1
$\begingroup$

In a spacetime diagram the spatialized time direction is the vertical $y$-axis and pure space direction is the horizontal $x$-axis, $ct$ and $x$, respectively.

The faster you go and therefore the more kinetic energy you have, you'll have a greater component of your spacetime vector in the $x$-direction. More of your energy and forward "motion" through spacetime is devoted to traveling through space than through time. A consequence is time dilation with which we are familiar.

My question arises because I am confused about how this relates to the energy-momentum 4-vector where the time component of this is $mc^2 + \frac{1}{2}mv^2$. The rest mass energy plus the kinetic energy. If the kinetic energy term is rather large, you have a large time component in the energy-momentum 4-vector, but if your kinetic energy is large shouldn't you be traveling "less" through time and "more" through space? There is some subtle disconnect here for me and I would appreciate it if someone could help me to think about this properly.

If the kinetic energy is zero we are left with simply $mc^2$, the energy that mass has on its own at rest. This tells me that it is the energy that the mass has as it moves through spacetime solely in the time direction. So if you add kinetic energy, the time component becomes larger, and more energy is devoted to travel in the time direction in spacetime. How do we reconcile that with larger energies and velocities means you travel more in the space-direction in the spacetime diagram where we consider the spacetime interval?

Furthermore, would this inability to reconcile this have to do with the hyperbolic geometry of Minkowski spacetime and how it changes the Euclidean Pythagorean relationship to a Non-Euclidean geometry?

$\endgroup$
4
$\begingroup$

Since your question involves varying the "relativistic kinetic energy", the following energy-momentum diagram might be helpful.

First, some definitions.

robphy-energyMomentum

  • We use rapidity $\theta$ [the Minkowski-angle between future-timelike vectors],
    where the velocity is $V=c\tanh\theta$ and
    the time-dilation factor is $\gamma=\displaystyle\frac{1}{\sqrt{1-(V/c)^2}}=\cosh\theta$.
  • In $(t,x)$ coordinates, the dimensionless 4-velocity $\hat u=(\cosh\theta,\sinh\theta)$,
    which has "slope" $(V/c)=\displaystyle\frac{u_x}{u_t}=\frac{\sinh\theta}{\cosh\theta}=\tanh\theta$ and
    unit square-magnitude $\hat u\cdot\hat u=(\cosh^2\theta - \sinh^2\theta)=1$.
  • The 4-momentum [or the momentum 4-vector] of a particle $$\tilde P= m\hat u = m(\cosh\theta,\sinh\theta)=(E,p),$$ which has a timelike component $P_t=E$ called the "relativistic energy"
    and a spacelike component $P_x=p$ called the "relativistic momentum".
    Its slope is also the velocity.

    The square-magnitude $\tilde P \cdot \tilde P=(E^2-p^2)=(m\cosh\theta)^2-(m\sinh\theta)^2=m^2$ is the square of the "mass", an invariant. On the energy-momentum diagram, the curve of constant mass $m$ (called the "mass-shell") is given by the hyperbola. (In this diagram, the tip of the 4-momentum $\tilde P$ is on the ``$m=4$ mass-shell.'')
  • The "relativistic kinetic energy" (with the $c$'s restored)
    $T=mc^2(\cosh\theta-1) \equiv mc^2\left(\displaystyle\frac{1}{\sqrt{1-(V/c)^2}}-1\right)\approx \left( \displaystyle\frac{1}{2}mV^2 + \frac{3}{8}\frac{m}{c^2}V^4+\frac{5}{16}\frac{m}{c^4}V^6 + \ldots \right)$.
    It can be visualized as the segment on the energy-axis from the mass-shell (at $mc^2$) to the tip of the timelike-leg (at $mc^2\cosh\theta$), which is physically the portion of the relativistic energy beyond the rest-energy (rest-mass energy)--hence, the relativistic kinetic energy.

Now consider various ways to increase the kinetic energy..
Note, it's important to keep track of how we are taking the limit... is anything being kept constant?

  • With fixed mass (staying on the mass shell),
    increasing the kinetic energy $T_{\rm faster}$ corresponds to increasing the rapidity, velocity, energy, and momentum ...toward the edge of the light-cone [but never reaching it].
    The tail of the kinetic-energy segment remains on the mass-shell [keeping $m$ constant] but the tip of the energy component increases.
  • With fixed energy (staying on the horizontal line of constant E),
    increasing the kinetic energy $T_{\rm impulsive}$ corresponds to increasing the rapidity, velocity, and momentum, but decreasing the mass (toward the photon mass).
    The tip of the kinetic-energy segment is unchanged [keeping $E$ constant], but the tail of the segment moves down to a lower-valued mass-shell.
  • With fixed momentum (staying on the vertical line of constant p),
    increasing the kinetic energy $T_{\rm B}$ corresponds to increasing the rapidity and velocity but decreasing the energy and the mass (toward the photon mass).
    The tip and tail of the kinetic-energy segment move down, while keeping $p$ constant.
  • With fixed velocity and rapidity (staying on the ray [of constant V]),
    increasing the kinetic energy $T_{\rm heavier}$ corresponds to increasing the mass, momentum, and energy... but not approaching the light cone.
    The tip and tail of the kinetic-energy segment move up, while keeping $V$ constant.
  • One can also use other trajectories for taking limits.

robphy-limits


update:
Using the relations in the first section, one could write various quantities as a function of the relativistic kinetic energy $T$ and $m$:

  • $E=m\cosh\theta=m(\cosh\theta-1)+m= T+m $
  • $p=m\sinh\theta=m\sqrt{\cosh^2\theta-1}=\sqrt{m^2(\cosh\theta-1)(\cosh\theta+1)}=\sqrt{T(T+2m)}$
  • $m=\sqrt{E^2-p^2}$
  • $v=\tanh\theta=\displaystyle\frac{p}{E}=\frac{\sqrt{T(T+2m)}}{T+m}$
$\endgroup$
1
  • $\begingroup$ This is all very interesting and much of it is quite new to me. It will take some time for me to study it and grasp it but I am rather intrigued. the diagrams are helpful. $\endgroup$ – MattGeo May 11 '20 at 23:33
1
$\begingroup$

Firstly, I would like to point out (as you might already know) that the time component of energy is not solely $mc^2 +\frac{1}{2}mv^2$. It is just the classical limit of energy upto first order. The total energy goes on increasing when you increase "kinetic energy". Now, to answer your question

if your kinetic energy is large shouldn't you be traveling "less" through time and "more" through space?

No, you should not. This is what intuitively can be stated as. "no massive particle can move at speeds greater than the speed of light in vacuum". Mathematically stated, you always stay within the light cone of Minkowski space.

This is what a lightcone looks like

As you can see from the image, the time component of position 4-vector is always greater than space component. Also, as you increase the space component of momentum 4-vector, you go on to more space component of position 4-vector. As momentum 3 vector is $m\frac{dx}{dt}$, so when you go increase space component of position 4-vector, you move nearer to the boundary of the light cone, but never leave it

$\endgroup$
2
  • $\begingroup$ Something with mass can never reach the 45 degree line which forms the boundary of the light cone, but regardles, it's the same as in a Minkowski spacetime diagram. When velocity increases by some amount the spacetime interval decreases by some amount in the time direction and increases by some amount in the space direction. $\endgroup$ – MattGeo May 11 '20 at 13:12
  • $\begingroup$ While the diagram is nice, it should be noted that: at each event, the tangent-vector to the worldline should always point into the future light-cone centered at that event [that is, have a slope that is less than 45 degrees from the vertical]. In the diagram, some segments of this roughly drawn worldline violate that... for example, at the sharp turns before and after the roughly straight segment. $\endgroup$ – robphy May 12 '20 at 12:50
0
$\begingroup$

Given a world line $x^{\mu}$, its derivate w.r.t to proper times is it's 4 velocity:

$$ \frac{dx^{\mu}}{d\tau} = u^{\mu} = (\gamma c, \gamma \vec v)$$

Note that:

$$ mu^{\mu} = (\gamma mc, \gamma m\vec v) = (E/c, \vec p) = p^{\mu}$$

So they are related by a scale factor, $m$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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