Tensor rank of acceleration in Newtonian mechanics Is the Wikipedia page on acceleration as of 3/16/2020 correct?
https://en.wikipedia.org/wiki/Acceleration
"Accelerations are vector quantities (in that they have magnitude and direction)[1][2], technically classified as a rank-2 tensor."
As I understand it, a vector is a rank-1 tensor.
 A: Yes, in Newtonian mechanics acceleration is the derivative of velocity with respect to time. Since velocity is a vector (i.e. a rank 1 tensor) then so is acceleration. Things are a little more complicated in general relativity, since we have to be careful how we define "time", but acceleration is still a vector.
I have reverted the change to the Wikipedia page.
A: Ok, let's try a different tact. Let's see if I can get my point across:
The tangent vectors to the coordinate axes form the basis of the vector space. They are often set in Euclidean space as orthogonal, but need not be. Which is why in Euclidean space the dual basis is generally the same as the tangent basis.
Contravariant components are expressed with respect to ordinary basis vectors. Covariant components are expressed with respect to dual basis vectors. Usually in 3-D Euclidean space, that's not an issue, since the dual and the ordinary bases are the same, and the contravariant and covariant components of a vector are equal.
So we need something like Minkowski space to map 4-D space-time to Euclidean space. The spacetime interval between any two events is independent of the inertial frame of reference in which they are observed. We accomplish this via reflections, rotations, translations, time translations and boosts (Galilean transformations). All Galilean transformations preserve the 3-dimensional Euclidean distance. This distance is purely spatial. The Minkowski inner product returns the spacetime interval when supplied the vector of coordinate differential as an argument. This preserves not just the spatial distance of Euclidean space, but space-time. As such, time is not Newtonian, it is Minkowskian (I'd call it Einsteinian, but Minkowski put forth the concept of a unified space-time first in 1908).
As it turns out, there is a close relationship between the dual basis and the vector derivative operator (usually denoted nabla, ∇). If dual basis vectors are written e^a (and ordinary basis vectors e_a), then we tend to say ∇ = ∑_a e^a ∂/ ∂x^a. Nabla, ∇, is defined in terms of dual vectors, not ordinary vectors
Force is a dual vector. A one-form is defined as a linear scalar function of a vector. The vector space of one-forms is called the dual vector (or cotangent)
space. The one-form adds one covariant component to the index of force, escalating its rank.
The Einstein summation rules dictate that repeated upper and lower indices are summed over their ranges, so in reality the dual vector of force is a rank-2 tensor, not rank-1.
F = m a
Force is a dual vector, rank 2; 
m is a scalar; 
a scalar times a rank 2 tensor is a rank 2 tensor; 
the equation balances.
As such the acceleration tensor is an antisymmetric tensor of rank 2, operating upon two vector fields... three of the six independent components of the acceleration tensor associated with the components of the acceleration field strength S, and the other three with the components of the acceleration solenoidal vector N.
The continuity equation for the mass 4-current ∇_A J^α=0 is a gauge condition that is used to derive the field equation from the principle of least action. Therefore, the contraction of the acceleration tensor and the Ricci tensor must be zero.
This is why the acceleration stress-energy tensor of a gravitational field (an accelerational field) is treated as a rank 2 tensor.
https://en.wikiversity.org/wiki/Acceleration_tensor
https://en.wikiversity.org/wiki/Gravitational_tensor
http://www.tapir.caltech.edu/~chirata/ph236/lec07.pdf
https://en.wikipedia.org/wiki/Einstein_tensor
http://sergf.ru/aten.htm
http://zen.uta.edu/me5312/03.pdf#page=3
A rank 1 tensor (a vector) can be described in terms of a uniform displacement per unit time (the magnitude of its velocity in the direction of motion). A body undergoing acceleration, however, cannot be described merely in terms of a uniform displacement per unit time, because the displacement per unit time is changing per unit time. If a body undergoing acceleration could be described in that manner, that would imply that it would fall at the same displacement per unit time in a gravitational field. There is a higher derivative (and thus its dual, the gradient) and thus a higher tensor.
Ok, how'd I do? LOL
