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  Equation of everything

+ 3 like - 3 dislike
4214 views

enter image description here

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

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Raisa
Closed as per community consensus as the post is a popular science question.
asked Sep 17, 2013 in Closed Questions by Raisa (0 points) [ no revision ]
recategorized Jan 8, 2015 by dimension10
Most voted comments show all comments
No measure of integration (both outer and in exp). And I think that adding gravity just like that inside functional integral is incorrect.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user user23660
@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.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Michael Brown
@Lacek: Um... No, it's correct. That's how the Dirac field looks like.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Dimensio1n0
@Dimension10 - ok. thanks

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Lacek

This question itself looks rather like a popular science question, even though it has obtained two good technical answers. To prevent it from attracting additional low-quality answers, I think we should close it now and therefore vote to close.

Most recent comments show all comments
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.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user dj_mummy
I do not like the "i" in the third term in the exponent. It seems incompatible with the i/hbar in front of the integral. Some terms seem to be taken in real time, some in the imaginary time. Also it does not specify over which fields the variables are taken.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Lacek

2 Answers

+ 7 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Mitchell Porter
answered Sep 18, 2013 by Mitchell Porter (1,950 points) [ no revision ]
+ 3 like - 0 dislike

It's wrong.

  1. The Path integral is missing a $\mathcal D\left[...\right]$.

  2. The Action integral is missing a $\mbox{d}x^4$.

  3. Even if one incorporates (1) and (2), it would not be the wavefunction, but the kernel/matrix element.

  4. The wavefunction is not specific to "Schrodinger".

  5. It's meaningless to label the constants that way, to say, "Hey, look! It's even about Planck and Newton!".

  6. Einstein - Hilbert is a classical action.

  7. One may not add the Einstein - Hilbert Lagrangian Density, $kR$, to the Standard Model Lagrangian Density, the rest of the wreckage.

  8. The Standard Model Lagrangian Density is on flat spacetime, for some reason, rendering the equation even meaningless at semi - classical gravity.

  9. It's not an equation with an elegant, maybe simple, gauge group, so it's a manmade theory. It may be the field - theory limit of another theory, like string theory, but it cannot be the theory of everything.

  10. It's missing an $\operatorname{h.c}$, hermitian conjugate.

  11. No $\Lambda$, cosmological constant.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Dimensio1n0

answered Sep 18, 2013 by dimension10 (1,985 points) [ revision history ]
You don't need empty space between the lines connecting a sigle term. You can write "semi-classical gravity", "Einstein-Hilbert action", etc.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user NiftyKitty95
@NickKidman: I know; it's just out of habit.

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user Dimensio1n0
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).

This post imported from StackExchange Physics at 2014-03-24 03:31 (UCT), posted by SE-user user23660




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