Energy conservation violation?

Question

Is energy in every possible inertial frame equal? If not, does is not violate energy conservation?

Answer 1

You are confusing between conserved and invariant quantities.

A quantity is said to be conserved if it is a constant of motion, i.e. does not depend on time.

A quantiy is said to be invariant if it remains the same when transformed between different frames of reference.

In Newtonian mechanics energy is not an invariant quantity, but a conserved one. It means that it might change when you move between reference frames. But if you choose a specific frame, then it will be the same in all times.

In special relativity there is, however, an invariant quantity that is related to the energy. In relativity energy and momentum come together in the form of the four-momentum $$p^{\mu}=\begin{pmatrix}\frac{E}{c}\\p_x\\p_y\\p_z \end{pmatrix}$$ The interval of this four-vector (in the case of signature $(1,-1,-1,-1)$ metric) is given by $$p^{\mu}p_{\mu}=m^2c^2$$ and is invariant under Lorentz transformations.

You can think on it in analogy to rotations. Rotations preserve the norm of a three-vector. Lorentz transformations are kind of rotations of space and time, and they preserve the 'norm' of four-vectors.

Answer 2

It is not necessary to think in the context of relativity. In newtonian mechanics it also happens; imagine an object (with mass $m$) moving at a constant velocity $v$, then the kinetic energy is $ \frac{1}{2} m v^2 $, but if you observe the same situation from the point of view of a bird which is flying beside the object at the same velocity, you are going to see a zero kinetic energy.

The value of the energy is frame-dependent, but it is constant for both observers (if the system is isolated and the observers are inertial).

Firstly, take into account that in relativity the energy depends on the inertial frame, because it is not invariant under Lorentz transformations (those that connect inertial frames). The energy is part of a four-vector called four-momentum:

$$ p^{\mu} = (E/c, p_{x}, p_{y}, p_{z}) $$

where $p_{x}, p_{y}, p_{z}$ are the components of the momentum $\vec{p}$. When you switch from one inertial frame to another, the components of this four-vector mix between them. The only thing that is completely independent of the frame is its module:

$$ \eta_{\mu \nu}p^{\mu}p^{\nu} = E^2/c^2 - \vec{p}·\vec{p} $$

which is what we call the rest mass $m$ of the object:

$$ E^2/c^2 - \vec{p}·\vec{p} = m^2c^2 $$

Secondly, if you choose a particular inertial frame in special relativity, the total four-momentum $p^{\mu}$ is a constant of motion in a (elastic) process with no external actions.

To sum up, in an elastic isolated process:

• What is conserved but frame-dependent: the components $p^{\mu}$.
• What is conserved and frame-independent: the module of $p^{\mu}$ (the rest mass, $m$).