In 3 spatial dimensions, $$[E] = [ML^2 T^{-2}]$$

Would it change in higher dimensions? If yes, then what would be the dimensions for 4 spatial dimensions?

  • 2
    $\begingroup$ In the 4-dimensional case, would the fourth dimension be pretty much the same thing as the first three spatial dimensions? $\endgroup$ – Nat Oct 11 '18 at 23:02
  • $\begingroup$ Note that in many problems and exercises on work and energy we limit the universe to lines / curves or planes / sufaces. This does not affect the interpretation or units of energy in any significant way. One could therefore empirically expect that it looks the same in higher dimensions as well. $\endgroup$ – Arthur Oct 12 '18 at 11:11

Suppose for a moment that we're specifically interested in the kinetic energy of a single, non-relativistic particle, so that $E=\frac{1}{2}m\vec{v}^2$. I include the vector notation for the velocity here because it's useful to keep in mind that $\vec{v}^2=\vec{v}\cdot\vec{v}$.

If the particle is moving in two dimensions, then $\vec{v}=v_x\hat{x}+v_y\hat{y}$ and $\vec{v}\cdot\vec{v}=v_x^2+v_y^2$, which has units of velocity squared. In three dimensions, $\vec{v} = v_x\hat{x}+v_y\hat{y}+v_z\hat{z}$ and $\vec{v}\cdot\vec{v} = v_x^2+v_y^2+v_z^2$, which still has the units of velocity squared. In four dimensions, $\vec{v} = v_x\hat{x}+v_y\hat{y}+v_z\hat{z}+v_w\hat{w}$, and $\vec{v}\cdot\vec{v}=v_x^2+v_y^2+v_z^2+v_w^2$, which, again, still has the units of squared velocity. This is because adding two quantities of the same units together does not change their units. Indeed, this holds for any number of dimensions, and suggests that the units of energy should be unchanged.

We can also see this more generally by the formal definition of work, which is the definition of change in energy:

$$W=\int_C \vec{F}\cdot d\vec{s}$$

for an object being acted on by a force $\vec{F}$ moving along a path $C$ with arclength parameter $d\vec{s}$. This is the formal, most general definition, and holds no matter how many dimensions of space you have. Once again, you can see that this definition involves a dot product. The dot product takes two vectors in any number of dimensions and outputs a one-dimensional quantity (i.e. a number). The work only cares about one component of the force, namely, the one pointing along the one-dimensional path that the particle is taking. No matter how many spatial dimensions this one-dimensional path is embedded in, the work only cares about things happening along one of those dimensions. As such, the units of work should always be the same, and, since work has the same units as energy (otherwise we wouldn't be able to add them together), the units of energy should also always be the same.


Classical mechanics is already (effectively) multidimensional.

Consider the equation of motion for one particle in one dimension: \begin{align} m \ddot x(t) = F(x). \end{align} Now consider the equation of motion for one particle in two dimensions, \begin{align} m \ddot x(t) & = F_x(x,y)\\ m \ddot y(t) & = F_y(x,y), \end{align} and in three dimensions \begin{align} m \ddot x(t) & = F_x(x,y,z)\\ m \ddot y(t) & = F_y(x,y,z)\\ m \ddot z(t) & = F_z(x,y,z). \end{align} Now consider the equation of motion for two particles in one dimension, \begin{align} m_1 \ddot x_1(t) & = F_{1}(x_1,x_2)\\ m_2 \ddot x_2(t) & = F_{2}(x_1,x_2), \end{align} or in twodimensions, \begin{align} m_1 \ddot x_1(t) & = F_{x,1}(x_1,y_1,x_2,y_2)\\ m_1 \ddot y_1(t) & = F_{y,1}(x_1,y_1,x_2,y_2)\\ m_2 \ddot x_2(t) & = F_{x,2}(x_1,y_1,x_2,y_2)\\ m_2 \ddot y_2(t) & = F_{y,2}(x_1,y_1,x_2,y_2), \end{align} or three, \begin{align} m_1 \ddot x_1(t) & = F_{x,1}(x_1,y_1,z_1,x_2,y_2,z_2)\\ m_1 \ddot y_1(t) & = F_{y,1}(x_1,y_1,z_1,x_2,y_2,z_2)\\ m_1 \ddot z_1(t) & = F_{z,1}(x_1,y_1,z_1,x_2,y_2,z_2)\\ m_2 \ddot x_2(t) & = F_{x,2}(x_1,y_1,z_1,x_2,y_2,z_2)\\ m_2 \ddot y_2(t) & = F_{y,2}(x_1,y_1,z_1,x_2,y_2,z_2)\\ m_2 \ddot z_2(t) & = F_{z,2}(x_1,y_1,z_1,x_2,y_2,z_2). \end{align}

The pattern should be pretty obvious: adding spatial dimensions is identical to adding particles, and classical mechanics is already perfectly equipped for handling additional dimensions, so that e.g. if you added a fourth spatial dimension, the equations of motion \begin{align} m \ddot x(t) & = F_x(x,y,z,w)\\ m \ddot y(t) & = F_y(x,y,z,w)\\ m \ddot z(t) & = F_z(x,y,z,w)\\ m \ddot w(t) & = F_z(x,y,z,w) \end{align} would have a form which is exactly identical to that of two particles in two dimensions.

So, what does that mean for energy? Well following the core principle that

The same equations have the same solutions and the same properties,

the dynamics in four spatial dimensions will have a conserved energy which is exactly analogous to that of two particles in two dimensions, $$ E = \frac12 m_1 \dot x_1^2 + \frac12 m_1 \dot y_1^2 + \frac12 m_2 \dot x_2^2 + \frac12 m_2 \dot y_2^2 + V(x_1,y_1,x_2,y_2) $$ i.e. a conserved energy of the form $$ E = \frac12 m \dot x^2 + \frac12 m \dot y^2 + \frac12 m \dot z^2 + \frac12 m \dot w^2 + V(x,y,z,w) $$ where $V(x,y,z,w)$ is a potential energy. This should make it clear that the physical dimensionality of energy, $$ [E] = [ML^2T^{-2}], $$ is not changed by the process - and its form from "normal" classical mechanics already fully incorporates effects spanning as many spatial dimensions as you want.


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