Download We don`t generally encounter forces, even in our theoretical studies

Of Particular Significance
Conversations About Science with Theoretical Physicist Matt Strassler
The Strengths of the Known Forces
Matt Strassler [May 31, 2013]
In this article I want to discuss basic properties of the forces that we know about — four that we’ve actually
observed, and a fifth — a new one — whose existence we infer from the discovery of the Higgs particle.
Specifically, I want to discuss what particle physicists mean in describing forces as being weak
or strong. It’s terminology that you’ll often see, but unless it’s been explained to you, there’s no
way to guess what’s actually meant. So here’s an explanation… a long one, but I hope it will
give you many insights into how nature works, as well as raising more questions I’ll have to
answer later…
“Weak” versus “Strong”
What do these terms mean? While in ordinary life you and I would think of a strong force as one
that can pull us off our feet and a weak force is one that we can counter by stiffening our muscles
a little, that’s not what particle physicists mean at all.
By “strong” and “weak”, particle physicists are not talking about whether a force is absolutely
strong, or absolutely weak. It’s not about whether the force could break a window, or hold up a
bar of gold. “Strong” and “weak” in this context are semi-absolute terms, in a sense that we
never use in daily life, or even in undergraduate physics classes. It’s a way of talking that only
emerged from a deep understanding of quantum field theory, the modern mathematical language
used to describe the known elementary particles and forces. But it’s fundamental to the way
particle physicists think about these issues today. So I’m going to start by explaining the
rationale behind this way of talking.
Take two objects of some type, perhaps elementary particles, and place them a distance r
apart. Suppose each exerts a force F on the other. Then we will say this force is weak if

F is much less than (h c / 2π r² )
where h is Planck’s quantum mechanics constant and c is the speed of light. Often it is
convenient in physics to use not h but

ℏ = h / (2π)
In short, for particle physicists,


a weak force has F r² much less than ℏ c
a strong force has F r² about as big as ℏ c
[We don’t generally encounter forces, even in our theoretical studies, that are much stronger
than (ℏ c / r² ); typically such strength makes them so complicated that we end up thinking about
them in a different way. Long story.] This is a measure not of whether the force is weak in an
absolute sense, but whether it is weak or strong compared to typical forces that are to be
found at a distance r. The question isn’t about the force; it’s about the force times the distancesquared, and whether that is smaller than or comparable to ℏ c.
To explain why this notion of strength is useful, I’ll illustrate the concept in the case of
electromagnetic forces acting on simple charged particles, such as electrons, anti-electrons
[“positrons“] and protons. Electrons have electric charge -e; protons and positrons have charge
+e.
First, imagine two stationary protons, each with mass m and electric charge +e, a distance r apart.
The electric force between them pushes them apart and has a strength given by the following
formula

F = ke2 / r2
[What is k? see below…] The same formula would apply for two electrons, which both have
electric charge -e. For an electron and a positron [an anti-electron, with charge +e], the force
would have the same size, but would pull them together instead of pushing them apart.
Now — what is k? It’s called Coulomb’s constant, and what its value is depends on how we
define e, the basic unit of charge. But it won’t matter, because in discussing electric forces
involving elementary particles, we’ll always see ke2 appearing as a group, together. So we don’t
need to know how big k is; we only need to know, how big is ke2?
It turns out that if r is larger than a millionth of a millionth of a meter, then ke2 is approximately
0.007 times (h c/2π), where h is Planck’s quantum mechanics constant and c is the speed of light.
So we may write the electric force, times r² as (approximately)

F r² = 0.007 ℏ c
Since 0.007 is much less than 1, electromagnetism is a weak force, and remains so at all
distances, down as far as we have measured.
It’s very mportant not to get confused here! Just because electromagnetism is a weak force in
this sense doesn’t mean that the force between two protons is weak in an absolute sense. In fact,
the electric force trying to push the two protons in a helium nucleus apart is comparable to the
weight of a truck! All that force acting on two tiny tiny particles!!! But still, this is, for such a
small distance, a rather weak force, and indeed a somewhat stronger force (the “residual strong
nuclear force”) counteracts that electromagnetic repulsion, and holds the protons and neutrons of
the helium nucleus together.
By the way, there’s a historical name for this number 0.007; it’s called the “fine structure
constant” (because it sets the size of little differences in the energies of various configurations of
atoms) and it’s normally called “α”:

α = ke2/ ℏ c = 0.007 297 352 57
It’s one of the most precisely measured quantities in nature. Often people write it as
approximately 1/137 (and many decades ago various people thought maybe there was something
special about the number 137), but if you’re going to do that really it should be written as
1/137.0359990…
Ok — so why is the fact that α is much less than 1 an indication that this force should be thought
of as weak rather than as strong?
Why α«1 means the electromagnetic force is weak
This is best illustrated in the case where the force is attractive, as it is for the electron and
positron, or the electron and proton. The electron and positron are a little easier to start with,
because they have equal mass m; they form an atom-like state called positronium, analogous to a
hydrogen atom formed by an electron and proton, but more symmetric, with the two particles
orbiting each other, rather than in hydrogen, where the electron orbits the nearly stationary
proton. In fact, the formulas for hydrogen, for those who know them, apply to positronium too,
with some minor changes (factors of 2) in a few places. (Yes, the electron and positron in
positronium eventually annihilate and turn into two or three photons, but only after the particles
have orbited each other many billions of times — which admittedly only takes a tiny fraction of a
second.) For positronium, in its lowest-energy state,





the typical speed of either particle is α/2 × c
the typical motion-energy (i.e. “kinetic” energy) of either particle is mc² × α²/8 .
the interaction-energy (or “potential” energy) of the two particles is –mc² × α²/2.
the binding-energy B of the positronium state (the sum of the motion-energy and the
interaction-energy) is mc² × α²/4.
the mass-energy of the positronium state 2 mc² – B; and since the latter is much smaller
than the former, the mass of the atom is just a tiny bit smaller than the sum of the mass of
the electron and of the positron.
In short, because α is much less than 1, there are three essential and related facts
1. The electron and positron move at speeds slow compared to the speed of light c.
2. Kinetic, potential and binding energy B are all small compared to the electron’s and
positron’s mass-energy, E = mc2.
3. The mass of positronium is very close to the sum of the masses of the electron and
positron.
All of these statements would be true no matter how large or small were the electron’s mass;
they depend only on α being small.
Together these facts mean that in describing this atom-like state, Einstein’s theory of special
relativity is not important; Newton’s laws of motion are good enough to make predictions, up to
details that are no larger than α, i.e. to about the 1% level or better. And as we’ll see in the next
section, this means the system is relatively simple. It can be described using quantum
mechanics, which has rather simple mathematics, without need for quantum field theory, which
would be necessary if Einstein’s relativity were important. The math for the hydrogen atom is
the same as for positronium, and it is so simple that physicists learn about it as undergraduates,
early in their first quantum mechanics class.
There’s another useful, though a bit less familiar, way to think about this. We should remember
that electrons, like all elementary “particles”, are really “quanta”, tiny ripples in quantum fields.
They are more like waves than they are like little balls. Consequently they vibrate, as all waves
do: they have a “frequency” of vibration. And the time it takes between one vibration and the
next — which I like, poetically, to call a particle’s “heartbeat” — is equal to hc/m. If α is small,
then the time it takes light to cross the atom-like state is much larger, by 1/α, than the heartbeat
of the particles that it contains. In this sense, positronium is relatively big. And since the
particles themselves travel much slower than light, it takes even longer for the particles
themselves to cross the atom-like state, something like 1/α² heartbeats.
Other things that could have been complicated are rather simple too, when α is small. For
instance, the force exerted by the positron on the electron can cause the electron to become,
sometimes and briefly, a virtual electron and a virtual photon. (Virtual “particles” aren’t
particles; a real particle is a well-behaved ripple in quantum fields, but a virtual particle is a more
generalized disturbance of those fields.) But that’s rare, if α is small. Even rarer, the virtual
photon will itself be disturbed into becoming a virtual electron and a positron. Since there isn’t
nearly enough energy around to make another real electron and positron, which would require
energy of 2mc2 to come from somewhere (recall the motion and interaction energy are much less
than this), it is very rare to make a virtual electron and positron. The fact that virtual particles are
rare is why we can say so simply that “a positronium atom consists of an electron and a positron”
— that’s indeed what it is, most of the time. Only in high-precision calculations do we need to be
more careful, and remember that’s not always quite true. The same is true of a hydrogen atom: it
is (almost all of the time) just one electron and one proton, held together by a simple electric
force.
You can read about atoms here, where my descriptions are relatively simple. In this article I
partially explained how the hydrogen atom’s size can be inferred from quantum mechanics
principles, and you can use that result to see why the speed comes out to be αc, and why the
motion-energy and interaction-energy come out to be ½ mc² α².
What would happen if α were closer to 1?
But now imagine making α gradually larger and larger, increasing toward 1. What happens to
positronium? [I must warn you what I’m about to describe isn’t a rigorous discussion! Those of
you who are planning to be experts someday need to be more careful than I’m about to be.]
Fig. 1: The simplicity of positronium (which closely resembles the simplicity of a hydrogen
atom) would become considerable complexity were α close to 1. Electrons are drawn in blue,
positrons in red, and photons in orange. Were electric forces much stronger, positronium would
be much smaller (not drawn to scale; it would shrink by a factor of 100 in radius), and its mass
quite different from 2m. The particles inside it would be faster, more energetic, more numerous,
changing in number, and less easily identifiable as “real particles” (as opposed to the more
general disturbances in fields called “virtual `particles’ “.)
As α increases, the force (at any given distance) between the electron and positron becomes
stronger, and since the force pulls harder, it pulls the particles in the atom-like state tighter
together. The particles move faster, approaching the speed of light. The motion-energy of each
particle becomes larger; and the magnitude of the interaction-energy becomes larger, so the
binding-energy becomes larger — becoming comparable to 2m itself. Conequently, the mass of
the atom-like state is no longer approximately equal to 2 m. The size of the atom-like state
becomes smaller, so that the time it takes light to cross the state, the time it takes the particles to
move from one side of the state to the other, and the time between heartbeats of the particles all
become about the same.
The stronger force between the electron and positron makes it more common for virtual photons
to be present; and the larger amount of energy flying around the atom makes it easier for the
virtual photon to itself become a virtual electron and positron. And in fact, when that happens, it
can become difficult to say which electron is real and which is virtual, because there are also
powerful forces between the two electrons, and between the electron and either positron, and
these can cause a particle that was real to become one that is virtual, and make the virtual one
real, and back again. Meanwhile, the virtual electrons and positrons can also release or absorb
photons, which may be virtual but may sometimes be real.
In fact the very distinction between real and virtual particles starts to become difficult to
decide. Real particles are supposed to be well-behaved ripples in a quantum field. But with the
atom-like-state so small, it really only takes a single heartbeat, more or less, for an electron or
positron to cross this atom, at which point powerful forces will already cause it to change
direction. How can we say that such a particle is a “well-behaved ripple”? A well-behaved
ripple should, well, ripple for quite a while — for many heartbeats — before being affected by
external influences. Here, our electron, while more like a real particle than the generic virtual
particle, is still highly distorted, and doesn’t fit the definition of “real particle” anymore either.
And the electron may not even be around for long. The production of a virtual electron-positron
pair can be followed by the annihilation of the formerly-real electron with the newly formed
positron, leaving the maybe-virtual maybe-real electron behind.
So instead of what we had at small α — a simple system of mass just below 2m, consisting of an
electron and a positron moving at speeds well below the speed of light — we find, as α
approaches 1, an exceedingly complex system, with multiple particles moving at or near lightspeed, with a mass that is very different from 2m. (See Figure 1) It’s impossible to say how
many particles are inside; do you count the real ones only? and if so, how do you precisely
distinguish the almost real ones from the mostly virtual ones? The number of real particles can
be constantly changing.
Those of you who’ve read about the proton may note some similarity. There are important
differences too, but yes, the similarities are not accidental.
Now this is what is characteristic of a truly strong force; the objects that it forms are vastly more
complex than atoms. Scientists were lucky, in a sense, that the first objects that were encountered
on the road to the theory of quantum field theory were atoms. These are held together by a weak
force — the electromagnetic force — and were easy to understand using the simpler mathematics
of quantum mechanics, in which the number of particles is held fixed. Protons, by contrast, are
held together by a strong force — the “strong nuclear force.” So it’s not that surprising that
protons are much, much more complex internally than atoms. (And I haven’t even attempted
here to tell you about some of the additional complications that arise there; those are covered in
this article.) Inside a proton, the number of particles is continuously changing — which
requires the much more complex mathematics of quantum field theory.
By the way, the electric force between two electrons is weak because α is small. The same is true
for forces between any two elementary particles, because the all of the known particles have
electric charges that lie between -e and e — for instance, top quarks have charge (2/3) e. You
might, however, wonder about the force between an electron and a uranium nucleus, since the
uranium nucleus has charge 92 e. Well, in that case the force does get pretty strong! But this only
has part of the effect that I’ve described for strong forces, because changing the charge on only
one of the objects involved (and in particular the heavy one) doesn’t increase the probability for
finding virtual electron-positron pairs. That would only change if the electron itself got a much
larger charge than e! So even a uranium atom remains rather simple compared to a proton…
How Weak is the Weak Nuclear Force? It’s Tricky…
How strong are the other known forces of nature? We’ve seen that electric forces have a strength
α, at least at macroscopic and even atomic and subatomic distances. And over those distances,
down to a millionth of a millionth of a meter, α is a constant; it doesn’t depend on r, which is
part of why it’s such a convenient quantity. But in fact, the strength of a force can change with
distance, which complicates matters. For electromagnetism, this isn’t so important; the effect is
very small. However, for other forces, it is a big deal.
The so-called weak nuclear force is, of course, weak. Well, it is weak at macroscopic, atomic,
and even nuclear distances. But in fact its strength isn’t constant. For distances large compared
with ℏ c/MW ~ 3 × 10-18 meters, i.e. (about 1/300 the radius of an proton), where MW = 80
GeV/c² is the mass of the W particle, its strength αweak is (roughly)

Fweak r² /ℏ c = αweak(r) = 0.02 e-Mwr/ℏc
The exponential e-Mwr/ℏc makes the force remarkably weak! Even at distances the size of a
proton, this factor gives a suppression of e-300, which means the force has already decreased by a
number so spectacular I can’t write it out here: it is a 1 with 130 zeroes after it. (That’s bigger
than a `googol’, a 1 with 100 zeros after it.) And the force becomes rapidly weaker from there.
Why? The same effect that gives the W particle (a ripple in the W field) a mass makes it
impossible for a particle to distort the W field over long distances, in contrast to the effect of an
electron or proton on the electric field. Consequently, the force generated by the W field is
completely ineffectual at long distances.
But for even shorter distances,

αweak(r) = 0.02
Notice this is several times stronger than the electromagnetic force! The weak force is not
intrinsically weak at all. See Figure 2. (Minor Caution: I’m leaving out a subtlety involving the
interplay of the weak and electromagnetic forces at such short distances, and a very slow change
in the force that becomes noticeable at vastly shorter distances.)
What makes the weak force so weak, when we observe it in the physics of nuclei, atoms, and
daily life is the large mass of the W particle. If the W particle were massless, effects of the
“weak” nuclear force would be stronger than those of the electric force! [This is another context
in which the Higgs field, which gives the W its mass, is really important to our lives!]
The Strong Nuclear Force
The strong nuclear force, which pulls and pushes quarks and gluons (but not electrons), does
something quite different. At the distances we were just discussing for the weak force — 3 × 1018
meters — the strong nuclear force is quite a bit stronger than both the weak nuclear force and
the electromagnetic force:

Fstrong r² /ℏ c= αstrong = 0.11 (at r ~ 3 × 10-18 meters)
That’s really not so strong; it’s about a tenth as strong as a really strong force, and only about ten
times stronger than electromagnetism. In fact, although they differ enormously at macroscopic
distances, the strong nuclear, weak nuclear, and electromagnetic forces differ in strength by
only about a factor of 10 at distances shorter than about 3 × 10-18 meters. This is remarkable,
and perhaps not accidental. It’s a small step from here to the notion of the “Grand Unification” of
these three forces — the idea that at much shorter distances, all three forces end up with the same
strength, and become part of a more universal force.
But at longer distances, the strong nuclear force gradually becomes (relatively!) stronger. [Again,
remember what we mean by “weak” and “strong” here; the force is actually becoming weaker in
absolute terms as r increases, but relative to, say, electromagnetic forces at the same distance r,
it’s becoming stronger.]

αstrong = 0.3 (at r ~ 10-16 meters)
That’s quite strong indeed! And by the time r reaches 10-15 meters, the radius of a proton, αstrong
is bigger than 1, and becomes impossible to define uniquely.
In short, the strong nuclear force, which is only moderately strong at distances far smaller than
the radius of a proton, grows (in relative terms) at larger distances, and becomes a truly strong
force at a distance of 10-15 meters. (This is shown in Figure 2.) It is this truly strong force that
creates the proton and the neutron, and a weaker residual effect of this strong force that combines
these objects into atomic nuclei. Other important effects of this force becoming so strong are the
conversion of high-energy quarks and gluons to jets (sprays of hadrons).
Why does the strong force become gradually stronger as r increases? That’s a story for another
day, but essentially it is a very subtle effect due to disturbances (“virtual particles”) in the very
gluon and quark fields that are affected by the strong force. Similar effects impact the weak and
electromagnetic forces, but have much less dramatic effects on those two forces, which is why I
haven’t mentioned them before. (For instance, at the distance of 3 × 10-18 meters, the
electromagnetic α is closer to 1/128 than its long-distance value of about 1/137.)
Given how strong the strong nuclear force is, why don’t we encounter it on a daily basis? It has
to do with important details of how the strong force binds quarks and gluons and anti-quarks so
tightly into protons and neutrons that we never observe them separately. This is in sharp contrast
to how the weaker electromagnetic force allows electrons to escape from atoms rather easily,
making the phenomena of static electricity (including lightning) and electric currents (including
those in electrical wires) possible.
The Strength of Gravity
What about gravity? Well, for the particles we know about, gravity is amazingly weak. For two
(stationary) particles of mass m, the force of gravity has a strength

αgravity = GN m2 / ℏ c
where GN is Newton’s gravitational constant. Compare this with the case of the electric force,
where α = ke2 / ℏ c; the role of k and e in electric forces is played by GN and m. (Here I am using
Newton’s formula for gravity, but as long as αgravity is small compared to 1, Einstein’s formula
for gravity between two objects is essentially identical.)
Now we can rewrite this in terms of what is called the Planck mass Mpl = 1019 GeV/c2, or about
the mass of 10 million million million protons, or of 20 thousand million million million
electrons. (This is about the mass of a . [OOPS! Proofread Failure. This is obviously
wrong. Speaking approximately, Mpl = 1019 GeV/c2 = the mass of 1019 protons = the mass of
1019 Hydrogen atoms = 0.00002 moles of Hydrogen = 0.02 milligrams = about a tenth of the
weight of a typical grain of salt. Thanks to reader Al Schwartz for pointing it out.]

αgravity = (m / Mpl )2
So for two protons, each of which has a mass of about 1 GeV/c2the gravitational force between
them has a relative strength of the square of (1/10 million million million), or (10-19)

αgravity = (10-19)2 = 10-38
which is a 1 preceded by 37 zeroes and a decimal point! Meanwhile for two electrons

αgravity = (10-19)2 = 3 × 10-46
which, since an electron has a mass almost 2000 times smaller than a proton, corresponds to a
force four million times weaker. Even for a pair of top quarks, nearly 200 times heavier than a
proton and with the largest masses of any known particles, the gravitational force has a strength
of only

αgravity = 10-34
That’s about 100,000,000,000,000,000,000,000,000,000,000 times smaller than the electric force
between two top quarks. That’s why gravity doesn’t show up in Figure 2.
If you think about it, this incredible weakness of gravity explains why you (using electric forces
that power your muscles and keep your body intact) can move so freely despite being pulled on
by the gravitational pull from the entire, enormous earth. In fact it even explains why the earth
can be so much larger than a single atom; gravity wants to crush the earth, but the integrity of
atoms, whose electrical forces resist this crushing, prevents this. If gravitational forces were
much stronger, or electric forces much weaker, gravity would crush the earth down to a much
smaller size and a much greater density.
Gravity is so weak that it’s amazing that we could discover it at all. So why was it the first force
we humans knew about? The reason is that it is the only force that survives to very large
distances in ordinary matter.




The weak nuclear force becomes extremely weak at long distances. (We’ll see in a
moment that the same is true of the Higgs force.)
Electromagnetism survives to larger distances, but though not very strong is still strong
enough to bind up most electrons and atomic nuclei into electrically-neutral
combinations, whose electric forces on other objects cancel. [For instance, a hydrogen
atom does not have pull on a distant electron, because the electron in the hydrogen atom
pushes and the proton in the hydrogen atom pulls that electron, with forces that
essentially cancel.]
The strong nuclear force is so strong that it binds quarks and gluons and anti-quarks
together into combinations that similarly have cancelling effects.
But gravity cannot be arranged to cancel in this way. There are no particles that generate
gravitational forces that push things apart, so you can’t combine two particles so that
their gravitational forces on all distant things cancel.
The Higgs Force!?
As of 2012, we have a new force to think about: the force between two particles induced by the
Higgs field! This is not to be confused with the effect by which the Higgs field gives the known
elementary particles their masses; the Higgs field can do this to a single, isolated particle. That’s
not a force; it doesn’t push or pull. But the Higgs field can also induce a force between two
particles; this happens in much the same way that electromagnetic forces are created. However,
as far as its effect on ordinary matter, this force is very, very hard to detect. At short distances,
for particles like electrons and the up and down quarks that dominate the proton, the Higgs force
is very weak (much weaker than electromagnetism, but much much stronger than gravity). At
long distances, like the weak nuclear force, the Higgs force becomes extremely weak, because
the Higgs particle, like the W particle, has a mass.
The Higgs field induces a force similar to the weak nuclear force in that it has a very short range,
becoming ineffectual at distances long compared to ℏ c / Mh ~ 2 × 10-18 meters (1/500 of a
proton’s radius), where Mh ≈ 125 GeV/c² is the Higgs particle’s mass. And at first glance the
formula is similar to that of gravity, in that it is an attractive force proportional to the masses m
of the two elementary particles being attracted.

αHiggs = (mc2 /4 π v)2 × e-Mhr/ℏc (for r » 2 × 10-18 meters)

αHiggs = (mc2 /4 π v)2
(for r « 2 × 10-18 meters)
where v = 246 GeV is the constant value of the Higgs field found throughout the universe.
[Actually, if one is careful, there is an extra square root of 2 in there, but let’s keep the formulas
simple-looking.]
Be cautious! The resemblance to gravity is misleading. This formula is only precise for the
known elementary particles — those objects that get their mass from the Higgs field. It works for
electrons and muons and quarks. It is not correct for protons, neutrons, atoms, or you! That’s
because a proton’s mass (and a neutrons, and therefore an atom’s, and therefore yours) does not
entirely come from the Higgs field. This is in contrast to the formula for gravity, which is
correct for all slow objects! Instead, for ordinary atomic matter, we’d have to replace the
formula with one that looks similar but has a different factor in front, slightly different for each
atom. Qualitatively, however, the dependence on the distance would remain similar.
Also, the formula I’ve written also assumes there is only one Higgs field and one Higgs particle
(which we don’t yet know to be true, but is the simplest possibility consistent with current data.)
If that’s not the case the formula will become more complicated, while remaining of a similar
form.
Fig. 2: The relative strengths (α) for the forces acting on a top quark and top anti-quark, with
mass almost 200 times larger than a proton, as a function of radius, between about 1/3000 of a
proton’s radius and 30 proton radii. The Higgs force and the weak nuclear force become
extremely weak at distances even smaller than the proton radius, while the strong nuclear force
becomes strong at the proton radius. Electromagnetism stays moderately weak at all
distances. There are subtleties, ignored here, involving electromagnetism and the weak nuclear
force at sufficiently short distances (grey box). Gravity is incredibly weak and lies far below this
graph. For the up and down quarks in the proton and neutron, the graph would be similar, except
that the Higgs force would be far too small to appear on the plot; the same would be true for two
electrons, though it is unaffected by the strong nuclear force.
How strong is this force? Well, at very short distances, shorter than 2 × 10-18 meters, the Higgs
force between two top quarks is comparable to the strong nuclear force at that distance (see
Figure 2)! But for electrons, which have a low mass because their interaction with the Higgs
field is small, the force even at short distances would be much weaker than electric forces —
more than a thousand million times weaker — though still thousands of millions of millions of
times stronger than gravitational forces between electrons. Yet if you consider two electrons in
an atom, which are about ten million or so times further apart than 2 × 10-18 meters, then the
Higgs force between them is much, much smaller even than the tiny gravitational force between
them, as it is suppressed by e-10,000,000. [Even if the Higgs field did give protons and neutrons all
their masses, the Higgs forces inside a nucleus would still be vastly smaller than those of gravity,
which in turn are incredibly small compared to the residual strong nuclear force that holds the
nuclei together.]
It is the incredible weakness of the Higgs force in the context of ordinary matter that makes it so
hard to discover. On the one hand, the Higgs force, like gravity, is always attractive and can’t be
cancelled. But on the other hand, that’s irrelevant, because, like the weak nuclear force, the
Higgs force does not survive to long distance, because the Higgs particle, like the W particle, has
a mass. The Higgs force at ultra-short distances is much stronger than gravity, but at nuclear and
atomic distances, it is much weaker, because of the Higgs particle’s mass. And for the low-mass
particles out of which we’re made, which interact weakly with the Higgs field, the Higgs force is
always thousands of millions of times weaker than electric forces, even at very short
distances. So even though every atom in the Earth exerts a Higgs force on every other atom in
the Earth, that force is so incredibly minuscule, even for neighboring atoms, and especially for
distant ones, that it has no detectable effect. This is why we had to go find the Higgs particle
directly to confirm the Higgs field exists; we couldn’t just go looking for the force it creates, the
way we can use the observation of electric and magnetic forces to confirm that the world has
electric and magnetic fields.
When might we actually observe this new force? It’s effects will first be observed either in the
scattering of W and Z particles off each other (which will eventually be done, indirectly, within
the proton-proton collisions at the Large Hadron Collider) or in the interaction between a top
quark and a top anti-quark (which can be observed at an electron-positron collider — in fact I
wrote my first particle physics paper [see in particular Figure 11 of the paper] about this very
phenomenon.)
Share this:




Twitter4
Facebook115
Google
104 responses to “The Strengths of the Known Forces”
1. Pingback: What is the “Strength” of a Force? | Of Particular Significance
2.
Kudzu | May 31, 2013 at 10:19 AM | Reply
An excellent and informative article as always. It answered a question i didn’t even know
I had. Two questions remain for me however:
1.) Is the weak force always attractive like gravity, or does it have opposing charges like
EM and the strong force? I’m assuming the former.
2.) I assume that while charged particles are not affected by atoms at a distance, as they
approach closer and closer to an atom or other electrically bound system that becomes
less so. If so, how would an electron within a hydrogen atom, closer than the bound
electron behave?
o
4gravitonsandagradstudent | May 31, 2013 at 12:39 PM | Reply
Matt can probably explain these better than I can, but I thought I’d give it a shot:
1. The weak force also has opposing charges. They’re more like the charges of the
strong force than the charges of EM, in that just like the three independent
“colors” of the strong force there are two independent sorts of charge in the weak
force.
2. Approach close enough to an atom and the electron and the proton indeed have
an effect. It’s a bit like holding a bar magnet at a distance from the north pole of
another magnet, in that as you get closer the two ends want to go different
directions.
o
Matt Strassler | May 31, 2013 at 3:33 PM | Reply
The weak force is like the strong nuclear and electromagnetic forces in that it can
pull and push, but it has other things that it can do — in particular, it can cause
decays. Requires a separate article.
The bound electron pushes on the second electron, while the proton pulls on it.
Which pushes/pulls harder depends on where precisely the second electron is.
When the second electron is far away from both particles, the forces nearly
cancel; when it is close in, then the forces don’t cancel at all. This is the usual
story — even from freshman year — about electric dipoles.
o
po8crg | June 23, 2013 at 3:21 PM | Reply
A complete answer to your question 2 would be the entire academic subject of
physical chemistry.
The interaction of non-atomic electrons with an atom is exactly how chemistry
happens.
3.
ablelawrence | May 31, 2013 at 10:22 AM | Reply
What would be the Higgs force acting between the virtual particles in empty space – the
sea of quarks and anti-quarks. Since Higgs force is always attractive and does not cancel,
out, the interactions between the constituents of QCD vacuum may have some effect. Can
we then think of it as the “surface tension” of vacuum space. If the vacuum energy (zero
point energy) is not constant between different parts of space time, what would be the
impact of this Higgs force between the virtual particles. Even though Higgs force does
not act at long distance, since the distance between virtual particles is rather small,
wouldn’t this have a cascading effect again similar to surface tension for the vacuum
space. We may not be able to perceive this because we cant feel the space because of
relativity but its effects may influence expansion of space itself in cosmology
o
Matt Strassler | May 31, 2013 at 3:36 PM | Reply
The Higgs force between virtual particles in empty space [remember virtual
particles aren’t really particles at all, http://profmattstrassler.com/articles-andposts/particle-physics-basics/virtual-particles-what-are-they/ ] is just one of many
important things going on in the very complex place we call “the vacuum of
empty space”. It’s not especially important or unimportant compared to other
things that are also going on.
4.
blerg | May 31, 2013 at 11:42 AM | Reply
One nagging question that remains for me: what is it about the weak force that causes it
to mediate a lot of particle decays? Is that just because of neutrinos and lepton number
conservation or is there something deeper?
o
Matt Strassler | May 31, 2013 at 3:37 PM | Reply
It’s deeper, and a long article is needed. It’s related to the Higgs field, the fact that
the W and Z have masses, the fact that up and down quarks have masses (not only
that, different masses), and a lot of other things. The buzzword is “symmetry
breaking”, but since there was no real symmetry and it really isn’t broken, it’s not
a good buzzword to hang one’s hat on… Stay tuned.
5.
aa. sh. | May 31, 2013 at 11:48 AM | Reply
Really great , thanks Matt.
You explained the Higgs force properties , but i could not find whereabout of its function
since all the micro-world was already described by QFT and QCD….. so what
ADDITIONAL function this force perform ?
o
Matt Strassler | May 31, 2013 at 3:39 PM | Reply
It doesn’t really play much of a role in daily life or in the structure of the universe
as we know it so far. That said, it might end up being how we discover dark
matter. Of course it may have some functions we don’t know about. But under
normal circumstances, everything that it could do is dwarfed by effects of other
forces.
6.
kashyap Vasavada | May 31, 2013 at 12:13 PM | Reply
Excellent article. I have two questions.(1) Your fig. 2 shows the strengths leveling and
not coming together as required in grand unified theory. Is it that they will come together
on left of y-axis eventually?
(2) Where do you put dark energy repulsion (lambda force)? It is hard to believe that
something affects only at the galactic scale and there is no trace of it at the microscopic
scale. Thanks.
o
Matt Strassler | May 31, 2013 at 3:42 PM | Reply
(1) on the one hand, yes, grand unification is far to the left of the plot. BUT also
there are subtleties hidden in that grey box. And the slow drift of the strengths of
the forces is essential to the idea. The Higgs force does not get unified with the
others in Grand Unification, nor is gravity. Any further unification would occur at
even shorter distances…
2) dark energy repulsion has no trace at the microscopic scale. It’s not a repulsion
between objects. It is causing space itself to slowly expand, and this is very hard
to detect on terrestrial scales.

kashyap Vasavada | June 1, 2013 at 10:30 AM | Reply
Thanks. But something like lambda, may be having much larger value
than the current small value of lambda (inflaton or whatever you call it)
generated exponential inflation in the beginning. Would that not have
microscopic implications in terms of forces?

Matt Strassler | June 2, 2013 at 4:33 PM | Reply
No. [I assume by “lambda” you mean the cosmological energy
density or cosmological “constant” (which isn’t always constant
but ok…). It’s Lambda, usually.]
If what you’re asking is whether the physics was that generated the
cosmological constant (or assured it was large enough in early
times to generate inflation and is now very small) might be
associated with microscopic forces, then the answer is yes, it
might, but there’s no guarantee. The cosmological constant itself is
not associated with any microscopic effects; it is a large-scale
effect, not a small-scale one.

kashyap Vasavada | June 2, 2013 at 5:44 PM |
Thanks . I see what you are saying. But I still find it
incredible that some very large field gave rise to
tremendous inflation, then turned off and no remnants left
for 13.8 B years!

Matt Strassler | June 2, 2013 at 5:53 PM |
The equations say that this can happen.
But first, you should not say there was no remnant of any
sort. The energy density from that period of inflation heats
up the universe, when inflation ends, and gets the hot-dense
period of the Big Bang going! So it’s not as though there
was no effect!! There’s just no microscopic force from that
effect.
However, if that energy density came from a standard
inflaton field, all that is required to end up with no
microscopic remnant is that the fields involved have
particles that are very heavy and unstable. That means there
aren’t any of those particles left, because they all decayed
away; we aren’t able to produce those particles at our
accelerators, because it would require too much energy per
collision; AND the forces they generate have too short a
range to have any effect on any current microscopic
measurement we can make.
7.
aa. sh. | May 31, 2013 at 12:39 PM | Reply
As far as i remember you previously told that the Higgs particle is not a force carrier ,
then what is the fifth force- force carrier ?
o
Matt Strassler | May 31, 2013 at 3:43 PM | Reply
The Higgs particle would be called the force carrier of the Higgs force, as the
photon is called the force carrier of the electromagnetic force. I don’t really like
the term “force carrier”; that article has yet to be written though.
8.
Thilo | May 31, 2013 at 2:41 PM | Reply
Excellent article.
Question: If one particle emits or absorbs a virtual force carrier particle, shouldn’t that
also affect its spin due to conservation laws? Or, in case of W boson exchange, always
result in both interacting particles being “changed” (as W is charged)?
o
Matt Strassler | May 31, 2013 at 3:49 PM | Reply
To really do this justice requires another article, which I do plan to write.
To your first question: No. The problem is that virtual particles aren’t particles
http://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtualparticles-what-are-they/ — they operate under more general rules — and it isn’t
really particles that carry forces, it’s disturbances in the corresponding fields. In
fact you already know this from freshman year study of electric fields, which are,
after all, spin-1 fields.
It is really the Z that you would say generates the weak force, and the W, which
really does change things, typically leads to decays.
In the case of W boson exchange, yes, both interacting particles are changed, and
typically that means that you can’t think of multiple exchanges as a push-or-pull
force.
In the case of gluons, both interacting particles are changed too, but in a way that
is unobservable, does not lead to decays, and just maintains the push-or-pull force
effect.

charge vs. charge | July 10, 2013 at 6:36 AM | Reply
“electric fields are spin-1 fields” -> what do you mean by this? Do you
mean, “electric fields” are made of spin-1 photons? with coherent quanta
(=real photons) or without coherent quanta(=virtual photons, or
disturbances?))? and why do you have to call “fields” spin-1, 0, 2 etc.?
Intrinsic “spin (0, 1/2, 1, 2 etc.)” is associated with “a particle” but you
can measure “spin of fields” without a real particle? or can you calculate?
talk about? “the spin of fields(?)” without any real particle in the field?
why this is so?(if “electric fields are spin-1 fields” is the bottom
property/current understanding of the field, then give a few specific
“simple(simplest)” examples that demonstrate why spin-1 field (but not
spin 1/2, 0, or 2) is required for (and also is consistent with) explaining
these examples(=physical phenomena in electric fields))? Massless virtual
photons also have spin-1? Do massive? photons in superconductor? also
still have spin-1? (“..It’s the same with electric charge in a
superconductor; charge is still conserved but it doesn’t look that way
because the vacuum is full of Cooper pairs, and the photon acts as a
massive particle.”)
9.
Derek | May 31, 2013 at 4:29 PM | Reply
Hi Matt,
Although I learned a great deal from your article, as I always do, I didn’t actually learn
the thing I was most hoping to! Basically, I’ve long wanted to understand why QFT uses
the concept of ‘force’ at all. In what sense is it helpful to think of an electron and a
positron pulling on each other, for example?
In Quantum Mechanics, one can formally recover Newtonian concepts because it uses a
Hamiltonian formulation (albeit with operators rather than numbers), but QFT uses
something totally different (and something I’m not very familiar with). As a specific
example, in QFT an electron and positron are supposed to attract each other by means of
an exchange of photons. Yet such an exchange, if taken literally, would lead to repulsion
rather than attraction, so Newtonian thinking seems totally misplaced here.
I’ve always preferred the term ‘fundamental interactions’ rather than ‘fundamental
forces’, because I just can’t see where their force-likeness comes from. This is especially
so in the case of the Weak Interaction, which seems to always manifest itself
transformatively rather than kinematically.
I guess another article explaining what attraction and repulsion (and hence ‘force’) mean,
and how they happen, in the context of QFT would really hit the nail on the head for me.
If you have already written such an article, I’d really appreciate you pointing it out!
Thanks again for all of your great work here.
o
Matt Strassler | June 2, 2013 at 5:29 PM | Reply
You are right that “interaction” is a much better term than “force” — and when
speaking to professionals, that’s what I use. But I haven’t yet written the article
that explains why professionals do this! I think the term “force” remains
necessary for communicating with non-expert audiences, because “force” has
some intutive reach and some connection with freshman physics and many books,
while “interaction” is too broad and vague a term to touch any intuitive concept.
With regard to this issue, I do indeed plan to write the article that you’ve just
suggested.
Meanwhile, don’t forget the weak nuclear (umm, interaction) does have a Z
boson, and that does give an honest-to-goodness force too, just the same as gluons
and photons! If we ever discovered a heavy weakly-interacting stable particle, and
could make a bound state out of it and its anti-particle, the Z-boson-induced force
would be measurable.

Kudzu | June 2, 2013 at 7:24 PM | Reply
Given that a popular model of dark matter is WIMPs, would the z-boson
mediated force have any consequences? Would it be strong enough to
cause disparate dark matter particles to collapse under certain
astronomical circumstances?

Matt Strassler | June 2, 2013 at 10:15 PM | Reply
1) the Z-boson mediated force could have been the one that made
WIMPs bounce off an atomic nucleus. But in this case WIMPs
would probably have been observed already. It appears something
weaker, like the Higgs force, may have to do the job – if anything.
2) No, a short-range force can’t affect dark matter particles that are
macroscopic distances apart. In fact it could barely affect them if
they were packed together. Think about how ineffectual
electromagnetism would be if its range were 10^{-18} meters.
10.
aa. sh. | May 31, 2013 at 5:13 PM | Reply
In fig.2 , what do you physically mean by the abrupt stop of the strong force curve ? i
guess it somehow must drop to meet the x-axis somewhere …..
o
Kudzu | June 1, 2013 at 2:07 AM | Reply
In that figure the strong force curve does not ‘stop’ or go to 0; quite the opposite.
It becomes so large that our way of describing it (In this article) breaks down.
This is because when a force passes the ‘1 mark’ on that graph things become
quite complicated.
o
Matt Strassler | June 2, 2013 at 5:20 PM | Reply
It simply stops being meaningful. I’ve addressed this in more detail when you
repeated the question in a later comment.
11.
PuzzledFan | May 31, 2013 at 6:07 PM | Reply
“the time it takes between one vibration and the next — which I like, poetically, to call a
particle’s “heartbeat” — is equal to hc/m”
but the dimensions of that expression are L^3.T^-2
Isn’t it t=h/mc^2? (from E=h/t=mc^2)
o
Matt Strassler | June 2, 2013 at 5:18 PM | Reply
Uh oh. Not sure how that happened! That goof would have been made a long time
back, and propagated through multiple articles. Thanks!
12.
Tony | May 31, 2013 at 6:25 PM | Reply
Great article, many thanks Professor.
Just a couple of questions, I always had problems trying to understand what a force really
is. I mean, what is the underlying mechanism that makes two particles affect each other.
Is the same mechanism for all forces? what’s the internal difference between a repulsive
and an attractive force? If gravity is “just” a distortion of space-time, what is a graviton?
Thanks again
o
Matt Strassler | June 2, 2013 at 5:17 PM | Reply
I’ll be addressing this in an article in the near future.
13.
Tienzen (Jeh-Tween) Gong | June 1, 2013 at 12:50 AM | Reply
Matt: “… Clearly there is a language difference here… as is often the case with words in
English and words in Physics-ese.
… to understand that this way of talking about the strengths of forces is the best one to
use.”
There is no issue about the accuracy of this article. Yet, indeed, the “language” can make
a big difference for the same story. In my view, the four known forces can be described
with a set completely different language. That is, we can view these forces in their
functionalities instead of their strengths. Then, these four forces can be classified into two
types.
a. Envelope constructing type:
i. Strong force (SF) — constructing the envelopes for proton or neutron. Then, the
residual of SF constructs the envelopes of nuclei.
ii. Electromagnetism force (EF) — constructing the envelopes of atoms and molecules.
Then, it also constructs the envelopes of the “causal universe”.
iii. Gravitational force (GF) — constructing the envelopes of large mass bodies. Then, it
also constructs the envelope of a non-causal universe.
b. Envelope penetrating type: It goes into an envelope or breaks out from one, the Weak
force (WF).
The above description can also be viewed in terms of an automobile, as the universe is
evolving “forward”. The constructing type is the “forward” gears while the penetrating
type is the “reverse” gear.
Then, we can go one more step; viewing these forces are building an amphitheater
(without ceiling and walls) with chairs and stages, similar to the universe with quarks,
nuclei, …, galaxies, etc.. The constructing force are “making” chairs, sections, …, stages.
The penetrating force “brings” audiences and actors from one chair to another or from
one section to a different one. That is, the WF is the usher of the theater.
In the case of unifying these forces, knowing their functionalities are much more
important than knowing their strengths. Even if their coupling strengths are exactly the
same, they can still be completely different forces. Yet, knowing their strengths converge
at some certain situations is still a great knowledge about them.
With this amphitheater model, the forces classification (degeneration) is the results of
that theater’s framework and functionalities. That is, there is a “fundamental/emergent”
relationship, or the force-genealogy as follow:
Force (degenerated) = K (degenerated) F(unified), K is the coupling constant.
F (unified) = ħ / (delta T * delta S) ; T, time; S, space.
This unified force equation shows the “sources” of all degenerated forces. The
degenerated K (couplings) is caused by different “charges” (job-types).
With different language, a story can be told differently. In this story, the alpha (α) is not
only the coupling constant for electromagnetism but is the “manager” for all other
charges, as it is a “lock” which locks the three most fundamental nature constants (ħ, C
and e).
14.
JollyJoker | June 1, 2013 at 7:11 AM | Reply
OK, I think I understand why strong forces are hard to deal with a lot better now.
However, this raises a bunch of questions I’m not really sure how to formulate. This
gives rise to 3-body or n-body problems, right? How then do the dualities linking the
strong regime of some theory to a weak one of another work? Linking something
complicated to something simple doesn’t generally sound like it could work. Presumably
this says something deep about one of the more surprising (to me) dualities, gravity /
Navier-Stokes. The ABC conjecture also comes to mind as a weird thing linking the
simple to the complicated.
This is probably pretty unclear post, but I really have too vague a sense of what puzzles
me to ask a sensible question.
o
Matt Strassler | June 2, 2013 at 5:08 PM | Reply
“Dualities” which allow you to write a very strong force’s behavior in terms of a
very weak one are remarkable, complex, quantum mechanical, hard to explain,
and in most cases are not well-enough understood even to state clearly.
I have not covered this issue on this website for a reason!
I’ll cover this someday, but not now. However, here’s a hint. Protons and
neutrons are incredibly complicated, with all sorts of internal structure. Yet the
interactions between protons and neutrons, at least if you keep all the motion- and
interaction-energies (i.e. kinetic and potential energies) below 1 GeV or so, are
pretty simple — which is why the residual nuclear force isn’t that complicated
and one can do a reasonable job with a semi-quantitative description of nuclear
structure. Now how did that happen? Shouldn’t it be the case that if something is
really complicated, everything you make out of it should be really complicated
too?
Apparently not.
Well, this isn’t enough of a real answer, but a proper answer would take much,
much longer, with many pictures.

JollyJoker | June 3, 2013 at 2:04 AM | Reply
Well, structures that make some quantity add up to zero is an obvious
example where you lose complexity. Or more accurately, hide some
complexity. Having an n-body problem on one side of a duality become
“something simple” on the other side wouldn’t hide anything but show
that the complexity of the whole thing is a mirage. (Side note; checking
Wikipedia for n-body problem, it seems a solution is known to be
impossible only for some specific method. I thought it was more general)
Looking forward to the explanation with many pictures. Linking to a N AH paper with a hundred weird diagrams is cheating ;)

JollyJoker | June 3, 2013 at 2:40 AM | Reply
One loophole that occurred to me is that since the situation is quantum, the
extra stuff probably has a smooth distribution of probabilities that could
add up to something simple although the classical case would be messy.
15.
aa. sh. | June 1, 2013 at 7:11 AM | Reply
What is the ” charge ” that is the Higgs force cause ?
What particles the Higgs force is acting on ?
o
Kudzu | June 1, 2013 at 9:09 AM | Reply
The Higgs force, like gravity has no charge. Anything that interacts with the
Higgs field feels the Higgs force. This includes the stuff we’re made of (Electrons
and quarks.) as well as many other particles like Tau leptons.

Tienzen (Jeh-Tween) Gong | June 1, 2013 at 12:58 PM | Reply
“Mass” is the charge for gravity.

Matt Strassler | June 2, 2013 at 5:14 PM | Reply
I don’t think I would say it this way. Energy and momentum are
gravitational charges; the energy-momentum tensor is the corresponding
conserved current.
The Higgs force you can think of, if you want, as being due to “Higgs
charge” (these are just the Yukawa couplings). But the main point is that,
unlike the charges in the electromagnetic, strong nuclear, and [to a limited
extent] weak nuclear forces, those charges are not conserved.

Kudzu | June 2, 2013 at 7:13 PM | Reply
Interesting. Is there a reason why the weak and Higgs forces’
charges are not conserved? Would it be possible to have a force
with two charges like electromagnetism not have conserved
charges with a physics similar to that of our universe?

Matt Strassler | June 2, 2013 at 10:18 PM |
This only has a technical answer; sorry for non-experts.
1) Higgs forces are from a spin-zero field; there’s no
associated conserved current. Spin-one fields always come
with a conserved current, for their very consistency; this is
true of all the other non-gravitational forces. The weak
force, however, is complicated by the fact that the W and Z
are massive; this means that charge is still conserved but in
a non-obvious way, because charge can disappear into the
vacuum. It’s the same with electric charge in a
superconductor; charge is still conserved but it doesn’t look
that way because the vacuum is full of Cooper pairs, and
the photon acts as a massive particle.
2) I’m confused by this question; not sure what you mean.
“Two charges”?

Kudzu | June 3, 2013 at 2:42 AM | Reply
Thankyou for your reply. The answer to my first question makes
sense and answers my second question as well.
o
Tienzen (Jeh-Tween) Gong | June 1, 2013 at 3:44 PM | Reply
What is charge? This is a language issue.
I was charged 10 dollars for the movie and 100 dollars for the dinner. “Money” is
a universal charge. Yet, in general, different action demands different charge. As
“no money, no fun”, it is the same that “no charge, no action”. If a force without a
charge, it can produce no action.
Yet, what is charge? Charge is the “measurement” for the action it has done. That
is, “charge” is in fact a “measuring ruler”.
What is e (electric charge)? Well, what is “e” measuring? “e”-charge measures
the “size” of the “causal universe”. The part of this universe beyond this causal
universe is called beyond the “event horizon”.
What is m (mass-charge) measuring? “m”-charge measures the “size” of the
internal “envelope”, such as the quark-envelope or the proton-envelope. If there
are envelopes beyond the event horizon, m-charge will measure them too. That is,
the m-charge has the power going over the event horizon. If there is an invisible
envelope, m-charge will measure it too.

Matt Strassler | June 2, 2013 at 4:45 PM | Reply
You wrote : ““e”-charge measures the “size” of the “causal universe”.”
No it doesn’t. Electric charge of a particle simply tells you how
powerfully a particle will distort the electric field in its vicinity. Just as
mass (actually, energy and momentum, in Einstein’s theory of gravity),
tells you how powerfully a particle will distort space-time (said another
way, distort the gravitational field) in its vicinity.

Tienzen (Jeh-Tween) Gong | June 2, 2013 at 10:11 PM |
Reply
Indeed, the “e” itself does only what you said. But the “causal
universe” is precisely defined by its squire (the photon). Thus, the
boss got the credit. This is the same as a king whose legs do not go
out of the palace takes the credit for everything happened in his
dominion.
o
Matt Strassler | June 2, 2013 at 5:12 PM | Reply
Umm… it’s the `Higgs charge’, but we generally don’t call it that for historical
reasons. And the Higgs force acts on all particles that interact with the Higgs
field… which also happens to be all of the known elementary particles that have
mass.
16.
johnmcAllison | June 1, 2013 at 10:19 AM | Reply
Matt, I’m looking for the fundamental difference between the electric and magnetic
force:
Would you agree that the acceleration on a moving charge in an electric field is
independent of the direction of its velocity, but not for a magnetic field?
o
Matt Strassler | June 2, 2013 at 4:33 PM | Reply
There is no “fundamental” difference; the right statement is that there is an
electromagnetic force, of which electric and magnetic forces are different parts…
and in particular, what one observer sees as a pure electric force will be viewed by
another observer, moving relative to the first one, as both electric and magnetic.
The force law that describes a particle moving in an electric and a magnetic field,
from the point of view of any (inertial) observer, is
F = q (E + v x B)
F is the force (a vector), E is the electric field (a vector), v is the velocity (a
vector), B is the magnetic field (a vector), and “x” is a “cross-product”.
Notice that velocity does not appear at all in the term involving the electric field.
So the acceleration of a charge, whether moving or not, FROM THE POINT OF
VIEW OF AN OBSERVER WHO VIEWS THE SITUATION AS
CONTAINING ONLY AN ELECTRIC FIELD AND NO MAGNETIC FIELD, is
entirely independent of its velocity, both direction and speed. A different
observer, moving relative to the first one, will see both electric and magnetic
fields, and draw a different conclusion.

johnmcAllison | June 17, 2013 at 2:36 PM | Reply
I don’t think this is correct:
“So the acceleration of a charge, whether moving or not, FROM THE
POINT OF VIEW OF AN OBSERVER WHO VIEWS THE SITUATION
AS CONTAINING ONLY AN ELECTRIC FIELD AND NO
MAGNETIC FIELD, is entirely independent of its velocity, both direction
and speed”.
Yes, on the RHS there is no velocity term associated with the electric
field, but the acceleration term on the LHS for the force does involve
velocity:
F = d/dt (m_0 \gamma vec_u) = (m_0 \gamma^3/c^2 vec_u_dot_vec_a) +
m_0 \gamma vec_a
Although the magnitude of acceleration vec_a is velocity dependent, it’s
direction is the same as that of the term on the RHS which is independent
of velocity for the electric field term.

Matt Strassler | June 19, 2013 at 10:54 AM | Reply
You are correct; I answered a question about the *force*, but as
you point out, your question was about the *acceleration*.
However, the issue you point out has nothing to do with electricity
or magnetism; it has to do with space and time, and is present both
for electric and for magnetic forces, as well as any other forces that
might be present, such as the weak nuclear force or Higgs force.
Newton’s law was F = m a; in this context both the F-side of the
equation (which has to do with electricty and magnetism) and the
“ma”-side of the equation (which has to do with space and time)
are modified.
I don’t think, therefore, I understood the nature of your question;
what is it that you are trying to understand?
17.
aa. sh. | June 1, 2013 at 11:43 AM | Reply
I am still quite unsatisfied with the curve of the strong force , any curve representing a
function cannot just stop nowhere , either it go to infinity in value /distance or it diminish
to zero , would you please clarify this point.
o
Kudzu | June 1, 2013 at 5:32 PM | Reply
Curves can and do stop, especially when what they are measuring breaks down. A
classic example of this is in the phase diagram of water where at the ‘critical
point’ the difference between liquid and gaseous water ‘vanishes’
http://serc.carleton.edu/images/research_education/equilibria/h2o_phase_diagram
_-_color.v2.jpg In the case of the strong force, when the value starts to exceed 1
its definition breaks down. You can’t say a curve goes to infinity if you can’t
measure or define it.
It is a simple diagram, it even lacks gravity. Some detail has been sacrificed for
clarity.
o
Matt Strassler | June 2, 2013 at 4:40 PM | Reply
The reason I wrote this article now is that I have to go back to the protons and
neutrons article, http://profmattstrassler.com/articles-and-posts/particle-physicsbasics/the-structure-of-matter/protons-and-neutrons/, and start to explain this
question. It does not have a short answer. The relevant buzzword is “quark
confinement”, though that’s not very meaningful by itself. What it means is that
you can’t ever pull quarks or gluons far enough apart to give meaning to
alpha_strong, but this is not because the force is infinitely strong. There are hints
of how this works in this article: http://profmattstrassler.com/articles-andposts/particle-physics-basics/the-known-apparently-elementary-particles/jets-themanifestation-of-quarks-and-gluons/
18.
aa. sh. | June 1, 2013 at 11:46 AM | Reply
For gravity as a FORCE with graviton as force carrier , does this represent exclusive
alternative to gravity as geometry ?
o
Kudzu | June 1, 2013 at 5:37 PM | Reply
At present they are both two equal interpretations, like the various ways of
looking at quantum mechanics. Neither one can currently claim to be ‘correct’
and the tension between relativity and quantum mechanics is interesting indeed.
Quantum gravity often tries to reconcile the two.
o
Matt Strassler | June 2, 2013 at 4:42 PM | Reply
No, these are complementary — two ways of viewing the same thing. The fact
that gravitational fields can be viewed as geometry (even before we worry about
gravitons or gravitational forces) is the meat of Einstein’s theory — but it doesn’t
change the relation between fields, forces and particles when those forces are
relatively weak. You have to be more careful when you start dealing with
complicated gravitational objects like black holes or the universe as a whole.
19.
andrew | June 1, 2013 at 1:12 PM | Reply
This is great! It’s terrific to have articles that are one step in detail above the usual
popular ones, without being actually mathematical.
I think there might be a typo: “If α is large, then the time it takes light to cross the atomlike state is much larger, by 1/α, than the heartbeat of the particles that it contains.”
Should be “If α is small”?
o
Matt Strassler | June 2, 2013 at 4:43 PM | Reply
Yes, thanks for catching that!
20.
Tienzen (Jeh-Tween) Gong | June 1, 2013 at 11:10 PM | Reply
Matt: “… the (still unobserved but surely present) Higgs force.”
This is a highly speculative issue. Yet, it can be discussed conceptually, with two stories
as below.
Story one:
Mother: Hi, baby, it is the time to hit the sack.
Baby: I am still working on my homework, the neutron decay. How can a three string
particle (u, d, d) become 5 strings (u, u, d, e, anti-e-neutrino), especially a “d” had a sexchange for becoming a “u”?
Mother: This is the most famous fairy tale. Here comes an angel who gave the “d” a kick
on its butt, and it comes the “u”.
Baby: Who is the angel?
Mother: Who? Who? The “W”.
Baby: Then, where are this e and the neutrino coming from?
Mother: Oh. The “W” angel flies away as the e and the neutrino.
Baby: Where was the “W” angel coming from?
Mother: There is a Higgs “field” which is omnipresent, and it has Higgs egg everywhere.
The “W” angel was hiding in the Higgs egg all this time.
Baby: Thanks Mon. This is a great fairy tale. I now can sleep in peace.
Story two:
Brother: Mon, your fairy tale is very interesting. But, I learned a different story about the
neutron decay.
Mother: There cannot be any different story. This fairy tale is the “Standard Model”. But,
go ahead tell me your story.
Brother: The (u, d1, d2) picks up a virtue d-pair (d3, -d3), the vacuum disturbance. And it
forms a 5-quark blob. Then,
Step 1: (d1, -d3) got together
Step 2: (d1, -d3) turns into (u1, -u3), this is a vacuum state transforms into another
vacuum state.
Step 3: (-d3, -u3) got together. Then, this blob has a genetic exchange and flies away as
the e and the neutrino.
So, (u, d1, d2) decay into as (u, u1, d2, e, neutrino).
Mon, you have mistaken this vacuum transformation blob as the Higgs egg.
Mother: Your story follows all physics laws, and it makes sense. But it is wrong because
that it is not the Standard Model. Only authority can be right. Higgs force has ensured
that you are wrong.
o
Tienzen (Jeh-Tween) Gong | June 3, 2013 at 11:42 PM | Reply
Note: there was a typo, (-d3, -u3) should be (d3, -u3).
This Higgs force speculation can go one step deeper, again, with a conversation.
Baby: I don’t like your “vacuum blob” story. It is too rational and too cold. I like
the fairy tale, not because of its being the Standard Model but because that it is
“Magical”. A Higgs chicken (field) is omnipresent and can lay Higgs egg any
time at any place at its choosing. Then, out it pops a “W”-angel when neutron got
out of from bondage. Furthermore, no gadget thus far (including the LHC) in this
world is able to distinguish your “vacuum blob” from the Higgs egg. That is, that
fairy egg can be as real as the “vacuum blob” reality, and there is no gadget data
can prove it otherwise.
Brother: Hi, baby, for the gadget testing concern, you might be right for the next
decade or two. But, in sports, any tie can be easily broken by “overtime”, the tiebreak-litmus test.
The Standard Model is absolutely correct in terms of that entire gadget testing
data. But, as soon as it goes out of its baby crib, SM fails from left to right, from
top to bottom. It (SM) cannot derive most of the “parameters” it used in the model
(such as, the Cabibbo and Weinberg angles, the Alpha, etc.). Furthermore, it fails
on all the following known physics facts;
i. Planck data — dark energy (accelerating expansion of this universe), dark
matter, Neff = 3 (minimum), etc. .
ii. Neutrino oscillations
iii. Proton’s stability
iv. Guidelines for SUSY (with s-particle)
v. About gravity
vi. Unification between quantum and determinism.
That is, the Standard Model is only the hodgepodge of the gadget data. And,
simply it has “no” theoretical “base” for deriving those parameters and for
connecting to those other physics facts.
Thus, those other facts above become the overtime litmus tests. Anyone who can
solve any one of those other facts or derive those free parameters is the winner of
this contest.
Baby: Yet, most of those issues are controversial and cannot be decided with the
current gadget testing.
Brother: Indeed. But, the Cabibbo – Weinberg angles and the Alpha are only three
“numbers”. The only gadgets needed for testing them are “paper and pencil”
which are in the hands of every first grader. There will be absolutely no
controversy about whether an equation for calculating one of them is correct or
not. Thus, we can simply choose Alpha as this overtime Litmus test, to be the
“first” criterion for the correct final physics.
Baby: That is great. But, verifying the “prediction” of a theory was the only
criterion for physics theory thus far. How can you convince physicists to accept
your overtime idea?
Brother: Most of theory is the “extension” of known physics. Thus, it must
“predict” something new in order to be credible. But, when a “system” is
constructed from a “base” which contains no “known” physics of any kind, it is
credible only if it can “reproduce” all known physics, not just any puny
“prediction”. That is, system has “outcomes (or consequences)” while theory has
“prediction”. And, there is a super strong requirement for “base”. It must not
contain any “known physics”, that is, no known physics is “put-in” into the base.
Thus, “reproduce all known physics from a base (without containing any known
physics)” is the second “Overtime test”, the second criterion.
Baby: Must a base be verified by gadget testing?
Brother: No, no, and no! The “base” needs no verification. If a “wrong” base
produces a system which can reproduce all known physics, it is a “good” base.
Baby: Just heard in the news that a “W”-einstein had discovered the final physics
theory. Do you know anything about it?
Brother: No, never heard about it. If he can pass two tests,
a. Litmus test — deriving the Alpha with a physics equation
b. A base is able to reproduce, at least, the known 48 elementary particles
then, he could be on the path to the final physics. Otherwise, even the “Z”einstein won’t do any good.
Baby: Obviously, the Higgs story failed the first overtime litmus test. Are you
saying that the Higgs game is over? Is there Higgs force?
Brother: Higgs force is “not” needed in the “vacuum blob” story which did pass
the first litmus test, and its calculation is available at another blog
(http://blog.vixra.org/2013/05/16/why-i-still-like-string-theory/ ).
21.
Edwin Steiner | June 2, 2013 at 11:52 AM | Reply
Dear Prof. Strassler,
Many thanks for another excellent article! I have a question that may tie in with the weak
vs. strong force distinction: It is about the concepts of “elementary” and “composite”
particles.
After getting a glimpse of QFT I find this distinction confusing. My understanding is:
The state that we call a physical electron is a kind of resonance that involves both the
(bare) electron field and the photon field (in the full SM we even need right- and lefthanded electron fields, the Higgs field and others). Although it can be treated
perturbatively due to the EM force being “weak”, the physical electron is not even
“close” to a bare electron quantum, as renormalization tells us that we have to ascribe
absurd infinite properties to the latter to end up with a realistic physical electron.
The state that we call a proton is a kind of resonance that involves the quark and gluon
fields (and others). Due to the strong force being “strong”, we cannot describe a proton
using perturbation theory.
As neither the electron nor the proton can be explained using a single field, what
distinguishes one of them as elementary and the other as composite? Is there any
fundamental difference that I am missing? Or is it that the quantitative difference (i.e.
weak vs. strong) is just so large that it makes sense to use different terminology for the
electron and the proton?
best regards
Edwin
o
Matt Strassler | June 2, 2013 at 4:53 PM | Reply
This is an excellent and very deep question. It’s the kind of question that I would
discuss extensively with graduate students, and I am reluctant to answer it here
precisely because requires a deep, technical discussion — beyond the scope of
this website.
What is the current status of your training? Are you a beginning graduate student?

Edwin Steiner | June 2, 2013 at 6:33 PM | Reply
Dear Prof. Strassler,
I wish I could say “yes”, but the status of my training is a bit complicated.
Many years ago I studied physics for two years, then I switched and made
my master in computer science. However, my love for physics and
mathematics has never died and I’ve been wrestling with Peskin &
Schroeder, etc. on and off in my spare time. Having found that I lack too
many of the fundamentals, I’m now working through more humble texts
like Dirac’s Principles of QM and I have taken up studying mathematics.
Alas, progress is slow next to my full-time job.
Your website is very valuable to me, as it fills in many insights and
intuitive explanations that are so hard to come by in text books. There is
such a gigantic stretch nowadays between the popular expositions and the
“real thing”, so any intermediate level of information is precious,
especially for someone like me who has no access to the “oral history”
passed on at universities.
best regards
Edwin

Matt Strassler | June 2, 2013 at 10:34 PM | Reply
Here’s the simplest way to answer, then.
The most obvious sign that an object is composite is that you can
excite it internally. Of course atoms can be excited. You can make
an excited version of the proton; the lowest-mass excited proton is
called the Delta. And then there is a large tower of additional, more
massive states. The mass scale for these excitations is a few
hundred MeV. And indeed, the size of the proton is hc/few
hundred MeV, or 10^(-15) meters. The reason the size and the
excitation energy are similar (which is not true for atoms) is that
the force that binds the system together is a strong one. If the force
is weaker, the excitation energy is smaller than the corresponding
inverse size.
There is no excited version of the electron with mass below 1 TeV.
[The muon and tau are not excited electrons; if they were, they
would rapidly decay as muon –> electron + photon, whereas in fact
this decay never happens.] This tells us that the electron is smaller
than 10^(-18) meters.
So even though an interacting particle is a complex object, there is
still a notion of size which affects whether there are excited
versions of that particle. Yes, an electron has, as part of its very
nature, a “cloud” of photons and electron-positron pairs around it;
but that cloud has no independent existence, and you can’t excite
the cloud into a higher state. That’s different from the “cloud” of
electrons around a nucleus inside an atom, and its different from
the “cloud” that is a proton.

Edwin Steiner | June 3, 2013 at 4:45 PM |
Thank you for your answer. It naturally raises the question:
When do we interpret two states as different states “of the
same object” (which is not a directly observable relation, to
my understanding)? I guess part of the answer is: If only
the spacetime-related eigenvalues differ between the states.
Maybe there is also a pragmatic part: When it makes
obvious sense to organize the states like that. – But then
this might head towards the discussion you do not want to
enter here.
Anyhow, the difference you point out between the
electron’s “cloud” and the “cloud” around the nucleus is an
interesting example and I think it will make a good test of
my understanding if I try to state this difference in precise
terms for myself (as soon as I’m up to that).

Matt Strassler | June 3, 2013 at 5:07 PM |
To some extent, interpreting two states of nature as
different states of the same object has to do with how we
think about that object’s properties. Why do we view the
ground state of Hydrogen and its first excited state as the
same object in two different states? Because the excited
states readily decay to the ground state emitting nothing but
a photon, which is a particle that we know is associated
with the same force that holds Hydrogen together. A
similar statement applies for the Delta, which decays back
to a proton by emitting a pion. But a muon decays to a
neutrino, an antineutrino and an electron, via a short-range
force that we know isn’t holding the electron together.
But I have a feeling there’s a better answer, and maybe I
can come up with it, if I think harder about it.

Oleg | June 26, 2014 at 5:36 PM |
Apologies for the following: replying at the wrong
conversation level, replying to a year-old thread, digging at
something that you probably can’t/don’t want to answer
without equations.
Concerning the question further down about a muon being
an excited form of an electron: could a particle (specific
concurrent ripples in many fields) be excited in more that
one of the fields that it interacts with? Could the muon be
an electron that carries an excited waveform in another
field that the electron normally interacts with, instead of the
electromagnetic field? Specifically, could the muon be an
excitation in the weak field? Note that the other particles
involved in the decay – neutrinos – interact with the weak
force, and that the ‘weak field’ is related to EM field
locally.
Thank you so much for being so involved with comments
and such a fountain of knowledge.
22.
veeramohan | June 2, 2013 at 3:35 PM | Reply
Professor, an excellent and informative article as always, with awesome neatness.
If strong nuclear force is really separated (splitted) will release incredible strength – more
than residual strong nuclear force ? (because in long distance it become more than 1 – fig
2).
We can interpret this, “there is already a Grand unification at very short distance much,
much lesser than 2x 10^-18 meters?” – there is already high energy there ?
The Higgs force was the force of Grand unification, which encapsulated, the quanta or
quantum of action ?
This “quanta” breaks the continuity (apeiron) at nuclear distances – like the appearance
of neutron and proton.
Then, Phenomenologically we say, this “quanta (mathematical)” as mass ?
The high energies created in LHC, will only break the Asymmetries and release the force
?
o
Matt Strassler | June 2, 2013 at 5:02 PM | Reply
The amount of energy you can get out of a system is not simply related to the
strength of the force that holds it together. It’s much more complex than that. So
the answer to your first question is no; it turns out you can’t split protons or
neutrons. Long story.
There’s no energy in a force, per se. If I have two objects held together at very
short distance, the energies involved in keeping them there will be very large.
That’s partly just because the distances are short, not just because the forces are
“strong” or “weak” in the physicists sense. Meanwhile, this does not mean I can
obtain large amounts of energy from such a source. I may have to put *in* lots of
energy, rather than drawing *out* lots of energy, in order get these objects to stay
together, or to pull them apart.
The Higgs force has nothing to do with grand unifaction; it isn’t part of the
unification force, and it doesn’t have any role in Grand Unification. In old-style
grand unification, there is another, Higgs-like field which plays a role, but it isn’t
the Higgs field that we’re talking about at the LHC. And there may be no such
field in newer versions of unification.
Energy, at the LHC or elsewhere, cannot “release a force”. And the LHC cannot
probe grand unification anyway; that would require a machine with collisions a
million million times more powerful.

veeramohan | June 3, 2013 at 3:37 AM | Reply
Thank you Professor,
it will take lifetime to get intuitively, the meanings of mass, force and
energy.
23.
jal | June 3, 2013 at 1:56 AM | Reply
Unanswered questions.
1. Where does the energy come from to create the disturbance in space that result in a
sphere of strong force, (proton)?
Why does this sphere of energy appear to be permanent/stable. It does not appear to need
a continuous input of energy to maintain itself. It does not appear to be “leaking any
energy”.
Does the existence of a proton violate Conservation of energy? It does appear to be a
perpetual machine.
o
Kudzu | June 3, 2013 at 3:00 AM | Reply
1.) Like most of the particles that make up the universe protons were
manufactured in the Big Bang, a hot, dense high-energy state. While not as simple
as just pulling protons from nothing it’s a quite intuitive explanation. If you want
to ask where the energy of the Big Bang came from things get more complicated.
(And head towards ‘Why is there something instead of nothing?’)
2.) The proton is stable in the same way a pile of ashes isn’t flammable. Anything
and everything that exists has energy (Including mass as energy.) Roughly, things
will only decay if they can break down into products that contain ‘less energy’*.
(A neutron decays quickly into a proton, electron and anti-neutrino. In fact nearly
all known particles decay in the blink of an eye.) The proton has energy, yes, but
it cannot have or break up into anything that has LESS energy. It is the ashes of a
fire that has burned out, a car without gas.
* ‘less energy’ is not accurate but a nice way of thinking about things.
3.) To violate the conservation of energy a machine of any kind must either
destroy or make energy from nothing. ‘Perpetual motion machines’ are not
impossible, an atom will have all its parts moving for an indefinite amount of
time. What IS impossible is a machine that gives you more energy out then you
put in. As protons are the lowest energy state they can be, as they cannot ‘leak
energy’ there is no way to make them produce energy from nothing.
24.
veeramohan | June 3, 2013 at 3:48 AM | Reply
Thank you Mr Kudzu.
25. Pingback: Courses, Forces, and (w)Einstein | Of Particular Significance
26.
jal | June 3, 2013 at 11:30 AM | Reply
I’ve read
http://en.wikipedia.org/wiki/Particle_in_a_box
and
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator
If the source of the energy is turned off then the wave structure disappear.
The wave structures in a box requires a box.
A continuous energy input is required to maintain the wave structures.
I’m still trying to get an understanding/explanation for the proton.
o
Matt Strassler | June 3, 2013 at 4:57 PM | Reply
I’m afraid (for reasons I can’t determine from your comment) that you’ve fully
misunderstood what you’ve read. You said “If the source of the energy is turned
off then the wave structure disappear.” That’s entirely false. You also wrote “The
wave structures in a box requires a box.” That is true, but has nothing to do with a
source of energy. The box neither adds nor removes any energy. So unfortunately
your statement “A continuous energy input is required to maintain the wave
structures” is completely wrong; NO energy input is required.
In the same way, no energy input is required to maintain the proton once it is
created. Indeed, the world would not have any protons in it, were continuous
energy input required. Protons are not like engines; they do not take energy in and
send exhaust back out. Once created, they will stay as they are, unless they are hit
hard by something else.
So I encourage you to go back and figure out how you misunderstood what you
read in Wikipedia. Of course it is possible that wikipedia has an error on this
point, but more likely you misunderstood it.

jal | June 4, 2013 at 11:38 AM | Reply
This article, “The Strengths of the Known Forces”, I hope has resulted in
achieving your objective.
Almost every comment has a version of
“I have a question”
and
“Thank You”
Put me on that list for encouraging me to continue searching for
knowledge.
Your answer,
“In the same way, no energy input is required to maintain the proton once
it is created. Indeed, the world would not have any protons in it, were
continuous energy input required. Protons are not like engines; they do not
take energy in and send exhaust back out.
Once created, they will stay as they are, unless they are hit hard by
something else.”
That is why I’m still trying to get an understanding/explanation for the
proton.
27.
Euan Tracey | June 3, 2013 at 12:00 PM | Reply
Hi, great article! Possible typo: shouldn’t the interaction energy of positronium be -mc² ×
α²/4 rather than -mc² × α²/2 ? (-2 x E.K.). On a related note, shouldn’t the binding energy
be negative too?
o
Matt Strassler | June 19, 2013 at 10:51 AM | Reply
The total kinetic energy is that of the electron AND that of the positron, so that
explains the factor of 2. For hydrogen, all the kinetic energy is essentially that of
just the electron.
The sign of the binding energy (B) is a matter of convention. I choose to state it as
a positive quantity, but that just means that the mass of the state is 2 m_e – B, not
+ B.
28.
veeramohan | June 10, 2013 at 10:58 AM | Reply
α = ke2/ ℏ c = 0.007 297 352 57. So, ke^2 = mass? – Here, mass of a partice also defined
by charge.
Because α is one of the most precisely measured quantities in nature, the perturbative
nature of eigenstates energy level differences in atom create a parallax of apeiron and
perion (quantas)- thus give mathematical authenticity? So the calculation of angular
momentum was possible using pythagorean axiom.
Because particles travel slower than light the Relativistic quantum mechanics was also
consistent.
This discreteness become not doable, when α reaching closer to 1. Eventhough
mathematics of relativity can solve the problems of particles reaching the speed of light,
the shrinking 1/α² heartbeats, will again make parallax between virtual particles and real
particles – thus making mathematics of Algebraic geometry possible – even at the
absence of apeiron and perion and axiom – duplicating consciousness ?
The amplitude that a virtual particle exists interferes with the amplitude for its nonexistence, whereas for an actual particle the cases of existence and non-existence cease to
be coherent with each other and do not interfere any more. So Virtual particles are
excitations of the underlying fields, but also duplicate measurable particles. Thus making
numerical production of concrete geometry of space possible – like hologram or quantum
3D print ?
29.
Steve | June 24, 2013 at 9:31 PM | Reply
Prof. Strassler, I hope I’m not too late to ask a question on this thread (I just discovered
your blog and I am thrilled that you are trying hard to convey these subtle concepts to the
public). My question: I’ve always had trouble making sense of the metaphor that forces
result from the exchange of virtual particles. There is an asymmetry here: forces can be
attractive or repulsive, but the “exchange” metaphor does not seem to accommodate this
asymmetry. Other commenters have asked this question, in somewhat different words,
but you punted. I’m hoping you can give us some intuitive understanding without feeling
you have to write a whole article.
30.
Steve | June 24, 2013 at 9:43 PM | Reply
Prof. Strassler, I apologize – you answered my question quite nicely in your “virtual
particles” post. Thanks a bunch. You have cleared up a mystery that has afflicted me for
many years!
31.
charge vs. charge | July 9, 2013 at 12:36 AM | Reply
How can you compare the force strengths with “which amount of charge”? For “electric
charge” the amount of elementary charge could be used for a typical(the smallest)?
amount of charge that an elementary particle(like an electron/positron etc.) possess but
what about gravity? and other forces(weak & strong)? “mass(=m)” in gravity in
(F=Gm^2 /r^2) looks like “e” in (F = ke2 / r^2) but m is not quantized(?) nor does it have
the smallest quantum(chunk)(?) Then, how can you compare the two? if “m” does not
come in a “chunk(quantum)/ or the smallest unit”, it seems impossible to compare the
two forces. Or all the charges (electric charge, color charge, and mass etc) are quantized
and comes in the smallest indivisible chunk/unit? Electric charge seems to have the
smallest amount/unit (an elementary charge or 1/3 of it for a quark?) but “mass” has the
smallest unit/quantum? What is the evidence? Or, they(physicists) just assume that it
comes in the smallest unit like the Planck mass, and calculate the relative strength of
force created by “e” and “m”? but the Planck mass is not the smallest possible mass
unit/quantum/chunk? So, why on earth the Planck mass appears in this article? And, are
“charges” for weak and strong force also come in the smallest chunk like “the elementary
(electric) charge”?
32.
Kudzu | July 9, 2013 at 3:19 AM | Reply
At small enough scales you reach a point where you can no longer meaningfully measure
something, be it mass or energy, time or even space. This gives us a certain limit to the
smallest mass, etc. (While not ‘clean’ like the fundamental unit of electric charge there
exists a certain amount of mass which, if two masses differ by less than it, we cannot
measure that difference in any meaningful way.)
But even if gravitational ‘charge’ were quantized how would we know that one
gravitational charge ‘equaled’ one electromagnetic one? It is like comparing apples and
oranges.
However in this case the situation is more prosaic; we are simply talking of the properties
of any given elementary particle. We can consider an electron or a proton as examples.
There is some variation in say mass-to-charge ratios but not significant enough to affect
our arguments here. In fact no elementary particle exists that would substantially alter the
arguments given here. It is possible to imagine a system (such as two charged black holes
orbiting each other.) where say the electric force begins to act differently but this is little
more than a side note to the arguments presented here.
33.
charge vs. charge | July 9, 2013 at 9:57 AM | Reply
“we are simply talking of the properties of any given elementary particle. We can
consider an electron or a proton as examples.” -> ok, so this means, just compare the
forces btw an electron and a proton (or electron and positron or any combination of two
particles) for example, and calculate the force by “e” and the force by “m” and compare
the strength of the two? Is that all this article is talking about? is that simple (that is all?)?
but, then, this means a coupling constant does not have a single value (assuming the
r(distance) is fixed)? because a coupling constant depends on the amount of charge like
“e” and “m.” I do not see where/how the each alpha=each coupling constant is coming
from(calculated) except “electric charge.” in this article.
o
Kudzu | July 9, 2013 at 7:01 PM | Reply
Some very good points. You have made me realize that your question is more
complex than it seemed and that a lot of my previous post was in error and
focused on the wrong ideas. Disregard it. (If you see this professor please delete
my previous comment.) Starting afresh:
The first thing to note is that ‘charge’ is simply a measure of how a field changes
around a certain point. The electric charge of an electron causes the
electromagnetic field’s value at various points from that charge to change in a
predictable manner. In itself this doesn’t mean much.
A force arises because there is an energy involved in this change of a field. When
talking about strength in the ‘usual’ manner this depends both on the rate of
change of the field per distance unit. That depends on both the ‘charges’ involved
and the number of distance units they are separated by. For the electromagnetic
force this is ke^2/r
What is being spoken of here however is a strength independent of distance and
charge. To find this we must remove both of those things. Charge can be removed
by setting e (Or m, etc) in ke^2 to the same value of charge used to calculate k (Or
G, etc.) This makes k and e related, changing the value of e used to calculate k
changes k. ke^2 thus becomes independent of ‘charge’ for the force.
By removing r and instead dividing by ℏc we remove the dependence on distance
and instead relate the force to a fundamental limit of the universe, the speed of
light. More importantly it relates the energies of the force to the energies at which
a number of effects (such as the creation and particle-antiparticle pairs) become
important.
A weak force is one that it is very difficult to create a system where these effects
are important while a strong force is one where such systems form easily. Under
this arrangement and definition the magnitude of the ‘charges’ involved are just a
variable that can be changed. Electromagnetism is weak because it is very
difficult indeed to create system where things like particle-antiparticle pairs are
important. (It could in theory be done with very high charges and/or short
distances, but it’s not easy.) The strong force is strong because it is very easy
indeed to create such a system, all protons are one.
There is probably a better and more accurate way to word this; your question
made me realize I still do not entirely understand everything laid out here. I hope
this helps though.
34.
charge vs. charge | July 10, 2013 at 5:29 AM | Reply
F r² = 0.007 ℏ c = ke^2 -> I got “approximately 0.007(297… etc.)”(=ratio of (Fr^2)/(ℏ c
), =unitless) by setting “e” as the value of elementary charge (=1.60217657… × 10-19
coulombs) but if the amount of “an elementary charge” is not used for “e”, (Fr^2)/(ℏ c )
is not 0.007(=the coupling constant). To get “a fixed value(=constant value)” of
“Fr^2″=”ke^2=(*1)” or (Fr^2)/(ℏ c )=(a coupling constant), you need to have a fixed
chunk(unit) of “a charge(of e, m, color charge etc.)” Is this not the case? It seems that to
compare any two fundamental forces by the values given by “Fr²”, or “(Fr^2)/(ℏ c )” you
need to have some kind of a “chunk(quantum)” in “charge(e, m, color etc.)” given by
nature or you just have to use the “m” value and the “e” value for existing/observed
(elementary) particles. So, the value of the coupling constant 0.007.. is calculated for two
point-like particles with “an elementary charge” for each? but any other two (point-like,
elementary) particles with greater or less “electric charge” than “an elementary charge”
has a coupling constant (for electric charge) of other values? For example, two “quarks”
can have either -1/3 or +2/3 electric charge, (−2/3 or +1/3 for anti-quarks), a coupling
constant (for e charge) for two quarks must be different from 0.007..? And what about a
charged object/system that acts like a (point-like?) particle such as a charged black hole?
If two charged black holes (with different amount of multiples of elementary charge) can
act like a point-like particles, they should have yet another a coupling constant (for
electric charge/force)?
btw,
unlike Fr² or ke^2 or (Fr^2)/(ℏ c ), Coulomb’s constant(k) or Gravitational constant(G)
seem to stay constant regardless of charge amount (“e” or “m”) or r(=distance)
numerically(=meaning if considered only within these equations.) but because they have
different units, it seems impossible to compare k and G. (does not make sense..? or? there
is more to them if considered beyond these equations that take the form of
F=k(“charge”^2)/(r^2)? or what?)
1)=this value seems to stay more or less constant regardless of r if “shielding effect”(or
the effect of virtual particle pair production at very close distance etc.) is ignored. I have
to check/calculate how exactly “shielding effect” etc. affects the force strength or
coupling constant though.. (and I am not sure how to do this precisely.)
o
Kudzu | July 10, 2013 at 4:35 PM | Reply
As you notice that e is NOT the smallest (and thus elementary) charge; down type
quarks have 1/3 this charge and up type quarks 2/3 of it. There’s no reason at all
that some exotic physics might throw up a particle with a tenth of this charge or
any amount of it at all. The electron’s charge is a useful unit but there’s nothing
‘special’ about it.
When we calculate α the first step is to multiply k * e * e. What we are doing is
setting ‘e’ as 1 in a sense so that 1 x 1 = 1 and it (charge) can be ignored. Now k
has units of Nm^2/C^2 and it’s the C^2 that’s important. e appears in it in that 1 C
= 6.241 × 10^18 e.
Imagine now that we are going to use 2e as out ‘1’ If we just plug in 2 into the
equation we will get an ‘α’ that is four times larger. But we didn’t set 2e as 1! our
answer is in fact 4α. In order to set 2e as 1 we must also alter k since k had 1e in it
hidden away. To do this the ‘C’ becomes half as large (since there is half as man
2e in 1C as there is e’s.) Since k has units of /C^2 it becomes four times smaller.
1/4 * 4 = 1 and α remains the same!
The same is true of all our αs for various forces, their k has in it somewhere *the
same unit we are setting as ‘1’* when working out α. The very fact that α must be
independent of charge (It cannot even be considered ‘per charge’) means you
have to pay attention not just to the units of e (or m, etc) but also those of k. k, G,
etc are constant ONLY when considered as ‘per’ [PER Coulomb, PER kilo].)
α is unitless and to make it so you must carefully cancel between the units of your
charge and the units of k (Or G, etc.)
35. Pingback: Final Day of SEARCH 2013 | Of Particular Significance
36.
dratman | September 2, 2013 at 12:26 AM | Reply
This is a wonderful article! I feel like I’m glimpsing a tiny bit of something an ordinary
mortal might someday understand. Thank you.
37. Pingback: Key mission of life | The Great Vindications
38. Pingback: Quantum Field Theory, String Theory and Predictions (Part 3) | Of Particular
Significance
39. Pingback: Quantum Field Theory, String Theory and Predictions (Part 4) | Of Particular
Significance
40.
warren d smith | October 10, 2013 at 9:45 PM | Reply
Where can we find the most precise and up to date measured values of the numbers you
get from nowhere such as 0.11 and 0.007? (Where did you get them?)
o
colin | March 21, 2014 at 11:49 AM | Reply
The electromagnetic fine structure constant is a very well-known fundamental
value published in lots of places. Here, for instance: http://physics.nist.gov/cgibin/cuu/Value?alph|search_for=abbr_in!
The strong force version is a function of radius not a simple constant, and so it’s
not documented as widely. Do particle physicists have a standard handbook of
these things, or is it a question of tracking down the right textbook?

Matt Strassler | March 21, 2014 at 12:14 PM | Reply
Here is a plot showing the strength of the strong nuclear force as a
function of Q, where Q is an energy scale.
http://inspirehep.net/record/896215/files/asq-2009.png To convert Q in
GeV to a distance scale L, do the following: the distance L is (h-bar c )/Q ,
where h-bar is Planck’s constant divided by 2 pi, and c is the speed of
light.
41.
Philip Maurone | December 6, 2013 at 4:15 PM | Reply
Even though the weak and strong interactions vary greatly over very small distances, is
there a fundamental source strength term analogous to the source terms for E&M and
gravity that one could use for the definitions of the “alphas” for these interactions that
could be separated out from their small distance varibility?
42. Pingback: Quantum Field Theory, String Theory and Predictions (Part 8) | Of Particular
Significance
43.
Paul Papke | January 31, 2014 at 5:09 PM | Reply
I found this article looking for a comparions between gravitational and electromechnical
forces and found it fascinating reading. Would you have a book you could recommend
that I could read more of your writings?
44. Pingback: Attractive higgs force and inflation | Question and Answer
45.
Siva P Kodukula | January 5, 2015 at 10:33 PM | Reply
I like this article. It triggered my imagination about Grand Unification in a right way.
Thanks to the author
Leave a Reply