Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

0

So basically, due to Wigner, determining the unitary, infinite dimensional irreps of Poincare group in a sense can be reduced to specifying the unitary irreps of its little groups, which in the massive case is so(3), and hence for the massive case the irreps are all finite-dimensional and given by the spin. Using Wigner's Little group one can classify ...

0

Each of those terms represent the price to pay, in terms if energy, to have some specific configuration of the field: configurations that change with time (price is estimated by time derivative) configurations that change with space (price is estimated by the gradient) magnitude of the field (mass plays a role in enhancing this effect) All those terms ...

0

Instead of $-e\Phi+e\mathbf{v}\cdot\mathbf{A}$ you will have something like $-j_\mu A^\mu$, where $j^\mu=i e(\psi^*\psi_{,\mu}-\psi^*_{,\mu})-2 e^2 A_\mu\psi^*\psi$.

1

You're going to learn that the $4$-potential is intimately connected with the process of taking derivatives. So, you begin with a Lagrangian that doesn't have an E&M interaction. For what you have written, the Lagrangian is $$L = \int \mathrm{d}x \ \frac{1}{2} (\partial_\mu \psi)^* \partial^\mu \psi,$$ assuming the metric signature is $(+,-,-,-)$. The ...

0

The reason is that the S matrix is written like $$S=\langle f|{\cal T}\exp\left(-\frac{i}{\hbar}\int dt'V(t')\right)|i\rangle$$ being $\cal T$ the time-ordering operator. This makes the integration with respect to time Lorentz-invariant.

3

In perturbation theory of QFT, we often (possibly implicitly) Taylor expand the action $$S[\phi]~=~S[\phi_0] ~+~\underbrace{\left. \frac{\delta S[\phi]}{\delta \phi^{\alpha}}\right|_{\phi_0}}_{=0}\eta^{\alpha}~+~\frac{1}{2}\left. \frac{\delta^2 S[\phi]}{\delta \phi^{\alpha}\delta \phi^{\beta}}\right|_{\phi_0}\eta^{\alpha}\eta^{\beta}~+~{\cal O}(\eta^3) \tag{... 0 If you write a general potential as a function of the field and you expand in Taylor series you have a linear term but if you want to have some global minimum of the potential at the field value zero then you have to ensure that there is no linear term. This is the best I can do. 0 This will only answer part of your question, namely "What is the physical motivation for wanting our lagrangian to be invariant wrt to local phase changes? What are some cases where we want local U(1) invariance(?)". I will try to explain why gauge invariance is forced upon us in a particular example. On a general note, you should usually think of local/... -1 You're on the right track. If you look at your third equation again and apply it to the scalar field \psi, you will see that because of the derivative in front, you will get a similar term for \psi as you got for \psi^*:$$ i \partial_t \psi^* = RHS $$Also on the RHS you should get the same minus as you got before, following the same reasoning as ... 0 Using c=\hbar=1 ... The original motive of Schroedinger was to find a wave analogue of particle dynamics. He actually started with the relativistic energy-momentum equation, and using an analogy to classical wave theory, found the equation (the Klein Gordon) he needed. In doing so, he came up with the idea of replacing E with i\frac{\partial}{\partial ... 1 This is easiest done not using index notation. Define the column vector$$ \vec{\phi} = \begin{pmatrix} \phi_1 \\ \phi_2 \\ \phi_3 \end{pmatrix} $$The derivative acts on the column vector as \mathbf{1}_3\partial_\mu where \mathbf{1}_3 is the identity matrix on the same vector space in which \vec{\phi} lives. That is,$$ \partial_\mu \vec{\phi} = \...

6

Momentum is defined via the Noether theorem. It is the conserved charge correspoding to translations. If you consider $\delta x^{\mu} = a^{\mu}$ where $a^{\mu}$ is just a constant four-vector with real components, you can run the Noether procedure to compute $$\delta S = \int d^{4} x \, a^{\mu} \partial^{\nu} T_{\mu \nu}$$ Now given a conserved current $... 0 The main reason is that one generally prefers to work with momentum eigenstates and these are complex exponentials. This is so because we have free fields and the relation$p^2=m^2$is satisfied anyway. One has $$p_\mu=i\partial_\mu$$ and then $$i\partial_\mu\phi_p=p_\mu\phi_p$$ yields complex solutions that are also solutions of the original equation. ... 1 If I got that right - the core of the question is really - why complex numbers. I am also not fond of canonical quantization, in particular because any complex field is representable as two real fields, anyway. So, my answer is that complex numbers come in because if you start with a classical field theory then the requirement to be Lorentz invariant means ... 0 This also caused me a deal of confusion because I too was trying to calculate the contributions$\int_{C_1}$,$\int_{C_2}$in order to determine the integral along the real axis (as is so often the target of exercises in complex variable courses). The key realisation is that the propagator$\Delta_F(x-y)$is defined (by Tong) as the entire integral \begin{... 1 All field entries with time component are non-dynamical; you can see this using the field equations :$\partial^{\alpha} \partial_{[\alpha} B_{\beta \gamma]} = 0$Therefore, use temporal gauge :$B_{0i} = 0$The dynamical variables are :$B_{ij}$. Similarly the gauge transformations :$B_{ij} \rightarrow B_{ij} +\partial_{[i} \epsilon_{j]}$Hence dof =... 1 The last term of (5), $$\hat{\phi}^a \hat{\phi}^b \epsilon^{bcd}\partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d= \delta^{ae} \epsilon^{bcd} \hat{\phi}^e \hat{\phi}^b \partial_\mu \hat{\phi}^c\partial_\nu \hat{\phi}^d ~.$$ Now recall the identity $$\delta^{e[a} \epsilon^{bcd]} =0,$$ since you cannot antisymmetrize four indices! Consequently $$\... 1 The term you computed is identically zero, since partial derivatives commute, and the \epsilon symbol is fully antisymmetric. In the language of differential forms, the extra term in the Lagrangian is \mathrm{d}A\wedge\mathrm{d}A, and the contribution to the equations of motion is simply \mathrm{d}(\mathrm{d}A)=0, since the exterior derivative is ... 3 By definition of the Levi-Civita symbol it is antisymmetric under permutations of its indices. So under permutation of \alpha and \mu it would obtain a minus sign. On the other hand the double derivative \partial_\alpha\partial_\mu is symmetric under this permutation. The contraction of a symmetric and an antisymmetric object is 0. 2 Monopoles are somewhat subtle, and there are different layers towards their classification, each being more correct than the previous one. Quick remark: what is usually called the \text{SO}(10) model should more properly be called the \text{Spin}(10) model, because it contains spinors. One has \pi_1\text{Spin}(n)=0 and \text{Spin}(2n)^\vee=\text{PSO}... 0 \nabla^{α}(\nabla_{α}φ\nabla_{β}φ) -\cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ\nabla_{ν}φ) - \nabla^{α}g_{αβ}V(φ) = 0 \Rightarrow \nabla^{α}\nabla_{α}φ\nabla_{β}φ +\nabla_{α}φ \nabla^{α}\nabla_{β}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla^{α}(\nabla_{μ}φ)\nabla_{ν}φ - \cfrac{1}{2}g_{αβ}g^{μν}\nabla_{μ}φ\nabla^{α}\nabla_{ν}φ - \nabla^{α}g_{αβ}V(φ) = 0 \... 1 Ok, this is the free Klein-Gordon Hamiltonian. Strictly, you should be writing this in terms of the momentum canonically conjugate to \phi. By convention, we use the symbol \pi for that. So, your original Hamiltonian density is \begin{array} \mathcal{H} = \frac{1}{2} \pi^2 + \frac{m^2}{2}\phi^2 + \frac{1}{2}(\nabla\phi)^2. \end{array} There's a good ... 0 I think it's more straightforward to vary the action with respect to \phi and cancel surface terms. Also, check this: Derivation of Klein-Gordon equation in General Relativity 2 From Ref. 1, one sees that the Euler-Lagrange equations generalize to$$\frac{\partial \mathcal{L}}{\partial\phi}=\nabla_\mu\left(\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}\right)$$In general, when one does these things, one must make sure of two things: that the index placement is the same, and that the indexes are NOT the same in the derivative ... 0 Hint: try to act with P_\epsilon^{\mu\nu} on k_\nu. 3 Applying the operator P^{\mu\nu} to a generic vector v_\nu,$$ -P^{\mu\nu} v_\nu = v^\mu - \frac{k^\nu v_\nu}{M^2} k_{\mu} . $$That is, up to a sign, P "removes" (projects out) from v its component parallel to k (assuming k_\mu k^\mu = M^2). Therefore, it projects any vector to the surface orthogonal to k. 0 Briefly, the action (Lagrangian^1 density) in the path integral are functionals (functions), respectively, as opposed to operators. This is a consequence of how the path-integral formalism is derived from the operator formulation (by inserting infinitely many completeness relations). The action & Lagrangian density usually don't depend on Planck's ... 2 You are just missing something very simple. There are two appearances of the variable \phi_i in that sum$$\frac{d}{d\phi_i}\sum_j(\phi_{j+1}-\phi_j)^2=\frac{d}{d\phi_i}\left((\phi_{i+1}-\phi_i)^2+(\phi_{i}-\phi_{i-1})^2\right)=-2(\phi_{i+1}-\phi_i)+2(\phi_{i}-\phi_{i-1})$$2 Hints: Recall that the action$S=\int \! \mathbb{L}$is an integral over the Lagrangian 4-form$\mathbb{L}~=~ d^4x~ {\cal L}$, where${\cal L}$is the Lagrangian density. It's a density in the sense that under a coordinate transformation$x\to x^{\prime}$, it transforms as${\cal L}^{\prime}={\cal L}/J$with the inverse Jacobian$J:=\det\frac{\partial x^{\...

0

My bad, I see now what happened, and as @Qmechanic noted above, Lagrangians are invariant under the addition of total derivatives.

2

There are at least 2 issues with OP's discussion (v2): One should properly distinguish between total and explicit spacetime derivatives, cf. e.g. my Phys.SE answer here. In particular, an infinitesimal quasisymmetry of the Lagrangian (density), means by definition that the infinitesimal variation is a total (space)time divergence. Note that not all terms ...

3

This isn't an answer, just an extended comment documenting a failed attempt, and why it fails in higher-dimensional space even though it works fine in 2- and 3-dimensional space. To keep this brief, I'll assume that the calculus of differential forms is familiar. Canonical form for the field Let $\mathbf{x}$ denote a point in flat $N$-dimensional space, ...

0

As you noted $A'^{\mu}(x') = \frac{\partial x'^{\mu}}{\partial x^{\lambda}}A^{\lambda}(x)$ is how a four-vector transform under Lorentz transformations, but a translation $x^\mu+\epsilon^\mu$ is not a Lorentz transformation. Some people call translations Galilean Transfromations. The group of Lorentz transformations + Translations is called Poincarè Group. ...

2

Spacetime isn’t a field. Spacetime is the arena where quantum fields live and interact. Each kind of elementary particle has a field, with values at every point in spacetime. For example, there is only one universal electron field, which fills spacetime, and all electrons and positrons are its quanta. The various quantum fields (for quarks, charged leptons, ...

0

In the kind of particle physics that is widely accepted, such as the Standard Model, quantum fields for quarks, charged leptons, neutrinos, photons, weak bosons, gluons, and Higgs exist in spacetime but are not part of the geometry of spacetime. Only gravitation currently has an accepted geometrical explanation. The electromagnetic, weak, and strong forces ...

1

What I could understand is that particles are different kinds of local fluctuation states of force fields. Particles are specific kinds of states (a certain asymptotic states defined according to the LSZ formalism) of the quantum fields--they do not necessarily correspond to the states of force fields. For example, the electromagnetic field is what can be ...

Top 50 recent answers are included