# Is kinetic energy relative to frame of reference?

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## Main Question or Discussion Point

According to special relativity, KE= ([PLAIN]https://upload.wikimedia.org/math/3/3/4/334de1ea38b615839e4ee6b65ee1b103.png-1)(mv^2)/2. [Broken] Since the velocity measured is dependent on a person's frame of reference, then does that mean energy too is dependent on frame of reference? For example, if an observer in Car A is moving at velocity, V, and a person in Car B is moving at velocity, V, also, then according to the observer in Car A, the person in Car B would have 0 J of kinetic energy?

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Drakkith
Staff Emeritus
Kinetic energy is indeed frame dependent.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
According to special relativity, KE= ([PLAIN]https://upload.wikimedia.org/math/3/3/4/334de1ea38b615839e4ee6b65ee1b103.png-1)(mv^2)/2. [Broken] Since the velocity measured is dependent on a person's frame of reference, then does that mean energy too is dependent on frame of reference? For example, if an observer in Car A is moving at velocity, V, and a person in Car B is moving at velocity, V, also, then according to the observer in Car A, the person in Car B would have 0 J of kinetic energy?
Energy is dependent on the frame of reference already in classical mechanics. In the rest frame of any object, the object has zero kinetic energy.

The kinetic energy in special relativity is given by $T = (\gamma - 1) m c^2$, nothing else. For small $v$, this is well approximated by the classical expression $T = mv^2/2$.

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Mister T
Gold Member
Since the velocity measured is dependent on a person's frame of reference, then does that mean energy too is dependent on frame of reference?
Yes. This is also true in newtonian physics.

In special relativity kinetic energy is $\gamma mc^2-mc^2$. The first term is the total energy, it's frame dependent. The second term is the rest energy, it's a relativistic invariant (has the same value in all reference frames).

vanhees71
Gold Member
2019 Award
One should note that the four-momentum
$$(p^{\mu})=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix}$$
is a four-vector if you define the energy such that it includes the rest energy, i.e., $E=E_0+E_{\text{kin}}=c \sqrt{m^2 c^2+\vec{p}^2}$. Then the covariant expression
$$p_{\mu} p^{\mu}=m^2 c^2$$
contains the relation between energy and momentum.

In terms of space-time coordinates along the particle's trajectory you have
$$p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} = m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m \gamma \begin{pmatrix} c \\ \mathrm{d} x/\mathrm{d} t \end{pmatrix}.$$
Thus the energy and momentum in terms of the three-velocity is
$$E=c p^0=m \gamma c^2, \quad \vec{p}=m \gamma \vec{v} \; \Rightarrow \; \vec{\beta}=\frac{\vec{v}}{c}=\frac{c \vec{p}}{E}.$$
If you need to transform from one to another frame via a Lorentz transformation, use covariant quantities like the four-momentum and then the just given expressions to derive the three velocity and other non-covariant quantity with its components in the new frame. That makes things much more convenient and much less dangerous to make a mistake. As any four-vector the four-momentum transforms as
$$p^{\prime \mu}={\Lambda^{\mu}}_{\nu} p^{\nu},$$
where ${\Lambda^{\mu}}_{\nu}$ is the Lorentz-transformation matrix.