# Tag Info

58

I'm going to go with a programmer metaphor for you. The mathematics (including "A field is a function that returns a value for a point in space") are the interface: they define for you exactly what you can expect from this object. The "what is it, really, when you get right down to it" is the implementation. Formally you don't care how it is implemented. ...

23

You say: she said to me that, if I wanted hardcore definitions, a field is a function that returns a value for a point in space. Now this finally makes a hell lot of sense to me but I still don't understand how mathematical functions can be a part of the Universe and shape the reality. You don't have to use super-complicated examples ...

19

The main distinction you want to make is between the Green function and the kernel. (I prefer the terminology "Green function" without the 's. Imagine a different name, say, Feynman. People would definitely say the Feynman function, not the Feynman's function. But I digress...) Start with a differential operator, call it $L$. E.g., in the case of ...

18

The higher the number of derivatives the more initial data you have to provide. If you have some Lagrangian that contains an infinite number of derivatives (or derivatives appearing non-polynomially, such as one over derivative) then you have to provide an infinite amount of initial data which amounts to non-local info, in the sense explained below. If you ...

17

1) yes, it basically will find a non-optimal solution. At every point, the top of the ray looks for the bigger potential gradient, the charge in the surrounding volume grows, polarizing surrounding material (air, in this case) until a bigger gradient shows up and the ray continues over that direction. This is why the lightining path looks like a jigsaw; its ...

16

Vladimir's answer has the right essence but it is also misleading, so let me clarify. The formula $$H = \sum_i p_i\dot q_i - L$$ relating the Hamiltonian and the Lagrangian is completely general. It holds in all theories that admit both Lagrangians and Hamiltonians, whether they're relativistic or not, whether or not they have any other symmetry aside ...

14

General approach First recall that Euler-Lagrange equations are conditions for the vanishing of the variation of action $S$. For a scalar field $\Phi$ with Lagrangian density $\mathcal L$ on some open subset U we have $$S[\Phi] = \int_U {\mathcal L}(\Phi(x), \partial^{\mu}\Phi(x)) {\rm d}^4 x$$ Consider a variation of the field in direction $\chi$ and ...

13

A field theory is a physical description of reality in which the fundamental entities are fields, i.e. objects having no definite spatial location but a certain value or intensity at each place. Examples of fields are the temperature in a room, for each location in the room, a temperature can be specified, although in most cases temperature will be pretty ...

13

Update to address new questions. The answer to this question is no. At least if you take the question purely formally. Only theories such as classical field theory, quantum field theory and continuum mechanics are field theories (you generally recognize them by having continuous degrees of freedom; also they usually have the word field in the title :-)). ...

12

First of all, it's not true that all important differential equations in physics are second-order. The Dirac equation is first-order. The number of derivatives in the equations is equal to the number of derivatives in the corresponding relevant term of the Lagrangian. These kinetic terms have the form $${\mathcal L}_{\rm Dirac} = \bar \Psi \gamma^\mu ... 11 Whether your current j^\mu is conserved off-shell depends on your definition of j^\mu. If you define it via the Dirac and other charged fields, it will only be conserved assuming the equations of motion. However, if you define j^\mu via$$ j^\mu = \partial^\nu F_{\mu\nu}, $$i.e. as a function of the electromagnetic field and its derivatives, then ... 10 The electric field itself is not accessible by experiments. We can only observe e.g. trajectories of charged particle, etc., to find the forces they are subjected to. It all comes down to the electric field just being a theoretical concept used to describe the phenomena covered by electrodynamics. Thus, we cannot make a definite statement on the nature of ... 9 One can rewrite any pde of any order as a system of first order pde's, hence the assumption behind question is somewhat questionable. Also there exist first order PDE's of relevance to physics (Dirac equation, Burgers equation, to name just two). However, it is common that quantities in physics appear in conjugate pairs of potential fields and their ... 9 In general, boundary conditons must be adapted to the real situation. Zero boundary conditions are just for the sake of simplicity. But they are realistic only when the field is really zero for some definite reason. If the boundary is at infinity, zero boundary conditions means that everything of interest happens in a finite domain and cannot be noticed ... 9 General Mumbo-Jumbo about Statistics When you have any Hamiltonian mechanical system, with degrees of freedom q_i, conjugate variables p_i, and Hamiltonian H(q_i,p_i) there is a conserved phase space volume, which is just the area in q,p space, defined by the volume element$$\prod_i dp_i dq_i$$The conservation of phase space volume is Liouville's ... 8 Clearly, an interaction involving \phi(x+h) deserved to be called nonlocal. But since \phi(x+h)=\sum_{k=0}^\infty \phi^{(k)}(x) h^k/k!, any nonlocal interaction can be expressed as a power series involving arbitrarily many derivatives. Therefore an action (or Lagrangian) is called nonlocal if it involves infinitely many derivatives. If there are only ... 8 As Lubos Motl and twistor59 explain, a necessary condition for unitarity is that the Yang Mills (YM) gauge group G with corresponding Lie algebra g should be real and have a positive (semi)definite associative/invariant bilinear form \kappa: g\times g \to \mathbb{R}, cf. the kinetic part of the Yang Mills action. The bilinear form \kappa is often ... 8 The trick is given in equation 4.4 of the attached article: First couple the theory to gravity, (by introducing a metric tensor in the integration measure and for each index raising) obtaining the action: S = \int_M d^4x \sqrt{-g} \mathcal{L} Then vary the action with respect to the metric tensor: T_{\alpha\beta} = \frac{1}{\sqrt{-g}} \frac{\delta ... 8 Just because F^{\mu\nu} has two indices does not mean that it represents a spin-2 particle. Note that the metric g^{\mu\nu} is a symmetric two indexed object while the EM field strength F^{\mu\nu} is antisymmetric. In fact, the metric g^{\mu\nu} is analogous to potential A^\mu in EM and the field strength of gravity is the four indexed Riemann ... 8 Lubos Motl and Vladimir Kalitvianski have already provided correct conventional answers concerning the Legendre transformation from Lagrangian to Hamiltonian formalism. Nevertheless, it seems appropriate to mention that OP's second equation(v2)$$\mathcal{H} ~=~ \pi_{\mu}\partial^{\mu} \phi - \mathcal{L}$$is precisely the starting point for De ... 8 I) It is worthwhile mentioning that there exists a basic approach well-suited to physics applications (where we usually assume locality) that avoids multiplying two distributions together. The idea is that the two inputs F and G in the Poisson bracket (PB)$$\tag{1}\{F,G\} ~=~ \int_M \!dx \left( \frac{\delta F}{\delta \phi(x)}\frac{\delta G}{\delta ...

8

The point is that eq. (1.35) should hold off-shell to have a symmetry, while eq. (1.37) may only hold on-shell. [The term on-shell (in this context) means that the Euler-Lagrange equations are satisfied. See also this Phys.SE post.] In other words: On-shell, the action will only change with at most a boundary term for any infinitesimal variation, whether ...

8

There is indeed a scalar field model of gravity, in fact Einstein originally tried that before settling on a spin 2 description. Scalar gravity is called Einstein-Nordstrom gravity, here is a link to wikipedia: http://en.wikipedia.org/wiki/Nordstr%C3%B6m%27s_theory_of_gravitation. At the nonlinear level it amounts to using $R$ in Einsteins equations instead ...

8

It is not. The correct identity is $$\frac{\delta}{\delta \Phi(y)} \Phi (x) = \delta(x-y)$$ where the derivative is the functional derivative. If $F : D(F)\ni \Phi \mapsto F(\Phi)\in \mathbb C$ is a function from a space of functions $D(F)$ to $\mathbb C$, the functional derivative of $F$, if it exists is the distribution $\frac{\delta F}{\delta \Phi}$ ...

7

Conformal field theories do not have a mass-gap, which is one of the assumptions [for the strong conclusions of non-mixing of Poincare spacetime symmetries vs internal symmetries] of the Coleman-Mandula no-go theorem. Similarly, for its superversion: the Haag-Lopuszanski-Sohnius no-go theorem. [In the supercase, the Poincare algebra is replaced with the ...

7

The actual paper by Haag, Łopuszański and Sohnius covers Conformal Supersymmetry, and it states explicitly that this extension is achieved by NOT assuming the mass gap.

7

Disillusionment with systems described by higher order Lagrangians harks back to a 1950 paper by Pais and Uhlenbeck, in which they showed that such systems were prone to pathologies, including states with negative energy and states with negative norm. There's a more recent discussion of this in arXiv:hep-th/0408104.

7

There are several inequivalent definitions, used in different contexts, which is the reason for your confusion. The word "Chiral" originally refered to chirality, or handedness of spin along the direction of motion. This is still the most often used definition. The spinor representations of the Lorentz group in even dimensions have components with a ...

7

Wigner always complained about people who used the word « invariant » (this was, of course, in the context of Special Relativity): he said one should say that the principle of relativity requires « covariance,» not invariance. Einstein's own papers on GR tend to carry out Wigner's request: the theory of GR (which is more general than Einstein's theory of ...

7

No, the statement is false even in the electric case. At the very beginning, the acceleration is $\vec a \sim \vec E$ so they have the same direction at $t=0$: the tangents agree. However, as soon as the particle reaches some nonzero velocity $\vec v \neq 0$, its acceleration is still $\vec a\sim \vec E$, in the direction of the field lines, however its ...

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