If we look at a particle moving in the positive $x$ direction from a frame that is accelerating in the positive $y$ direction then its acceleration will be different from that in an inertial frame and hence the force on the particle must also be different in this frame.

Is this right?

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    $\begingroup$ Does this answer your question? Does force definition depend on frame of reference? $\endgroup$ – Guru Vishnu Mar 28 at 8:18
  • $\begingroup$ I have already read that but the answer does not explain why force is independent of frame. I get it that if we see from a non inertial frame then we have to apply a pseudo force to proceed normally though it is not real and is acting only this frame and since it is acting it must add up with other real forces to make up for net force in this frame. $\endgroup$ – WantToLearnPhysics Mar 28 at 8:26
  • $\begingroup$ Are we talking Newtonian mechanics here or special relativity? $\endgroup$ – Harry Johnston Mar 28 at 23:27
  • $\begingroup$ Newtonian mechanics, I am not a major a physics and have almost zero knowledge about relativity $\endgroup$ – WantToLearnPhysics Mar 29 at 11:10

As described in the question, you have only said that the frame is moving. But if motion of the frame is uniform, then acceleration is not changed and therefore the force is not changed (Newton II).

If the frame is accelerating relative to an inertial frame, then there will be an inertial force (or pseudo force). Inertial forces are distinct from active forces (inertial literally means not acting). The inertial force accounts for the difference in the acceleration of the particle when seen from the accelerating frame.

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  • $\begingroup$ Sorry I meant to write accelerating frame, I have edited it $\endgroup$ – WantToLearnPhysics Mar 28 at 9:15
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    $\begingroup$ @WantToLearnPhysics I think Charles answer still sufficiently answers your edited question? See for further reference en.wikipedia.org/wiki/Fictitious_force $\endgroup$ – TheoreticalMinimum Mar 28 at 9:46
  • $\begingroup$ Thanks, I got it :) $\endgroup$ – WantToLearnPhysics Mar 28 at 12:53
  • $\begingroup$ How is acceleration not changed? If a rocket has an acceleration that's constant in its own frame, its velocity will asymptotically approach c, so its acceleration will approach 0. Moreover, its mass will change, so even if the acceleration is constant, the force will change. $\endgroup$ – Acccumulation Mar 28 at 18:39
  • $\begingroup$ @Acccumulation, That is not how it goes. It may have a proper acceleration, but its own frame is also accelerating. Its velocity is always 0 in its own frame. Its rest mass does not change. An inertial observer should calculate using vector quantities, which are the same for the same for all observers. He should not naively try to apply Newtonian formulae in his own coordinates, as one all too frequently finds in low grade descriptions. $\endgroup$ – Charles Francis Mar 28 at 19:05

Note: this answer is based on Newtonian mechanics not special or general relativity.

If we start in an inertial frame and consider a particle acted on by a real force in the $x$ direction then the motion of the particle is given by

$m \ddot x=F$

$m \ddot y=0$

$m \ddot z=0$

Now, consider a reference frame initially at rest with respect to the inertial frame and accelerating at $a$ in the $y$ direction then to transform between coordinates we can use


$Y= y -\frac{1}{2}a t^2$


So in this frame the motion of the particle is given by

$m \ddot X =m \ddot x = F$

$m \ddot Y =m \ddot y -ma=-ma$

$m\ddot Z=m\ddot z=0$

So in the accelerated frame the same real force $F$ acts in the $X$ direction but in addition there is a term that looks like a force $-ma$ acting along $Y$ which is the direction of the non-inertial frame’s acceleration.

This term is often called a fictitious force or an inertial force, but although it is called a force it is an artifact of the reference frame and not an interaction between objects, meaning that it does not follow Newton’s 3rd law.

So in a non inertial reference frame the real force still exists but in addition there is an inertial force present.

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All forces can be categorised under the four fundamental forces(or a combination of them).



3.Stong Nuclear force

4.Weak Nuclear force

Let us consider an example - Gravitational force. $$F=\frac{Gm_1m_2}{r^2}$$ G is an universal constant(independent of frame of reference) $m_1$ and $m_2$ are also frame independent. r-the separation between the 2 point masses is also independent(displacement is frame dependent, but not separation). Hence the net force will also be independent of the frame of reference. Similarly one can argue for the other forces. Hence we consider Force to be independent of frame of reference.

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  • $\begingroup$ I'd like to remind that this discussion hasn't included relativity. Separation does change when relativity is taken into account. This answer is to give you an idea, not an accurate one to be honest. $\endgroup$ – Vilvanesh Mar 28 at 9:27
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    $\begingroup$ That was a very good method of explanation, thanks :) $\endgroup$ – WantToLearnPhysics Mar 28 at 12:54
  • $\begingroup$ Separation and mass are both frame dependent. $\endgroup$ – Acccumulation Mar 28 at 18:35

I will answer this with Galilean relativity for Newtonian mechanics. These are elementary boosts with $\vec x\rightarrow \vec x+\vec ut$ and velocity change $\vec v\rightarrow \vec v+\vec u$. Newtonian force is $\vec F=\frac{d\vec p}{dt}$, such that momentum is $\vec p=m\vec v$. Momentum is the frame dependent, but clearly because $\frac{d\vec u}{dt}=0$ force is not frame dependent. It also means that momentum, while frame dependent, does not increase or decrease with time due to a change of frame, and is thus conserved.

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  • $\begingroup$ Why is du/dt= 0, particle does have acceleration $\endgroup$ – WantToLearnPhysics Mar 28 at 12:45
  • $\begingroup$ Newton's first law of motion implies that one observes physics in an inertial or nonaccelerated frame. So for $\vec u$ the boosted velocity of a frame it is chosen so that it is constant or the frame is inertial. $\endgroup$ – Lawrence B. Crowell Mar 30 at 14:29

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