Position formula in a non-inertial frame

[kent davidge]

The equation $x = x_0 + vt + at^2/2$ is derived assuming a constant acceleration $a$. My question is , from what frame is this path $x(t)$ described? Can I use it from a non inertial frame?

 

[Nugatory]

Can I use it from a non inertial frame?

Only in special cases and with great care. The initial position coordinate $x_0$, coordinate velocity $v$ (which you should have written as $v_0$ to indicate that it is also an initial constant and not a function of $t$) and coordinate acceleration $a$ are frame-dependent, so must be correctly represented in your non-inertial frame; that's the "great care" part. If $a$ is not constant in this frame, then of course the formula will fail; that's the "special case" part.

 

[kent davidge]

Ok. So let us consider that the acceleration is constant, what should we do next?

Suppose the frame is that of an elevator going up in Earths gravitational field. The accelerator itself has an acceleration $a$ and gravity gives an acceleration $g$ (assume this acceleration is constant too). Can we use a "resultant" acceleration in this case? That is, would we use $a - g$ in the formula for the particle's path?

 

[Nugatory]

Ok. So let us consider that the acceleration is constant, what should we do next?

Suppose the frame is that of an elevator going up in Earths gravitational field. The accelerator itself has an acceleration $a$ and gravity gives an acceleration $g$ (assume this acceleration is constant too). Can we use a "resultant" acceleration in this case? That is, would we use $a - g$ in the formula for the particle's path?

You would use $a+g$ if the elevator is accelerating upwards - the fictitious force needed to make $F=ma$ work in your non-inertial frame is working in the same direction as gravity, pulling the dropped object towards the floor more strongly.

It would be a good exercise to write down the coordinate transformations between the inertial frame in which the surface of the Earth is at rest (you posted this in classical physics so we're considering gravity to be a real force and an object resting on the surface of the Earth to be at rest with zero net force on it) and the non-inertial frame in which the floor of the elevator is at rest, then transform the formula for the trajectory of the object from one frame to the other.

 

[kent davidge]

It would be a good exercise to write down the coordinate transformations between the inertial frame in which the surface of the Earth is at rest (you posted this in classical physics so we're considering gravity to be a real force and an object resting on the surface of the Earth to be at rest with zero net force on it) and the non-inertial frame in which the floor of the elevator is at rest, then transform the formula for the trajectory of the object from one frame to the other.

I don't know how to do it?

May be $$x' = x + v_e t \\ v' = v + v_e + a_e t \\ a' = a + 2 a_e$$ where primes denote the particle's coordinates in the elevator frame and unprimed ones denoted the particle's coordinates in the inertial frame of the Earth. And $a_e$ is the elevator's acceleration.

In the case in consideration, the particle is seen having an acceleration $g$ (from gravity) when it's dropped from the elevator's ceilling, so $a' = g + 2 a_e$.

But that nasty factor of 2 in front of $a_e$ seems to be telling me that I'm wrong in all of this.

 

[PeroK]

I don't know how to do it?

May be $$x' = x + v_e t \\ v' = v + v_e + a_e t \\ a' = a + 2 a_e$$ where primes denote the particle's coordinates in the elevator frame and unprimed ones denoted the particle's coordinates in the inertial frame of the Earth. And $a_e$ is the elevator's acceleration.

In the case in consideration, the particle is seen having an acceleration $g$ (from gravity) when it's dropped from the elevator's ceilling, so $a' = g + 2 a_e$.

But that nasty factor of 2 in front of $a_e$ seems to be telling me that I'm wrong in all of this.

Everything can be deduced from the position of the origin in the accelerating frame. Let's assume a particle moves with initial position $x_0$, initial velocity $v_0$ and constant accleration $a$ in an inertial frame.

Further, let's assume that the origin of an accelerating frame is at $\frac12 a_e t^2$. You can generalise more by giving the acclerating frame an initial non-zero position and velocity, but that won't affect the equation for acceleration.

Now, we can write down the position of the particle in the accelerating frame:

$x' = x_0 + v_0 t + \frac12 at^2 - \frac12 a_e t^2$

And, twice differentiating wrt time gives:

$a' = a - a_e$

In the special case where $a = -g$ (where upwards is positive and $g$ is taken to be positive), then:

$a' = -g - a_e$

I.e. if the elevator is accelerating up, then the dropped object accelerates down in the elevator frame by more than $g$ etc.

 

[Nugatory]

I don't know how to do it?

Here's the general process you'll go through. (For this particular problem it's overkill; once you understand the process you can skip the intermediate steps where you formally write down the coordinate transformations, skip straight to writing the trajectory in the primed coordinates as @PeroK just did).

First we write down the coordinate transformations between the primed and the unprimed frames. These have nothing to do with the motion of any particles anything else; they just tell you how to calculate the primed coordinates of a given point from the unprimed coordinates and vice versa. The transformations will be four functions$$x'=x'(x,y,z,t) \\y'=y'(x,y,z,t)\\z'=z'(x,y,z,t)\\t'=t'(x,y,z,t)$$ and their inverses will give you the coordinates in the unprimed frame of a point, given its coordinates in the primed frame. In this problem with the elevator accelerating at $a_e$ we can choose the origins of the two frames to coincide and be at rest relative to one another at time zero; then the transformations are straightforward:$$x'(x,y,z,t)=x-a_et^2/2 \\y'(x,y,z,t)=y\\z'(x,y,z,t)=z\\t'(x,y,z,t)=t$$and say that everything is the same except that the the origin of the primed frame is displaced by $a_et^2/2$ from the origin of the unprimed frame. The only non-trivial inverse is $x(x',y',z',t')=x'+a_et'^2/2$.

Ok, so we have two coordinate systems and we know how to transform between them. Now we're ready to take on the problem of transforming the trajectory of a moving particle from one frame to the other. The trajectory is described by three functions giving the coordinates of the position of the particle at time $t$:$$x_p=x_p(t)\\y_p=y_p(t)\\z_p=z_p(t)$$We need to find the corresponding functions $x_p'(t')$, $y_p'(t')$, and $z_p'(t')$.
In this particular problem we have $x_p(t)=x_0+v_0t-gt^2/2$ because that's the trajectory you gave when you started this thread ($y_p(t)$ and $z_p(t)$ are zero of course, and $-g$ is the downwards gravitational acceleration).

From here it's just algebra: we use the transformation equations to get rid of all the unprimed coordinates to get the description of the trajectory using only primed coordinates. For this problem it's pretty easy:$$x'_p(t)=x'(x_p(t), y_p(t), z_p(t))\\=x_0+v_0t-gt^2/2-a_et^2/2\\=x_0+v_0t-gt^2/2-a_et^2/2\\=x_0+v_0t'-(a_e+g)t'^2/2$$where we used $t'=t$ in the last step.

That's the equation that @PeroK supplied above for the trajectory of the particle in the primed frame. Differentiate once with respect to $t'$ to get the speed in that frame, differentiate again to get the acceleration.

Note also that that the formula you asked about in the first post of this thread works in both the primed and the unprimed frame, although the speed, acceleration and distance traveled are frame-dependent and different in the two frames. That won't be true for all transformations, but it works for this particularly simple case.

 

[kent davidge]

Thanks, but how would I know that I should use $x'(x,y,z,t) = x - a_e t^2 / 2$ besides the fact that we are using the convention that the systems origins coincide at $t = 0$. I mean, in principle I could use $x'(x,y,z,t) = x + \alpha t$ where $\alpha$ is a constant giving $\alpha t$ dimensions of length, and the two origins would still coincide at $t = 0$.