# Mass-Energy derivation

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Ziang
SR does not only deal with inertial motion. It can handle acceleration just fine.

So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?

SiennaTheGr8
So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?

SR is fundamentally inconsistent with Newtonian gravity, so in general it's best to neglect gravity altogether in SR (though there may be circumstances in which you can overlook the inconsistency and get some "good enough" results that aren't quite right; I'm not sure whether your question falls into this category).

To deal with gravity in a way that's consistent with SR requires GR.

2022 Award
So is it ok to say that according to SR, the potential energy of an object m at a height h is PE = mgh, because g is as same as acceleration?
No. As Sienna says, you can't handle gravity in SR, you need GR.

Gravity is usually used as a straightforward example of a force in Newtonian physics, but it is extremely complicated in relativity. You can handle electromagnetism within the framework of SR if you want an example of a force.

Ziang
SR is fundamentally inconsistent with Newtonian gravity, so in general it's best to neglect gravity altogether in SR (though there may be circumstances in which you can overlook the inconsistency and get some "good enough" results that aren't quite right; I'm not sure whether your question falls into this category).

Yes, I would like to see how SR potential energy look like in simple cases like constant gravitational acceleration. "Good enough" is good enough. :)

Staff Emeritus
Yes, I would like to see how SR potential energy look like in simple cases like constant gravitational acceleration. "Good enough" is good enough. :)

I think you'll need tensor methods usually associated with GR to work this out. I'll take a stab at how it goes. Unfortunately I can't gurantee I won't make errors.

If we take "constant gravitational acceleration" to be the Rindler metric, then there is at least a conserved notion of total energy. There are several variants of the RIndler metric one might use, I would use

*CORRECTION*

$$-\left(1+gz\right)^2 \left(\frac{dt}{d\tau}\right)^2 + \left(\frac{dx}{d\tau}\right)^2 + \left(\frac{dy}{d\tau}\right)^2+\left(\frac{dz}{d\tau}\right)^2$$

This replaces the Minkoskii metric for an inertial observer

$$-dt^2 + dx^2 + dy^2 + dz^2$$

I've used geometric units for both the Rindler and Minkowskii metrics, which is how I avoid having the constant c (the speed of light) appear in the metric tensor.

If we let ##t=x^0, x=x^1, y=x^2, z=x^3## we have the energy momentum 4-vector ##E^i## and it's co-vector ##E_i##. They are related by the usual method of raising and lowering indices with the metric tensor, i.e. ## E^i = g^{ij} E_j ## and ##E_i = g_{ij} E^j##.

The metric tensor, and the associated need for vectors and coverctors, make the formulation a bit more complicated if one isn't familiar with tensor methods already.

The only component of the metric tensor that's neither zero nor unity is ##g_{00}## which is a negative number with the chosen sign convention, ##g_{00} = -(1+gz)^2##. The corresponding inverse metric has ##g^{00} = -1/(1+gz)^2##.

We have a static space-time, so we can use the same techniques to define various conserved quantities, of which there is the conserved energy ##E_0##, the covector component of the energy-momentum 4-vector, and two momenta, ##E_1## and ##E_2##. ##E_3## won't be conserved. By "conserved" I mean that ##E_0, E_1, E_2## are all constant for an object with no forces acting on it, i.e. an object following a space-time geodesic.

There may be some sign issues remaining here, but regardless of the best choice of sign, the above are constants of motion.

We can then find that ##E^0 = m \frac{dt}{d\tau} = g^{00} E_0## so we wind up with ##E^0 = -\frac{E_0}{(1+gz)^2}## with ##E_0## being some constant of motion that we call "energy". I'm not aware of any scheme to break up total energy into "kinetic energy" and "potential energy" in this formulation, but the expression for total energy is of some interest.

The other formula we mainly need is that ##E^i E_i = -m^2##, the tensor version of ##E^2 = (pc)^2 + m^2##. Using this relationship, we can find, for instance the value of ##dz/d\tau## given the conserved constants of motion of the object representing "energy" ##E_0##, "x-momentum" ##E_1## and "y-momentum" ##E_2##. "z-momentum", isn't conserved, but we can calculate ##\frac{dz}{d\tau}## knowing the other conserved quantities ##E_0, E_1, E_2##, the invariant mass ##E^i E_i##, and the z-coordinate of the object much as we might calculate the velocity of a falling object knowing it's energy, x & y momentum, and it's current height in the Newtonian case.

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Mentor
We have a static space-time, so we can use the same techniques to define various conserved quantities

[Note: corrections made corresponding to those made by @pervect in his post.]

I think it's worth discussing a bit the details of how this is done. I'll run through the details with the "energy" conserved quantity, since that's the one of interest here.

The "energy" conserved quantity exists because we have a timelike Killing vector field ##K^\nu##, which in the coordinate chart you've chosen is simply the vector ##(1, 0, 0, 0)##. The corresponding conserved quantity ##E## (no index because it's a Lorentz scalar), by Noether's Theorem, is then ##E = - g_{\mu \nu} P^{\mu} K^{\nu}##, where ##P^{\mu}## is the 4-momentum of an object which is undergoing geodesic motion, and the minus sign is because of the ##- + + +## metric signature convention you're using. Writing this out, we get

$$E = \left( 1 + g z \right)^2 P^0 = \left( 1 + g z \right)^2 m \frac{dt}{d\tau}$$

We can invert this to get

$$P^0 = \frac{E}{\left( 1 + g z \right)^2}$$

Lowering the index then gives ##P_0 = - E##. However, I think it's still worth keeping these two quantities conceptually distinct--one is the 0 component of the 4-momentum, which will change if you change coordinate charts, while the other is a conserved Lorentz scalar that remains the same in any chart. For example, it's a good exercise to re-do the computations I did above in the Minkowski chart; you will find that ##P^0## and ##P_0## look very different in terms of ##E## along the same geodesic.

I'm not aware of any scheme to break up total energy into "kinetic energy" and "potential energy" in this formulation

The way to do it is to pick a "zero point" for potential energy. For this case, an obvious such point is ##z = 0##--which is actually a hyperplane. The reason this is an obvious choice is that, throughout this hyperplane, the metric is actually Minkowski, which is a nice property for the "zero point" of potential energy to have. Then the conserved energy ##E## of an object on a geodesic that is momentarily at rest at some ##z## in these coordinates will be ##m \left( 1 + g z \right)^2##, and since the object is momentarily at rest, all of this energy is potential energy; you can then trace the object's motion along the geodesic and see how the potential energy is converted to kinetic energy as it falls. (Note that this computation also makes it clear that in this formulation, the rest energy ##m## is part of the potential energy.)

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Staff Emeritus
I think it's worth discussing a bit the details of how this is done. I'll run through the details with the "energy" conserved quantity, since that's the one of interest here.

The "energy" conserved quantity exists because we have a timelike Killing vector field ##K^\nu##, which in the coordinate chart you've chosen is simply the vector ##(1, 0, 0, 0)##. The corresponding conserved quantity ##E## (no index because it's a Lorentz scalar), by Noether's Theorem, is then ##E = - g_{\mu \nu} P^{\mu} K^{\nu}##, where ##P^{\mu}## is the 4-momentum of an object which is undergoing geodesic motion, and the minus sign is because of the ##- + + +## metric signature convention you're using. Writing this out, we get

$$E = \left( 1 + g z \right) P^0 = \left( 1 + g z \right) m \frac{dt}{d\tau}$$

We can invert this to get

$$P^0 = \frac{E}{1 + g z}$$

Lowering the index then gives ##P_0 = - E##. However, I think it's still worth keeping these two quantities conceptually distinct--one is the 0 component of the 4-momentum, which will change if you change coordinate charts, while the other is a conserved Lorentz scalar that remains the same in any chart. For example, it's a good exercise to re-do the computations I did above in the Minkowski chart; you will find that ##P^0## and ##P_0## look very different in terms of ##E## along the same geodesic.

The way to do it is to pick a "zero point" for potential energy. For this case, an obvious such point is ##z = 0##--which is actually a hyperplane. The reason this is an obvious choice is that, throughout this hyperplane, the metric is actually Minkowski, which is a nice property for the "zero point" of potential energy to have. Then the conserved energy ##E## of an object on a geodesic that is momentarily at rest at some ##z## in these coordinates will be ##m \left( 1 + g z \right)##, and since the object is momentarily at rest, all of this energy is potential energy; you can then trace the object's motion along the geodesic and see how the potential energy is converted to kinetic energy as it falls. (Note that this computation also makes it clear that in this formulation, the rest energy ##m## is part of the potential energy.)

Note that I made an unfortunate typo for the RIndler metric in my original post, which I've since corrected.

Ziang
Thank you all,
I know GR is the best to deal with gravity. But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

SiennaTheGr8
Thank you all,
I know GR is the best to deal with gravity. But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

It sounds like you want an answer that reconciles SR with Newton's theory of gravitation, but no such thing exists.

You can just use the ##mgh## calculation from Newtonian physics -- as far as I know, SR doesn't "modify" that at all. But what if the object is traveling so fast relative to the Earth that SR is needed to describe the relative motion? does that make the Newtonian calculation less accurate?

I'm going to guess that the answer is yes, and that you then need GR to get a good handle on the physics. But I'm out of my depth here. (And it's probably worth mentioning that the very concept of energy conservation is complicated in GR, so for all I know the question might be ill-posed).

Mentor
Note that I made an unfortunate typo for the RIndler metric in my original post, which I've since corrected.

Ah, yes, good catch. I'll update my post accordingly.

Gold Member
But Sienna said SR can deal fine with acceleration. So let us assume constant g is as same as constant acceleration, to figure out what is the potential energy of an object m at a height h, using SR, please.

Think about what you're asking here. The premise is that SR can be used to analyze and understand acceleration. You cannot conclude that it's therefore possible to use SR to analyze and understand everything that involves an acceleration.

In other words, the quantity ##mgh## contains the factor ##g## which is the free fall acceleration. SR involves acceleration. It doesn't follow that ##mgh## can be understood using SR.

In fact, calling ##mgh## the potential energy is an approximation, valid only in the Newtonian realm, and even then valid only when the gravitational field is constant.

Ziang
You can just use the ##mgh## calculation from Newtonian physics -- as far as I know, SR doesn't "modify" that at all. But what if the object is traveling so fast relative to the Earth that SR is needed to describe the relative motion? does that make the Newtonian calculation less accurate?
The potential energy may not be mgh.
Let us consider the simplest case only. The object m is dropped freely from a height h.
dE = fdx (as they do in the derivation of mass-energy relation in textbooks)
but for now, x = h, and f = mγ3a = mγ3g
So PE = mgI3dh)
This PE is larger than mgh.

Ziang
Mentor
It is as same as it is pulled by a force f = mγ3a.

No, it isn't, because the frame you are working in is not an inertial frame. In an inertial frame there is no "g"; in order to "simulate" gravity in SR, you have to use a non-inertial frame. An earlier post by @pervect, with some additional details added in a post by me, give the basics of how the simplest version of such a frame--Rindler coordinates--would work, including how potential energy is defined. I suggest reading those posts carefully.

Staff Emeritus
I can only guess what Ziag's background is. It appears to me he's struggling with the Lorentz transform, and has some issues understanding the relativity of simultaneity. Getting sidetracked into the more advanced methods needed to handle accelerating frames just isn't going to be productive for anyone who can't handle the much simpler case of special relativity in inertial non-accelerating frames.

I'm not sure if it well help Ziang, but a paper by Scherr , et al, <<link>>, entitled "The challenge of changing deeply held student beliefs about the relativity of simultaneity" at least illustrates the fact that he is not alone in having trouble with understanding the relativity of simultaneity. The Scherr et al paper is written for teachers and not students, still it might be helpful.

Ziang
To derive the mass-momentum relation, SR defined momentum ##\mathbf{p} = \gamma m \mathbf{v}##.
After the relation derived, SR defined mass ##(mc^2)^2 = E^2 - (pc)^2##.
Thus, the definitions are circular. The definition of mass is not clear.

2022 Award
Not really. You define the four velocity, ##U##, which is normalised to ##c^2##, then you observe (from experiment) that there is an object-specific quantity that we'll call ##k^{(i)}## for the ##i##th body such that the sum of ##k^{(i)}U^{(i)}## over all bodies is conserved in collisions. Then you do a Taylor series expansion and show that ##k^{(i)}## is the Newtonian mass of the ##i##th body

Science is empty if you don't ground it in experiment.

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SiennaTheGr8
SiennaTheGr8
@Ziang

I think you're confusing yourself by imagining that there's only one "path" through all of this material. Really, there are many possible starting points that will all lead to the same set of equations.

Yes, most people in the know today would probably agree that mass in SR is best defined as the magnitude of the four-momentum. But Einstein developed SR without recourse to the four-vector formalism at all! My point is that there's more than one way to skin a cat. Don't get hung up on seemingly contradictory approaches.

I suggest getting yourself a good book on special relativity and starting from the beginning. Get an understanding of the basic kinematics first. Then you'll be able to easily grasp the dynamics you're grappling with.

Ibix
Ziang
You define the four velocity, ##U##, which is normalised to ##c^2##, then you observe (from experiment) that there is an object-specific quantity that we'll call ##k^{(i)}## for the ##i##th body such that the sum of ##k^{(i)}U^{(i)}## over all bodies is conserved in collisions. Then you do a Taylor series expansion and show that ##k^{(i)}## is the Newtonian mass of the ##i##th body

Science is empty if you don't ground it in experiment.

So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?

2022 Award
So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?
Same as in Newton, except you can't use gravitational experiments - you need GR for that. You could collide them with something (probably a sequence of identical asteroid-sized bullets at a range of velocities). Or make them acquire a known charge and use electromagnetism.

For something like a planet none of the above is really practical. You really need GR, then you can study orbits. Or show that Newtonian gravity is the weak-field low-speed approximation and just use the masses implied by Newtonian physics - errors will be tiny.

Gold Member
To derive the mass-momentum relation, SR defined momentum ##\mathbf{p} = \gamma m \mathbf{v}##.

No, you can't both define and derive something. You can take an approach where you define momentum, or you can take an approach where you derive momentum based on definitions of other things. But however you do it, you will have to start with some definitions of things, things you define rather than derive.

And by the way, ##\gamma m \vec{v}## is a valid expression for the momentum of only massive, as opposed to massless, particles. ##\left( \frac{E}{c^2} \right) \vec{v}## is a more general expression that's valid for all particles.

So what experiments could be to get the masses for the Sun, Earth, and Moon, based on definitions of mass in SR?

You can't do experiments with these objects, we don't have the technology to manipulate them. What you can do is observe, and based on those observations compare their masses with each other. By determining the value of the gravitational constant ##G## you compare Earth's mass with the mass of the standard body. All measurements of mass are comparisons of one object's mass with another.

The norm of the energy-momentum four-vector equals what we measure when we measure mass. All we're doing is saying the momentum of these objects is zero, so their energy is the norm of their energy-momentum four-vector, which in turn is equivalent to their mass.

Ibix and SiennaTheGr8
Sorcerer
In SR, fundamental concepts velocity, acceleration, force, momentum were defined as four-vectors.
https://en.wikipedia.org/wiki/Four-vector
Does someone show me a derivation of the equation E = mc2 from those definition?
Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work).

Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

SiennaTheGr8
Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work).

Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

Not following you at the end there.

Sorcerer
Not following you at the end there.
You’ll get an expression for kinetic energy that has two terms separated by a minus sign.

T = E - U.

The U, the “potential energy” (that is, the “rest” energy that remains when speed u=0),” will end up being mc2.

I probably should have mentioned that.

SiennaTheGr8
Ah, that's not usually referred to as "potential energy." It's "rest energy" / "invariant energy" / "proper energy" / "mass-energy."

Sorcerer
Ah, that's not usually referred to as "potential energy." It's "rest energy" / "invariant energy" / "proper energy" / "mass-energy."
True. I used that phrasing more as an analogy since it works out to have the same form. Ambiguity aside, the derivation works like a charm. Set speed to zero and you get what OP wanted. *shrugs*

Ziang
The GR could lead to the relation too?

Sorcerer
The GR could lead to the relation too?
For small intervals of spacetime, spacetime is approximately flat. So if it's true in SR it's true locally in GR, if I understand it correctly.

2022 Award
The GR could lead to the relation too?
GR is a generalisation of SR, so all results in SR can be derived using the tools of GR.

In this particular case, you'd probably start from the notion of a worldline and do a slightly more rigorous derivation of the four-velocity as the tangent to the worldline. The rest is the same.

Ziang
Could Newtonian mechanics concepts/absolute space and time lead to the relation as well?

SiennaTheGr8
Here’s my favorite way. (Ibknow the discussion has moved on, but OP asked for a quick and easy derivation, so here is one way to do it that seems to work).

Start with a spatial differential dx. Then divide by proper time (so that everyone agree with what you do going foward).

From there, use the usual kinemetics. Multiply by mass to get momentum, take the time derivative to get force, integrate over distance to use the work-energy theorem to get an expression for kinetic energy, then notice that kinetic energy = total energy minus potential energy, and you’re pretty much done.

I'm sure I've seen this done for the rectilinear case (though I can't remember where), but I thought I'd give it a crack for the general case. The idea is that one has already obtained the equation for relativistic momentum ##\vec p = \gamma m \vec v## and plugs it into the "work" side of the work-energy theorem to see what pops out on the "energy" side. Using ##\vec f = \dot{\vec p} = mc^2 \, d(\gamma \vec \beta)/d(ct)##:
$$\begin{split} \int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec r_f}_{\vec r_i} \dfrac{d (\gamma \vec \beta)}{c \, dt} \cdot d \vec r \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left( \gamma \, d \vec \beta + d \gamma \, \vec \beta \right), \end{split}$$
where
$$\begin{split} d \gamma &= d \left[ \left( 1 - \vec \beta \cdot \vec \beta \right)^{-1/2} \right] \\[3pt] &= \dfrac{1}{2} \left(1 - \vec \beta \cdot \vec \beta \right)^{-3/2} \, d ( \vec \beta \cdot \vec \beta ) \\[3pt] &= \gamma^3 ( \vec \beta \cdot d \vec \beta ) , \end{split}$$
so
$$\begin{split} \int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left[ \gamma \, d \vec \beta + \gamma^3 ( \vec \beta \cdot d \vec \beta ) \vec \beta \right] \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \Bigl[ \gamma \left( 1 + \gamma^2 \beta^2 \right) \Bigr] ( \vec \beta \cdot d \vec \beta ) \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \gamma^3 ( \vec \beta \cdot d \vec \beta ) \\[3pt] & = mc^2 \int^{\gamma_f}_{\gamma_i} d \gamma \\[3pt] &= mc^2 \Delta \gamma \\[3pt] &= mc^2 + mc^2 \Delta \left( \dfrac{1}{2}\,\beta^2 + \frac {3}{8} \, \beta^4 + \frac{5}{16} \, \beta^6 + \frac{35}{128} \, \beta^8 + \dots \right), \end{split}$$
where the ##\beta^2## term suggests calling everything in the parentheses "kinetic energy" ##E_k## and the invariant first term "rest energy" ##E_0##. Then it makes sense to define ##\gamma mc^2 = \gamma E_0 \equiv E## as the total energy, and:
$$E^2 - (pc)^2 = (\gamma E_0)^2 - (\gamma \beta E_0)^2 = E_0^2,$$
though I suppose one might like to know whether ##E## is indeed conserved before calling it "energy."

I think I prefer to "justify" the existence of rest energy by Einstein's own reasoning (from his original ##E = mc^2## paper): if a body at rest emits identical light waves in opposite directions, then its kinetic energy remains the same in this frame (zero), but it loses energy nevertheless.

Sorcerer
I'm sure I've seen this done for the rectilinear case (though I can't remember where), but I thought I'd give it a crack for the general case. The idea is that one has already obtained the equation for relativistic momentum ##\vec p = \gamma m \vec v## and plugs it into the "work" side of the work-energy theorem to see what pops out on the "energy" side. Using ##\vec f = \dot{\vec p} = mc^2 \, d(\gamma \vec \beta)/d(ct)##:
$$\begin{split} \int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec r_f}_{\vec r_i} \dfrac{d (\gamma \vec \beta)}{c \, dt} \cdot d \vec r \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left( \gamma \, d \vec \beta + d \gamma \, \vec \beta \right), \end{split}$$
where
$$\begin{split} d \gamma &= d \left[ \left( 1 - \vec \beta \cdot \vec \beta \right)^{-1/2} \right] \\[3pt] &= \dfrac{1}{2} \left(1 - \vec \beta \cdot \vec \beta \right)^{-3/2} \, d ( \vec \beta \cdot \vec \beta ) \\[3pt] &= \gamma^3 ( \vec \beta \cdot d \vec \beta ) , \end{split}$$
so
$$\begin{split} \int^{\vec r_f}_{\vec r_i} \vec f \cdot d \vec r &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \vec \beta \cdot \left[ \gamma \, d \vec \beta + \gamma^3 ( \vec \beta \cdot d \vec \beta ) \vec \beta \right] \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \Bigl[ \gamma \left( 1 + \gamma^2 \beta^2 \right) \Bigr] ( \vec \beta \cdot d \vec \beta ) \\[3pt] &= mc^2 \int^{\vec \beta_f}_{\vec \beta_i} \gamma^3 ( \vec \beta \cdot d \vec \beta ) \\[3pt] & = mc^2 \int^{\gamma_f}_{\gamma_i} d \gamma \\[3pt] &= mc^2 \Delta \gamma \\[3pt] &= mc^2 + mc^2 \Delta \left( \dfrac{1}{2}\,\beta^2 + \frac {3}{8} \, \beta^4 + \frac{5}{16} \, \beta^6 + \frac{35}{128} \, \beta^8 + \dots \right), \end{split}$$
where the ##\beta^2## term suggests calling everything in the parentheses "kinetic energy" ##E_k## and the invariant first term "rest energy" ##E_0##. Then it makes sense to define ##\gamma mc^2 = \gamma E_0 \equiv E## as the total energy, and:
$$E^2 - (pc)^2 = (\gamma E_0)^2 - (\gamma \beta E_0)^2 = E_0^2,$$
though I suppose one might like to know whether ##E## is indeed conserved before calling it "energy."

I think I prefer to "justify" the existence of rest energy by Einstein's own reasoning (from his original ##E = mc^2## paper): if a body at rest emits identical light waves in opposite directions, then its kinetic energy remains the same in this frame (zero), but it loses energy nevertheless.
I never noticed that you can do a change of variable to dγ. All the times I’ve done this I’ve had to do trig substitution. Such an obvious thing but I missed it.

SiennaTheGr8
I didn't find that obvious, but in hindsight I guess I should have (since I already knew the energy equations I was working toward). Anyway, fun exercise!

Ziang
Is there any way to derive the relation based on Newtonian mechanics clearly and convincedly?