TE
TechEcho
Home24h TopNewestBestAskShowJobs
GitHubTwitter
Home

TechEcho

A tech news platform built with Next.js, providing global tech news and discussions.

GitHubTwitter

Home

HomeNewestBestAskShowJobs

Resources

HackerNews APIOriginal HackerNewsNext.js

© 2025 TechEcho. All rights reserved.

Why is symmetry so important in particle physics?

121 pointsby mfnover 2 years ago

10 comments

moringover 2 years ago
I find it incredibly frustrating that again and again, the Lagrangian gets introduced and then said it should be minimized without ever explaining the motivation behind doing so. What <i>is</i> the Lagrangian and <i>why</i> should it be minimized?<p>I totally get how L gets defined mathematically, how it is derived from Newton&#x27;s laws (this part is typically well explained by textbooks), and why in the case of point particles, a curve that violates Newton&#x27;s laws does not minimize L. But there is no understanding at all, just saying &quot;okay, it checks out&quot; on a math level.<p>It doesn&#x27;t help at all that on a math level, L isn&#x27;t actually minimized but its derivative set to 0, which isn&#x27;t even equivalent to &quot;minimized or maximized&quot;. Why doesn&#x27;t a single textbook explain why maximizing L is also okay when they first stated &quot;minimized&quot;? Or why derivative=0 is sufficient? As a reader, I always get the impression that &quot;well, of course they cannot explain that, because they don&#x27;t even know why L should be minimized in the first place&quot;. It&#x27;s just all formulas that are easy to verify but don&#x27;t convey a single bit of understanding.<p>Just for comparison, I found quantum mechanics based on the Schrödinger equation and the Hamiltonian rather easy to grasp, because every piece of it has an easy-to-understand meaning, that also gets explained really well. Why is this seemingly impossible for the Lagrangian?
评论 #33403863 未加载
评论 #33406132 未加载
评论 #33403557 未加载
评论 #33404104 未加载
评论 #33408137 未加载
评论 #33407550 未加载
评论 #33403359 未加载
评论 #33409928 未加载
评论 #33404307 未加载
评论 #33403634 未加载
评论 #33404222 未加载
评论 #33403814 未加载
评论 #33404394 未加载
评论 #33403773 未加载
评论 #33411806 未加载
评论 #33405163 未加载
teleforceover 2 years ago
This is an excellent book on the subject by Jakob Schwichtenberg [1].<p>[1]Physics From Symmetry:<p><a href="http:&#x2F;&#x2F;physicsfromsymmetry.com&#x2F;" rel="nofollow">http:&#x2F;&#x2F;physicsfromsymmetry.com&#x2F;</a>
评论 #33402466 未加载
adamnemecekover 2 years ago
This idea shows up in essentially all scientific fields. It’s the idea of adjointness. Together with norm, they give you the idea of fixed points, (invariants, spectra, embeddings, braids etc).I&#x27;m<p>Lawvere&#x27;s fixed point theorem is I think the best formulation of the idea<p><a href="https:&#x2F;&#x2F;ncatlab.org&#x2F;nlab&#x2F;show&#x2F;Lawvere%27s+fixed+point+theorem" rel="nofollow">https:&#x2F;&#x2F;ncatlab.org&#x2F;nlab&#x2F;show&#x2F;Lawvere%27s+fixed+point+theore...</a><p>I&#x27;ve been putting together a brain dump on the topic<p><a href="https:&#x2F;&#x2F;github.com&#x2F;adamnemecek&#x2F;adjoint&#x2F;" rel="nofollow">https:&#x2F;&#x2F;github.com&#x2F;adamnemecek&#x2F;adjoint&#x2F;</a><p>Join the discord <a href="https:&#x2F;&#x2F;discord.gg&#x2F;mr9TAhpyBW" rel="nofollow">https:&#x2F;&#x2F;discord.gg&#x2F;mr9TAhpyBW</a>
评论 #33401991 未加载
xeonmcover 2 years ago
When being &quot;here&quot; and &quot;there&quot; is exactly the same, you move spontaneously.<p>When being &quot;here&quot; and &quot;there&quot; is almost-but-not-quite the same, you move easily.<p>The degree of &quot;not-the-same-ness&quot; is called the Lagrangian.<p>In other words, symmetry is &quot;fungibility of states&quot;.<p>Things happen because the before and after is not very different, the transaction of different-ness is the energy involved.<p>In a classical system it&#x27;s nigh impossible to encounter an &quot;exactly the same&quot; situation, because there are just too damn many participants to rule out every possible interaction.<p>In a quantum system you encounter &quot;exactly the same&quot; situations frequently because there is only a tiny number of participants interacting.
misja111over 2 years ago
After reading to half way into the article, there is this disappointing text:<p>&gt; It turns out that there are a range of particles in nature that exhibit this kind of symmetry. There isn’t any easy way to argue why this kind of symmetry must exist, but it does.<p>And then the article continues to show some implications and predictions that follow when one assume that this symmetry must exist. But I opened the article expecting to find the why ..
bigbacaloaover 2 years ago
When we don&#x27;t have symmetry we don&#x27;t know what to do and can&#x27;t compute anything. Fortunately many situations can be modeled as near a symmetric one. We solve the symmetric one and study it&#x27;s asymmetric perturbations.<p>The prevalence of symmetry reflects our inability to do anything in its absence.
评论 #33403657 未加载
ranger207over 2 years ago
I don&#x27;t like to be critical, but I&#x27;ve been wanting to understand symmetry in physics for a while, so here&#x27;s a few points of confusion I have<p>&gt; The principle, then, is that the particles and fields that were used to build up the theory will move in a way that minimizes or maximizes the sum of L over the path taken by the system<p>The next couple of examples only minimize the Lagrangian; are there any systems in this article that maximize it?<p>&gt; a collection of objects and a recipe to build a Lagrangian from those objects, with the movement of those objects determined by a path that minimizes the Lagrangian<p>What in this case are the objects? Just particles? Is mass in the first example (of classical motion) an object? I&#x27;m trying to figure out what kind of objects to use to build an equation, or basically, what type (in the programming sense) an object is<p>&gt; This is a surprising fact<p>Why? Are there other theories, maybe from earlier in the development of physics, that used a different approach?<p>&gt; We could use particles as the building blocks, and represent each particle by its position and velocity. However, fields turn out to be a much more useful way of representing the way particles behave. A field ϕ(x,t) is a function that takes a point in spacetime and spits something out for each point.<p>I assume that in this passage &quot;building blocks&quot; is equivalent to &quot;objects&quot; in the last passage? Why are fields more useful? Is there an example of what using particles as objects would look like? In particular, a field looks to me like a function; if you used particles as an object, would you represent a particle as a function using its position and velocity? Would that function have time as a parameter like fields do? Typing that out I can kind of see why you&#x27;d use fields<p>&gt; For example, a field could take a position (in spacetime) and spit out a number (which could be real or complex)<p>What does the output represent? Anything in particular? If not, then it seems like you could define the field function to be anything since the output doesn&#x27;t represent anything, then when you feed the field function into the Lagrangian eventually you&#x27;d get massively different results<p>&gt; The simplest Lagrangian we can write for this field is: L=δtϕ⋆δtϕ<p>How is the Lagrangian constructed? This &quot;simplest&quot; Lagrangian is the derivative of the field with respect to time, along with the derivative of the field&#x27;s complex conjugate with respect to time, but how&#x27;d you know to do that? What makes this the simplest possible Lagrangian? Calling this the &quot;simplest Lagrangian&quot; hints that there are other equally valid ways to create a Lagrangian; is that correct? What are the rules for that? Why would you make a more complex Lagrangian?<p>&gt; One interesting observation about Lagrangians is that any term of the form V(ϕ) represents potential energy.<p>What is V(ϕ)? My initial assumption would be velocity, but how do you take velocity of a field? Actually, I can see what they&#x27;re doing: velocity of a particle is the derivative of it&#x27;s position with respect to time, so I guess V(ϕ) aka velocity of the field is the derivative of the field with respect to time. That could&#x27;ve stood to have been spelled out<p>&gt; Since there are no time or space derivatives involved<p>Ok I guess V(ϕ) doesn&#x27;t represent the velocity of the field. I&#x27;ve got no clue what it is<p>&gt; Now let’s take this a step further. In the previous example, we rotated the field at all points by some angle. But why do we need to rotate the field the same way everywhere? What if we measure things at one point with one coordinate system, but measure them at a different point using a different coordinate system that’s rotated. Although it’s hard to imagine why anyone would want to do this, one would still expect that this shouldn’t affect the actual physics predicted by the theory... Multiplying these together, we see that the Lagrangian is different. This is not what we wanted - rotating the complex plane in has affected the results of our theory.<p>I have no idea what&#x27;s going on here. Why would you measure different points of the field with different coordinate systems and expect sensical results? I&#x27;m imagining a surveyor walking in a line starting from the origin: he takes a measurement at the origin, then at (1,0), then at (2,0), then at (3,0), etc. (Imagine that the underlying field is frozen in time so we aren&#x27;t dealing with the Lagrangian yet.) Since we know the field equations we can predict what he&#x27;ll measure at each of those points in the line.<p>But if the coordinate system changes with every step, he&#x27;s still moving in a straight line as seen from a bird flying overhead, but at his first step he&#x27;s at (1,0), then at the next step (2,0) turns into (5,1), then at the next step (6,1) (aka (3,0)) turns into (12,-3), etc, because the coordinate system changes each step. It&#x27;s still (1,0),(2,0),(3,0) if you measure in the original coordinate system. But the underlying field wouldn&#x27;t change in that case. Sure, if you put (5,1) into the field equation you&#x27;ll get a different result than if you put in (2,0), but if you&#x27;re only changing the coordinate system then that has to be compensated for in the field equation itself and you&#x27;re not going to get different results for the same physical point. I mean, you should get the same result if you do f((2,0), coordinate system a) as if you did f((5,1), coordinate system b)<p>Edit: I think the core of my confusion is that in order for f((2,0), coordinate system a) to equal f((5,1), coordinate system b) then you need knowledge of how the coordinate system changes, and I don&#x27;t see how that gets incorporated into the function<p>&gt; Note that the issue here is that when we take the derivative of the field, we get an extra iϕδθδt term proportional to the derivative<p>Proportional to the derivative of what?<p>&gt; This property - that a theory is not affected by changing some symmetry parameter throughout spacetime - is called gauge invariance<p>Why is it called that? I assume someone chose that name because it made sense to them for good reasons<p>&gt; Also note that this new term, iAϕ, looks like a potential from the perspective of our field, with V(ϕ)=iAϕ<p>There&#x27;s V(ϕ) again. I still don&#x27;t know what it represents<p>&gt; our theory now predicts some type of force involved in the interaction between our field and this new field<p>Wait, &quot;our field&quot; and &quot;the new field&quot;? What fields are those? We were talking about a field defined by the function ϕ(x,t) and thinking about its Lagrangian. We added a term A to the Lagrangian and that was it. What&#x27;s the &quot;new field&quot;? Why does ϕ(x,t) have an interaction with it? Is A the new field?<p>&gt; This mechanism of introducing an additional field to make an existing theory gauge invariant is exactly what gives rise to photons in the Standard Model! ‘Rotations’ of the electron field correspond to an additional field, called a gauge field, that behaves exactly the way photons do.<p>Ok, I guess A is a new field. I can see how it arises, but I&#x27;m not sure how you actually get to it<p>&gt; This derivative doesn’t really make sense if we aren’t using the same coordinate system everywhere<p>you don&#x27;t say<p>&gt; The way we measure our field at x is different than the way we measure it at x+δx, so to get the actual difference, we need to make the field comparable by fixing it up before subtracting it<p>What is &quot;the way we&#x27;re measuring it&quot;? I think it&#x27;s, basically, the coordinate system of the surveyor changes each step, so that&#x27;s a different &quot;way&quot; of measuring it? I still don&#x27;t see why changing the coordinate system makes new stuff pop out of the equation<p>&gt; As a first step, we can expand it: W(x,x+δx)=1−iδxA(x)+O(δx2)<p>How do you expand it? Are you giving the definition of W(x, x+δx)? How&#x27;d you get that? What&#x27;s O(δx^2)?<p>&gt; A group is a set of elements associated with some operation... What’s important here is that these two sets - the set of rotations, with the operation being composition, and the set of 1×1 complex matrices with the operation being multiplication - have the exact same behavior<p>Where&#x27;d the second set come from? Wait, I see, it&#x27;s just saying that you can say that &quot;multiplying a number by a 1x1 matrix&quot; is the same thing as saying &quot;you can compose a number with a rotation&quot;. It&#x27;s literally the same thing, just said in a less clear manner. Does the new terminology get us anything useful?<p>&gt; U(1) invariance of the electron field gives rise to the photon field. &gt; SU(2) invariance across lepton fields (such as the electron and electron neutrino) leads to W+, W−, and Z bosons. SU(2) has two generators, so there are three gauge bosons. &gt; SU(3) invariance across quark fields leads to eight gluons, since SU(3) has eight generators.<p>What? How do you know how many generators there are? Why does SU(2) have two generators but three gauge bosons?<p>&gt; It’s remarkable how the observation that an equation doesn’t change under some operation, which seems quite trivial, can have deep consequences, dictating the nature of forces and interactions in the theory.<p>Yeah, I think the part I&#x27;m not getting is how changing coordinate systems affects the equation. I think I can see that if you insist on doing something ridiculous like this you&#x27;d need some math to correct for it and if the new correction functions are fields then it looks like new particles popping out, but I don&#x27;t see how that doesn&#x27;t result in an infinite number of new particles. Like, I can add a function f(x) = x^2 to the Lagrangian, then a g(x) = -x^2 to compensate for it, but those don&#x27;t represent new particles do they? Why do those cancel out but A doesn&#x27;t? I just don&#x27;t see how changing coordinate systems results in different results<p>Despite my questions, I think I have a better idea of what&#x27;s going on. You have a function; it should spit out the same numbers when you rotate it; you need a function to correct for the rotation; in physics the new function looks like a particle. I can kinda sorta see how it works now. Thanks for the article!<p>Edit: Ok, I think I&#x27;ve narrowed down my confusion to the θ(t). I can see that if you want to measure the same (x,y) over time as the coordinate systems change even though that (x,y) represents a different physical point every t, then you&#x27;d need to take the change in coordinate system over time into account in the derivative. But I&#x27;m not sure how that would be useful, nor how that would result in new physics over the case of a fixed θ
评论 #33403048 未加载
评论 #33404107 未加载
评论 #33403052 未加载
nyc111over 2 years ago
“Let’s say we have some complex field [. . .] Just to keep things as simple as possible, let’s say that the field is only a function of time.”<p>This is not simplifying things. First they assume that spacetime is not space and time separately but a different entity then reduce this entity to time only. Then we are not dealing with spacetime. This is like assuming a cube then reducing the cube to one dimension but calling the line a cube. It’s not.
评论 #33402445 未加载
评论 #33402506 未加载
psychoslaveover 2 years ago
Let me guess, because symmetry is far more easier to handle from a human cognitive point of view?
评论 #33404589 未加载
behnamohover 2 years ago
I wonder how much of this can be ELI5&#x27;ed with diagrams and intuitions instead of algebra.
评论 #33402044 未加载
评论 #33402390 未加载
评论 #33402061 未加载