Energy definition in special relativity

I'm going through the early homework assignments for my special relativity course and I've got myself a little confused about energy. I've got a basic understanding of what the 4-momentum is, having defined it as $m\dfrac{dx^{\mu}}{d\tau}$, and shown that this is equal to $m\gamma (|\vec{v} |)(c,\vec{v})$ where $\vec{v}$ is the classical velocity in the inertial reference frame associated to the above Cartesian coordinates.

Now one of my assignments has began by saying

Denote the components of $p^{\mu}$ as $(E/c, \vec{p})$...

I don't understand why the time component of the 4-vector is being denoted as $E/c$. You can't just go arbitrarily denoting quantities by other quantities! So I'm left wondering whether this is a definition of relativistic energy, or if it follows from some other result in special relativity?

Thanks for any clarification.

• Well, $\gamma m c^2$ is the relativistic energy. This can be proven in a number of ways, depending on your level. – Javier Jul 3 '14 at 1:28
• Now think about the zeroth (or fourth depending on the convention your book uses) of the vector that you working with. Is the interpretation clear yet? If not have you been shown the energy needed to accelerate a mass yet? – dmckee Jul 3 '14 at 1:28
• You might also want to consider units. – user41976 Jul 3 '14 at 22:46
• You can also consider the way momentum has to be defined in special relativity so that it becomes a conserved quantity. Then you can consider the way this transforms under Lorentz transformations, and those transformations can be written in a covariant way in terms of the 4-momentum such that the last 3 components are the components of the momentum. It then follows immediately that the zeroth component is also conserved. – Count Iblis Jul 4 '14 at 0:31

Your book may be treating things a little backwards from the way they are usually done. The usual way is to define the momentum four-vector as the combination $(E/c, \vec{p})$, where $E$ is already known to be the total energy (the thing that reduces to $mc^2 + \frac{1}{2}mv^2$ for $v\ll c$) and then go on to show that it satisfies the properties expected of a four-vector. But it sounds like your book defines the four-momentum via $m\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}$, so you know it satisfies the properties expected of a four-vector from the start, and then they're going to have you prove that the time component of this four-vector has a low-velocity limit of $mc^2 + \frac{1}{2}mv^2$.

They're probably using the notation $E/c$ to be suggestive, but at this stage, it's just an arbitrary notation - that is, they don't intend $E$ to mean energy yet. If you prefer, you could use $p_t$ or something instead, until you actually do show that it's equal to the total energy divided by $c$.

• Thanks for the reply! You're right: I looked ahead and that's exactly what they've done. After showing that $E\cong mc^2 + \frac{1}{2}mv^2$ in the limit $v<<c$, they write "we have found that $E$ is the energy." But aren't there a whole bunch of functions whose low velocity limit is the classical energy? There must be a better reason for defining the energy to be $mc^2\gamma$ – James Machin Jul 3 '14 at 21:36
• Yeah, I glossed over that detail for simplicity, but showing that $E$ has that low-velocity limit is not sufficient to prove that it is the energy. But a real proof gets a little tricky; what it entails will depend on what you take to be the fundamental definition of energy. For example you could require it to be a conserved quantity for a certain class of physical systems, or you could define it as the value of the Hamiltonian, or you could define it as the thing that transforms as the time component of a four-vector and is equal to $mc^2$ in the rest frame, etc. – David Z Jul 4 '14 at 3:51
• All these definitions are equivalent, by the way (except maybe for some very weird physical systems), but how you do your proof depends on which definition you choose. – David Z Jul 4 '14 at 3:52
• @DavidZ: "[...] or define it as the thing that transforms as the time component of a four-vector and is equal to $mc^2$ in the rest frame" -- Would this provide a unique definition of "energy"? Would there be a reason (besides "habit") to reject for instance $$\tilde E[ \mathbf v ] := \frac{m~c^4}{\sqrt{ c^4 - | \mathbf v |^4 }},$$ along with some suitable "space part" $$\tilde{ \mathbf p}[ \mathbf v ] := \frac{m~c^2~| \mathbf v |~\mathbf v}{\sqrt{ c^4 - | \mathbf v |^4 }}$$ ? (OTOH, do these components even constitute a legitimate 4-vector ? ...) – user12262 Jul 4 '14 at 14:45
• @user12262 true, it wouldn't be unique. What I had in mind was that the spatial components would also have to reduce to nonrelativistic momentum in the low-velocity limit. That may be enough to uniquely define the energy (I'm not sure offhand though). – David Z Jul 4 '14 at 19:39

I don't understand why the time component of the 4-vector [ $m~\gamma(|\vec v|)~(c, \vec v)$ ] is being denoted as $E/c$.

So the underlying question is two-fold:
Why is "energy" considered the time component of some 4-vector at all?, and Why this specific time component expression, among time components of all different 4-vectors imaginable?
(Where we're obviously referring to "energy of something which is characterized by $m$", "with respect to the system or reference frame which determined the value $|\vec v|$ of that something".)

A suitably general and readily applicable definition of (how to measure) "energy" seems to be as "time component of the generator of translations"
(or "generator of succession"; alongside the definition of how to measure corresponding space components, namely of momentum as "generator of translations"):

$$\hat E :\simeq \frac{d}{dt}\!\!\big[ ~ \big].$$

Applying this operator to $\tau(\vec v)$ (to what else?) yields (by my naive calculation):

$$\hat E\big[ \tau(\vec v) \big] :\simeq \frac{d}{dt}\!\!\big[ t~\sqrt{ 1 - \left( \frac{|\vec v|}{c} \right)^2 } \big] := \frac{d}{dt}\!\!\big[ t~\sqrt{ 1 - \left( \frac{|\vec x|}{c~t} \right)^2 } \big] =$$ $$= \sqrt{ 1 - \left( \frac{|\vec x|}{c~t} \right)^2 } - \frac{\frac{t}{(-t^3)}~\left(\frac{|\vec x|}{c^2}\right)^2}{\sqrt{ 1 - \left( \frac{|\vec x|}{c~t} \right)^2 } } = \frac{1}{\sqrt{ 1 - \left( \frac{|\vec x|}{c~t} \right)^2 } } = \frac{1}{\sqrt{ 1 - \left( \frac{|\vec v|}{c} \right)^2 } },$$

where obviously $|\vec v| := |\vec x| / t$.

A similar excercise with one component of the momentum operator $\hat p_x :\simeq \frac{d}{dx}\!\!\big[ ~ \big]$ results in:

$$\hat p_x\!\big[ \tau(\vec v) \big] :\simeq \frac{d}{dt}\!\!\big[ t~\sqrt{ 1 - \left( \frac{|\vec v|}{c} \right)^2 } \big] := \frac{d}{dt}\!\!\big[ t~\sqrt{ 1 - \frac{x^2 + y^2 + z^2}{(c~t)^2} } \big] =$$ $$= -\frac{t~x}{(c~t)^2} \frac{1}{\sqrt{ 1 - \left( \frac{x^2 + y^2 + z^2}{c~t} \right)^2 } } = \frac{-x}{t~c^2} \frac{1}{\sqrt{ 1 - \left( \frac{x^2 + y^2 + z^2}{c~t} \right)^2 } } = -\frac{v_x}{c^2} \frac{1}{\sqrt{ 1 - \left( \frac{|\vec v|}{c} \right)^2 } )},$$

with $x$, $y$, $z$ denoting distances in three orthogonal directions, in a flat space, of course.

With suitable proportionality constants

$$\hat E := m~c^2~ \frac{d}{dt}\!\!\big[ ~ \big]$$ and

$$\hat p_x := m~c^2~ \frac{d}{dx}\!\!\big[ ~ \big], \,\,\, \hat p_y := m~c^2~ \frac{d}{dy}\!\!\big[ ~ \big], \,\,\, \hat p_z := m~c^2~ \frac{d}{dz}\!\!\big[ ~ \big]$$

then together

$$\left( \frac{1}{c^2} (\hat E)^2 - (\hat p_x)^2 - (\hat p_y)^2 - (\hat p_z)^2 \right)\!\big[ \tau(\vec v) \big] = (m~c)^2$$

which is a result evidently independent of $\vec v$, therefore an invariant characteristic of the "something" whose energy and momentum components were being determined; and $(\frac{E}{c}, \vec p)$ is a corresponding 4-vector expression.

All this applies in the simplest case that the "something" which is characterized by the invariant $m$ is "free". If instead a "potential" enters the consideration then the invariant is rather expressed as

$$\left( \frac{1}{c^2} (\hat E - q~A_t)^2 - (\hat p_x - q~A_x)^2 - (\hat p_y - q~A_y)^2 - (\hat p_z - q~A_z)^2 \right)\!\big[ \tau(\vec v) \big] = (m~c)^2,$$

where $\mathbf A := (\frac{A_t}{c}, \vec A)$ is a suitable 4-vector potential (whose components may in turn be expressed as derivatives of a suitable "phase function" $\alpha( \mathbf x )$), and $q$ represents a "charge".

• Thanks for the reply! I knew that the energy and momentum were operators in Quantum Mechanics, but I've not yet seen them as operators in anything else -it might take me a while to get my head around this – James Machin Jul 3 '14 at 21:30
• @James Machin: "Thanks for the reply!" -- Sure; thank you in turn for asking. "[...] it might take me a while to get my head around this" -- Me, too ... – user12262 Jul 4 '14 at 14:52

Without calculus, Special Relativity is treated in the fully discrete theory of "causal sets." Einstein suggested that a discrete theory of space-time could provide an inherent metric, while a continuum requires that a metric be imposed as an accessory to space-time. In causal set theory, the manifold is formulated in terms of time alone, primitive spatial relations being excluded from the theory. (If successful, that would constitute the first reduction of parameters in physics since Newton's original reduction.) I noticed, in a causal set diagram of 3 arrows, that frequency ratios are formed, useful for defining energy ratios in accord with Planck's E=hf. The "causal link," or discrete temporal transition, is implicated as the quantum of energy ratios, or simply the quantum of energy. We have here a structural definition of energy and its quantum, in terms of time alone, illustrated in the simplest case by a time diagram of 3 arrows. Thus we have the prospect of reducing space, time, and energy to causal sets, which are simply formations generated by sheer temporal succession. See "Causal Set Theory and the Origin of Mass-ratio." http://vixra.org/pdf/1006.0070v1.pdf

• Full disclosure: you're the author of the linked paper. – Brandon Enright Sep 13 '16 at 16:27

protected by Qmechanic♦Sep 13 '16 at 15:54

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?