# Definitions of relativistic kinetic energy and momentum [duplicate]

I can not solve this "loop" in my learning:

Relativistic kinetic energy is usually defined as (see by example here):

$$T = \int \mathbf{v} d \mathbf{p} = \int \mathbf{v} d(\gamma m \mathbf{v}) = (\gamma-1) m c^2$$ where $$\mathbf{p} = \gamma m \mathbf{v}$$ is used as premise.

(Four) momentum is defined as:

a) from mass multiplied by position derivative respect to proper time (something usually deprecated due to the mix of reference frames for $$\mathbf{x}$$ and $$\tau$$):

$$\mathbf{p}=m\frac{d\mathbf{x}}{d\tau}=m\frac{d\mathbf{x}}{dt}\frac{dt}{d\tau}=m\mathbf{v}\gamma$$.

b) from action related to Lagrangian of the free particle (see here):

$$\mathbf{p} = \frac{\partial S}{\partial \mathbf{q}} = \frac{ \partial L}{\partial \dot{\mathbf{q}}}$$

where $$S=\int L dt$$ and the Lagrangian $$L = - \frac{ 1 }{\gamma} m c^2$$ is obtained from the total energy $$E=\gamma m c^2$$ and:

b1) total energy from kinetic (loop!) $$T=(\gamma-1)mc^2$$ plus rest energy $$mc^2$$ or,

b2) from energy-momentum relation $$E^2 = p^2c^2+m^2c^4$$ and energy-momentum relation (see here) from four-momentum norm (loop!).

I can not find a linear (non-circular) sequence of definitions that allows me to define and evaluate (four) momentum and kinetic energy in relativistic context.

• Energy is the time component of the 4-vector. The invariant mass is the norm in all frames of reference. You need a metric for taking inner products, e.g. (-1, 1, 1, 1). Then, $-E^2 + p^2 = -m^2$, where I use $c = 1$. Then the above relation comes out. I think you have some typos in your question, like $p^2c^2$ and I don't understand how you define momentum in terms of position.
– user196418
Dec 15 '20 at 13:03
• @ggcg: fixed typo and "a)" clarified. Dec 15 '20 at 13:21
• Ask yourself what is $\gamma$, and how that relates to the Lorentz transform. Many of your definitions are all self consistent. I am not sure what you mean by "as linear as possible" in your comment. $T = E - mc^2$, total energy minus rest mass energy. Your comments b1 and b2 are self
– user196418
Dec 15 '20 at 13:33
• Don't use Wikipedia for learning. I check out that link and it's just a long list of factoids, which may all be correct but don't illuminate what is going on. All of your definitions seem self consistent. I think the biggest issue is understanding that energy is not a scalar in relativity, like time and space mix, energy and momentum mix.
– user196418
Dec 15 '20 at 13:36
• @ggcg: if I obtain value $T$ from value of $\mathbf{p}$, value of $\mathbf{p}$ from $E$ and $E$ from $T$ is a loop. If obtain $p$ from $L$, $L$ from $E$ and $E$ from $p$, it is a loop. By linear I mean a sequence of definitions and inferences without loops. Dec 15 '20 at 13:36

For your specific case, you may simply define the four-momentum as $$p^{\mu}=mu^{\mu}$$ where $$u^{\mu}=\frac{d}{d \tau}x^{\mu}$$ is the four-velocity. Then everything else can be derived. But be aware that other authors may have taken a different definition and then derived the above expression. That is not a problem.
In particular, here it can be shown from the above that in an inertial frame $$p^{\mu} = (E/c,p_x,p_y,p_z)$$. Then $$p^{\mu}p_{\mu}=E^2/c^2-\vec p^2 = m^2 c^2$$. In the rest frame this simplifies to $$E_{rest}=mc^2$$. So the kinetic energy is simply the total energy minus the rest energy $$E-mc^2$$, which simplifies to the usual expression.