## Why isn't kinetic energy a vector instead of a scalar?

Kinetic energy is a scalar because it is

The reason why scientists define certain words to mean certain things is that some definitions are more useful than others. In particular, if a quantity is useful, it is likely to have a name. The quantity \(T = \frac{1}{2}mv^2\), a scalar,

Now, there are at least three different reasons why kinetic energy, as a scalar, is a useful quantity in physics:

*defined*to be a scalar.The reason why scientists define certain words to mean certain things is that some definitions are more useful than others. In particular, if a quantity is useful, it is likely to have a name. The quantity \(T = \frac{1}{2}mv^2\), a scalar,

**is useful**, so it gets a name:*kinetic energy*. The quantity \(\frac{1}{2}mv^2 \hat{\mathbf{v}}\), which is a vector whose magnitude is the kinetic energy and whose direction is the direction of motion,**is not useful**, so it doesn't have a name.Now, there are at least three different reasons why kinetic energy, as a scalar, is a useful quantity in physics:

- It's part of the total energy—kinetic plus potential—which is very important because of conservation of energy. \(E = T + V\) would have to be changed to \(E = \|\mathbf{T}\| + V\) if we defined kinetic energy as a vector. That is, we'd just throw away the direction. There is no vector conservation law for energy, the way there is for momentum.
- It's a term in the Hamiltonian, which tells you how a physical system in general evolves in time. The Hamiltonian's value is usually the same as the total energy, \(H = T + V\), although for some systems it's different. Again, if we defined kinetic energy as a vector, we'd just have to throw away the direction, \(H = \|\mathbf{T}\| + V\).
- It's a term in the Lagrangian, which, like the Hamiltonian, governs the behaviour of a physical system, and is very important theoretically because its mathematical form makes certain properties of physical systems obvious. The Lagrangian for a particle is \(L = T - V\). Again, if we defined kinetic energy as a vector, we'd have to throw away the direction, and write \(L = \|\mathbf{T}\| - V\).