Where did the 1/2 go in E=mc^2?

[PeterDonis]

what is the proof that the speed of light is a constant?

It's not a proof, it's a definition: the SI definition of the meter makes the speed of light a universal constant.

The physical reason why this definition works well is the geometry of spacetime.

 

[vanhees71]

Out of interest, what is the proof that the speed of light is a constant?

There is of course never a "proof" of a fundamental natural law as in mathematics, which is based on some axioms from which everything can be derived by logical thought. The difference between math and theoretical physics is that math creates its own universe and doesn't care about whether it describes the real universe. Theoretical physics is the opposite: It tries to describe objective quantitative observations about the universe as we find it using our senses and all kinds of devices to enhance them with as few as possible "fundamental laws", from which all known observed facts can be, more or less accurately, deduced. The constancy of the speed of light in our observable universe is one of the most fundamental laws. It cannot be proven, but always tested with more accurate observations. So far, there's no hint at a violation of this fundamental law, which is the basis of the relativistic description of space and time (or rather spacetime). The same holds for Planck's constant $h$, which is the fundamental constant entering physics through quantum theory, the other most fundamental theory we have.

As you see, the physical edifice is not complete. It consists of two fundaments, the spacetime model (General Relativity and Special Relativity as an approximation, when gravity can be neglected, which fortunately is the case for almost everything concerning the description of matter) on the one hand and relativistic local quantum field theory on the other. Although intimately connected (e.g., the very fundamental properties all quantum descriptions must follow are derived from the mathematical structure of spacetime in terms of its symmetries), a completely consistent theory is still missing, i.e., there's no quantum theory of spacetime or, again intimately related, gravitation, i.e., a unified theory of "everything" (and we can't even be sure whether this really is everything)!

 

[Mister T]

Out of interest, what is the proof that the speed of light is a constant?

The problem with proofs is that their conclusions don't contain any information that wasn't already present in the assumptions used to construct the proof. So, for example, if you assume the validity of Maxwell's equations you can prove that the speed of light is independent of the motion of the source or observer (which is what I assume you mean by constant in this context).

The issue is not one of proof, but of observation. We always observe that the speed of light is independent of the speed of the source or observer.

The fact that the speed of light can be measured more precisely than the length of a meter stick resulted in the metrologists setting an exact value for the speed of light rather than defining the length of a stick to be exactly one meter.

 

[metastable]

So, assuming my thoughts are correct, what happened to the

1/21/2​

1/2? Einstein's formula doesn't contain it.

Is it anything to do with pair creation vs black holes? Suppose 1.022 MeV photon near a nucleus produces a electron-positron pair. I now drop the positron from infinity into a black hole. What is the rest energy + kinetic energy of the positron (with respect to a similar particle at its original position) just before the event horizon?

 

[PeterDonis]

What is the rest energy + kinetic energy of the positron (with respect to a similar particle at its original position)

This is not well-defined. To compare energies in a curved spacetime, the objects must be close to each other.

 

[PeterDonis]

Is it anything to do with pair creation vs black holes?

No, the question under discussion in this thread has nothing to do with that.

 

[metastable]

Mass/energy and length/time still have inverse physical dimensions to each other, and GGG still has units of length squared/inverse mass squared.

What is the physical meaning of the inverse of photon frequency. I’ve heard here before here it’s not emission duration.

 

[Dale]

What is the physical meaning of the inverse of photon frequency.

The inverse of the frequency is the period.

 

[vanhees71]

It has the usual physical meaning of any wave phenomenon the inverse frequency is the temporal period of the wave, $T=2 \pi/\omega$, while the inverse wave number is the wave length, i.e., the spatial period, $\lambda=2 \pi/|\vec{k}|$.

 

[metastable]

Is there any appropriate mathematical conversion from mass to energy to frequency to time to length?

 

[PeterDonis]

Is there any appropriate mathematical conversion from mass to energy

Of course, just multiply by $c^2$ (which in natural units means $1$).

energy to frequency

Sure, just divide by Planck's constant (which in natural units means $1$).

frequency to time

Um, just take the reciprocal?

time to length?

Multiply by $c$.

 

[Federation 2005]

Get ready to have your eyes opened! I'm going to show you where the 1/2 went (and where it still resides). This will be new to you (no matter who you are). In fact, we're going to make this entertaining for you. I will pose the answer as an exercise. You will have to solve it to find what you need.

Exercise:
Assume αv² < 1 and m ≠ 0, and let u = v²/(1 + √(1 - αv²)).

Prove that if L = mu, H = Mu and M = m + αH, then M = m/√(1 - αv²), H = pv - L, where p = Mv, H = p²/(m+M) and p² - 2MH + αH² = 0

Calculus bonus: Prove that p = ∂L/∂v.

Comments:
Interpretations should be obvious, if you are familiar with the letter names. L,H,m,M,p,v are respectively named Lagrangian, Hamiltonian, rest mass, moving mass, momentum, velocity. The extra coordinate u has no name; but plays a central role in all of this.

The magic parameter α distinguishes different types of kinematics from one another. The geometry associated with it has, as its invariants, the following:
dt² - α(dx² + dy² + dz²), ∇² - α (∂/∂t)²

For α > 0, you have relativity with an underlying Minkowski geometry in which the invariant speed is given as c = 1/√α. Upon substitution into the formulae above, you arrive at the usual expressions for the associated terms that any Relativist should be familiar with.

For α = 0, you have an underlying non-relativistic framework (associated with a flat spacetime Newton-Cartan geometry). As you can see, the extra coordinate u reduces to your familiar ½ v². That's where it is.

For α < 0, these would be the expressions that apply in a Euclidean form of the kinematics, where the underlying geometry has 4 dimensions of space and none of time.

Heuristically, you can think of the Legendre transform from L to H as a kind of "renormalization" in which one first adds the mass equivalent αmu of the "bare" kinetic energy mu to the mass m; then the mass equivalent
α(αmu)u of the bare kinetic energy (αmu)u of that extra mass; then the mass equivalent α(α(αmu)u)u of the bare kinetic energy (α(αmu)u)u of the second contribution to the mass, and so on ad infinitum, to arrive at the total moving mass M:
m + αmu + α(αmu)u + α(α(αmu)mu)mu + ⋯ = m/(1 - αu) = m/√(1 - αv²) = M
(the last two equalities are what the exercise was, ultimately, to show!)

The corresponding kinetic energy Mu is the Hamiltonian H.

Underlying geometry;
Notice the mass shell invariant has 5 components (if you treat the momentum p as a 3-vector), not 4. The associated signature is 4+1, independent of what α is. Divide out M² from p² - 2MH + αH², and you get the invariant v² - 2u + αu². In addition, you also have the linear invariants M - αH and, associated with this, 1 - αu.

If, in place of u, you write -du/dt (I prefer it with the other sign), and express the velocity v as dr/dt, then you may (upon multiplying by dt²) rewrite this as a line element: dx² + dy² + dz² + 2dtdu - αdu²; and linear invariant ds = dt + αdu. For relativity, setting the line element to 0 (that is, confining one to the 5D light cone for this geometry) reduces ds to the proper time and the line element to the Minkowski metric.

This still has meaning, even when α = 0 ... and is closely linked to the natural 5-dimensional representation of the Galile group and its central extension, the Bargmann group. For α ≠ 0, it gives you the relativistic version of this interpretation.

The unification of Mikowski, Newton-Cartan and Euclidean geometries here is entirely analogous to the unification of Euclidean, spherical and hyperbolic geometries into projective geometry.

Symmetries:
Treating the momentum, now, as a vector p↑, then under infinitesimal rotations ω↑, infinitesimal boosts υ↑, the momentum p↑, moving mass M and kinetic energy H transform together as a 5-vector Δ(H,p↑,M) = (-υ↑·p↑, ω↑×p↑ - υ↑ M, -αυ↑·p↑). From this, you can derive a finite transform by exponentiating it (H,p↑,M) → exp(sΔ) (H,p↑,M) = (1 + sΔ + s²/2! Δ² + s³/3! Δ³ + ⋯)(H,p↑,M).

This leads to another exercise that generalizes this a bit further still...

Exercise:
Define the following operations
Δ (dr↑, dt, du) = (ω↑×dr↑ - β υ↑ dt, -αυ↑·dr↑, υ↑·dr↑)
Δ (H, p↑, M) = (-αυ↑·p↑, ω↑×p↑ - υ↑ M, -αβυ↑·p↑)
on 3-vectors p↑ = (p₁,p₂,p₃), dr↑ = (dx,dy,dz) and scalars dt, where ()×() and ()·() respectively denote the cross product and dot product.

Express, in closed form, exp(sΔ)(dr↑, dt, du) (where exp(sΔ) = 1 + sΔ + s²/2! Δ² + s³/3! Δ³ + ⋯) in terms of θ↑ = sω↑ and w↑ = sυ↑.

You may use the following functions
C(λ,x) = 1 + λ x²/2! + λ² x⁴/4! + ⋯
S(λ,x) = x + λ x³/3! + λ² x⁵/5! + ⋯
E(λ,x) = x²/2! + λ x⁴/4! + λ² x⁶/6! + ⋯
in your answer [note: the E function is ultimately where the v²/2 and its relativistic analogues stem from]; and assume the following properties
C(λ,0) = 1, S(λ,0) = 0, E(λ,0) = 0
C(λ,x) - λE(λ,x) = 1
S(λ,x)² - 2E(λ,x) C(λ,x) + λE(λ,x)² = 0
C(λ,x)² - λS(λ,x)² = 1
Prove the operator identity
Δ²(Δ² + ω² - αβυ²) = αβ(ω↑·υ↑)²
For αβ ≠ 0, express the answer also in terms of
v↑ = w↑/w S(αβ,w)/C(αβ,w)
(where w = |w↑|) restricted to vectors v↑ where αβv² < 1, and show that
C(λ,x)² = 1/(1 - αβv²).

Prove that the following are invariants under these transforms:
dx² + dy² + dz² + 2βdtdu + αβdu², dt + α du, β(p₁² + p₂² + p₃²) - 2MH + αH², M - αH
and, from these, derive the reduced 4D invariants
α(dx² + dy² + dz²) - β dt², M² - αβ(p₁² + p₂² + p₃²)

Discuss each of the cases αβ > 0, αβ < 0, (α = 0 & β ≠ 0), β = 0. For αβ > 0 what's the invariant velocity in terms of α and β; and what does the case β = 0 correspond to?

Related References:
Bargmann structures and Newton-Cartan Theory
C. Duval, G. Burdet, H. P. Künzle, and M. Perrin
Physical Review D, Volume 31, Number 8, 15 April 1985
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.31.1841
Although their 5D geometric formulation (corresponding to my β = 1 cases above) is tailored for Newton-Cartan geometry and non-relativistic gravity; the underlying framework is general and provides a nearly seamless unifying framework that also includes 4D General Relativity (embedded in the 5D geometric representation just described) as a special case. The "almost" part is that there is still an obstruction that prevents a smooth continuous transform from α > 0 → α = 0.

Possible Kinematics
Henri Bacry, Jean-Marc Lévy-Leblond
Journal of Mathematical Physics 9, 1605 (1968);
https://aip.scitation.org/doi/10.1063/1.1664490
They provide a general classification of all possibilities according with general assumptions ... but arbitrarily exclude the Euclideanized cases (αβ < 0) - despite the utility that they have in field theory! Their notation's different than mine, because everything I laid out here was independently developed (and much cleaner) and only later cross-fitted with their work.

Kinematical Lie algebras via deformation theory
José M. Figueroa-O'Farrill
https://arxiv.org/abs/1711.06111
An even bigger classification that derives everything as a deformation of the "static" group (which corresponds to the α = 0, β = 0 above). I did the same thing several years before this, but didn't take it as far as they did (they allow the central charge -- the invariant M - αH in my notation above) to have non-zero Lie brackets -- something I specifically excluded in my version.

 

[Dale]

@Federation 2005 please use LaTeX for posting mathematical expressions

 

[olgerm]

if m is restmass, then $E_{rest}=m*c^2$, but in classical physics $E_{kinetic}=\frac{m*v^2}{2}$ so these 2 are just not the same quantities.

but reltivistic kinetic energy approches classical relativistic energy if speed of light is very big.
$E_{kinetic}=E-E_{rest}$
$E_{rest}=m*c^2$
$E=\frac{m*c^2}{\sqrt{1-\frac{v^2}{c^2}}}$

therefore
$E_{kinetic}=\frac{m*c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m*c^2$
to get classical limit where c>>v
$E_{kinetic\ classic}=\lim_{c \to \infty}(\frac{m*c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m*c^2)=\frac{m*v^2}{2}$

 

[DrGreg]

$E_{kinetic\ classic}=\lim_{c \to 0}(\frac{m*c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m*c^2)=\frac{m*v^2}{2}$

The limit is effectively $c \to \infty$, not $c \to 0$, or, perhaps better, $(v/c) \to 0$

 

[olgerm]

The limit is effectively $c \to \infty$, not $c \to 0$, or, perhaps better, $(v/c) \to 0$

Yes. I corrected it now.

 

[olgerm]

If m is relativistic mass in the original post:
K and E are still not same quanities, beacuse $E=m_{relativistic}*c^2$ is rest energy plus kinetic energy and $K=\frac{m_0*v^2}{2}$ is kinetic energy.

 

[cmb]

I suggested a while back that, of course, E=mc^2 is not an equation with 'c' as a variable, it is really a conversion factor. At least in the rest frame.

I have seen the following additional explanation, which describes the connection with Newtonian mechanics. If one writes the Lorenz factor as y = 1/SQRT(1-b^2), where b = v/c (sorry no latex) then the equation in special relativity, E = ymc^2, can be expanded as a Maclaurin series.

y = 1/SQRT(1-b^2) = 1 + 1/2.b^2 + 3/8.b^4 + 5/16.b^6 ... etc

As v (i.e. b) tends to small values then the end terms of the series vanish.

The first two terms of the expansion are then

E = mc^2 + 1/2.mv^2

which is therefore the Newtonian approximation for small v.

So I suppose an alternative answer for the OP is that the "1/2" can be made to "reappear" by a Maclaurin expansion of the Lorentz factor.

 

[Ibix]

So I suppose an alternative answer for the OP is that the "1/2" can be made to "reappear" by a Maclaurin expansion of the Lorentz factor.

This is a standard way of showing that the low speed limit of relativistic kinetic energy is Newtonian kinetic energy, yes.

 

[Mister T]

The kinetic energy of a particle can be expressed as $\frac{\gamma^2}{\gamma+1}mv^2$. In the low speed limit $\gamma \approx 1$ so the fraction $\frac{\gamma^2}{\gamma+1}\approx \frac{1}{2}$. So you can see that the $\frac{1}{2}$ didn't go anywhere. It's still there!