1
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ 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. $\endgroup$
    – user196418
    Dec 15, 2020 at 13:03
  • $\begingroup$ @ggcg: fixed typo and "a)" clarified. $\endgroup$ Dec 15, 2020 at 13:21
  • $\begingroup$ 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 $\endgroup$
    – user196418
    Dec 15, 2020 at 13:33
  • $\begingroup$ 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. $\endgroup$
    – user196418
    Dec 15, 2020 at 13:36
  • $\begingroup$ @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. $\endgroup$ Dec 15, 2020 at 13:36

1 Answer 1

2
$\begingroup$

First, it is a mistake to worry about "loops". A loop simply says that a set of concepts is self-consistent. There is nothing wrong with self-consistency, in fact, it is pretty important. Once you have a self-consistent set of statements then you can usually arbitrarily label one set as "definitions/assumptions/axioms" and the rest as "results/conclusions/theorems". The choice of which statements belong in which category is inherently somewhat arbitrary and different authors are free to change them.

For example, Einstein assumed his two postulates from which he derived the Lorentz transform. Lorentz postulated the Lorentz transform from which the two postulates can be derived. Other authors have derived the Lorentz transforms starting from assumptions about isotropy and homogeneity, which Einstein would have derived from the Lorentz transform. All of those approaches are valid.

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.