It is known that total linear momentum of a system is conserved in an inertial frame in which net force is zero. I tried extending it to non inertial frames by taking into account the pseudo forces. The system I chose was two particles whose mass are m1 and m2 of opposite charges (+1C, -1C respectively) revolving around their centre of mass and the distance between them and the centre of mass is r1 and r2.

Let the first inertial frame be A . I chose a non inertial frame (let it be B) in which the positive charge is at rest.

Now, since: $\bar F_{B2} = \bar F_{A2} + \bar F_{fictitious} $ -----(1)
(where $\bar F_{A2}$ and $\bar F_{B2}$ are the net forces on the negative charge in the first frame(inertial) and second frame(non inertial)).

Here: $\bar F_{A2} = \frac{K}{(r_1 + r_2)^2}\ \hat r $ ----(2) (where $\hat r$ is the unit vector always facing towards the positive charge from the com)


cited from wikipedia

click here to see the derivation of this formula -----(3) (here m is m2, $ \bar a_{AB}$ is the acceleration of the frame B with respect to frame A and the $ \bar v_j \ and\ \bar x_j \ $ are the velocity and position vectors of negative charge with respect to frame B, and $\bar u_j$ is the unit vector of frame B.)

Here the second term and third term of the equation of pseudo force will become zero as our frame is not rotating (it is revolving around the com).Now the $ \bar a_{AB} = -\frac{K}{(r_1 + r_2)^2\times{m_1}}\ \hat r$ (where $\hat r$ is the unit vector always facing towards the positive charge from the com).

Now $ m_2\bar a_{AB} =-\frac{K\times{m_2}}{(r_1 + r_2)^2\times{m_1}}\ \hat r $ ----(4)

Now using eqn (1), (2), (3) and (4):

$\bar F_{B2} = \frac{K}{(r_1 + r_2)^2}\ \hat r - \frac{K\times{m_2}}{(r_1 + r_2)^2\times{m_1}}\ \hat r $

$\bar F_{B2} = \frac{K\times{(m_1 - m_2)}}{(r_1 + r_2)^2\times{m_1}}\ \hat r \ \neq 0 $

But , $\bar F_{B1} = 0 $ ,since we chose the frame B in which net force on positive particle is zero. ((where $\bar F_{B1}$ is the net force on the positive charge in the second frame(non inertial))

Since, $\bar F_{ net} = \bar F_{B1} + \bar F_{B2}$ (where $F_{net}$ is the net force on the system in frame B)

But since $\bar F_{B1} = 0 $ and $\bar F_{B2}\neq 0 $, this implies that , $\bar F_{ net} \neq 0 $, which in turn implies that law of conservation of linear momentum is not valid in this frame.

Can somebody please tell me where I went wrong or missed something or that what I am trying is not possible? It will be really helpful.


It is really hard to follow all of your work, and there are some assumptions I am not entirely sure about, but we can circumvent all of this by noting that momentum is not conserved in non-inertial frames with fictitious forces.

The reason this is the case is because fictitious forces do not obey Newton's third law. What this means is that if a fictitious force acts on an object and changes its momentum, there is no "equal but opposite" force acting on something else that will change its momentum in the opposite way to keep momentum conserved. Therefore, momentum is not conserved with fictitious forces.

  • $\begingroup$ Does this mean that there is no possible way to show that linear momentum is conserved in non inertial frames? $\endgroup$ – Dhrxv Apr 15 '20 at 14:03
  • $\begingroup$ @DhruvDhangar What part of my answer could be explained better? $\endgroup$ – BioPhysicist Apr 15 '20 at 14:05
  • $\begingroup$ I understood what you were trying to convey, but I read somewhere that if pseudo force were introduced in a non inertial frame then it would counter the changes in the momentum resulting in conservation of it. Link to that source:google.com/url?sa=t&source=web&rct=j&url=https://… $\endgroup$ – Dhrxv Apr 15 '20 at 14:14
  • $\begingroup$ @DhruvDhangar That link says nothing about non-inertial frames $\endgroup$ – BioPhysicist Apr 15 '20 at 14:17
  • $\begingroup$ Changes in the total momentum of the system due to us choosing a non inertial frame of reference. $\endgroup$ – Dhrxv Apr 15 '20 at 14:17

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