What I said earlier is correct if you think $$x^\mu$$ as a particular coordinate direction ($$\mu$$ is a fixed index). But if you think in a coordinate invariant way ($$\mu$$ is an abstract index, merely expressing the type of the object), $$\partial_\mu\phi:TM\rightarrow\mathbb{R}$$ is the derivative map of $$\phi:M\rightarrow\mathbb{R}$$, where $$M$$ is the underlying (spacetime?) manifold. So $$\partial_\mu\phi$$ is a section of the cotangent bundle, $$\partial^\mu\phi$$ is a section of the tangent bundle, and $$\partial_\mu\phi\partial^\mu\phi$$ is the 'inner product' between them.
The QH QM QA thread.
- Thread starter QuarkHead
- Start date
Well, temur originally said what I said. He then said that the $$\partial_{\mu},\;\partial^{\mu}$$ are vector fields, and I think he must be right. The fact that the $$\partial_{\mu}, \; \partial^{\mu}$$ are interacting with scalar fields, rather than elements of these fields strongly implies they are vector fields.
temur Let's not get bogged down in the niceties of manifold theory - we can do that in the other thread if you like.
Anyway, to continue.
This is a nice choice of transformation, since $$e^{-i \alpha}e^{i \alpha} = e^0 = 1$$, this implies that $$e^{-i \alpha}\phi e^{i \alpha}\phi^* = \phi \phi^*$$. This transformation is an isometry!
They are? As far as I know the Lorentz transformations are concerned with bringing NON-local coordinate functions into register. Or are you perhaps using the term "local" in a way I don't understand?
So what's with the centre dot? Surely not scalar product? Or is it? I am confused Ok, I shan't attempt to follow the math that follows this, but I have one more question
Well I think I see this bit, but in future we may need to talk in more general terms, as I don't really understand your math.
So, here is my final question. Why did you select the $$\partial_{\mu}\alpha(x) = A_{\mu}$$. Why could you not have set $$\partial^{\mu}\alpha(x) = A^{\mu}$$? Moreover, I am not happy about the "local spacetime coordinate" $$ x$$. As far as I am aware (I am happy to be corrected), each and every point in spacetime has a minimum of 4 local spacetime coordinates.
I'm going back to my knitting, I find this stuff too hard.
Absolutely. You just showed that the mass term and the $$\lambda$$ term are invariant under the same transformation. This is why I chose it
Well, ``local'' means ``a function of a space-time coordinate''. So the Lorentz transformation tells you how to link one coordinate frame with another coordinate frame, as you said, which require you to know something about the separation between the two frames, which requires you to set some coordinates on your space-time.
Nothing magical, I think.
Hmm. Would this help? $$\phi$$ transforms as a tensor of rank 0 under the Lorentz group, whereas $$A_{\mu}$$ transforms as a tensor of rank 1.
More abuse of notation, I'm affraid. Firstly, I should have warned you, we're working in a Minkowski space time, which is a bit boring. The metric is $$\eta_{\mu\nu} = diag(+,-,-,-)$$. The raising and lowering of indices is just accomplished with the metric:
$$A_{\mu} = \eta_{\mu\nu}A^{\nu}$$.
There's probably a fancy math word that would make you understand all of this, but sadly I don't know it.
The space-time coordinate $$x$$ is an abbreviation for $$x_{\mu}$$, which I was pretty careless about. So you are correct, there ARE four coordinates hiding in $$x_{\mu}$$.
Yes you should!! I briefly entertained the idea we were in phase space!
Yes you do, it's called tensor contraction (recall that $$\eta_{\mu\nu}$$ is a type 0,2 tensor)
Aargh, phucking physicists!! You explicitly stated that $$\phi$$ is a scalar field and that $$A_{\mu}$$ is a vector field. When will you guys learn that a) a tensor is not the same as a tensor field and that b) tensors can be satisfactorily defined as objects without reference to their transformation properties.
Grrr. Didn't I go to some length to explain this in another thread? (PS, I love you....)
Anyhoo, before I go off my head, how about this..
Now, I have a ton of questions about spin lined up, but I would prefer to defer them. But where does the "charge" come from. As I dimly recall, there is a theorem of Noether to the effect that for every symmetry there is a conserved current, or some such. I didn't understand it then, and I don't now.
Is it relevant here ?
If anyone chooses to answer, no physicists math, please, it confuses me!!
I also would like to know what is difference between "charge" and "flavour". How these are reflected in Lagrangian for example?
There was a thread opened by Ben that answers these kind of questions but I could not find. I posted here because if I open one for myself there will be many similar threads. If this is not the place for questions by me please guide me to an appropriate place.
There was a thread opened by Ben that answers these kind of questions but I could not find. I posted here because if I open one for myself there will be many similar threads. If this is not the place for questions by me please guide me to an appropriate place.
Ehh...Yes, Noether's theorem is important. We could have used Noether's theorem to tell us that the action was invariant under the U(1) symmetry which we invoked earlier.
Noether's theorem will give you the conserved current, which is a (Lorentz) vector (tensor of rank 1).
Also, you can think of spin as the transformation properties of the object under the Lorentz group.
Ahh... Well, the idea of defining a tensor without referring to its transformation properties more or less goes against the grain of how we do physics. So something that transforms as a tensor (field) of rank 0 under the Lorentz group is CALLED a scalar (read: Lorentz scalar), and has spin 0. Further, the scalar (field) can be a representation of other symmetries, under which it ISN'T a scalar.
Plus, it is now aparent to me that you mathematicians have monopolized the vocabulary... After spending a few minutes on Wikipedia, I DO think that we mean the same thing when we say ``field'', and more properly, $$\phi$$ and $$A_{\mu}$$ are scalar (type (0,0) tensor) fields and vector (type (0,1) tensor) fields, respectively. ``Scalar'' and ``Vector'' refer to the field's transformation properties under the Lorentz group.
Perhaps it is best to go back and clarify some of the vocabulary, which I took for granted. I am (clearly) not used to being strict in defining things, so we should work very slowly.
Well...Flavor is just a word that distinguishes quarks (i.e. we've six flavors of quarks), whereas charge is a more general idea that generally dictates how a specific field transforms under some U(1).
So $$\phi$$ has charge +1 and $$\phi*$$ has charge -1, because
$$\phi \rightarrow e^{i\alpha(x)}\phi$$
$$\phi* \rightarrow e^{-i\alpha(x)}\phi*$$.
Thanks for the replies. I am just trying to understand these things. So flavor is like "weak charge", comparable to color in the strong interaction, and electric charge in electrodynamics right? If you compare electrodynamics with QCD, do positive and negative charges correspond to a color and its anti-color, or to two different colors? What would be magnetic charge in terms of colors?
Ben, I hope you weren't offended by my remark about physicists; it was intended as a friendly interdisciplinary ribbing. In general, when I get anal like that I am best ignored - of course we mean the same thing when we talk about fields, vectors, scalars etc.
And if it suits you to define scalars and vectors in terms of their Lorentz transformations, go ahead. It wouldn't suit me, as I would like to be able to use these terms in different contexts, but that's neither here nor there
And if it suits you to define scalars and vectors in terms of their Lorentz transformations, go ahead. It wouldn't suit me, as I would like to be able to use these terms in different contexts, but that's neither here nor there
Heh...no worries man. I've heard it all before from mathematicians.
You are correct though---in general, $$\phi$$ will transform under a set of symmetries, not JUST the Lorentz symmetry. But physics is very much a work in progress, and it takes generations to change things---so the fact that we call $$\phi$$ a scalar field is just because someone, long ago, didn't think that symmetries besides the Lorentz symmetry were important.
To make matters worse, we also use an outdated nomenclature for $$\phi$$'s transformation properties under other symmetries. If $$\phi$$ transforms as a scalar under some other symmetry, it is called a ``singlet'' under that symmetry. This is old notation from the days when people did spectroscopy.
So between the three of us, I think we should be able to understand things a bit better.
You are correct though---in general, $$\phi$$ will transform under a set of symmetries, not JUST the Lorentz symmetry. But physics is very much a work in progress, and it takes generations to change things---so the fact that we call $$\phi$$ a scalar field is just because someone, long ago, didn't think that symmetries besides the Lorentz symmetry were important.
To make matters worse, we also use an outdated nomenclature for $$\phi$$'s transformation properties under other symmetries. If $$\phi$$ transforms as a scalar under some other symmetry, it is called a ``singlet'' under that symmetry. This is old notation from the days when people did spectroscopy.
So between the three of us, I think we should be able to understand things a bit better.
Well, the thing that I've been dancing around is that QED is described by U(1), and QCD is described by SU(3). Further, the generators of the group play the role of the bosons.
So if you look above, U(1) is generated by some phase $$\alpha$$---if I'm being sloppy with definitions again, forgive me! But essentially we identified $$\alpha(x)$$ (some U(1) rotation) with the photon. Likewise, we can do the same exercize with SU(3), and find not one generator but 8. So instead of having one gluon, we have 8.
So to answer your question, it's a generalization. The electron has negative charge, but quarks can be red, green, or blue. To obfuscate matters further, sometimes this is called ``color charge''. Likewise, flavor can be thought of as ``weak charge''.
Find a basis for it's algebra! I confess I am still slightly confused by Ben's terminology, but it looks like this what he is getting at - the algebra of U(1) has a single basis vector, the algebra of SU(2) has three, the algebra of SU(3) has 8 basis vectors. These will define the space of all actions on these groups (I think - correct me someone if I misunderstood the question)
That's exactly correct. We call the basis is called the adjoint representation.
============= \begin{Invisible to QuarkHead} ===================
Usually, the way we think about representations is in terms of matrices, although this tends to drive mathematicians nuts.
for example, SU(3) is a group of 3x3 unitary matrices (complex conjugate + transpose = inverse), which have determinant 1. It turns out that you can write any member in the group as a linear combination of 8 basis vectors, typically called the Gell-Mann matrices. These matrices act on a three component vector---in this case, the 3x3 matrix is thought of as a gluon, and the three component vector describes a quark.
============= \end{Invisible to QuarkHead} ====================
Well, I've been tugging my beard furiously over spin, and while I still don't fully understand it, I discovered something that may be relevant.
So. I dug out an old exercise from 3 years ago, where I had found the basis for the algebra of SU(2), let's call it su(2), to be the set of traceless, anti-hermitian matrices
$$ X = \begin{pmatrix}0 & i\\
i & 0 \end{pmatrix},\quad Y= \begin{pmatrix}i & 0\\
0 & -i \end{pmatrix}, \quad Z= \begin{pmatrix}0 & 1 \\
-1 & 0 \end{pmatrix} $$.
I then looked up the Pauli spin matrices, let's use the standard physicists notation here, and saw straight away that $$X= i\sigma_x,\; Y = i\sigma_y,\; Z = i\sigma_z$$.
Now such coincidences do crop up from time-to-time in Lie theory, and most of them have no real significance.
Is this one of them?
So. I dug out an old exercise from 3 years ago, where I had found the basis for the algebra of SU(2), let's call it su(2), to be the set of traceless, anti-hermitian matrices
$$ X = \begin{pmatrix}0 & i\\
i & 0 \end{pmatrix},\quad Y= \begin{pmatrix}i & 0\\
0 & -i \end{pmatrix}, \quad Z= \begin{pmatrix}0 & 1 \\
-1 & 0 \end{pmatrix} $$.
I then looked up the Pauli spin matrices, let's use the standard physicists notation here, and saw straight away that $$X= i\sigma_x,\; Y = i\sigma_y,\; Z = i\sigma_z$$.
Now such coincidences do crop up from time-to-time in Lie theory, and most of them have no real significance.
Is this one of them?
No coincidence! Bear with me, and remember to breathe.
The Pauli matrices, as you've written them, act on a two component vector---more generally, you'd probably say that the fundamental (or smallest) rep of su(2) is 2 dimensional and real. This means that the 2x2 matrices act on a 2 component, real vector. This means that we have 2 degrees of freedom---let's call them ``spin up'' and ``spin down'', and write them as
$$ \left|\uparrow\rangle, \left|\downarrow\rangle$$
I'm using the bra-ket notation of Dirac, which I think you hate. If we're good so far, then we can keep going, but if not then we should clear up any issues.
The Pauli matrices, as you've written them, act on a two component vector---more generally, you'd probably say that the fundamental (or smallest) rep of su(2) is 2 dimensional and real. This means that the 2x2 matrices act on a 2 component, real vector. This means that we have 2 degrees of freedom---let's call them ``spin up'' and ``spin down'', and write them as
$$ \left|\uparrow\rangle, \left|\downarrow\rangle$$
I'm using the bra-ket notation of Dirac, which I think you hate. If we're good so far, then we can keep going, but if not then we should clear up any issues.