Energy and momentum & the relation between them

Question

I am trying to wrap my head around energy, mass and momentum, especially in the more general scope of special relativity where massless objects moving at the speed of light also have momentum. So I am not particularly interested in how much speeds change with forces or anything. I am just wondering if there is a way to relate the three quantities and then go on to derive the energy-momentum relation and I find the way these quantities are usually defined circular or confusing.

So I thought to myself: From Newtonian mechanics, we have $$\mathbf F=\frac{d\mathbf p}{dt}$$ and at the same time $$\Delta E=\int_{x_1}^{x_2}\mathbf F\cdot d\mathbf x$$ which is equivalent to $$F=\frac{dE}{dx}=\frac{dE/dt}{dx/dt}.$$ Or we could say, $$\frac{dE}{dt}=\frac{dp}{dt}v\\ \frac{dE}{dp}=v.$$

My questions are:

1. Is the last equation true in special relativity?
2. If so, and we use it to define momentum, assuming we already know what energy is (which I am not sure about), what more do we need to arrive at the energy-momentum relation?

Answer 1

The easiest way to derive the energy momentum relation of the special theory is to build things up using the notion of 4-vectors. This notion is useful for deriving quantities in the special theory, because relativity is based on the covariance of electrodynamics which requires that space and time be treated equally.

A position vector or point in Minkowski space can be represented by the following 4-vector:$$r=(ct(\tau),x(\tau),y(\tau),z(\tau))=(ct(\tau),\vec r(\tau)),$$ where $\tau$ is the proper time (time as recorded in the object's rest frame) used to parametrize paths in 4-space and $\vec r$ is the usual thee space position. Further, one must calculate inner-products according to the Minkowski metric so that:$$r^2=r\cdot r=c^2t^2-x^2-y^2-z^2,$$ and similarly for any other vector in 4-space. Now the obvious way to proceed for calculating the 4-velocity is to compute the time derivative of $x$. A look at the metric for Minkowski space-time, $ds^2=(cdt)^2-dx^2-dy^2-dz^2=(cd\tau)^2-d{x^\prime}^2-d{y^\prime}^2-d{z^\prime}^2$ allows one to compute the following relation for time dilation:$$d\tau=dt/\gamma;\quad \gamma=(1-v^2/c^2)^{-1/2}.$$ Hence, $${dct\over d\tau}=\gamma c;\quad and\quad{dx\over d\tau}=\gamma {dx\over dt}$$ Extrapolating, one can find the 4-velocity,$u$ to be :$$u={dr\over d\tau}=(\gamma c,\gamma \vec v)$$With the momentum given by $$p=mu=(\gamma mc,\gamma\vec p).$$ Now if one pays attention to the units of $\gamma mc$ one can conclude that the units are such that they correspond to $E/c$, thus one can write the 4-momentum as $$p=(E/c,\gamma\vec p).$$ Now this argument from units is motivating, however, the real sign-post that the temporal component of the 4-momentum is the total energy comes from an understanding of the electromagnetic stress-energy tensor, the first column of which $T^{\mu 0}$, is the energy-momentum flux density. Thus, by analogy to electrodynamics, the 4-momentum is the energy momentum vector of an object traveling along a path in Minkowski space-time. Now for the energy momentum relation of the special theory one has but to calculate the magnitude of the 4-momentum vector:$$p^2=p\cdot p=m^2c^2{\gamma}^2-m^2|{\vec v}|^2{\gamma}^2={E^2\over c^2}-|{\vec p}|^2,$$ which after a little algebra yields:$$E^2=m^2c^4+|{\vec p}|^2c^2.$$

Answer 2

The most transparent way to determine the expressions for energy, momentum and angular momentum of a certain (mechanical) system (described by its action integral $S$) is to study its behaviour under time translations, spatial translations and spatial rotations. If the system is time-translation invariant, the Noether theorem tells us that there exists a conserved quantity being defined as the energy $E$ of the system (up to some factor). If the system is invariant under translations in all three spatial directions, the corresponding three independent conserved quantities are defined as the three component of the momentum vector $\vec{p}$ of the system. If the system is invariant under rotations around a certain axis (say, the $z$-axis) the corresponding constant of motion is defined as the $z$-component $L_z$ of the angular momentum vector.

The general form of the action integral of a system with $f$ degrees of freedom with generalized coordinates $q_1, \ldots q_f$ has the form $$S=\int\limits_{t_1}^{t_2} \! dt \, L(q_1, \ldots, q_f, \dot{q}_1, \ldots, \dot{q}_f,t), \tag{1} \label{eq1}$$ where $L(q,\dot{q},t)$ is the Lagrangian function, characterizing the system under investigation. The equations of motion (Euler-Lagrange equations) $$ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i}, \quad i=1,\ldots f \tag{2} \label{eq2},$$ derived from the requirement that the action integral \eqref{eq1} of the system attains an extremum (usually a minimum) for the actual motion of the system between the fixed initial and final points $q_i(t_1)$, $q_i(t_2)$ for $i=1,\ldots f$.

Time-translation invariance of a system ist reflected by the fact that in this case the Lagrangian function $L$ does not depend (explicitly) on the time variable $t$, i.e. $L=L(q,\dot{q})$. As a consequence, the expression $$ E:=\sum\limits_{i=1}^f \dot{q}_i \frac{\partial L}{\partial \dot{q}_i} -L \tag{3} \label{eq3}$$ is a constant of motion, as can easily be seen by computing the time derivative of $E$, $$ \frac{dE}{dt}=\frac{d}{dt}\left(\sum\limits_{i=1}^f \dot{q}_i \frac{\partial L}{\partial \dot{q}_i}-L \right)= \sum\limits_{i=1}^f \left( \ddot{q}_i\frac{\partial L}{\partial \dot{q}_i} +\dot{q}_i \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} -\frac{\partial L}{\partial q_i} \dot{q}_i -\frac{\partial L}{\partial \dot{q}_i} \ddot{q}_i \right)=0,$$ where the equations of motion \eqref{eq2} were used to arrive at the final result. The quantity $E$ defined in \eqref{eq3} is refered to as the energy of the (time-translation invariant) system under investigation.

In the case, where $L$ happens to be independent of one of the coordinates (say $q_k$), we have $\partial L /\partial q_k=0$ and \eqref{eq2} for $i=k$ implies $$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} =0,$$ such that the (generalized) momentum $$p_k:= \frac{\partial L}{\partial \dot{q}_i} \tag{4} \label{eq4}$$ is a constant of motion.

As an illustration, let us apply this general formalism to a free, nonrelativistic particle with mass $m$ in an inertial frame. Choosing cartesian coordinates $x,y,z$, the form of the corresponding Lagrangian function $$L=m (\dot{x}^2+\dot{y}^2+ \dot{z}^2)/2 = m \vec{v}^2/2 \tag{5} \label{eq5} $$ is dictated by homogeneity of time ($L$ has no explicit time dependence) and space ($L$ is independent of the coordinates $x,y,z$) and isotropy of space ($L$ depends only on $|\vec{v}|$) in an inertial frame together with the requirement of Galilei invariance. Employing our general cooking recipe, the momenta are given by $$ p_x= \frac{\partial L}{\partial \dot{x}}= m\dot{x}, \quad p_y= \frac{\partial L}{\partial \dot{y}}=m \dot{y}, \quad p_z= \frac{\partial L}{\partial \dot{z}} = m\dot{z} \tag{6} \label{eq6}$$ and the energy becomes $$ E= \dot{x} p_x + \dot{y} p_y + \dot{z} p_z-L = m \vec{v}^2/2. \tag{7} \label{eq7} $$

In the relativistic case, the Lagrangian function of a free particle with mass $m$ (in an inertial frame) takes the form $$L= -mc^2 \sqrt{1- \vec{v}^2/c^2}\tag{8} \label{eq8} $$ ensuring the Lorentz invariance of the action integral (remember that $c^2 dt^2 - d\vec{x}^2$ is Lorentz invariant). The prefactor $-mc^2$ in \eqref{eq8} is required by consistency with the nonrelativistic Lagrangian function \eqref{eq7} in the nonrelativistic limit $|\vec{v}| \lt \! \lt c$ as $-mc^2 \sqrt{1-\vec{v}^2/c^2} \to -mc^2 +m \vec{v}^2/2+ \ldots$ (the additional constant term $-mc^2$ does not affect the equations of motion). The further steps are now straightforward. Applying the general formalism, one finds $$\vec{p} = \frac{m \vec{v}}{\sqrt{1-\vec{v}^2/c^2}}, \quad E=\frac{mc^2}{\sqrt{1-\vec{v}^2/c^2}} \tag{9} \label{eq9}$$ for the relativistic momentum vector and the relativistic energy. Expressing $E$ in terms of $\vec{p}$, we obtain the relativistic energy-momentum relation $$E=c\sqrt{m^2 c^2 + \vec{p}^2 } \tag{10} \label{eq10}. $$ Note that \eqref{eq10} is also applicable for massless particles (like the photon), where $E= c |\vec{p}|$.

The methods desribed here can also be generalized to systems with an infinite number of degrees of freedom (in particular fields). As an example, in electrodynamics one finds expressions $\sim \vec{E}^2+ \vec{B}^2$ for the energy density and $\sim \vec{E} \times \vec{B}$ for the momentum density of the electromagnetic field.

Answer 3

Or we could say... $$ \frac{dE}{dp}=v.$$

The above equation is generally true when the symbol $E$ represents the Hamiltonian energy.

The Hamiltonian equations of motion are well known to be: $$ -\frac{\partial E}{\partial \vec q}=\dot{\vec p} $$ and $$ \frac{\partial E}{\partial \vec p}=\dot {\vec q} \equiv \vec v $$

My questions are:

1. Is the last equation true in special relativity?

Yes.

2. If so, and we use it to define momentum, assuming we already know what energy is (which I am not sure about), what more do we need to arrive at the energy-momentum relation?

In special relativity, we know that the energy is given in terms of the momentum as: $$ E={\left(p^2 c^2+m^2c^4\right)}^{1/2}\;.\tag{A} $$

So, $$ \vec v=\frac{\partial E}{\partial \vec p} = \frac{1}{2}{\left(p^2c^2 + m^2 c^4\right)}^{-1/2}2\vec p c^2 =\frac{c^2\vec p}{E}\;. $$

Or, solving for $p^2$ in terms of $v^2$, we have: $$ p^2 = \frac{m^2 v^2}{(1-v^2/c^2)}\;. $$

Or, solving for the energy in terms of v, we have: $$ E=mc^2\frac{1}{\sqrt{1-v^2/c^2}}\;.\tag{B} $$

Both Equations (A) and (B) are well known expression for the energy in special relativity. The former gives the energy of a free particle in terms of its momentum. The latter gives the energy of a free particle in terms of its velocity.