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If my layperson's understanding is correct (probably vastly generalized), this should be considered exciting news. A few years ago the existence of the Higgs B

• cshimmin 8y ago
  ATLAS physicist here -- you're mostly right! The existence of a Higgs boson has been confirmed with extremely high confidence. And this news... it's exciting in the sense that it's an important result, but it's a bit of a damper on those of us who were hoping to discover new secrets about the universe. Interestingly, the primary production mechanism for the Higgs boson at the LHC proceeds via the fusion of gluon particles from the colliding protons. This seems strange at first glance, since the Higgs interacts with particles proportionally to their mass, while the gluons are massless. In the standard model, there is a higher-order process wherein the gluons annihilate into a virtual "loop" of top quarks (which are very massive), which then "lend" their mass to produce the Higgs. Because previous measurements have been consistent with the Standard Model prediction, we already had some evidence that the coupling to top quarks was nominal. However, there are other possible scenarios involving exotic physics that could mimic this production mechanism, which could imply the existence of new forces/particles, and that the Higgs boson does not couple to "normal" particles in the predicted fashion. Because of the nature of these heavy virtual loop processes, it turns out that it's pretty hard to distinguish between different such scenarios by examining the kinematic properties of the events. Hence, the main observable is the rate, and even that can be made to match a wide range of values with or without standard couplings, given a clever enough theory. This recent result demonstrates directly that the Higgs boson does indeed couple to the top quark in a matter roughly consistent with the Standard Model expectation (the CMS measurement shows an upward fluctuation, indicating even stronger coupling than expected, although this is probably just a statistical fluctuation). Therefore, no fancy/exotic physics (such as string theory, strong gravity, etc) are required to explain the observed production of Higgs bosons so far at the LHC.
  • dspillett 8y ago
    > The existence of a Higgs boson has been confirmed with extremely high confidence. And this news... it's exciting in the sense that it's an important result, but it's a bit of a damper on those of us who were hoping to discover new secrets about the universe. There is a certain beauty in not only accepting that you might be wrong, but actively wanting to be wrong because that could be so much more interesting! And a certain sadness in the disappointment some obviously have when they accept that the less interesting outcome is the one that they receive.
    • cshimmin 8y ago
      Well, it's a bit selfish really. It's a terribly interesting outcome as it is, but the particle physicists in the 70's and 80's got to have all the fun.
      • deleted 8y ago
        [deleted]
      • da_chicken 8y ago
        Sort of how the chemists in the 18th and 19th centuries got to have all the fun. "I'll call it 'Aluminum.' No, wait! 'Aluminium!' Uh... ship that first print run of my book with the wrong spelling to North America. Nobody will notice!" "I'll call it Potassium, because it's found in potash." "That's not very creative." "Fine, I'll call it Kalium and give it a symbol of K." "Where'd you get Kalium from?" "Well, it's very alkaline, you see. It's from potash, after all." "That's still not very creative!" "Fine. We'll call it Potassium, but give it a symbol of K. Happy now?"
        • vidarh 8y ago
          It's still called Kalium in a bunch of languages.
          • staticautomatic 8y ago
• pacala 8y ago
  As a layperson, this is so confusing. On one hand, we have Higgs Bosons "giving" particles mass, presumably in a nondeterministic quantum manner, on the other hand we have relativity theory that says mass "warps" space-time itself. What is the relation between the two theories?!
  • etatoby 8y ago
    Nobody knows. It's like knowing how electronics work, from transistors downwards; and how software works, from machine language upwards; but completely missing the entire concept of a CPU, how it's built, how complex (or not) it is, and how to study it and tinker with it. Quantum gravity is one of the biggest remaining mysteries or unknowns in Physics. (IANAP)
  • raattgift 8y ago
    The short answer is that when doing quantum mechanics on curved spacetime, we tend to do it semiclassically: the quantum stuff is approximated or averaged, and then that is taken as the source of classical spacetime curvature. The averaged (or whatever) value of the Higgs field at a point in spacetime, like the value of all the other fields at the same point, is then just a contribution to the stress-energy tensor at that point in spacetime. The Einstein Field Equations [1], dropping constants and indices, and with a vanishing cosmological constant, can be written as G = T. Here G is the Einstein tensor, which describes curvature at each point, and T is the stress-energy tensor, which describes the flux of momentum-energy through each point. Below, for ease of understanding T, let's apply Cartesian-like coordinates (really, Minkowski coordinates, which are spatial Cartesian coordinates x,y,z and time coordinate t, such that when calculating distances between two Minkowski points the two sets of coordinates differ by a sign and a constant: ds = sqrt(dx^2 + dy^2 + dz^2 - c^2dt^2) is one way of writing out the line element in a Minkowskian fashion, but below let's use ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2, which is the "mostly-minus" or "+,-,-,-" metric signature; below we'll refer to this line element as "the metric". The metric is a component of the Einstein tensor G, and typically in the Einstein Field Equations you can see the metric exploded out as its own tensor g [2]. Let's look shallowly at the stress-energy tensor. We have made a deliberate choice of metric, and of coordinate basis, and will lean a bit on the natural spacetime slicing into space and time that these particular choices give us. T can be written as a 4 x 4 matrix with rows and columns starting at 0 and running to 3. T_{00} or shorter T_00, two zeroes subscripting the letter T, means the 0th row and 0th column; 0 in our metric above means the time direction; 1 means the x direction; 2 means the y direction; 3 means the z direction. With our set of choices, T_00 corresponds to \gamma m_0 c^2: it is the "matter" at a point in spacetime that has come from the past and is going to the future, and is not moving in the x, y, or z direction. "Matter" is all the contributions to energy-momentum. Here that's some expectation value for each of the quantum fields at the point. T_ii means we look at T_00, T_11, T_22, T_33. Let's look at the he lower three of these diagonals, T_ii, i != 0, so we are not looking at T_00. With all these choices made, T_11 is the flux of x momentum in the x direction. If x is "left" and "right" then we are thinking about momentum going from left to right entering the point from the right and exiting the point to the right. Again, that momentum can be photons or any other quantum field content. T_0i and T_i0 can be thought of momentum-energy which originates or terminates at the point, arriving or departing in the i direction; more precisely, if you know some special relativity, T_00 here looks like \gamma m_0 c^2 and T_0i looks like \gamma m_0 \vec{v}_{i} c. For completeness, the non-diagonals, T_ij, i,j ! =0, i != j are the fluxes of i momentum the j direction, or equivalently with our choices, the total momentum times the velocity in the j direction. So, considering the middle space of three 3d spaces at time coordinates t-1, t, and t+1, at a point p the stress-energy tensor T encodes the all the quantum field values and changes thereof that contribute to the energy-momentum that is constant at p (in T_00) or which arrive and depart p in the three spacelike directions. The momentum is deposited into the "matter" at the point and adjacent (but in the future light cone) points in spacetime we are looking at, and arrives from the matter the adjacent points in spacetime in the past light cone. [4] When taking this sort of semiclassical approach we usually write G = <T> where the angle brackets mean the expectation value for what they enclose. <T> is the source for the curvature in G, and thus the curvature does not depend on the details of what number of electron neutrinos vs photons are at the point. This approach works very well in practice except (a) in regions that our technology has no hope of probing any time soon: the deep interiors of black holes, and the very hottest densest phase of the universe in our past and (b) where there (under our choices above) is a superposition of position of the quantum field values where the spatial separation is significant and the superposition lasts long enough to "measure" gravitationally (e.g. with a Cavendish apparatus). We will probably probe (b) directly within the next few years. Unfortunately, the Minkowski metric we were using above is an extremely special case of no curvature. No curvature and no cosmological constant means, strictly, no matter. Fully classically, G = T, we then have T = 0 : our curvature-free spacetime is vacuum. However, semiclassically G = <T>, is only guaranteed a vacuum on average; there could be some configuration of quantum field values that are non-vacuum, and in principle they could condense, exposing us to a problem similar to (b) in the previous paragraph. We can also attack this class of problem with perturbation theory, where our metric is a background, and we perturb around it. Quantum Field Theories we're interested in are linear, so we can just sum perturbations of a quantum field; we can do the same in weak gravity -- this is how linearized gravity[3] works -- but because the Einstein Field Equations are nonlinear, we stop being able to add a perturbation of a metric to the background metric; so we run into the (a) problem where the perturbations of the quantum field are so hard to exactly match with corresponding perturbations of the background metric that nobody knows how to do it yet. But if we look again at the "exploded" Einstein Field Equations [2], then we can say g = \eta + h + O(h^2) + ..., where \eta is the background (in this case, flat spacetime) metric, and h is a field perturbations on it, O(h^n) are higher-order perturbations. As long as we can essentially ignore O(h^n), n > 1, semiclassical gravity linearizes beautifully accurately with supporting evidence from solar-system scales to binary pulsars. Finally, when we use a different metric or a different set of coordinates, or both, the interpretations of the components of the stress-energy tensor at each point must be varied accordingly, however, in all cases the tensor value T itself determines the Einstein Tensor in all systems of coordinates and with all metrics. More details if you like at : https://en.wikipedia.org/wiki/Semiclassical_gravity https://en.wikipedia.org/wiki/Semiclassical_gravity which has a good References section (Birrell and Davies is former gold standard, but Leonard Parker & David Toms _Quantum Field Theory in Curved Spacetime_ is a bit more up to date). - -- [1] https://en.wikipedia.org/wiki/Einstein_field_equations https://en.wikipedia.org/wiki/Einstein_field_equations [2] metric exploded out form (without cosmological constant): https://wikimedia.org/api/rest_v1/media/math/render/svg/7da0def1c2c8d85120b36307ccbab4ee5a4766bf https://wikimedia.org/api/rest_v1/media/math/render/svg/7da0... versus the Einstein Tensor https://wikimedia.org/api/rest_v1/media/math/render/svg/2174f519deba2c9198068e9d7fc6b22e262d5b73 https://wikimedia.org/api/rest_v1/media/math/render/svg/2174... [3] https://en.wikipedia.org/wiki/Linearized_gravity https://en.wikipedia.org/wiki/Linearized_gravity [4] With a different metric, the momentum-energy flux through a point in spacetime can be gravitational too, and that is a critical source of nonlinearity; if we slightly modify our choices above to allow for it, a gravitational wave moving in the x direction will influence the T_i1 and T_1j components of the stress-energy tensor, for instance. Returning to the quantum question, that means that a field value at that point might be excited by the passing wave; with a strong enough excitation, perhaps a particle might appear there that would not have if the gravitational wave had not been coincident at that point.