How is the law of conservation of energy valid in all frames?

Question

I am aware there are similar questions already asked, however, I find none of the answers satisfactory, they either do not contain any mathematics at all, or mathematics of a level I am not capable of comprehending.

My doubt is that, if the change in kinetic energy is dependent on frame, then how is the law of conservation of energy valid in all frames? One possible explanation I have thought of is that the change of potential energy is also frame dependent, but I cannot think of a suitable example to justify the same.

Please answer only using newtonian mechanics I do not know statistical or lagrangian mechanics.

Answer 1

In classical mechanics, Galilean transformations (we'll assume a non-relativistic problem) relate the coordinates of events as measured in two different inertial reference frames that are related by a constant relative velocity $v$ along a straight line. The Galilean transformations for space and time coordinates are given by:

$$x′ = x−vt$$ $$y′ =y,$$ $$z′=z,$$ $$t′=t.$$

If you consider a system with a particle of mass m and velocity u, the kinetic energy K in one frame is given by:

$$K= \frac{1}{2} m u^2$$

Let's find the kinetic energy $K′$ in a frame moving with a velocity $v$ relative to the original frame. In the new frame, the velocity of the particle $u′$ would be: $u′ =u−v$, so

$$u′^2 =(u−v)^2 =u^2 + v^2 − 2u⋅v.$$ The kinetic energy $K'$ in the moving frame is:

$$K′ = \frac{1}{2}mu′^2 = \frac{1}{2}m(u^2+v^2−2u⋅v).$$ We can see that $K′$ is not equal to $K$ unless $v =0$ or $u⋅v=0$. Therefore, kinetic energy is not invariant under a Galilean transformation.

For systems involving not just kinetic but also potential energy, the story is similar. Potential energy often depends on the relative positions of particles, so for many systems of interest (e.g., gravitational or electrostatic systems), the potential energy would be the same in both frames. However, the total mechanical energy, being the sum of kinetic and potential energy, would still transform like the kinetic energy under a Galilean transformation.

Answer 2

I don't know to what level you want a response, but if you are comfortable with analytical mechanics then perhaps the following will be of interest to you. The system of euler lagrange equations corresponding to the functional \begin{equation} J[y_1...y_n]=\int_{a}^{b}\mathcal{L}(x,y_1,...y_n,y'_1,...y'_n)dx \end{equation} has the first integral

\begin{equation} \mathcal{H} = -\mathcal{L} + \sum_{i=1}^{n}y'_ip_i \end{equation}

then you just need Noethers theorem to see how conservation laws emerge; if $J$ does not depend on x explicitly it stands to reason that it should not matter if we make the change of coordinates \begin{equation} x*=x+\epsilon \end{equation}

where $\epsilon \in \mathbb{R}$ . Therefore $\mathcal{H}$ is the first integral of the euler lagrange equations corresponding to the functional if and only if it is invariant under the coordinate change and as we shall see there is a connection between certain first integrals of the EL equations and the invariance of the action under certain transformations or \textit{symmetries}. Consider the following transformations;

\begin{equation} x*= K(x,y,y';\epsilon), y_i*=G_i(x,y,y';\epsilon) \end{equation}

The functional $\int_{x_0}^{x_1}\mathcal{L}(x,y_1,...y_n,y'_1,...y'_n)dx$ is said to be invariant or symmetric under the transformation iff $\int_{x_0}^{x_1}\mathcal{L}(x,y_1,...y_n,y'_1,...y'_n)dx = \int_{x_0*}^{x_1*}\mathcal{L}(x*,y*_1,...y*_n,y'*_1,...y'*_n)dx*$.

Noethers theorem says If the action $J[y]=\int_{x_0}^{x_1}\mathcal{L}(x,y,y')dx$ is invariant under the family of transformations for arbitrary $[x_0,x_1]$ then

\begin{equation} \sum_{i=1}^{n}p_ik_i+ \mathcal{H}g = c \end{equation}

where $k_i$ and g are as follows

\begin{equation} \begin{cases} g(x,y,y')=\frac{d{G}}{d\epsilon}|_{\epsilon=0}\\ k_i(x,y,y')=\frac{d{K_i}}{d\epsilon}|_{\epsilon=0} \end{cases} \end{equation}

In plain terms, every one parameter family of transformations leaving the action invariant leads to a first integral of the Euler Lagrange equations and this first integral is constant and that is the conserved quantity corresponding to the symmetry of the action. You can show that Lorentz transformations also have conserved quantities and this is the whole idea of a Lorentz scalar or relativistically invariant quantities.