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TL;DL

In an inertial frame of reference, one fundamental law that always holds is the conservation of momentum. If you take the reference frame of one of the interacting objects the conservation of momentum no longer holds, so it is possible to detect accelleration (or a force). If this is true then why (in general reletivity) can we consider the motion of a falling object to be inertial?

If the answer requires the mathematics of general relativity, I am comfortable with it as I have studied it before. Even though I'm comfortable with it, it's easier to understand my question from a Newtonian perspective (However the principles should stay the same when upgrading to special and then general reletivity)

My working

If we assume that $\gamma$ is a list of all properties needed to work out the force on an object (i.e. position, mass, etc..) we can write the force on an object as $$\vec{F}(\gamma)$$

Let's also assume that we observe two particles interacting from a distance of infinity as to not interfere with the particles and to assume that we are in an inertial frame. We observe two particles with initial velocities $\vec{v_1} = \vec{0}, \vec{v_2} = \vec{0}$ and therefore can calculate the momentum of each particle to be $\vec{p_1} = \vec{0}$ and $\vec{p_2} = \vec{0}$; therefore the total initial momentum of the $\vec{P} = \vec{0}$. Since we are in an inertial frame this quantity should not change. We can calculate the new momentum after $dt$ seconds using the force $\vec{F}(\gamma)$. We can write the force in terms of momentum $$\frac{\vec{dp_1}}{dt} = \vec{F}(\gamma)$$ Solving for $dp$ we get $$\vec{dp_1} = \vec{F}(\gamma) dt$$ Adding this onto the momentum of the particle gives $$\vec{p_1} = \vec{F}(\gamma) dt$$ This means that the total momentum is now $$\vec{P} = \vec{F}(\gamma) dt$$ To counteract this we can bring in the conservation of momentum to make sure the other object has equal but opposite momentum $$\vec{p_2} = -\vec{p_1}$$ Which means that $$\vec{p_2} = - \vec{F}(\gamma) dt$$ This makes the total momentum equal to $0$. However, if we consider the new velocities we get $$\vec{v_1} = \frac {\vec{F}(\gamma)}{m_1} dt$$ $$\vec{v_2} = -\frac {\vec{F}(\gamma)}{m_2} dt$$ We can now observe the interaction from the frame of reference of the first particle by doing the transformation $x_2' = x_2 - x_1$, $v_2' = v_2 - v_1$ this implies that the new velocity is $$v_2' = - \frac {\vec{F}(\gamma)}{m_2} dt - \frac {\vec{F}(\gamma)}{m_1} dt$$ $$v_2' = - \vec{F}(\gamma) ( \frac {1}{m_2} + \frac {1}{m_1}) dt$$ This implies that the momentum for the second object is now $$p_2' = - \vec{F}(\gamma) ( 1 + \frac {m_2}{m_1}) dt$$ Since we are in the frame of reference of the first object, it is not moving reletive to it's self, therefore has $0$ momentum this means that there is now a global change of momentum from the perspective of the falling mass.

So how can it be that the falling observer is inertial? How does general reletivity account for the break in conservation of momentum when calling the falling observer inertial?

A more intuitive way of explaining this is to think about falling towards the earth, the earth is coming up to you gaining momentum without another object to interact with to move in the opposite direction to keep momentum constant so, the only logical thing to conclude is that you are that person that is accellerating. However, this breaks General Reletivity's postulate that a falling observer is an inertial.

Thanks Josh.

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  • $\begingroup$ Newtonian and gr has different intuition for force $\endgroup$ Nov 12 '20 at 18:46
  • $\begingroup$ @PrateekMourya I know that gravity isn't really a force, however if the definition of a inertial frame is such that momentum is conserved. Then it would imply that gravity is a force if when free falling momentum is not conserved. If momentum is conserved from the falling perspective then what am i missing? $\endgroup$ Nov 12 '20 at 19:15
  • $\begingroup$ You should ask john rennie sir , he is very good at these things chat.stackexchange.com/users/37155/john-rennie $\endgroup$ Nov 13 '20 at 2:26
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The first issue is that when spacetime is curved inertial frames are only local. They are local free falling frames, but they are only local. So your setup with infinitely separated objects is a non-starter. We must first bring your particles in close to each other.

The second issue is that if there is a force $\vec F(\gamma)$ acting on the object in a local inertial frame then the object is not inertial. Therefore when you transform into its rest frame the rest frame is also non inertial. We do not expect the momentum to be conserved in a non inertial frame.

The third issue is with your comment about the earth gaining momentum. That is related to the first issue. Remember, the inertial frame is local, so it doesn’t cover the whole earth, just the nearby part of it. That nearby section of ground experiences a large upward pressure force from the earth underneath it. Since gravity is absent in a local inertial frame there is no downward force to counteract the upward pressure. The section of ground therefore accelerates upward as predicted by the unbalanced upward force. Indeed, this upward acceleration is easily measurable by an accelerometer placed on the ground.

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