How can Higgs field explain proton's inertial resistance to acceleration?by johne1618 Tags: acceleration, explain, field, higgs, inertial, proton, resistance 

#1
Dec1911, 11:16 AM

P: 355

If most of the mass/energy of a proton is due to the kinetic energy of its quarks and gluons, rather than interaction with the Higgs field, then how can we explain its inertial mass, i.e. its resistance to acceleration, as being due to a drag induced by the Higgs field?
Alternatively imagine a body made up of particles whose mass/energy is provided by the Higgs field. Now let us spin that body very fast. Its mass/energy will increase due to relativistic mass increase of the moving particles. We know that its inertial mass will increase  it will be harder to accelerate the whole body linearly with a given force. But the Higgs field is Lorentz invariant so that if the inertial resistance force is due to the Higgs field then it shouldn't be any harder to accelerate the body whether it is spinning or not. 



#2
Dec1911, 09:26 PM

P: 270





#3
Dec2011, 03:24 AM

P: 355





#4
Dec2011, 03:43 AM

P: 2,892

How can Higgs field explain proton's inertial resistance to acceleration? 



#5
Dec2011, 04:32 AM

P: 640

It is not scientifically justifiable to introduce more and more dynamic degrees of freedom in that way if there is no indication or need for such a construction from an experimentally point of view. On the other hand. there was/is a scientifically justifiable reason for introduction of the Higgs mechanism/potential, because gauge theories seemed fine in most respects apart from not at first being able to explain the origin of the masses of fundamental particles. 



#6
Dec2011, 04:35 AM

P: 748





#7
Dec2011, 04:37 AM

P: 640

I have no idea what kind of accuracy you get if you attempt a calculation like this. If you start from experimentall observed masses of baryons, instead of quarks and gluons, you'll get very good results, though. 



#8
Dec2011, 04:48 AM

P: 748

One reason this is a nice topic, is that the mass of the proton should be much the same, even if up and down quarks are massless! So clearly the Higgs field doesn't "explain proton's inertial resistance to acceleration". You could have QCD with two massless flavors of quark, coupled to electromagnetism and gravity, and the behavior of the proton would be the same.
So a coherent "philosophy of inertia and acceleration" has to be able to deal with at least two cases  situations without a Higgs field at all, like the one I just described, and also situations where the Higgs field is at work  e.g. instead of a proton, we have an electron which does get its mass from the Higgs mechanism. I am used to just saying that gravity couples universally to energymomentum, but it should be enlightening to dig into the details of the two cases. 



#9
Dec2011, 05:51 AM

P: 2,892

So according to your philosophy no money should be spent in the LHC,etc no? 



#10
Dec2011, 06:03 AM

P: 640

So I want to think about the origin of the Higgs mass when experiments determine that the current Higgs mechanism doesn't perfectly conform to experiments. Of course, everybody are free to think and research whatever they want. That's just as important as the scientific method itself... :) 



#11
Dec2011, 06:16 AM

P: 2,892

I was not referring to "existence" in that deep sense, I asked about the mechanism that allows the Higgs particle to acquire his own mass, just hinting at the problem the very Higgs mechanism can trigger while apparently resolving , say, the origin of the mass of the electron. 



#12
Dec2011, 06:26 AM

P: 640

I guess my original point was only the somewhat naive observation that IF the current theoretical Higgs mechanism turns out to be fine from a theoretical and experimental standpoint, then there is no need for another "underlying explanation", since the mass of the Higgs field would simply be explained by the fact that its potential has a certain quadratic term. 



#13
Dec2011, 07:45 AM

P: 2,892





#14
Dec2011, 08:14 AM

P: 640

[tex] L = \frac1{2}(\partial \phi)^2  \frac1{2}m^2\phi^2 [/tex] The Higgs mechanism is just a field that is a Lorentzscalar (+ some nontrivial gaugegroup properties which I skip here) which has a Lagrangian function similar to (a bit simplified to avoid unnecessary details): [tex] L = \frac1{2}(\partial \phi)^2  V(\phi) [/tex] where [itex]V(\phi)[/itex] has a continuum of global minima, away from [itex]\phi=0[/itex]. The standard Higgsmechanism uses something like: [tex] V(\phi) = \mu^2\phi^2 + \lambda\phi^4 [/tex] Therefore, at low energy, the Higgs field [itex]\phi[/itex] settles into a nonzero global minimum [itex]\phi_0[/itex] which gives it a nonzero vacuum expectation value (VEV). Around this minimum, the potential has the form: [tex] V(h) = a + \frac1{2}m^2h^2 + ch^3 + dh^4 [/tex] where [itex]h := \phi  \phi_0[/itex], and where the Higgs mass [itex]m[/itex] is determined from the two parameters [itex]\mu,\lambda[/itex] that define the Higgs potential. So the introduction of the Higgs potential defines the mass of the small fluctuations around the potential mininum, which is by definition the Higgs mass. As a reference with more details: http://en.wikipedia.org/wiki/Higgs_mechanism 



#15
Dec2011, 01:32 PM

P: 2,892

Thanks, it must be said the Higgs potential and that quadratic term has its own problems, like the hierarchy problem and vacuum related instabilities depending on the Higgs mass that are not solved at all as of now. But anyway this only shows that the Higgs mas is introduced "by hand" in that potential and doesn't explain what I was referring to, wich is an "underlying" problem: how does a Lorentz invariant field gives itself an invariant mass? How can something that is out to break a symmetry respect itself that symmetry?




#16
Dec2011, 02:38 PM

P: 640

Yes there are some theoretical problems related to renormalisation, and that's not a surprise considering the weak mathematical grounding for the whole field of quantum gauge theories. As far as I know, the perturbation expansion that is used even in quite simple QFTs doesn't even converge. 


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