Formula for the Bekenstein bound S &#x2264;<!-- ≤ --> 2 &#x03C0;<!-- π -->

This is the common problem of a charged particle moving in a static electric and magnetic field.
Say $\mathbf{\text{E}}=(E_x,0,0)$ and $\mathbf{\text{B}}=(0,0,B_z)$
In the inertial frame of reference, the equation of motion is (1):
$\frac{d\mathbf{\text{v}}}{dt}=-\frac{q\mathbf{\text{B}}}{m}\times \mathbf{\text{v}}+\frac{q}{m}\mathbf{\text{E}}$
We can find equations for $v_x$ an $v_y$ and see that the resulting motion is a circular orbit with a constant drift velocity $v_d=\frac{E_x}{B_z}$.
Surely I should get the same answer if I solve the problem in a rotating frame of reference?
I know that (2):
$\frac{d\mathbf{\text{v}}}{dt}{|}_{Inertial}=\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}+\mathit{\omega }\times \mathbf{\text{v}};$
If I use Eq. (1) as the LHS of Eq. (2), and choose $\mathit{\omega }=-\frac{q\mathbf{\text{B}}}{m}$, then I get (3):
$\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}=\frac{q}{m}\mathbf{\text{E}};$
How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field E=$\mathbf{\text{E}}=(E_x,0,0)$ change form in the rotating frame?

In physics problems, the earth is usually considered to be an inertial frame. The earth has a gravitational field and the second postulate of the general theory of relativity says:
"In the vicinity of any point, a gravitational field is equivalent to an accelerated frame of reference in gravity-free space (the principle of equivalence)."
Does this mean that accelerating frames of reference can be inertial?

"You are traveling in a car going at a constant speed of 100 km/hr down a long, straight highway. You pass another car going in the same direction which is traveling at a constant speed of 80 km/hr. As measured from your car’s reference frame this other car is traveling at -20 km/hr. What is the acceleration of your car as measured from the other car’s reference frame? What is the acceleration of the other car as measured from your car’s reference frame?"

Shouldn't they both appear to have an acceleration of zero, because both velocities are constant? I can imagine sitting in the faster car and watching the slower car, its speed would not appear to change, only its position?

For a car that is accelerating linearly, the non-inertial frame of reference is the driver in the car where from his reference frame, the car is stationary. It is so called stationary because the non-inertial frame of reference has the same acceleration as the car. Is like the car's acceleration "transform" the driver frame of reference into a non-inertial. That's why in the non-inertial frame of reference, there is no force acting on the car.
But when the car is driving in circles at a constant speed, in the non-inertial frame of reference there is a force acting on the car, which is the centripetal force. Why isn't this frame of reference like the above, not having the acceleration found in their each respective inertial reference frame? Why can't we have a non-inertial reference frame(due to rotation) whereby there is no centripetal force, subsequently eradicating the need for a centrifugal force?

"Strictly speaking, Newton’s laws of motion are valid only in a coordinate system at rest with respect to the 'fixed' stars. Such a system is known as a Newtonian, or inertial reference frame. The laws are also valid in any set of rigid axes moving with constant velocity and without rotation relative to the inertial frame; this concept is known as the principle of Newtonian or Galilean relativity."
Why should the inertial frame of reference not spin?

"If the frame of reference is translated or rotated, the vector doesn't change."
Although the length of the vector won't change, the angle that this vector makes with the positive direction of the newly defined x-axis changes, no? Hence, how is it possible to state that the vector doesn't change? Have I misunderstood the basic definition of a vector?

The Fourier rate equation of heat conduction states:
$\overrightarrow{q}=-k\mathrm{\nabla }T$
But I'm wondering if this is valid in every frame of reference, because the heat flux $\overrightarrow{q}$ does obviously change when the area under consideration is moving. The equation is often applied to moving fluids, and because there is no objective way to determine a non-moving frame of reference in a moving substance, I guess that the equation is valid for every inertial frame of reference, but I just wanted to be sure.