rests on the equality of gravitational and inertial mass as demonstrated by Galileo, Huygens, Newton, Bessel, and Eötvös. Einstein reflected that, as a consequence, no external static homogeneous gravitational field could be detected in a freely falling elevator, for observers, their test bodies, and the elevator itself would respond to the field with the same acceleration. This can be easily proved for a system of particles $N$, moving with nonrelativistic velocities under the influences of forces $\mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)$ (e.g. electrostatic or gravitational forces) and an external gravitational field $\mathbf{g}$. The equations of moation are \begin{equation} m_N \frac{d^2\mathbf{x}_N}{dt^2} = m_N\mathbf{g} + \sum_M \mathbf{F}(\mathbf{x}_N-\mathbf{x}_M) \end{equation} Suppose we perform a non-Galilean space-time coordinate transformation \begin{equation} \mathbf{x}'=\mathbf{x}-\tfrac{1}2{}\mathbf{g}t^2 \phantom{XXXX} \mathbf{t}'=\mathbf{t} \end{equation} Then $\mathbf{g}$ will be canceled by an inertial "force," and the equation of motion will become \begin{equation} m_N \frac{d^2\mathbf{x}'_N}{dt'^2} = \sum_M \mathbf{F}(\mathbf{x}'_N-\mathbf{x}'_M) \end{equation} Hence the original observer O who uses coordinates $\mathbf{x}t$, and his freely falling friend O' who uses $\mathbf{x}'t'$, will detect no difference in the laws of mechanics, except that O will say that he feels a gravitational field and O' will say that he does not.

Remark on My Confusion

To me, the first equation looks like it has both a gravitational force $m_N\mathbf{g}$ and some other force $\sum_M \mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)$. The third equation then looks like it only has $\sum_M \mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)$. If I am understanding him correctly, what he is saying is that if we pick a coordinate transformation such as $\mathbf{x}'=\mathbf{x}-\tfrac{1}2{}\mathbf{g}t^2$, then the force $m_N\mathbf{g}$ will seemingly disappear in the $\mathbf{x}'$ reference frame.

But I don't understand where it goes. Does this mean the term $\frac{d^2\mathbf{x}_N}{dt^2}$ changes from the $\mathbf{x}'$ frame to the $\mathbf{x}$ frame? In other words, if $m_N \frac{d^2\mathbf{x}_N}{dt^2}=m_N \frac{d^2\mathbf{x}'_N}{dt'^2}$ and $\sum_M \mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)=\sum_M \mathbf{F}(\mathbf{x}'_N-\mathbf{x}'_M)$, then the equations don't make sense. That would be like saying $a = b + c$ and $a = b$ but $c \neq 0$.

My Question:

(1) Does $m_N \frac{d^2\mathbf{x}_N}{dt^2}=m_N \frac{d^2\mathbf{x}'_N}{dt'^2}$?

(2) Does $\sum_M \mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)=\sum_M \mathbf{F}(\mathbf{x}'_N-\mathbf{x}'_M)$?

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  1. No.

  2. Yes, because the forces only depend on the separations. (Translation symmetry.)

Elaboration of 1.: We have $\mathbf{x}'=\mathbf{x}-\tfrac{1}{2}\mathbf{g}t^2$. Let us calculate the acceleration: $$\frac{d^2\mathbf{x}'}{dt^2}=\frac{d^2}{dt^2}(\mathbf{x}-\tfrac{1}{2}\mathbf{g}t^2)=\frac{d^2\mathbf{x}}{dt^2}-\mathbf{g}$$ Thus the force is $$m\frac{d^2\mathbf{x}'}{dt^2}=m\frac{d^2\mathbf{x}}{dt^2}-m\mathbf{g}$$ Since $\mathbf{F}=\mathbf{F}'$ we have $$m\frac{d^2\mathbf{x}'}{dt^2}=\mathbf{F}'=m\frac{d^2\mathbf{x}}{dt^2}-m\mathbf{g}=\mathbf{F}$$ Thus $$m\frac{d^2\mathbf{x}'}{dt^2}=\mathbf{F}'$$ $$m\frac{d^2\mathbf{x}}{dt^2}=m\mathbf{g}+\mathbf{F}$$ the unprimed frame experiences a gravitational force $m\mathbf{g}$ whereas the primed frame does not.

Elaboration of 2.: We have $\mathbf{x}'=\mathbf{x}-\tfrac{1}{2}\mathbf{g}t^2$. Thus $$\mathbf{x}_N'-\mathbf{x}_M'=(\mathbf{x}_N-\tfrac{1}{2}\mathbf{g}t^2)-(\mathbf{x}_M-\tfrac{1}{2}\mathbf{g}t^2)=\mathbf{x}_N-\mathbf{x}_M$$ implies $$\mathbf{F}(\mathbf{x}_N'-\mathbf{x}_M')=\mathbf{F}(\mathbf{x}_N-\mathbf{x}_M)$$

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  • $\begingroup$ That was much simpler than I expected. What do $\mathbf{x}_N$ and $\mathbf{x}_M$ stand for? $\endgroup$ – Stan Shunpike Feb 10 '15 at 16:39
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    $\begingroup$ I take it $M$ is an index set denoting the various bodies in the gravitational field. $\mathbf{x}_N$ is the $N$th body. $\endgroup$ – Ryan Unger Feb 10 '15 at 16:48
  • $\begingroup$ By translation symmetry, you mean 2 is invariant under transformation of $\mathbf{x}$ to $\mathbf{x}'$ and you demonstrate that in your second elaboration, correct? Is this spatial translation? $\endgroup$ – Stan Shunpike Feb 10 '15 at 18:31
  • $\begingroup$ Translation symmetry means that if the positions of both particle $N$ and $M$ changes by the same amount, then the force between them does not change. The coordinate transformation to the accelerated frame is a time-dependent translation. Thus the force is invariant. $\endgroup$ – Ryan Unger Feb 10 '15 at 19:03
  • $\begingroup$ I get it. In the first case, there is a force but not an accelerating frame. In the second, there is an accelerating frame but no experienced gravitational force. Yet the laws or mechanics remain the same because the new equation still holds. $\endgroup$ – Stan Shunpike Feb 10 '15 at 20:00

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