# Doubt in fictitious forces chapter in Morin

The question is this -

I know 2 is what the non-inertial frame measures, but isn't $$\frac{d\mathbf{A}}{dt}$$ the real thing, the physical thing? And you can write that too in terms of the unit vectors of the non-inertial frame.

Similarly here, after $$\mathbf{A}$$ is set to be $$\mathbf{r}$$. Isn't $$\frac{d^2\mathbf{r}}{dt^2}$$ the real acceleration? What's stopping one from taking its components in the second frame? Again, I do understand why a is what's measured in the non-inertial frame, but what about $$\frac{d^2\mathbf{r}}{dt^2}$$?

• Please simply ask your physics question without discussing how you think about physics. I suggest removing the first paragraph. Commented Sep 23, 2023 at 17:06
• @Ghoster Done. Sigh.. Commented Sep 23, 2023 at 17:19

but isn't dA/dt the real thing, the physical thing?

Let's write $$\frac{d\mathbf{A}}{dt}=\frac{d}{dt}\mathbf{A}(t)$$. $$\mathbf{A}(t)$$ is just a symbol, we use it to represent the physical vector (which in this context can be viewed as a literal arrow from an origin in space at a given time).

I know 2 is what the non-inertial frame measures

Sure, let's look at a primitive example, say you are dropping a ball on the equator of an asteroid rotating at angular velocity $$\omega$$. An inertial observer measures the position of the ball as $$\mathbf{r}(t)=\left(h_0-\frac{1}{2}g_at^2\right)\mathbf{\hat{y}}'$$, where the primes denote the inertial observer.

In the asteroid's frame we have $$\mathbf{\hat{x}}=\cos(\omega t)\mathbf{\hat{x}}'+\sin(\omega t)\mathbf{\hat{y}}'$$ and $$\mathbf{\hat{y}}=-\sin(\omega t)\mathbf{\hat{x}}'+\cos(\omega t)\mathbf{\hat{y}}'$$. So in the asteroid's frame it sees: $$\mathbf{r}(t)=\sin(\omega t)\left(h_0-\frac{1}{2}g_at^2\right)\mathbf{\hat{x}}+\cos(\omega t)\left(h_0-\frac{1}{2}g_at^2\right)\mathbf{\hat{y}}$$

See this visual. The asteroid's frame is the solid black lines that the dashed lines are projecting to, and the inertial observer's frame is the global stationary gridlines.

You can see that to find $$\frac{d}{dt}$$ of $$\mathbf{r}(t)$$ we need to see how our $$\mathbf{\hat{x}}$$ and $$\mathbf{\hat{y}}$$ change with time, so we use the chain rule.

• Thank you for taking the time to write the answer. But I did understand that the unit vectors on the asteroid change, and we need to use the chain rule. My question is why $\frac{d\mathbf{r}(t)}{dt}$ can't be found, and then projected on to the asteroid axes? The book instead finds $\frac{d\mathbf{r}(t)}{dt}$, but separates it into two parts - with the first part being called the velocity wrt the asteroid. Further in the next comment, as character limit being near. Commented Sep 23, 2023 at 18:32
• $\frac{d\mathbf{r}(t)}{dt} = (\frac{dr_x}{dt}\hat{x} + \frac{dr_y}{dt}\hat{y} + \frac{dr_z}{dt}\hat{z}) + (r_x\frac{d\hat{x}}{dt} + r_y\frac{d\hat{y}}{dt} + r_z\frac{d\hat{z}}{dt} )$ The first term here is called the velocity wrt to the asteroid frame, while I say the total can also be the velocity, and we can just project that onto the axes of the asteroid frame. Commented Sep 23, 2023 at 18:40
• Emphasizing what @NeeladriReddy commented, if we set $g_a=0$ we have this. An inertial observer sees the ball doesn't move, since it feels no force, but in the rotating frame, we (at the green point) see it rotating (because our frame is rotating). If the rotating observer believes they are at rest (the same way we don't feel the rotation of the earth) then since the object is seen rotating, the rotating observer must (falsely) conclude a force acts on the ball (hence the introduction of the fictitious forces). Commented Sep 24, 2023 at 14:25
• Thanks for this example @QPhysl. I see your point - I was saying $\frac{d\mathbf{r}}{dt}$ can be the velocity in the new frame, but you easily disproved it by showing how when $\frac{d\mathbf{r^{'}}}{dt}$ = $\frac{d\mathbf{r}}{dt}$ - clearly the velocity in the non-inertial frame isn't zero. I came up with an example of the ball rotating in an orbit, so velocity in the non-inertial frame would be zero(orbit angular velocity equal to the rotational frame's), but clearly then $\frac{d\mathbf{r}}{dt}$ is not zero. So rotation needs to be taken into account. Commented Sep 25, 2023 at 13:26