Is kinetic energy a scalar?

Question

Is it correct to say that kinetic energy is a scalar?

Answer 1

Just to add some explicitness to the above answers: take an isolated particle at rest; it's KE is zero. Now switch to a reference frame with relative velocity $\beta$ wrt the particle. In this frame, it's KE is $$KE' = E - mc^2 = (\gamma - 1)mc^2 = \frac{1}{2}mv^2 + O(\beta^4).$$ We see that $KE' \neq KE,$ thus it's not a scalar.

Answer 2

I would say it depends on the context. In an Euclidean 3 space it is a scaler (provided its total energy is kinetic) for non relativistic case. In 4D relativistic case, it is a component of a 4 vector.

Answer 3

A scalar is something that doesn't transform under coordinate transformations. A vector is something that transforms "like a vector," in other words, its coordinate transforms are realized as a local multiplication by a linear operator (e.g., a matrix).

In Newtonian mechanics, it's simple to see that kinetic energy, being proportional to the square of a vector (a length) doesn't change under the allowable coordinate transformations (rotations and translations) in Newtonian mechanics, so it is a scalar.

In relativity, we mix up space and time coordinates with coordinate transformations, so there, it transforms as the time component of a 4-vector. This is because energy comes from symmetries associated to time translation. Symmetries associated with space translations are associated with conservation of momentum. Because coordinate transforms in relativity mix up space and time coordinates, we have a single 4-vector whose components are (energy, momentum in x direction, momentum in y direction, momentum in z direction) instead of separate notions of energy and momentum.