相对论能量&动量

Note: This article assumes some previous knowledge on special relativity to understand.

To develop the idea of relativistic momentum, we can proceed by analogy to classical mechanics: $p=mv$ Consider the spatial components of the 4-momentum. In an inertial reference frame, Newton's Laws are still valid, implying that (classical) momentum in conserved in without outside forces on the system. To show that this is still valid in the relativistic regime, we will consider the relativistic form under Galilean and Lorentz transformations.

Consider a particle which moves $v_{x}, v_{y}, v_{z}$ in frame $S$. In frame $S'$ moving at speed $u$ along the $x$ direction with respect frame $S$. Then we have, $$p'_x=p_x-mu$$ $$p'_y=p_y$$ $$p'_z=p_z$$

This shows the classical result that momentum is conserved in $S$ and $S'$ since $m$ and $u$ are constants. However, consider this form under a Lorentz transformation, $$x'=\gamma (x-ut)$$ $$t'=\gamma(t-\frac{ux}{c^2})\rightarrow v'_x=\frac{x'}{t'}=\frac{x-ut}{t-\frac{ux}{c^2}}=\frac{\frac{x}{t}-u}{1-\frac{u}{c^2}\frac{x}{t}}=\frac{v-u}{1-\frac{uv}{c^2}}$$ $$y'=y\rightarrow v'_y=\frac{y'}{t'}=\frac{y}{t'}=\frac{y}{\gamma(1-\frac{ux}{c^2})}=\frac{v_y}{\gamma(1-\frac{uv_x}{c^2})}$$ $y\to z$ applies as well. Since the primed velocity in the y and z direction depend upon the speed in the (unprimed) x direction, we see that the classical definition of momentum will not be conserved under Lorentz transformations.

If we examine a collision in one frame and find the total 3-momentum of the system to be $mv$ before and after the collision, then in another inertial frame of reference (IRF) the 3-momentum will be conserved in that frame as well. Why? The transformation from one frame to the other is simply the momentum multiplied by $\gamma$, a constant for constant relative speeds. So in the other frame the total momentum is simply $\gamma mv$ before and after the collision. Momentum is conserved in both frames, however the values of total momentum in the two frames will not agree.

The spatial part of the 4-momentum is clear, but what about the temporal part? What is the meaning of $mv^{ct}$? We have, $mv^{ct}=m\frac{\Delta t}{\Delta \tau}=m\gamma$. For the spatial part, we have a classical analogy, the 3-momentum. What is the classical analog of $m\gamma$, if there exists one? To explore this take the low velocity limit, $v\ll c$. To do this recall the binomial expansion, $(1+x)^n\cong 1+nx+n(n-1)\frac{x^2}{2!}+...$ if $x\ll 1$.

If $v\ll c$, or $v/c\ll 1$, we can expand $\gamma$ in this manner, $$\gamma=(1-\frac{v^2}{c^2})^{-1/2}\cong 1+(-\frac{1}{2})(-(\frac{v}{c})^2)+...\approx1+\frac{1}{2}(\frac{v^2}{c^2})$$ and, $$p^{ct}\approx_{v\to 1} mc+\frac{1}{2}mc(\frac{v^2}{c^2})\to mc+\frac{1}{2}mv^2\frac{1}{c}$$

The last term is simply classical kinetic energy divided by c. Hence, by taking the classical limit of the time component of out 4-momentum is $mc+KE/c$. This leads us to interpret that this component is related to energy.

Passing from the classical regime to the relativistic regime, the ct component of the 4-momentum picks up the extra terms in the expansion. This is not very significant since both m and c are constants. Recall that classically we can define the zero of energy as we wish since it is only the difference in eregy which matters. Hence, classically we can absorb this constat into the definition of energy by resetting the zero.

Now we define the relativistic energy, $E$, as $E=p^{ct}c=mc^2\gamma$. In the low velocity limit, we regain the classical form of energy (ignoring the other forms for now).

Note that when $v=0$ we are led to the most famous equation of all physics! $v\to 0$ , the$\gamma\to 1$ and $E_0 =mc^2$. This equation can be viewed as arising from the motion of an object through spacetime, along the ct direction.

So we have defined 4-momentum as the following:

$$p^\mu:(mc\frac{\Delta t}{\Delta \tau},m\frac{\Delta x}{\Delta \tau},m\frac{\Delta y}{\Delta \tau},m\frac{\Delta x}{\Delta \tau})=(\frac{E}{c},p^x,p^y,p^z)$$

and we have found the relativistic energy to be $$E=p^{ct}c=mc^2\gamma$$

and we found that the 4-momentum's magnitude is constant, $$(p^\mu)^2=p^\mu p_\mu = (p^{ct})^2-(p^x)^2-(p^y)^2-(p^z)^2$$

We can also define another related 4-vector which we can call the energy 4-vector, or usinf Taylor's terminology "momenergy". This is simply the 4-momentum multiplied by c, $q^\mu=cp^\mu$, $$q^\mu:(E,cp^x,cp^y,cp^z)$$ with $$(q^\mu)^2=E^2-c^2p^{x2}-c^2p^{y2}-c^2p^{z2}=m^2c^4=(mc^2)^2$$

The magnitude of this vector is $mc^2$ and is a constant. The momentum-energy 4 vector is conserved in collision as well. We can represent relativistic energy in another manner now, $$E=\sqrt{(mc^2)^2+(p^xc)^2+(p^yc)^2+(p^zc)^2}$$ This form makes it easier to determine an object's energy givenits mass and momentum.

This energy and total energy of the particle, KE plus all else. What this is saying is that a hot object will have more mass than a cold one, although the addition is very small ($\approx\frac{1}{c^2}$). From this definition of the relativistic energy, we define the kinetic energy, K, as the difference between the total energy and the rest energy.

We know the expressions for rest energy, total energy and kinetic energy following the answer above. But how do we verify these expressions to check if they are correct?

One way is to use the Work-Energy Theorem. This will be left as an exercise for the reader.