2
$\begingroup$

An inertial frame according to wikipedia

"is a frame of reference that is not undergoing any acceleration. It is a frame in which an isolated physical object—an object with zero net force acting on it—is perceived to move with a constant velocity (it might be a zero velocity) or, equivalently, it is a frame of reference in which Newton's first law of motion holds."

I am having trouble reconciling this first part of the definition "A frame that has zero acceleration", with "a frame of reference under which newtons first law of motion holds"

Family of inertial frames:

Consider a frame, that measures position vectors as $\vec{x}$, and that this frame is assumed to have "zero acceleration" ( What ever that means)

In this frame of reference, the force on an object is $\vec{F} = m\vec{a}$

A family of frames that leave newtons first law of motion unchanged are

$$x' = x -\vec{v}t$$

$$\frac{d\vec{x'}}{dt}=\frac{d\vec{x}}{dt} - \vec{v}$$ $$\frac{d^2\vec{x'}}{dt^2}=\frac{d^2\vec{x}}{dt^2}$$ $$\vec{F'} = \vec{F}$$

And hence this family of frames for all constant vectors v is said to be inertial.

"Non inertial frames:"

Now lets consider a frame transformation with respect to the "zero acceleration" frame, such that:

$$x_{\tau} = x -\vec{v}t^2$$ $$\frac{d\vec{x}_{\tau}}{dt} = \frac{d\vec{x}}{dt} - 2\vec{v}t$$ $$\frac{d^2\vec{x}_{\tau}}{dt^2} = \frac{d^2\vec{x}}{dt^2} - 2\vec{v}$$ $$\vec{F}_{\tau} = \vec{F} -2\vec{v}$$

From here we can see that because this frame is accelerating, newtons laws are not frame invariant with respect to $x$ and $x_{\tau}$. However I wouldnt really dismiss the use of newtons laws in this frame so easily, since this frame is accelerating only WITH RESPECT to frame x,

Consider a frame transformation from frame $\vec{x}_{\tau}$ to a frame $\vec{x}_{\lambda}$:

$$\vec{x_{\lambda}} = \vec{x}_{\tau} - vt$$ $$\frac{d\vec{x}_{\lambda}}{dt}=\frac{d\vec{x}_{\tau}}{dt} - \vec{v}$$ $$\frac{d^2\vec{x'}_{\lambda}}{dt^2}=\frac{d^2\vec{x}_{\tau}}{dt^2}$$ $$\vec{F}_{\lambda} = \vec{F}_{\tau}$$

Hence, it would appear that newtons laws are invariant with respect to a frame transformation between $\vec{x}_{\tau}$ to a frame $\vec{x}_{\lambda}$, I can identify a whole family of frames related by the same transformation, for all values of the constant vector v, under which newtons laws all hold, and thus fit the definition of an inertial frame.

These second family of frames seem "Inertial" respect to each other, dispite being non inertial with respect to the first family of "inertial" frames. So does it actually make sense talking about an inertial frame, over a family of inertial frames?

Acceleration apparently is meant to be absolute, however given the above analysis, I could have just as easily started with the second family of frames, reverse the transformation and say that the first family of frames are non inertial as they have a relative acceleration with the second family of frames.

Is acceleration absolute? I cant really see why people say so, given we measure acceleration with RESPECT to a specific "starting frame". i can only make sense of talking about a SET of inertial frames, over "This is inertial, this is not".

Imagine a vector field in space that was constantly accelerating everything with respect to some rest frame, in the frame of reference of those accelerating objects, everyone would have a net zero relative acceleration, and thus in that universe, if the creatures there devised newtons laws of motion, they would claim that THEY are inertial since they belong to the Set of inertial frames described by their motion with zero relative acceleration, that Is infact accelerating with respect to "some frame".

I would prefer answers not refering to special relativity, since the concept of absolute acceleration was introduced way before einstein.

$\endgroup$
4
  • 1
    $\begingroup$ $x_{\tau} = x -\vec{v}t^2$ this equation is invalid, units does not match in terms. $\endgroup$ Commented Jan 31, 2023 at 21:44
  • $\begingroup$ $\vec{v}$ is not velocity in this context though, just some constant vector v, corresponding to a velocity of $2\vec{v} t [m/s]$, $2\vec{v}$ having units $m/s^2$ (its acceleration) giving the total units $m/s^2 * s^2 = m.$ $\endgroup$ Commented Jan 31, 2023 at 22:54
  • 2
    $\begingroup$ You can measure acceleration with an accelerometer without referring to any other frame. $\endgroup$
    – John Doty
    Commented Feb 1, 2023 at 0:55
  • $\begingroup$ Accelerometers measure changes in acceleration throughout the accelerometer, in freefall when all parts are accelerating at the same time the accelerometer reads zero for example in freefall $\endgroup$ Commented Feb 1, 2023 at 11:28

1 Answer 1

4
$\begingroup$

Newton's first law of motion says that an object that starts moving at a constant velocity, and isn't acted on by an external force, will continue moving at that velocity forever.

Let me define two frames based on the wording in your question:

  • The "zero acceleration frame"
  • The "$\tau$ frame" (which in your notation is accelerating at a constant acceleration $-2v$ relative to the zero acceleration frame).

Let me consider doing a thought experiment in these two frames. In both frames, I sit in a spaceship out in space (so we can ignore the effect of gravity). Then I let go of an apple with zero velocity relative to teh walls of the ship.

In the zero acceleration frame, the apple will not hit any of the walls of the ship. It will remain moving at the same exactly velocity I released it at.

In the $\tau$ frame, the apple will immediately begin accelerating in the direction of $-\vec{v}$.

Now, what I understand your objection to be, is that we can simply say that Newton's first law applies in the $\tau$ frame, but there is also a constant force $-2m\vec{v}$ that acts "by default".

To some extent this is semantics. However, I take the point of Newton's laws to be that an applied force is one that an experimenter can choose to apply, or not, at least in principle. For example, we can remove an applied gravitational force by going into space. However, this fictitious $-2\vec{v}$ constant force is one that applies to all objects, everywhere, in the $\tau$ frame. It cannot be removed by moving your experiment somewhere else, or changing conditions of the experiment (while remaining in the same family of inertial frames).

Now I should emphasize that it is not the "constant-ness" of the force that makes it fictitious. You can also generate non-inertial forces that depend on space, such as the Coriolis effect. However, these fictitious forces are all have a certain strangeness when you try to think of them as being fundamental; for example, the force acting on each particle is proportional to the mass, so that the mass cancels out in Newton's second law. Additionally, there are no "experimental knobs" you can use to tune these forces to zero.

This property is the one that makes the zero-acceleration frame special; there are no "fictitious" forces.

$\endgroup$
14
  • $\begingroup$ Yes you knew what my objection would be to this answer, continuing on from this objection, if there were a force that is always present in this frame (relative the the force in the original frame) I could simply instead re write this as $\vec{F}_{\tau} + 2v = \vec{F}$ to say the force in the original frame is the same as the other frame "plus some constant acceleration" which then leading from your arguement is a force i "cannot turn off" hence the original frame is fictitious. The notion of placing an object and it automatically follows the frame acceleration may seem weird, until you look- $\endgroup$ Commented Feb 1, 2023 at 11:35
  • $\begingroup$ Look At it from the other perspective, and see that this is the same thing that could be shown to happen to the original frame, as viewed by the second. I cannot shake the idea that is it all relative since the math can be rearranged to make either frame the original frame and the other the secondary, which your conclusions can also be applied to $\endgroup$ Commented Feb 1, 2023 at 11:36
  • $\begingroup$ (1) Your equation relating the forces in the different frames should read $F_\tau + 2m\vec{v} = \vec{F}$ -- you need an extra factor of $m$ in the fictitious force (this is a clue that it is fictitious). (2) "The notion of placing an object and it automatically follows the frame acceleration may seem weird, until you look Look At it from the other perspective, and see that this is the same thing that could be shown to happen to the original frame, as viewed by the second." -- I do not know what this means. In the zero acceleration frame, an object you put down will not accelerate. $\endgroup$
    – Andrew
    Commented Feb 1, 2023 at 13:25
  • $\begingroup$ $\vec{F}_{\tau} + 2m\vec{v} = \vec{F}$ By definition, if we assume that relative to frame $\vec{x}$, there are no forces acting on it. Moving into the frame described by $ x_{\tau} = \vec{x} - \vec{v}t^2$ we will see in this new frame, that the force in frame $\vec{x}_{\tau}$ is : $$\vec{F}_{\tau} + 2m\vec{v} = 0$$ $$\vec{F}_{\tau} = - 2m\vec{v} $$ Hence, If now, I assume that $\vec{x}_{\tau}$ is my preferred original frame, This force is in frame $\vec{x}$ is, although zero, different from the "new original" frame hence there is still a ficticious force in this frame, causing aZeroForce - $\endgroup$ Commented Feb 1, 2023 at 13:59
  • 1
    $\begingroup$ "Imagine that my new preffered frame 𝑥⃗ 𝜏, has an object placed in it that has a constant velocity relative to this frame." -- Your argument hinges on this being a symmetrical situation with the zero acceleration frame. But it is not, in order for the particle to have a constant velocity in the $\tau$ frame, you need to apply forces which cancel the fictitious ones. In the zero acceleration frame, you do not need to apply any forces to the particle for it to move at a constant velocity. $\endgroup$
    – Andrew
    Commented Feb 1, 2023 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.