# Tag Info

10

Putting space and time on the same footing means to treat time as another dimension in addition to the other three physical dimensions. In the context of relativity, time is treated as another dimension (but within this idea of Spacetime space and time are not the same). In classical Newtonian physics, space is treated within the ideas of three dimensional ...

8

After some thought, this is what I understand: In Newtonian physics, a particle's path can be specified by $x^i(t)$ where the time $t$ can be seen as an independent parameter. The space coordinates $x^i(t)$ are dependent variables that depend on $t$. We thus say that space and time are not treated on an equal footing. In relativity, a particle's worldline ...

8

We have to be careful here about what we mean by "field." A field is a mathematical object that has been found to be very useful in making physical theories. I don't want to get bogged down in definitions, so I'll take a more conceptual approach. An example could be basic quantum mechanics, where the state of a quantum system is represented by the ...

5

Here is the action written with explicit sums over the indices $$S = \int {\rm d}^4 x \sum_{\dot{\alpha}=1}^2 \sum_{\beta=1}^2 \sum_{\mu=0}^3 \chi^\dagger_{\dot{\alpha}} i \sigma_{\dot{\alpha} \beta}^\mu \partial_\mu \chi_\beta$$ $\chi$ is a two component spinor and $\partial_\mu \chi$ is the gradient of a spinor. At a deeper ...

4

Consider the following picture. We have a field which is large in the red rectangle and small elsewhere. The function which tells us the field value at some point at coordinates $\mathbf x$ is $\phi$; that is, $\phi(\mathbf x)$ is the value of the field at the point labeled by coordinates $\mathbf x=(x^1,x^2)$. Now we perform an active transformation ...

3

In this case you can demonstrate that this Lagrangian represents a massive field. You do this by looking at the exponential term $e^{m\phi}$ in $\mathcal{L} = \frac{1}{2} (\partial_\mu \phi )^2 - e^{m \phi} + \lambda \phi^4$ and performing a Taylor expansion. That is $e^{m\phi} = 1 + m \phi + m^2 \large \frac{\phi^2}{2} + ....$ This means the Lagrangian can ...

3

Fundamentally, I guess my confusion is this: In classical E&M I know that picking q>0 or q<0 will lead to different motion for positive and negative charges (for example via the Lorentz force law). This is not true. It's certainly true that in a given electromagnetic field, the motion of a positively charged particle will be different than the ...

3

One can get extended SUSY by combining two supersymmetric multiplets into one with the appropriate choice of coupling constant. For example - in order to get $\mathcal{N} = 2$ SYM (Super Yang-Mills) one combines the gauge multiplet with gluon $A_\mu$ and gluino $\lambda_\alpha$ with the chiral multiplet with a fermion $\psi_\alpha$ and a scalar $\phi$, all ...

3

OP's question is not about the Klein-Gordon equation per se, but rather about the very definition of the functional/variational derivative (FD). OP's eq. (4) is just plain wrong. To complicate matters, OP's eqs. (8) refers to a misleading 'same-space' FD notation, which when applied naively, leads to a mismatch of integral signs in OP's eqs. (3), (7) & (...

2

How do particles know how to interact with other particles / fields ? Interactions depend on the properties of particles. How do we know that, for example, the charge on an electron never changes ? This is an empirical fact - we have never seen this happen. How do we know that all electrons have the same charge ? Once again, experimental evidence. Could we ...

2

I suspect you are unfamiliar with the (mercifully) loose language of physicists. Strictly matrix realizations of a Lie algebra, and hence group, are termed Representations, like (3.18); but anything else, including (3.16) and its SU(2) antecedent, are just termed Realizations: versatile maps (linear, in this case) which satisfy the Lie Algebra (3.17). In ...

2

TL;DR: The main rule is that kinetic terms should be positive, cf. above comment by G. Smith. Examples: The Lagrangian density for a scalar field is $${\cal L}~=~\pm \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi -{\cal V}(\phi),$$ with EL equations $$\mp\Box\phi~=~{\cal V}^{\prime}(\phi),$$ if the signature Minkowski metric is $(\pm,\mp,\mp,\mp)$, ...

2

You are most likely confused because methods from functional calculus are often obscured by hacks to make the calculations easier to understands but it also hides a lot of the intuition. So starting with functionals. Functionals are objects that take in functions and spit out a real (complex) number. They are denoted by square brackets to indicate that they ...

2

You simply derivate every component. You can write (sum over repeated indices implied, where $\alpha,\beta$ run from $1$ to $4$) $$L = \psi^*_{\alpha} \gamma_0 (i (\gamma_{\mu})_{\alpha \beta} \partial^{\mu} -m) \psi_{\beta}.$$ Now the components $\psi_{\alpha}$ are just complex numbers (or operators in the quantum theory) and so is $$\pi_{\alpha} \equiv \... 2 I often read from textbooks that in relativity, space and time are treated on an equal footing. What do authors mean when they say this? I actually give a brilliant aid to understand what does it mean? It's called surveyors parable introduced by Tayloe and Wheeler. Suppose a town has daytime surveyors, who have the North star. These notions differ, of ... 2 It's actually self-explanatory, but you have made a hash of terminology and illustrations to concoct a mystery. You are talking about the same object, really. Your unitary V s are group elements, exponentials of elements L in the Lie algebra (in which you also take ψ to be: in your setup, it is traceless hermitian). The dimensionality of the matrices you ... 1 \chi_{\sigma \mu \nu} = \frac{1}{2}(\chi_{\sigma \mu \nu} + \chi_{\sigma \mu \nu}) = \frac{1}{2}(\chi_{\sigma \mu \nu} + \chi_{ \mu \sigma \nu}) = 0 due to the anti-symmetry property of \chi, \chi_{\sigma \mu \nu} = -\chi_{\mu \sigma \nu}. The minus sign does not need to be where you think it should be since you can also write \chi_{\mu \sigma \nu} = -... 1 It is a generic feature of differential equations in second order partial derivatives that knowing the values of fields on a closed boundary establishes uniquely the solution in the bulk region. Some caveats are: Some equations are degenerate and admit multiple bulk solutions, hence invalidating the claim above. In physics, such situations are usually ... 1 The division of the total angular momentum into orbital and spin parts is unphysical, a bit like how the division of energy into potential energy and kinetic energy is unphysical. (Are static electric and magnetic fields purely potential energy while electromagnetic waves are purely kinetic? This is not a physically meaningful question, so we don't ask it.) ... 1 Your observation is correct. The total angular momentum, integrated over volume, contains both contributions as pointed out in the other answer. However it takes the overall for of an orbital, position dependent, angular momentum while clearly the electromagnetic potential has internal AM as well. My answer is written down in a peer reviewed, published ... 1 The symmetric EM tensor does contain the spin. There is a discussion by Michael Berry about this, and also some in Wikipedia here 1 Do not confuse multiplication with the idea of a dot product. The problem at hand, of constructing the kinetic term in components, is not equivalent to multiplying (a+b)(c_d). Rather, we have two vectors whose components are \langle \partial_0,\nabla\psi\rangle (the spatial components could be written out as well) and we are computing a dot product as ... 1 In Special Relativity, there is the invariant interval defined as$$\Delta s^2=c^2\Delta t^2-\Delta x^2$$(for relative motion in the x-direction only). Here \Delta t and \Delta x are the difference in t and x for two events in some frame of reference. It has the same value in any other inertial reference frame using that frame's coordinates t' and x' to ... 1 One possible approach is writing the sum of logs as a log of product and using the formulas for infinite products in Gradshtein and Ryzhik. 1 S=\sum_n\ln(-i\omega_n+\epsilon)=\sum_n\ln(i\omega_n-\epsilon)+C (usually this is an action and the constant is irrelevant. This transform is unneccessary just for convenient) =\mathrm{Res}\left\{\ln(z-\epsilon)g(z)\right\} when g(z) is what you've mentioned. Then the problem is to evaluate this integral. We could select the branch cut as (-\infty,\... 1 For this kind of stuff you usually integrate by parts. First change your sum to an integral:$$\sum_n \log\left(-i\omega_n +\frac{k^2}{2m}+\mu\right) \Rightarrow \int \mathrm{d}\omega \, \log(f - \mathrm{i}\omega), $$where f here is k^2/2m+\mu which I am assuming are not functions of \omega. Then integrate by parts:$$\int \mathrm{d}\omega \, \...

1

Yes each point of the space have some field energy located there. The potential energy is another expression of it in a simpler form. This equality can be seen by integration by part. $$\mathcal{E}_{field}=\iiint (\nabla \phi)^2d^3x=-\iiint \phi\Delta\phi d^3x = \iiint\phi \rho d^3x = \mathcal{E}_{potential}$$ where $\phi$ is the potential and $\rho$ the ...

1

It is the difference between a discrete and a continuum approach. In classical discrete mechanics, there are particles moving in 3D space along the time. The action is the sum of all integrals of their Lagrangians between $t_1$ to $t_2$. But in continuum classical mechanics for example, the displacement field of the material have to be dealt with, and ...

1

Your evaluation of the functional derivative in your third equation is wrong. It should read $$\dot \phi(x)= \frac{\delta H}{\delta \pi(x)}= \pi(x)$$ There is no integral. The rest of what you have wirtten has the same problem. Read up on how to define functional derivatives. Let's actually evaluate the functional derivatives:  \delta H = \int d^dx\left (...

1

Your three questions are three different ways of recasting a single problem: what is a field? Let me start with what a field is not. It cannot be "a region of space" equipped with special properties. The reason is already evident in your comment, but I would add that there is no way to make sense of a negative or even complex region of space. ...

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