# Equation of everything

Is this equation in the image true? Can you give some topics that I can cover the equation?

Similar equation from http://www.preposterousuniverse.com:

• Can you provide a link to where you got the image. It looks to me like a probability derived from a path integral, but the left hand side would not be a wavefunction as the picture implies. – John Rennie Sep 17 '13 at 16:58
• – John Rennie Sep 17 '13 at 17:02
• Interesting. There are a few terms missing like the cosmological constant and some other omissions and inaccuracies. This is an oversimplification. One does not simply shove the Einstein Hilbert action into a path integral. – dj_mummy Sep 17 '13 at 17:05
• No measure of integration (both outer and in exp). And I think that adding gravity just like that inside functional integral is incorrect. – user23660 Sep 18 '13 at 3:00
• @Lacek The $i$ in the Dirac lagrangian is just fine. Recall that the momentum operator is $i \partial_\mu$, so the kinetic term is Hermitian (up to a total derivative) as it should be. – Michael Brown Sep 18 '13 at 4:03

The equation is not literally correct. The single terms labeled Maxwell-Yang-Mills, Dirac, and Yukawa, are standing in for whole families of terms from the standard model lagrangian, a version of which you can see on page 1 here.

The "F^2" term, which comes from electrodynamics, should really be more like "trace(G^2) + trace(W^2) + B^2", where G is for gluons, and W and B are for the weak and hypercharge gauge fields before symmetry is broken by the Higgs. (The paper contains a further "trace(G G~)", a "theta term" which ought to exist according to the rules of lagrangian construction, but whose coefficient appears to be very close to zero in the real world.)

There is a single Dirac term in the picture, but in the paper I cite, you will see an analogous term for each of Q (left-handed quarks), U (right-handed up-type quarks), D (right-handed down-type quarks), L (left-handed leptons), E (right-handed charged leptons). In Q and L, the left-handed quarks and the left-handed leptons of each generation are treated as a single object that interacts with the weak bosons, whereas the right-handed quarks (e.g. up and down) and right-handed leptons (e.g. electron and electron-neutrino) are regarded as separate.

[For clarity I will emphasize that, in the standard model, there is a left-handed part and a right-handed part to each fermion, except for the neutrinos, which in the original standard model are purely left-handed. (That would make the neutrinos massless, so this aspect of the SM needs to be extended but we don't know exactly how.) So e.g. up quark has a left-handed part and a right-handed part, but the left-handed part of the up quark is bundled with the left-handed part of the down quark, into the combined left-handed quark field Q, whereas the right-handed parts remain alone as U and D.]

Also, there is a subscript i that runs from 1 to 3, because there is a Q,U,D,L,E in each particle generation. Also, the original Dirac term is just a kinetic term for the fermion, but the D-with-a-slash means that the so-called "long derivative" or "covariant derivative" is being used, which also includes the interactions of the fermions with the G,W,B fields.

Similarly, the single Yukawa term in the picture really stands for the three interaction terms QUH, QDH, LEH (as they are written in the paper), and the "lambda" coefficient actually stands for a 3x3 matrix of "yukawa couplings", which form the link between the left-handed and right-handed fermions (coupling Q to U and Q to D, and also L to E), thereby generating the masses and the weak-force mixings (the latter, because the Higgs field has an electroweak charge).

R in the picture is the Ricci scalar, which appears in the Einstein-Hilbert action for general relativity. So that part introduces gravity. The final two terms are the kinetic and potential terms for the Higgs field.

It's wrong to add a classical action to a quantum one, since the two are treated differently in the physics. The Standard Model lagrangian has its variation set to zero, not calculated for its kernel. It's also kind of silly, since we know what each of the component Lagrangians are about, but their sum (even if it were right) isn't exactly an elegant unification of the theories. To take a famous Feynman example -- you could as well say the "equation of everything" is $U=0$, where $U$ is the "unphysicality", containing terms like $(F-Gm_1m_2/r^2)^2+...$, but it won't tell us anything interesting about the physics. It's not exactly useful just to declare something an "equation of everything" if it doesn't make new predictions.

Some typos/minor errors --

• The left-hand-side is the Kernel, not a wavefunction. Even otherwise, it's odd to label the wavefunction as "Schrodinger" -- this seems more like a silly attempt to insert the names of a bunch of physicists to look unifying-ish, instead of bothering with the actual physics. Same with the ridiculous labelling of the constants, like Newton's and Planck's.
• Infinitesimals in the integral are missing, like $\mathcal{D}[...]$ and $dx^4$.
• Where's the Cosmological Constant? Also, only one component of the Standard Model Lagrangian is included, its complex conjugate needs to be added too.
• You don't need empty space between the lines connecting a sigle term. You can write "semi-classical gravity", "Einstein-Hilbert action", etc. – Nikolaj-K Sep 18 '13 at 7:38
• Cosmological constant could be hiding inside $V(\phi)$. Also, obviously not hermitian but complex conjugate (and I would think this should be minor not major flaw). – user23660 Sep 18 '13 at 14:16
• What the heck is this targeted downvoting whenever I edit my old answers? How does this not count as serial downvoting? – Abhimanyu Pallavi Sudhir Jul 25 '18 at 19:01

It's not meant to be taken as a literal equation. This is an attempt to summarize everything we know in Physics. I have seen it in material by Neil Turok at Perimeter Institute, he refers to it as "All Known Physics".

## protected by Qmechanic♦Jan 4 '16 at 1:12

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