Einstein’s Most Famous Equation, E = mc2
Einstein’s Theory of Relativity is simply a trivial extension, to 4-dimensions, of the famous Theorem of Pythagoras.
The Pythagorean theorem itself can easily be seen to be true, just by glancing at these two (equal-area) big squares:
Obviously, the square on the hypotenuse equals the sum of the squares on the other two sides of (any) red triangle.
The invention of algebra allowed us to write this famous Pythagorean theorem as an equation: ds2 = dx2 + dy2
Relativity asserts that Pythagoras’s theorem—but in our 4-dimensional universe—is: ds2 = dx2 + dy2 + dz2 − dt2
where ds is the separation in space-and-time of any two events (for example, your snapping your fingers twice, as
you wave your arm around)—and asserts that ds is invariant: that is, is the same for any observer of the two events.
Note the critical minus sign: this is the only way that time differs from being merely one more space dimension!
How do we know that relativity is true? Because—as we will shortly see—it successfully predicts the atomic bomb!
Newton taught us that, for a moving mass m, the important quantities are momentum mv and kinetic energy 1/2 mv2
So!—we will be keeping a sharp eye out for Newton’s mv and 1/2 mv2 as we explore the implications of Einstein’s
remarkable pseudo-Pythagorean assertions.
What Einstein specifically claimed was that ds (the separation of two events) is the same for a stationary observer,
and a moving observer. In the following two diagrams, the stationary observer is you, with your x-y-z coordinate
system, while the moving observer is your sister (redheart, with her x′-y′-z′ moving coordinate system): 𝑑𝜏 ≡ 𝑑𝑡 !
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Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
We are going to discover that redheart's velocity u (the red arrow) is predicted, from Einstein's Pythagorean claim, to
be greater than the velocity v (the black arrow) that you find for her : dx divided by your clock's change, dt.
In your stationary coordinate system redheart moves a distance dx, in time dt (by your two synchronized clocks).
The two events that we are considering are redheart’s snapping her fingers at one moment in her motion (our first
panel), and then snapping her fingers again at some later moment in her motion (our second panel):
those two events are what is illustrated in the two panels.
Einstein's claim is that
dx2 + dy2 + dz2 − dt2 = ds2 = dx′ 2 + dy′ 2 + dz′ 2 − dt′ 2
i.e, ds is invariant.
In our example, with the motion being entirely in the x direction, dy′= dy = 0 and dz′= dz = 0, and so we can write
Einstein’s claim as that:
dx2 - (c dt )2 = ds2 = dx′ 2 - (c dt′ )2
The c’s that I inserted (multiplying the dt’s) are to convert the units of time from seconds to meters—required so
that the units will be the same for everything in the equation: but the c’s have no physics significance at all. The
required value of c was long ago determined from experiment to be about 299,792,458 meters per second—very
close to the the speed of light—and this number is now defined to be the exact value (just as there are exactly
3 feet in a yard). Light moves at that speed because light is massless—all massless particles, including for example
gluons and gravitons, move at that identical speed, which is set by the remarkable geometry of spacetime. [If we
don’t use different units for time (compared with space) then c = 1 and can be omitted from all equations entirely.]
Snap test: what is the value of dx′ ?
Right! dx′ = zero! For dx′ is the spatial separation of those two finger snaps in redheart’s own coordinate
system—which moves right along with her!
Since dx′ is zero only in redheart's coordinate system and in no other, redheart's coordinate system is special
indeed—so, instead of writing the time separation (in her special coordinate system) as dt′ we choose to write (from
now on) instead of dt′, dτ (the Greek letter tau). So, instead of dx2 - c2 dt2 = ds2 = dx′ 2 - c2 dt′ 2 we now have
c2 dτ2 = c2 dt2 - dx2
and this equation will now expose us to the shocking true nature of photons of light. For consider the case that
redheart is a photon of light created on the sun, and absorbed (in your retina) on Earth. Distance equals velocity
times time: dx = v dt, and so for light, dx = (c dt) and so c2 dτ2 = c2 dt2 - c2 dt2 = 0 and Einstein predicts that dτ
which is the time, by the photon's own watch, of the photon's trip—from its creation on the sun to its ceasing to
exist in your retina—is: zero! By its own clock, the photon never exists! So what did Albert Einstein think of this?
Einstein (1954): "All these fifty years of conscious brooding have brought me no nearer to the answer to the
question, 'What are light quanta?' Nowadays every Tom, Dick, and Harry thinks he knows it, but he is mistaken."
Ready for one more major consequence of our 'trivial' postulate? Suppose our "photon" from the sun were some
kind of "magical" photon that could go at speed C—just a bit faster than 299,792,458 meters per second. Well, look
at our latest equation, which, for this case, would state that c2 dτ2 = c2 dt2 - C2 dt2 . On the left hand side we have
c2 dτ2 and since any number squared must be positive, the left hand side of our equation is positive, yet the right
hand side would be negative for any such magical photons! We not only conclude that there could be no magical
photons, we also conclude that no velocities greater than that of light can ever occur: if anyone says that they can,
that would be closely analogous to someone—absurdly—announcing that they can go north of the north pole. So!
Speed limit is one—that’s why we introduce 299792458: otherwise, your car would go at 0.000000089 (= 60 mph).
Now remember that we are seeking the famous equation E = mc2. To get that, we must launch a bunch of algebra.
But it is nothing but algebra: no additional physics, of any kind. (But how does Mother Nature know algebra,
which we humans (e.g. Omar Khayyam) developed? No one has any idea: but, know it Mother Nature does!)
Redheart's rocket’s speed is 𝑣 = 𝑑𝑥/𝑑𝑡 as measured by us on Earth [and 𝑢 = 𝑑𝑥/𝑑𝜏 as measured by redheart
herself—the very same distance dx of course, but (as we shall bring out) a quite different time interval dτ]. So,
we have:
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Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
𝑐!
𝑑𝑡
𝑑𝑡 𝑑𝑥 𝑢
=
∙
=
=
𝑑𝜏
𝑑𝑥 𝑑𝜏 𝑣
!" !
1
1−
= 𝑐 ! − 𝑣 ! , so
!"
𝑣!
𝑐!
𝑣!
≡𝛾 = 1− !
𝑐
!
!
!
=1+
1 𝑣! 3 𝑣!
+
+⋯
2 𝑐! 8 𝑐!
That last step is a binomial expansion (invented by Isaac Newton). Since frequently the velocity v is very much
smaller than c, that last term (plus all the higher-order terms: + ⋯) can often be ignored. (Be sure to memorize the
above definition of γ! Also, note that when v = 0, γ is one—but that as v approaches c, γ approaches infinity!)
Now, return to our equation 𝑚 ! 𝑐 ! 𝑑𝜏 ! = 𝑚 ! 𝑐 ! 𝑑𝑡 ! − 𝑚 ! 𝑑𝑥 ! where I have now multiplied both sides of the
equation by m2 (we seek E = m c2, so m, the mass of redheart (or of redheart plus her rocket) must be introduced
somehow!) From that equation,
𝑚! 𝑐 ! = 𝑚! 𝑐 !
!" !
!"
!" !
− 𝑚!
or 𝑚 ! 𝑐 ! = 𝑚𝑐𝛾
!"
!
− 𝑚! 𝑢!
Notice that u = γ v [and also that dt = γ dτ] — if you insert u = γ v into the last equation, it dissolves. Try it!
𝑚 ! 𝑐 ! = 𝑚𝑐 1 +
𝑚! 𝑐 ! =
!
!
𝟏
!
!!
!
!!
𝑚𝑐 ! + 𝒎𝒗𝟐 + 𝑚
𝟐
+⋯
!
1 𝑣! 3 𝑣!
+
+⋯
2 𝑐! 8 𝑐!
− 𝛾 𝒎𝒗
!
!
− 𝑚! 𝑢!
or, if we now abandon Newton and, with Albert Einstein,
!
!
!!
we redefine momentum to be: p = mu = γmv and total energy to be: 𝑬 = 𝑚𝑐 ! + 𝑚𝑣 ! + 𝑚 ! + ⋯ ≡ 𝑚𝑐 ! + 𝐾
!
!
!
(with K being the kinetic energy) then
𝐸 ! = 𝑐𝑝
!
+ 𝑚𝑐 !
!
or
𝐸 ! = 𝑝! + 𝑚!
You verified all of that? OK! If redheart is not moving at all, her momentum p is zero, and we have, at long last:
𝐸 = 𝑚𝑐 !
( or, E = m )
Simply our deriving E = mc2—the most famous formula in all of physics—from a tiny tweak of the Pythagorean
theorem—is not enough to have us also understand how the atomic bomb results. So let’s go through that historic
application of this most famous equation, so as to fully understand what E = mc2 really means in practice.
If a neutron hits a nucleus of 235U (mass mU) that nucleus decays into a nucleus of 92Kr , plus a nucleus of 141Ba , plus
3 neutrons—and those 3 neutrons can then go on to hit 3 more 235U’s, producing a chain reaction—and thus a bomb.
The Uranium, if stationary, had total energy E = mU c2. The resulting Kr and Ba nuclei are observed to each have, in
addition to their own internal rest energies mKr c2 and mBa c2 huge kinetic energies K. The total energy after the
decay, ignoring the low-mass neutrons, is, according to our above-expounded special relativity,
EKr + EBa = (mKr c2 + KKr) + ( mBa c2 + KBa)
Conservation of energy tells us that energy E = mU c2 = EKr + EBa = (mKr + mBa) c2 + (KKr + KBa)
so the total kinetic energy is K = KKr + KBa = c2 × [mU - (mKr + mBa)]
It is an experimentally observed fact that the sum of the masses of the Krypton and the Barium nuclei, plus those
neutrons that are also produced, is significantly LESS than the mass of the original Uranium nucleus: and when this
difference is put into Einstein’s equation and so is multiplied by the enormous number c2 well:—that atomic bomb!
And all that, from just these three steps:
a) the pseudo-Pythagorean Theorem of Albert Einstein: : ds2 = dx2 + dy2 + dz2 − dt2
b) a small bunch of algebra—in other words, nothing new: no additional facts, or ideas—just algebra!
c) one observed fact regarding the masses of the U, Ba, and Kr nuclei
The most amazing item on the list: our human development of algebra: so, thank you Omar Khayyam (et al!)
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Einstein’s Most Famous Equation, E = mc2
Epilogue
An extract from Einstein's 2nd 1905 paper on relativity—the journal paper in which he obtained the famous E = mc2 :
“It is not impossible that with bodies whose energy-content is variable to a high degree (e.g. with radium salts) the
theory may be successfully put to the test." My friend, that entire paper by Einstein was only 2 pages long—and yet
its application—in 1945—vaporized Hiroshima and Nagasaki. Over a brief 40 years, one tiny idea led humankind
to one horrendous weapon.
Lewis Carroll Epstein: “Algebra is a wonderful invention. It enables fools to do physics without understanding.”
Let me name some:
Wolfgang Rindler, eminent General Relativist, in his book “Essential Relativity,” talks about mass increasing with
velocity, which it does not (as the reader has, above, clearly seen—mass m is a constant) and then incredibly says,
“We should not be too surprised at this, since there must be some process in nature to prevent particles from being
accelerated beyond the speed of light.” Rindler simply does not understand what relativity means.
I cannot be totally surprised at this lack of understanding of what Rindler otherwise does so superbly well, because I
have read Walter Isaacson's biography of Einstein in which he notes that Einstein's bosom pal, the elderly Hendrik
Lorentz, discoverer of the Lorentz transformation 𝑑𝑥 ! = 𝛾 𝑑𝑥 − 𝑣 𝑑𝑡 ; 𝑐 𝑑𝑡′ = 𝛾( 𝑐 𝑑𝑡 − 𝑣/𝑐 𝑑𝑥) well before
Einstein's discovery of Relativity, never understood Special Relativity. (Please Lorentz-transform (dx′)2 - (c dt′)2.)
Richard Feynman, in his book “Surely You're Joking, Mr. Feynman,” recounts his shocked discovery that an
eminent colleague had defective understanding of elementary physics.
And physics Nobelist Steven Weinberg revealed in a Physical Review Letter, on 30 January 1989, his fundamental
lack of understanding of the use of Hilbert space in Quantum Mechanics—as was pointed out to all the world (in
PRL exactly one year later, 28 January 1990) by his colleague—just down the hall from him—Joseph Polchinski.
Folks: we are all human, and we should all pull together to get all of us humans grasping what our situation really
is—and that is the purpose of the present paper. We are all in this together, so let's get our act together!
General Relativity is a simple adaptation of (the above) Special Relativity to curved spacetime (conserving energy)
and GR explains gravitation— three cheers for LIGO's recently reaffirming that!
Quantum mechanics, widely regarded as deeply mysterious, is in fact trivial (you see that I do like that word): given
the fact that observations all ultimately have the character of numbers, and given the fact that in all our experience
there is invariance of physics under translation through space—under translation through time—under rotation in
space—then quantum mechanics follows as a necessity: and so quantum mechanics is not even slightly mysterious:
American Journal of Physics 58, 1087, 1990 and
J. Physics A: Mathematical & Theoretical, 41, 175303
So! When you add all of this up, what does it mean? It means (brace yourself) that what we perceive as a universe is
a scam—it is all in our minds. We proved that photons do not exist—and neither does anything else: Nature, 436, 29
Yes, my friend—life is but a dream!
PART TWO—PHYSICS: THE ORIGINS
!
I'm afraid I baited you—with the famous E = mc2 (but then I showed you that actually, 𝐸 = 𝑚𝑐 ! + 𝑚𝑣 ! + ⋯) and
!
I did it again with Newton's mv and 1/2 mv2 (so familiar to you—but: why are those two items conserved? And why
that 1/2? ) So let’s now go back to pre-Einstein days, to identify these two quantities’ geometrical significance.
Emmy Noether taught us that conservation laws result from the (unexplained) existence of symmetries in our world:
Suppose that we are somewhere in intergalactic space, and we let go of a brick. Nothing happens. But—if we do
the identical thing here on Earth, the brick falls, and kinetic energy T appears, seemingly out of nowhere. (We will
pretend we have no idea how T might depend on either the mass of the brick, m, or on its velocity, v : that is
something that we want to deduce from symmetry.)
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Einstein’s Most Famous Equation, E = mc2
With our intergalactic experience under our belts, we recognize that energy must be hiding, somehow, in the Earth’s
gravitational field: so we label that mysterious energy “potential energy,” and we symbolize it V. We then
confidently write E = T + V, and we insist, with Emmy Noether (because of symmetry under translation through
time) that E be conserved.
Our total energy E depends on the kinetic energy T of the brick (which clearly depends, somehow—but we don't yet
know exactly how—on the brick’s mass m and on its velocity v) and on the potential energy V of the brick, which
depends, somehow, only on the height of the brick in the Earth’s gravitational field (for if we were to go high
enough, there would be no potential energy: we’d have returned to outer space where the potential energy is zero).
The total energy being conserved means that there must be no change in the total energy E from the moment we let
go of the brick to when the brick is below us by distance d𝑦, where it is then moving downward with velocity v.
That is, the sum of the changes in energy E due to: a) change (with time t) of position y, and b) change (with time t)
of velocity v, must add up to zero:
𝜕𝐸 𝑑𝑦 𝜕𝐸 𝑑𝑣
+
=0
𝜕𝑦 𝑑𝑡 𝜕𝑣 𝑑𝑡
𝐸 = 𝐸 𝑦, 𝑣
An equation like this one applies to each and every one of the m atoms of which the brick is composed, so there will
be a total of m such equations, each identical to all the others. If we add all these equations up, of course we get
𝜕𝐸 𝑑𝑦 𝜕𝐸 𝑑𝑣
𝑚
+
𝑚
=0
𝜕𝑦 𝑑𝑡 𝜕𝑣 𝑑𝑡
𝑜𝑟
𝜕𝐸 𝑑𝑦
𝜕𝐸 𝑑(𝑚𝑣)
+
=0
𝜕𝑦 𝑑𝑡 𝜕(𝑚𝑣) 𝑑𝑡
We recognize mv as the momentum of the brick. You might ask—Emmy says that momentum, too, is conserved?
But there was none—the brick drops, and now there is some? Where did it come from? Answer: the Earth itself
moved upward in response to the fall of the brick, conserving momentum, which is a vector. Earth’s upward
velocity is miniscule, because Earth’s mass is so enormous: it is the vector sum of the two mv’s that is conserved.
Energy, in contrast, is a scalar, and is, as we shall discover, always of the same sign.
Sometimes (and in fact often) it is more convenient to use a system of coordinates different from the Cartesian x, y, z
with which we are most familiar—for example, spherical polar coordinates. To allow such generalized coordinates,
even though we will not be using any particular set in our discussion, we now switch to the symbol q for position y,
and to the symbol p for momentum mv. In that notation:
𝜕𝐸 𝑑𝑞 𝜕𝐸 𝑑𝑝
+
=0
𝜕𝑞 𝑑𝑡 𝜕𝑝 𝑑𝑡
We use the partial-derivative symbol 𝜕 with E (instead of d ) to remind us that E is a function of more than one
variable (position q, and what we have recognized as momentum p = mv). And now let’s switch to Newton’s “dot”
to indicate differentiation with respect to time:
𝜕𝐸
𝜕𝐸
𝑞+
𝑝=0
𝜕𝑞
𝜕𝑝
𝑜𝑟
𝜕𝐸
𝜕𝐸
𝑝=−
𝑞
𝜕𝑝
𝜕𝑞
Of course the minus sign could be placed on either side of the equals sign: what we show is the conventional
choice—which will result in mass being positive.
But changes in position with time (that is 𝑞) and changes in momentum with time (that is 𝑝) are quite independent
of one another! Just try it: if you change the position of a marble, that does NOT, by itself, change its momentum!
If you change the momentum of a marble, that does NOT by itself change its position (at that same moment, of
course).
But, to satisfy Emmy Noether (or, rather, to comply with the great underlying truth, to which Emmy drew our
attention) our last equation must hold true for any pair of values 𝑞, 𝑝. There’s only one way to accomplish that: it
must be true that
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Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
!"
!"
= 𝑞 and −
!"
!"
=𝑝
← Hamilton's
Equations!
for in that case our equation reads 𝑞𝑝 = 𝑝𝑞 — which is necessarily true!
We have derived Hamilton’s Equations from nothing but conservation of energy: and we are on our way to F = ma!
We assumed the conservation of energy in producing Hamilton’s equations. Let’s now see how symmetry can
require conservation of momentum mv, and can also produce Newton's T = 1/2 m v2 . Suppose that we are again
out in intergalactic space. We find that translational symmetry prevails—the energy E of our brick does not change
(indeed nothing changes) if our brick is displaced, say, several meters to the right.
That is, we firstly find in those circumstances that the second of Hamilton's two equations reads
−
𝜕𝐸
=𝑝=0
𝜕𝑞
𝑜𝑟
𝑑 𝑚𝑣
=0
𝑑𝑡
or mv = constant. We have discovered, with Noether, that, in these circumstances, momentum is conserved.
Now let's once again consider the second of Hamilton's equations, −
!"
!"
= 𝑝 𝑜𝑟 −
! !!!
!!
=𝑝
Kinetic energy does not depend on position q, it depends only on velocity. (If you move a marble to another
position, that does not change its kinetic energy—whereas, in contrast, if you change the marble’s velocity, you do
change its kinetic energy.) So,
!"
!!
= 0 and we have
−
𝜕𝑉
= 𝑝 = 𝑚𝑣 = 𝑚𝑎 ≡ 𝐹
𝜕𝑞
We have derived Newton’s assumed F = ma, from Emmy Noether’s symmetry! Newton’s famous equation is a
mere consequence of symmetry—it is not a fundamental law at all! In fact (please note my use of ≡ instead of =)
we have here invented the actually inessential concept of “force” F.
Notice, by the way, that Newton’s “force” is a vector (the negative gradient of scalar V ), and therefore is much
more difficult for beginning students to use than are Hamilton’s scalar quantities E, T, and V.
Finally, let's now see what Hamilton’s first equation gives to us!
!"
!!
= 𝑞 or
!(!!!)
!!
=𝑞
Potential energy V , by experiment, does not depend on velocity v, or on momentum p = mv , it depends only on
!"
position: how high above the Earth are you located? So,
= 0 and we have
!!
𝜕𝑇
=𝑞
𝜕𝑝
so
𝜕𝑇
𝑑𝑦
=
=𝑣
𝜕(𝑚𝑣) 𝑑𝑡
so
𝜕𝑇
= 𝑚𝑣
𝜕𝑣
so:
𝑇=
𝑚𝑣 𝑑𝑣 = 𝑚
𝑣 𝑑𝑣 =
1
𝑚𝑣 !
2
We have deduced from symmetry the pre-Einstein dependence of kinetic energy on mass and velocity. (Hey, you
always wondered where that 1/2 in 1/2 mv2 came from, now didn't you!) We could, of course, also have deduced this
from symmetry under translation through time, just as we did deduce that p = mv from symmetry under translation
through space. (Notice that if we had chosen to put the minus sign on the other of Hamilton’s two equations, mass
would invariably be negative. That is why, throughout the universe—and unlike electric charge—mass is always of
the same sign.) Anyway, all this 'pre-Einstein' is replaced, today, with the symmetry of the Lorentz transformation.
I could continue, but, I am not trying to teach you physics—all I am trying to do is teach you philosophy— and what
I have done, in this tract, is, I hope, quite enough to answer your most basic philosophical questions, about:
our experience in this, our so-called "universe"— a playpen for our human minds and spirits.
AND NOW! Onward, to Einstein's General Relativity
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Einstein’s Most Famous Equation, E = mc2
Part One (of 3) : Einstein’s Theory of General Relativity
… is one of the greatest intellectual achievements of the human race.
Let’s see how easy it was for Albert Einstein to discover General Relativity.
Einstein’s truly brilliant idea was that the presence of mass-energy (e.g., our massive Sun) somehow produces
curvature of spacetime, which results in planets moving on curved paths around the Sun—instead of the planets
simply moving at constant speed in straight lines (as Newton said happens if no “force” exists). That is, Einstein
postulated that:
Curvature of Spacetime = Mass-Energy
Mass-Energy
≡ Tµν which is the energy-momentum tensor, with 4 × 4 = 16 components
Example: for a perfect gas (e.g., the “gas” of galaxies in the universe)
⎛
⎜
ν
Tµ = ⎜
⎜
⎜
⎜⎝
⎞
⎟
p 0
0 ⎟
0 p
0 ⎟
⎟
0 0 − ρc 2 ⎟⎠
p 0
0
0
0
0
where p is the pressure (for galaxies, ~ zero), and ρ is the density. (Isaac
Newton did not know Special Relativity, and thus did not know that p, tiny as
it is, has to be included. So, Newton’s law of gravity is only one equation
(dealing with the –ρc2 ) and not the 16 equations (well, in this simple
example, two independent equations) of General Relativity!)
0
Now, what do we place on the “curvature of spacetime” side of our equation? Fortunately, Einstein had available
the work of the brilliant mathematicians Riemann and Ricci.
α
Riemann characterized geometrical curvature as R β γ δ which has 4 × 4 × 4 × 4 = 256 components! We can’t
ν
set it equal to Tµ because the latter has only 4 × 4 = 16 components. But! Ricci to the rescue: the purely
α
mathematical operation of contraction gives us: R β α δ = Rβ δ where the right-hand side is named the Ricci tensor.
ν
It has only TWO indices, so we are in business! Writing it in mixed covariant-contravariant form (like Tµ ) gives
us
⎛
⎜
G⎜
ν
Einstein’s First Guess: Rµ = 4
c ⎜
⎜
⎜⎝
⎞
⎟
p 0
0 ⎟
0 p
0 ⎟
⎟
0 0 − ρc 2 ⎟⎠
p 0
0
0
0
0
0
(G is Newton’s G, the
c 4 fixes units)
Looks good, but there is a problem: mass-energy is conserved, but the Ricci tensor is not.
“Conserved” means that its derivative is zero. So, Einstein needed to patch the left-hand side so that its derivative
would also be zero. Hey, it looks like a kluge, but here it is:
⎛ ν 1
⎛ G ν ⎞ where R = Rνν is the contraction of Ricci’s tensor, and gµν is the
ν⎞
ν
R
−
R
g
+
Λg
=
8
π
⎜⎝ µ
⎜⎝ 4 Tµ ⎟⎠
µ ⎟
µ
⎠
2
c
metric tensor (just 1’s on the diagonal— trivial). Since we are
requiring derivatives to be zero, naturally we can toss in an arbitrary constant, Λ (the famous Cosmological
Constant): and we have done that!
If we also stick in the 8π —we’ve done that, too—our equation easily reduces, in the first approximation (and
ignoring Λ) to Newton’s Law of gravity! We have derived the…
Field Equations of General Relativity!
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Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
Part Two (of 3) : The Cosmological Constant: Dark Energy
Rµν −
1
8π G
ν
R gµν + Λgµν = 4 Tµν , and the cosmological constant Λgµ must be structured like
2
c
⎛
⎜
ν
Tµ = ⎜
⎜
⎜
⎜⎝
⎞
⎟
p 0
0 ⎟
0 p
0 ⎟
⎟
0 0 − ρc 2 ⎟⎠
p 0
0
0
0
0
0
except that, since we want to represent a cosmological constant, all
four entries on the diagonal must be identical, and must be constant.
What does that take?
Well, we will do what Einstein did, and consider the possibility that the vacuum is not vacuous, as we’d always
thought, but rather has some constant mass density ρV (subscript V means vacuum). A simple thought
experiment then gives us the pressure pV for the vacuum:
nothing
vacuum
pV
ρV
(not even
vacuum)
Suppose the above were a conventional piston-and-cylinder, with gas in it, not vacuum, and we added heat ΔQ .
There would be an increase in the temperature (that is, in the energy E) of the gas, and the piston would move.
Conservation of energy, of course, applies:
ΔQ = ΔE + pΔV , where V is the volume. This is also called the first law of thermodynamics, but it is just
conservation of energy. pΔV is the work done as the piston moves (high school physics). Now apply the equation
to our thought experiment above, for the case ΔQ = 0 , that is, no heat added. Then we have
0 = ρV c 2 ΔV + pV ΔV
Any outward motion of the piston (“expansion of the universe”) spreads further apart any galaxies that are in the
cylinder (thus reducing their mutual gravitational attraction) but (by Einstein’s remarkable hypothesis) does not
diminish the vacuum’s mass density ρV at any location. ρV ΔV is the change in the mass contained in the cylinder
(remember that density is mass divided by volume V). OK! Cancel the ΔV 's and we get our answer:
pV = − ρV c 2 (which is negative; and which is enormous, because of the c2 ) and
⎛
⎜
Λgµν = − ρV c 2 ⎜
⎜
⎜⎝
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
⎞
⎟
⎟ = − ρV c 2 gµν
⎟
⎟⎠
Dark Energy!
We’ll see that it repels, instead of attracting!
8
Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
Part Three (of 3) : Finally, let us Compare with What Newton Thought
We want to compare our result with what Newton would have asserted, using his
Gravity,
F=−
) and his Law of
F = ma ( ≡ m R
GM m
(the minus sign indicating an attractive force). Our galaxy, having mass m,
R2
is attracted by everything in an arbitrary huge sphere, of radius R, containing total mass M.
ρM =
The density for that vast sphere is, of course:
Combining our 3 “Newton equations” gives us:
M
.
4π 3
R
3
R m
M
 = − 4π G ρ M R
acceleration = R
3
It is negative! The expansion of the Universe is decelerating! But, what does Einstein say?
Rµν −
1
8π G
R gµν + Λgµν = 4 Tµν
2
c
The above is a compact way of writing 16 separate equations (μ and ν each go from 1 to 4, since there are
4 dimensions: 3 space, and 1 time). You saw that for a gas of galaxies, only the 4 diagonal elements of the energymomentum tensor were non-zero (and also that 3 elements were the same, p). All 4 diagonal elements are 1’s for
the metric tensor. As for the Ricci tensor, if you assume the simplest possible isotropic geometry (called RobertsonWalker) for the expanding universe (such simplicity is strongly supported by observations of the famous 3K
background radiation), the first three elements are again the same. We’ll use RW for a flat universe, since
observations indicate that our universe is flat, to a high degree. So, instead of the fancy equation above, with its
raised and lowered indices (the reason for “raised/lowered” is too boring to discuss, and we don’t need it anyway),
our fancy equation reduces to just two equations, both at high-school-level:
−
 8π G
R 2 2 R
+
= 2 p
2
R
R
c
and
R 2 8π G
=
ρ
R2
3
The first equation, the one with p on the right, occurs 3 times—but of course, we don’t need 3 copies! We have
omitted Λ—we will see it appear as dark energy, from Einstein’s ρV idea (and his deduction that
that
ρ
pV = − ρV c 2 ), so
and p each have two contributors:
1) matter M (including dark matter), and 2) dark energy (subscript V). So
ρ = ρ M + ρV and p = pM + pV ≈ pV
(since the pressure of the galaxies, as well as of the only-weakly-interacting dark matter, is ~ zero, in drastic
contrast to the enormous pressure of the dark energy). In our equations, R is an arbitrary distance proportional to the
size of the universe at any time, R with a dot on it (Isaac Newton’s notation) is the universe’s expansion speed at that
time, and R with two dots on it is the acceleration of the expansion of the universe—which is what interests us most.
R from our two equations (easy). Result:
 = − 4π G ( ρ M c 2 + 3pV + ρV c 2 ) R = − 4π G ( ρ M c 2 − 2 ρV c 2 ) R ≈ + 8π G ρV R = R

acceleration = R
3c 2
3c 2
3
Now eliminate
If we ignore the dark energy, the first part gives our negative Newtonian result! But, if the universe has expanded
enough (diminishing ρM to the point that it is negligible, as in our present universe) we get our final expression
above for the acceleration of the universe:
—which is positive! So, instead of living in a universe that is decelerating (negative 𝑅), we find ourselves in a
universe that is accelerating its expansion!
And so:
Dark Energy now rules the Universe!
9
Richard Conn Henry 2017 April 7
Einstein’s Most Famous Equation, E = mc2
Rµν −
1
8π G
R gµν + Λgµν = 4 Tµν
2
c
… but, in the vacuum, and with cosmological constant zero, these
Field Equations of General Relativity reduce to
Rµν = 0
The photo shows Albert Einstein writing this equation (in covariant form; that is,
with two subscripts). Einstein followed his equation with a question mark!
Anyone reading my four-page account of GR might wonder why, General Relativity being so
simple and so apparently arbitrary (especially regarding dark energy), it is held in such high
esteem? It is because this theory makes quantitative predictions, for example of the evolution
of the orbits of cosmic binary neutron stars, that have always turned out to be highly accurate!
The Global Positioning System (GPS) could not function without our use of General Relativity;
Special Relativity (with time as the 4th dimension, but: with a MINUS sign in the Pythagorean
Theorem) led directly to atomic power and — to the hydrogen bomb.
10
Richard Conn Henry 2017 April 7
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