• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,079 questions , 2,229 unanswered
5,348 answers , 22,758 comments
1,470 users with positive rep
819 active unimported users
More ...

  How can there be a quantum field theory that predicts all particle masses?

+ 7 like - 0 dislike

Say I have a theory with only one (energy) scale, e.g. one given by the fundamental constants

$$\epsilon=\sqrt{\dfrac{\hbar c^5}{G}}.$$

In this case, where I can't compare to something else, is there a way to argue that

$$\epsilon<\epsilon^2<\epsilon^3<\dots\ ?$$

By that reasoning, can there be a (field?) theory, where values are obtained from some expansion like in a path integral (which needs a hierarchy of that sort)?

If you really only need/have a theory with $\hbar, c, G$, how can energies like particle masses be deduced from the theory (instead of being experimental input)? And then if, at best, the theory predicts some mass of a particle $\phi$ to be $m_\phi=a_\phi\dfrac{\epsilon\ }{\ c^2}$, then the number $a_\phi$ must have some geometrical meaning, right?

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user NiftyKitty95
asked May 27, 2012 in Theoretical Physics by NiftyKitty95 (95 points) [ no revision ]
I just want to note that, for the record, the Planck mass of $\epsilon / c^2$ is equal to $10^{22}$ times the mass of the electron. Obviously this is problematic since your proposal now requires $a_{\phi} \ll 1$. I feel better when my geometry factors are between 1 and 10.

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user AlanSE

4 Answers

+ 4 like - 0 dislike

You can't do it for real in quantum field theory, there are always adjustible parameters. The reason is that quantum field theory doesn't have a fundamental length, it is defined on the continuum, so it can always be rescaled. But if you have a quantum gravity theory that reduces to quantum field theory at energies less than some large energy, you can get exponentially small masses out without putting them in using coupling constant running. If you say that the coupling of a confining theory is something small (but not absurdly small) at the Planck scale, the mass of the confined particles is extremely small, and the confinement radius is enormous. This is how the proton mass is determined.

In principle, you could have one confining theory generate condensates that then break other symmetries, and so on, and give masses to all particles using only such a mechanism. This is called "technicolor", and it would predict the masses of the low-lying particles from dimensionless coupling constants. But the scaling that produces a continuum means that the full relations between all constants can't be predicted without a theory of quantum gravity, which breaks the continuum limit in field theory.

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user Ron Maimon
answered May 27, 2012 by Ron Maimon (7,730 points) [ no revision ]
+ 3 like - 0 dislike

This is a very good question. I think there is no quantum field theory which predicts all particle masses. The following is my personal opinion.

Masses (measured in Planck unit) are real numbers. The real numbers are NOT predictable, just like the radius of the orbit of Earth moving around Sun (measured in Planck unit) is not predictable. So the real fundamental constants are NOT predictable, and have to be inputted into the theory by hand.

However, those unpredictable real quantities have a predictable property: they are time dependent (but may change very very slowly). This is also just like the average radius of the orbit of Earth moving around Sun. So masses (measured in Planck unit) are not predictable, but may be time dependent.

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user Xiao-Gang Wen

answered May 27, 2012 by Xiao-Gang Wen (3,485 points) [ revision history ]
edited Apr 5, 2014 by Xiao-Gang Wen
+ 2 like - 0 dislike

Theories don't predict units unless you put units in. A theory which predicts the masses of the fundamental particles would actually only predict the mass ratios $a_\phi$. Presumably they would emerge as eigenvalues of some operator, or perhaps as the zeros of some complicated function.

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user David Z
answered May 27, 2012 by David Z (660 points) [ no revision ]
+ 0 like - 0 dislike

Here is my take on it from "history" point of view:

  • With $c$ we understood that there is "no difference" between space and time and we started to use same units to measure them. Same thing for energy and momentum, magnetic and electric fields, e.t.c.

  • Then $\hbar$ appeared and we realized that we could measure energies, momenta, distances and time intervals with same units (like $GeV$s). That the inly fundamental unit we have, right? All other units are just introduced for convenience and can be reduced to $GeV$s.

  • Now we want to include $G$ into the picture. But that would mean that we will leave the last unit we had -- we will measure all the energies in Planck units. Meaning that we will work just with "pure" dimensionless numbers.

So your theory with only $c,\hbar,G$ parameters should be essentially a pure mathematical construction, that gives you dimensionless numbers "in Planck units".

These numbers can:

  1. have some geometric nature (as you said),

  2. be solutions or eigenvalues of some functions or operators (as David Zaslavsky said),

  3. be just "accidental" numbers like Earth radius and even drift with time (as Xiao-Gang Wen said)

Or some combination (or none) of the above.

This post imported from StackExchange Physics at 2014-04-05 04:28 (UCT), posted by SE-user Kostya
answered Jul 5, 2012 by Kostya (320 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights