Understanding Mass and Vectors in Quantum Mechanics: The Role of the Higgs Field

[kye]

Before measurements, quantum states are vectors in Hilbert space and there may not even be momentum or position chosen before wave function collapse or measurement. I'd like to understand how mass or the higgs field couple to wave function. Can higgs home into the specific ray or vector in Hilbert space or does the higgs field only couple to it after collapse of the wave function (or measurement). If it is prior, then higgs has capability to navigate the vectors in Hilbert space and find the specific basis? (perhaps explaining why there is mass in the atoms even without being measured?) I understand that in gauge theory, masses can't be forced directly into the equations of the equation or the symmetry is destroyed. So higgs field were proposed to give masses to the properties, but how they exactly interact in Hilbert configuration space.

 

[dauto]

The Higgs field doesn't couple to the wavefunction. It couples to the other fields - the electron fields for instance - giving the particles their masses. When applying Quantum Mechanics to field theory to obtain Quantum Field Theory, the fields play the role of the observable degrees of freedom (equivalent to the position and/or momenta in QM). In other words, the fields are operators, they are not states. Operators don't collapse - states do.

 

[kye]

If so, why does QM has mass if the higgs field can't couple to the wave function.. unless QM is not complete model to integrate mass and one needs quantum field theory even for a single particle in double slit experiment? If so, then since the observable is not position but fields, then how do you treat a one body problem and mass in ordinary qm and its relationship to the double slit experiment for instance?
So before the collapse of the wave function (or related mechanism like decoherence),

 

[strangerep]

If so, why does QM has mass [...]

You're confusing a general mathematical framework, i.e., quantum theory, with its application to a particular physical scenario (e.g., Newtonian dynamics obeying Galilean symmetry, or Special Relativity obeying Poincare symmetry).

So... general quantum theory does not "have mass". But quantum theory, applied to model realistic dynamical situations does, because both Newtonian and Einsteinian dynamics have a concept of mass. We're simply representing the dynamical variables as operators on some Hilbert space, following the prescription from general quantum theory.

BTW, which QM book(s) have you studied? Ballentine's textbook develops quite well the relationship I summarized above.

since the observable is not position but fields, then how do you treat a one body problem and mass in ordinary qm and its relationship to the double slit experiment for instance?

Treating that in ordinary QM (without fields) always involves some hand waving and interpretation. I don't like those treatments.

To do it properly, with fields instead of particles, requires proficiency in quantum field theory. I could tell you it's described in the book by Mandel & Wolf on quantum optics and optical coherence, but I'm reasonably sure you're not yet at that level. The basic idea is that events registered on a spatially extended detector (like a photo-sensitive screen) obey a probabilistic formula. I.e., the math of quantum field theory, applied to this case, predicts a certain probability for a given area $\delta A$ on the screen to register an event within a given time interval $\delta t$.

 

[kye]

You're confusing a general mathematical framework, i.e., quantum theory, with its application to a particular physical scenario (e.g., Newtonian dynamics obeying Galilean symmetry, or Special Relativity obeying Poincare symmetry).

So... general quantum theory does not "have mass". But quantum theory, applied to model realistic dynamical situations does, because both Newtonian and Einsteinian dynamics have a concept of mass. We're simply representing the dynamical variables as operators on some Hilbert space, following the prescription from general quantum theory.

I read that mass in QM enters as a parameter that relates the momentum operator to the kinetic energy part of the Hamiltonian. Higgs field coupling to electron field details not necessary. Are you saying that it is not possible to apply QM to an electron in a hydrogen atom without using QFT and higgs field?

I know that QFT is applied when there is no fixed numbers of particles and it is moving close to speed of light, etc.

 

[strangerep]

Are you saying that it is not possible to apply QM to an electron in a hydrogen atom without using QFT and higgs field?

I have no idea how you got that from what I actually said.

The case of hydrogen (electron in a spherically symmetric potential) is discussed in every QM book I know of (and I note that you ignored my question about which QM books you've read).

QFT is of course necessary to achieve the best accuracy (e.g., Lamb shift).

 

[kye]

I have no idea how you got that from what I actually said.

The case of hydrogen (electron in a spherically symmetric potential) is discussed in every QM book I know of (and I note that you ignored my question about which QM books you've read).

QFT is of course necessary to achieve the best accuracy (e.g., Lamb shift).

I own dozens of pop-sci quantum and qft books and also have access to many qm/qft textbooks. You said that "So... general quantum theory does not "have mass". But quantum theory, applied to model realistic dynamical situations does, because both Newtonian and Einsteinian dynamics have a concept of mass.". I thought you meant that general qm didn't use mass, but dynamic situation like particle creation/annihilation, near light speed, etc. did. So what you meant was that qm theory doesn't have mass in the sense that geometry or algebra doesn't have mass. Of course we knew this. You continued "We're simply representing the dynamical variables as operators on some Hilbert space, following the prescription from general quantum theory." The reason I'm asking all this is because with the discovery of higgs.. it is certainty that mass came from the higgs field so I refocus my attention to how QM interact with it. So it doesn't, only QFT did. Or maybe we just don't focus on the detailed interactions of it or coupling in QM and just make mass as given like time. In other words, QM doesn't handle coupling details or dynamics (or just effective theory).

 

[BruceW]

In simple QED I think the electron actually does have mass. And QED is a field theory. So it's not like an absolute rule that in QM particles have mass, but in QFT's particles don't have their own mass. Anyway, (I think) the problem comes along when we try to make a gauge-invariant, renormalisable electroweak field theory, it is not possible for the electron to have mass. And so we need this Higgs model to explain how the electron acquires its mass.

 

[kye]

I understand the Higgs mechanisms and the doublet thing gauge invariance of it and not questioning it.

Isn't it that over 90% of electron mass came from special relativity (energy-mass equivalence), 10% or fewer involves the higgs. Now I'd like to know is how the higgs interact with an electron in between the emitter and detector in a double slit experiment. Before detection, the wave function as a ray or vector in Hilbert space is not yet collapsed so in conventional formalism, they fundamentally don't have positions (except Many world or bohmian's which let's ignore for now). But we can detect the mass at all times. So even though an electron doesn't have position before measurement yet it has mass, so could the higgs field somehow betray their presence by the screening effect giving mass to them or does the higgs field possesses the property of superpositions too or maybe it doesn't obey it hence there is always mass in a quantum system before measurement unlike position. What is your comment.

 

[BruceW]

I'm very new to this whole quantum field theory stuff. So bear with me. But I'm reading an amazing book - Quantum field theory by Mandl and Shaw. It's really good, introduces the subject gently. I'd recommend it if you are looking for good books.

So, anyway. I'm not sure what you mean 90% of electron mass comes from special relativity and 10% comes from the higgs... I'm pretty sure that 100% of the invariant mass of the electron comes from the higgs. But if we're talking about relativistic mass, then that is $p^2/c^2+m^2$ (where m is the invariant mass of the electron, and p is the relativistic momentum). So yes, the relativistic mass of the electron is not all from the higgs, some of it comes from the momentum of the electron.

About the higgs interacting with the electron. There's two separate concepts that are important (as far as I can tell). 1) the fact that there is a higgs field, and that it couples with the lepton field, means that the electrons have a specifically defined mass. 2) perturbations of the higgs field are higgs bosons. These are particles that can collide with the electron, just like how a photon can collide with an electron.

So due to 1) yes, even though the electron does not have a specific position, it does have a specific mass. Also, as the electron travels between emitter and detector, (I think) that it is good to imagine Feynman's sum-over-states, so yes there is some probability that the electron will collide with a higgs particle at some place along its path, but since we don't observe exactly what goes on, we must sum over all possible collisions that could have happened with higgs bosons, along all possible paths that the electron could have taken. So even though there is a higgs field, this does not allow us to know the precise position of the electron as if it was moving through some kind of bubble chamber. Also, important to note that the higgs field couples weakly with the electron (and most other particles I think), which is why it's presence is hard to detect. And for this reason, it does not collide with the electron very often.

 

[vanhees71]

The electron mass is due to the Higgs mechanism. That's not true for hadrons, whose mass is to a large amound dynamically generated by the strong interaction.

Further you should not use the idea of an "relativistic mass". For very good reasons in HEP we use the mass always in the sense of invariant mass. What was known as "relativistic mass" in the very early days of relativity is simply the energy of the particle (divided by $c^2$).

 

[kye]

I have questions about this "yes, even though the electron does not have a specific position, it does have a specific mass". When you weight molecules or atoms.. the electron has no position because they are smeared out probability yet they have mass. Can anyone give me other example where something has no positions yet have mass? And how does this get mapped in spacetime? How can spacetime locate no position but can detect mass? Please share how this is viewed or understood by the mainstream in the conceptual and mathematical aspect. Thanks.

 

[BruceW]

I'm not totally sure what you mean. If we measure the position of 4 identically prepared electrons, we can get an answer like 1cm, 11cm, 7cm, 2cm, e.t.c. So even though each of these electrons was identically prepared, we get a different answer for their position each time. And that is why the electron is said to be spread out over space. But, if we take 4 identically prepared electrons and measure their mass, we will get 0.51MeV/c² every single time. And this is true for any kind of state we prepare that electron to be in. So therefore, in our theory we specify that the electron has a definite mass, so that our theory matches experiment. For example, in the Dirac equation $(i \gamma^\mu \partial_\mu -m)\psi =0$ Automatically specifies that our electrons and positrons have a definite mass. Or, in the field-theoretic interpretation, that our field is made up of electrons and positrons which each have a definite mass.

 

[kye]

I'm not totally sure what you mean. If we measure the position of 4 identically prepared electrons, we can get an answer like 1cm, 11cm, 7cm, 2cm, e.t.c. So even though each of these electrons was identically prepared, we get a different answer for their position each time. And that is why the electron is said to be spread out over space. But, if we take 4 identically prepared electrons and measure their mass, we will get 0.51MeV/c² every single time. And this is true for any kind of state we prepare that electron to be in. So therefore, in our theory we specify that the electron has a definite mass, so that our theory matches experiment. For example, in the Dirac equation $(i \gamma^\mu \partial_\mu -m)\psi =0$ Automatically specifies that our electrons and positrons have a definite mass. Or, in the field-theoretic interpretation, that our field is made up of electrons and positrons which each have a definite mass.

in the dirac equations, mass is a parameter and didn't include interactions with the higgs field, isn't it?

anyway, my main question is that in conventional belief. Electrons don't have trajectories in the atoms, because if they do, they may lose electromagnetic energy and fall down to the nucleus, so the consensus is they are probability clouds and we can only describe them as probabilities. Hence there is no position in principle because no trajectories. This means at certain time, the electron is not a particle in the atom, now I'd like to understand how the mass can still exist in the times it is not a particle.. unless you mean the mass becomes energy (from e=mc^2) and this become part of the Hamiltonian of the wave, is this the explanation, pls. elaborate, thank you.

 

[atyy]

I read that mass in QM enters as a parameter that relates the momentum operator to the kinetic energy part of the Hamiltonian.

And how does this get mapped in spacetime? How can spacetime locate no position but can detect mass? Please share how this is viewed or understood by the mainstream in the conceptual and mathematical aspect. Thanks.

Since mass is a parameter relating "momentum" and "energy", in quantum theory momentum is spatial frequency (1/wavelength) and energy is temporal frequency, so "mass" of a wave is a parameter in the relationship between frequency and wavelength. This is what the mass of a wave is, since in quantum theory a "particle" is a wave.

 

[kye]

Since mass is a parameter relating "momentum" and "energy", in quantum theory momentum is spatial frequency (1/wavelength) and energy is temporal frequency, so "mass" of a wave is a parameter in the relationship between frequency and wavelength. This is what the mass of a wave is, since in quantum theory a "particle" is a wave.

But they said you can't force the mass terms in the Schroedinger equations or it can destroy the gauge symmetry, that is the precise reason they proposed the higgs mechanism.. so if mass came from the higgs mechanics.. what is its relationship to the "mass" of a wave as "a parameter in the relationship between frequency and wavelength"? How do you put the two together or relate them?

 

[atyy]

But they said you can't force the mass terms in the Schroedinger equations or it can destroy the gauge symmetry, that is the precise reason they proposed the higgs mechanism.. so if mass came from the higgs mechanics.. what is its relationship to the "mass" of a wave as "a parameter in the relationship between frequency and wavelength"? How do you put the two together or relate them?

With the Higgs mechanism, you can have mass as a parameter between frequency and wavelength in the wave equation for a gauge field.

 

[BruceW]

... Hence there is no position in principle because no trajectories. This means at certain time, the electron is not a particle in the atom, now I'd like to understand how the mass can still exist in the times it is not a particle.. unless you mean the mass becomes energy (from e=mc^2) and this become part of the Hamiltonian of the wave, is this the explanation, pls. elaborate, thank you.

I am super-confused. I agree, the electron is spread out over space. but it still has mass. If you measure the position of a ground state electron, then in each experiment you will get a different value of the position. But if you measure the mass of the electron, then you always get the same value. Also I don't know what it means "at a certain time, the electron is not a particle in the atom" why not? for example, an electron in the ground state is just going to stay in that ground state, so it is a particle in the atom for a long time.

edit: yes, I'm saying the electron is a particle, even though it is spread out over space. This terminology makes sense in that you can measure the position of the electron as accurately as you want, and you will find it to be a point particle.

 

[Nugatory]

Hence there is no position in principle because no trajectories. This means at certain time, the electron is not a particle in the atom, now I'd like to understand how the mass can still exist in the times it is not a particle

It means no such thing. The electron is in a bound state and therefore part of the atom even though its position is uncertain (it would be more accurate to say that its position is undefined until we measure it).

The Schrodinger equation that we use to calculate the probability of finding the electron at various locations starts with the electron in a bound state; the probabilities that it spits outs are the probabilities for an electron that is bound into the atom.

The bad news is that there's no way of really seeing this except to set up and solve the equations, and that requires paying some fairly serious mathematical dues - a year or so of study beyond introductory differential and integral calculus. The good news is that it's worth the effort - it's the difference between reading about a delicious meal and cooking and eating a delicious meal.

 

[kye]

I am super-confused. I agree, the electron is spread out over space. but it still has mass. If you measure the position of a ground state electron, then in each experiment you will get a different value of the position. But if you measure the mass of the electron, then you always get the same value. Also I don't know what it means "at a certain time, the electron is not a particle in the atom" why not? for example, an electron in the ground state is just going to stay in that ground state, so it is a particle in the atom for a long time.

edit: yes, I'm saying the electron is a particle, even though it is spread out over space. This terminology makes sense in that you can measure the position of the electron as accurately as you want, and you will find it to be a point particle.

If you agree the electron is spread out over space, it doesn't have trajectory from a to b, it's like it's simultaneous in a and b.. a wave... note qm only gives probabilities of its positions (it doesn't show it has position before measurement as Bohm emphasized).. see this thread here

https://www.physicsforums.com/showthread.php?t=98413

"Thus the closest quantum mechanical equivalent of "motion"" is momentum. It's a mistake to think of quantum phenomenon as being classical - Bell's theorem shows the pitfalls in assuming that quantum particles are classical ones with "hidden variables". If it weren't for the pitfalls involved with hidden variables, one could assume that the electron always had a definite position, but that it wasn't known. Unfortunately this application of hidden variables leads to contradictions as Bell's theorem illustrates. Therfore one cannot assume an electron has a definite position, and neither can one assume that it has a velocity".

 

[kye]

With the Higgs mechanism, you can have mass as a parameter between frequency and wavelength in the wave equation for a gauge field.

Since the higgs field only interact with the lepton fields and there is no concept of field in QM, then the mass in QM is only given and do you think non-relativistic QM is somewhat ad hoc?

 

[atyy]

Yes, non-relativistic QM as used in condensed matter and chemistry is only an effective theory that does not work at high energies. However, it is important to know that even the standard model of particle physics may be an effective theory that fails at high energies, near the Planck scale where quantum gravity is important.

You should also read http://arxiv.org/abs/1206.7114 about how most mass is not produced by the Higgs mechanism.

 

[kye]

Yes, non-relativistic QM as used in condensed matter and chemistry is only an effective theory that does not work at high energies. However, it is important to know that even the standard model of particle physics may be an effective theory that fails at high energies, near the Planck scale where quantum gravity is important.

They don't work at high energies, no problem about that. But there is still mass that doesn't come from the higgs field since there is no higgs field in QM. So how does non-relativistic QM treat the mass coming from the higgs contribution? Maybe you were saying QM just treat the mass coming from higgs and kinetic-mass contribution as parameter or just input them ad hocly without saying how they did it? And the phase invariance of U(1) is a QED thing and not basic QM?

You should also read http://arxiv.org/abs/1206.7114 about how most mass is not produced by the Higgs mechanism.

I know kinetic energy of the particles produced the mass.

 

[atyy]

Maybe you were saying QM just treat the mass coming from higgs and kinetic-mass contribution as parameter or just input them ad hocly without saying how they did it?

Yes.

 

[kye]

Yes.

Then it doesn't make sense to contemplate how electrons don't have positions (before measurements) in the atoms yet have fixed mass? This is also ad hoc and the complete analysis can only be done in the quantum field theory interactions between the higgs field and leptics field via the higgs mechanism?

 

[BruceW]

yeah, and if you do the complete analysis, then you get electrons which have a single definite mass. It is still a bit ad hoc. We still have parameters which we have to choose to fit experimental evidence. The idea is that they are trying to always reduce how many of these parameters are in our theory. So it's an ongoing thing.

 

[kye]

yeah, and if you do the complete analysis, then you get electrons which have a single definite mass. It is still a bit ad hoc. We still have parameters which we have to choose to fit experimental evidence. The idea is that they are trying to always reduce how many of these parameters are in our theory. So it's an ongoing thing.

Are you saying that in QFT, the parameter is still there only it's the strength of the higgs interaction with the electron? At least we are treating the interaction already instead of just putting the value of mass to QM and changing it to wave equivalent.

Now I'm imagining how higgs field and mass interact with space that is related to position. But spacetime or time in QFT is fixed.. we have fields on fixed background. Smolin said a real theory must be background independent.. so the fields must not be on a fixed spacetime background but they must interact dynamically.. so this means in QFT we can't even know how mass and spacetime interact (and hopeless for QM to model it.. ).. what do you think?

 

[BruceW]

ah man. I just wrote a flippin' long answer, but then I lost internet connection and I lost it all. darn. I'll try to write down the rough idea.

ah, I wrote, the free Dirac field for massive particles is also a QFT, so we have particles with a specific, definite mass, even without the Higgs interaction. The problem is that they wanted to make an electroweak interaction that obeyed Gauge-invariance. Therefore the mass of electrons (and gauge bosons) must be zero. But experimentally, we find that such particles do have mass. So, they introduced the Higgs particle and Higgs mechanism to explain how the particles obtained their mass. And luckily, they have found Higgs bosons, which validates the Higgs mechanism. Also, spacetime does not interact. You can just think of it as a fixed background (since we are talking about special relativity here, not general relativity). It is the various different fields which interact with each other. Spacetime does not interact with the fields.

 

[kye]

ah man. I just wrote a flippin' long answer, but then I lost internet connection and I lost it all. darn. I'll try to write down the rough idea.

ah, I wrote, the free Dirac field for massive particles is also a QFT, so we have particles with a specific, definite mass, even without the Higgs interaction. The problem is that they wanted to

Isn't it Stephen Hawking said information can't be lost even inside the event horizon?

You mean the free Dirac field also use mass as parameter (from the kinetic-mass), without needing the Higgs? Won't this make it weigh less? But even for the Dirac equation, won't the symmetry would be lost if you introduce mass by force...

make an electroweak interaction that obeyed Gauge-invariance. Therefore the mass of electrons (and gauge bosons) must be zero. But experimentally, we find that such particles do have mass. So, they introduced the Higgs particle and Higgs mechanism to explain how the particles obtained their mass. And luckily, they have found Higgs bosons, which validates the Higgs mechanism. Also, spacetime does not interact. You can just think of it as a fixed background (since we are talking about special relativity here, not general relativity). It is the various different fields which interact with each other. Spacetime does not interact with the fields.

It's said that "mass tells spacetime how to curve... spacetime tells mass how to move".. so how do fields interact with geometry, what mathematical language do you use for the interface besides the General Relativity formula that makes it as given without giving the interaction details?

 

[BruceW]

You mean the free Dirac field also use mass as parameter (from the kinetic-mass), without needing the Higgs? Won't this make it weigh less? But even for the Dirac equation, won't the symmetry would be lost if you introduce mass by force...

yeah, the free Dirac field just has mass as a parameter, without needing the Higgs field. So in this way, the electron just has its own mass. When we use a Higgs field, we say the electron has zero mass, but acquires mass by the Higgs mechanism. So in both cases, we can say the electron effectively ends up having the same mass (which is determined experimentally).

If you only have the free Dirac field, then that is not a gauge theory. So there is not any symmetry to be lost anyway. Once we introduce photons, then we have QED which is a gauge theory. And it is invariant under gauge transformations. So the symmetry is not lost, even though we are using mass as a parameter (and without any Higgs). The reason we need the Higgs mechanism is when we try to make an electroweak theory, if we let the electron have nonzero mass, then the symmetry is lost, in the sense that our theory is no longer gauge invariant. Therefore, they say the electron has zero mass, but acquires mass through the Higgs mechanism, so that the electroweak theory is gauge invariant. So if you don't care about the weak interaction, you can just use QED, where the electron does have mass and there is no Higgs particle. This is fine for a lot of cases, for example Compton scattering, you don't really care about the weak interaction, since you just want to find out what happens when an electron and photon collide.

It's said that "mass tells spacetime how to curve... spacetime tells mass how to move".. so how do fields interact with geometry, what mathematical language do you use for the interface besides the General Relativity formula that makes it as given without giving the interaction details?

I'm not sure what you mean by "the General relativity formula that makes it as given without giving the interaction details". I think the standard thing to do is if you have several fields, then you work out how they interact with each other, and 'plug in' the resulting energy density and pressure into the Friedmann equations, then solve as if it was a general relativity equation which involved normal energy and pressure. I don't know a lot about this stuff, but it seems pretty interesting.

 

[BruceW]

Isn't it Stephen Hawking said information can't be lost even inside the event horizon?

haha, yeah not lost, but probably not worth the effort of collecting the information together again :)

 

[kye]

yeah, the free Dirac field just has mass as a parameter, without needing the Higgs field. So in this way, the electron just has its own mass. When we use a Higgs field, we say the electron has zero mass, but acquires mass by the Higgs mechanism. So in both cases, we can say the electron effectively ends up having the same mass (which is determined experimentally).

If you only have the free Dirac field, then that is not a gauge theory. So there is not any symmetry to be lost anyway. Once we introduce photons, then we have QED which is a gauge theory. And it is invariant under gauge transformations. So the symmetry is not lost, even though we are using mass as a parameter (and without any Higgs). The reason we need the Higgs mechanism is when we try to make an electroweak theory, if we let the electron have nonzero mass, then the symmetry is lost, in the sense that our theory is no longer gauge invariant. Therefore, they say the electron has zero mass, but acquires mass through the Higgs mechanism, so that the electroweak theory is gauge invariant. So if you don't care about the weak interaction, you can just use QED, where the electron does have mass and there is no Higgs particle. This is fine for a lot of cases, for example Compton scattering, you don't really care about the weak interaction, since you just want to find out what happens when an electron and photon collide

Thanks for this illuminating piece. I'll memorize this word for word. :)

I'm not sure what you mean by "the General relativity formula that makes it as given without giving the interaction details". I think the standard thing to do is if you have several fields, then you work out how they interact with each other, and 'plug in' the resulting energy density and pressure into the Friedmann equations, then solve as if it was a general relativity equation which involved normal energy and pressure. I don't know a lot about this stuff, but it seems pretty interesting.

Some kind of duality? Remember General Relativity is pure geometry, and you can't talk of interaction fields because if you do, you are not talking about GR where gravity is literally geometry curvature and gravity fields don't exist... but maybe they can be tied up using some sort of duality where you can transform field and geometry, is this what you are saying?

 

[BruceW]

Thanks for this illuminating piece. I'll memorize this word for word. :)

hehe, I might have said already, but you should definitely try to get your hands on the book 'quantum field theory' by Mandl and Shaw. There might be a lot better books out there, I don't know, I haven't searched very hard. But this one is a really good basic introduction for physics fans like me, who want to get a basic idea of some of the physics involved, without going into too much detail. It does have some good physics (it has equations as well as word explanations). And in the equations, it will omit the longer proofs and extra detail that I just wouldn't find as interesting.

Some kind of duality? Remember General Relativity is pure geometry, and you can't talk of interaction fields because if you do, you are not talking about GR where gravity is literally geometry curvature and gravity fields don't exist... but maybe they can be tied up using some sort of duality where you can transform field and geometry, is this what you are saying?

In stuff I've seen, it looks like they keep General Relativity as pure geometry, but the stress-energy tensor results from the various fields. In Einstein's field equations:
$$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = \frac{8\pi G}{c^4} T_{\mu \nu}$$
all the stuff on the left-hand-side is related to the curvature e.t.c. like normal. And the stuff inside $T_{\mu \nu}$ contains all the stuff from the energy density and pressure due to the fields. And they usually mention that this only holds true in regions where the energy density is not too great and the curvature is not too great, otherwise we would need a theory of quantum gravity to describe what is going on in that region. The book 'particle physics and inflationary cosmology' by Andrei Linde is really good. But I know even less about this kind of stuff, so yeah, I can't really say much for certain. P.S. he put his book on the internet for free. You can find it at arxiv if you're interested. I definitely recommend reading a bit about quantum field theory first though.

 

[kye]

hehe, I might have said already, but you should definitely try to get your hands on the book 'quantum field theory' by Mandl and Shaw. There might be a lot better books out there, I don't know, I haven't searched very hard. But this one is a really good basic introduction for physics fans like me, who want to get a basic idea of some of the physics involved, without going into too much detail. It does have some good physics (it has equations as well as word explanations). And in the equations, it will omit the longer proofs and extra detail that I just wouldn't find as interesting.


In stuff I've seen, it looks like they keep General Relativity as pure geometry, but the stress-energy tensor results from the various fields. In Einstein's field equations:
$$R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R = \frac{8\pi G}{c^4} T_{\mu \nu}$$
all the stuff on the left-hand-side is related to the curvature e.t.c. like normal. And the stuff inside $T_{\mu \nu}$ contains all the stuff from the energy density and pressure due to the fields. And they usually mention that this only holds true in regions where the energy density is not too great and the curvature is not too great, otherwise we would need a theory of quantum gravity to describe what is going on in that region. The book 'particle physics and inflationary cosmology' by Andrei Linde is really good. But I know even less about this kind of stuff, so yeah, I can't really say much for certain. P.S. he put his book on the internet for free. You can find it at arxiv if you're interested. I definitely recommend reading a bit about quantum field theory first though.

Real forces are vector, the GR geometry is described by the metric, a rank-2 tensor. So GR is pure geometry where there are really no forces. Some people get confused with this. Do you think GR is some kind of duality to the forces or do you treat the tensor as like wave function just to represent measurements?

 

[BruceW]

$T_{\mu \nu}$ is the stress-energy tensor, so this is where forces enter the equation (well actually energy, momentum and stress, but they are related to forces). In GR, there is no gravitational force, but there are other forces.

edit: uh, I guess forces don't directly affect the metric. But they do affect it indirectly. For example, if there was two separate dust clouds, both positively charged, then they would have an electromagnetic force that tries to push them away from each other, and if this causes them to move apart, then energy, stress and momentum are affected, so $T_{\mu \nu}$ is affected, and so the metric is affected.