If there is a non-zero expectation value for the Higgs boson even in "vacuum", since the Higgs boson

Kolten Conrad 2022-07-14 Answered
If there is a non-zero expectation value for the Higgs boson even in "vacuum", since the Higgs boson has a mass unlike photons, then I would expect it to have a rest frame.
So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Mateo Carson
Answered 2022-07-15 Author has 15 answers
The potential energy energy for the Higgs field looks like
V ( h ) = m H 2 2 ( h v ) 2 + O ( ( h v ) 3 )
for h near v, so it is minimized by h ( x , y , z , t ) = v. Note that V ( h ) is a Lorentz scalar so when added to the Lagrangian, it preserves the Lorentz symmetry.
The nonzero value of h at each point is what breaks the electroweak symmetry because h is a component of a doublet whose only symmetry-invariant point is h = 0. But because this configuration is constant, it doesn't pick any preferred reference frame. There are no physical particles in the vacuum state.
Physical Higgs bosons may be added and they're energy quanta that allow h to deviate away from h = v - more than the uncertainty principle requires. Those quanta carry energy and momentum ( E , p ) that satisfies E 2 p 2 = m H 2 . That's the energy one has to add to the energy of the vacuum state; the latter state is free of particles and its energy may be chosen to be zero.
When it comes to the difference between the vacuum, which has no particles, and states with particles (Higgs bosons), you may define a new field, h = h v, and this field will be expanded around zero just like the electromagnetic fields or other fields. The vacuum will be at h = 0 and it may perhaps make it easier to understand why there's no Lorentz-symmetry breaking in the vacuum. The S U ( 2 ) × U ( 1 ) symmetry will act "non-linearly" on h - in a way that can be easily deduced from h = h v.

We have step-by-step solutions for your answer!

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-05-18
Environmentally induced decoherence makes wave function collapse unnecessary. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) So, doesn't then the decoherence time depend on the motion of the prepared state relative to the environment? And, doesn't the ultimate environment, all particles in the universe, introduce a preferred frame into quantum mechanics in the sense that the decoherence time is relative to this frame? And would this be measureable, at least in principle
asked 2022-08-24
Is there any way in which a bound state could consist only of massless particles? If yes, would this "atom" of massless particles travel on a light-like trajectory, or would the interaction energy cause it to travel on a time-like trajectory?
asked 2022-08-11
Manipulating an n × n metric where n is often > 4, depending on the model. The 00 component is always τ*constant, as in the Minkowski metric, but the signs on all components might be + or , depending on the model. Can I call this metric a Minkowski metric? Or what should I call it?
asked 2022-09-16
Since frequency of a wave is a function of time, then for a particular ray of the wave, why would the frequency remain the same when observed from a moving reference frame? The frequency should change according to time dilation. No?
asked 2022-09-25
consider a particle, A receiving energy from a second one,particle B in a one dimensional collision.
E 2 = p 2 + m 0 2
E d E = p d p
For particle A:
E A d E A = p A d p A ( 1 )
For Particle B:
E B d E B = p B d p B ( 2 )
d E A ∣=∣ d E B ( 3 )
From Conservation of linear momentum we have:
p A + p B = K
where k is a constant vector. Now,
d p A + d p B = 0
d p B = d p A
d p B ∣=∣ d p A ( 4 )
Applying relations ( 3 ) and ( 4 ) to ( 1 ) and ( 2 ) we have:
E A E B = p A p B ( 5 )
A pair of particles cannot interact unless relation ( 5 ) is satisfied. Can we conclude that that relation ( 5 ) to be a restriction for 1 D collisions?
Now let's move to the general type of 3d collisions between a pair of particles A and B.
A frame is chosen where the particle B is initially at rest in it.
E B d E B = p B d p B ( 6 )
If the particle B is initially at rest the RHS of ( 6 ) is zero. But the LHS cannot be zero unless d E B = 0.
How does one get round this problem?
asked 2022-05-19
Is there a time + two spatial dimension representation of a Minkowski-space surface which could be constructed within our own (assumed Euclidean) 3D space such that geometric movement within the surface would intuitively demonstrate the “strange" effects of the Lorentz transformation (length contraction, time dilation)?
asked 2022-09-04
Let's say a body with m=2kg falls from 100 meters. Obviously it's speed would be far lower than the speed of light so the change in mass (if it exists) would be very tiny. However, if the speed increases, its mass would increase too. That's because its kinetic energy would become bigger. On the other hand, its potential energy would decrease in the same amount that KE has increased.
Does this suggest that either the mass is not going to change (due to the conservation of energy) or it would become slightly bigger, but never less?

New questions

1) The parallel axis theorem provides a useful way to calculate the moment of inertia I about an arbitrary axis. The theorem states that I = I c m + M h 2 , where Icm is the moment of inertia of the object relative to an axis that passes through the center of mass and is parallel to the axis of interest, M is the total mass of the object, and h is the perpendicular distance between the two axes. Use this theorem and information to determine the moment of inertia ( k g m 2 ) of a solid cylinder of mass M = 2.50 kg and radius R = 2.50 m relative to an axis that lies on the surface of the cylinder and is perpendicular to the circular ends.
2) A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provide a means for storing energy in the form of rotational kinetic energy and are being considered as a possible alternative to batteries in electric cars. The gasoline burned in a 231-mile trip in a typical midsize car produces about 2.97 × 109 J of energy. How fast would a 45.3-kg flywheel with a radius of 0.209 m have to rotate to store this much energy? Give your answer in rev/min.
3) Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.5 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?
4) A playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of 1.69 m, an initial angular speed of 3.22 rad/s, and a moment of inertia of 121 k g m 2 . The mass of the person is 45.0 kg. Find the final angular speed of the carousel after the person climbs aboard.
Ive tried figuring these out but i need help