# Tag Info

3

There is really no complication in arriving at equation (5) given equation (5). We have: $$\frac{d}{d\rho}\left[\frac{\rho^3}{\sqrt{1+\left(\frac{dy_6}{d\rho}\right)^2}}\frac{dy_6}{d\rho}\right]=0.$$ We solve this differential equation. $$\frac{\rho^3}{\sqrt{1+\left(\frac{dy_6}{d\rho}\right)^2}}\frac{dy_6}{d\rho}=\tilde{c}$$ $\tilde{c}$ being a ...

1

Well, technically, both newtonian physics, relativity and QM work in tandem, all the time. However, some of the abstractions break down - for example, when you're dealing with an electron in isolation, it behaves well in accord with newtonian physics. The same way, even if that electron moves at half the speed of light, from the POV of the electron, it sill ...

2

No other thresholds comparable to quantum-classical are known, nor is there any reason known at present to suspect them. The precise threshold between quantum and classical physics is actually rather simply: it is ignorance (quantum) vs knowledge (classical). More precisely, regardless of the sizes or masses or scales involved, quantum rules always apply ...

2

The answer depends on the thermodynamic temperature of the environment of these objects, the interaction strength with which they couple to this environment and their lifetime. The spatially largest and consequently longest lived observed quantum effects, that I am aware of, are interference fringes of light that came from galaxies that are millions of ...

10

Newtonian physics is generally a good approximation in a problem as long as any significant differences in the action involved in the problem are much larger than Planck's constant (if not, quantum mechanics will be needed), the speeds involved in the problem are much less than the speed of light (if not, special relativity will be needed), and as long as ...

1

This type of calculation can be done in Mathematica using the xTensor package (its free). There is a bit of a learning curve, but the documentation is great and they have a very active google-group. Typically you will need to write the Lagrangian explicitly as a polynomial in the Riemann Tensor. Once you do this, the VarD command (xTensor) can handle the ...

1

Classical fields emerge when there is a large (but not definite) number of particles in a coherent state. For a simple example, for a scalar field $\phi(x)$ we can write a state that describes a classical configuration as something like $$\exp\left(\int d^D p\; \tilde\phi(p) a^\dagger_p\right)|0\rangle.$$ Note that this isn't an eigenstate of particle ...

3

This is a situation where knowing the history of the terminology can be helpful. The QFT/string theory terminology comes from algebraic geometry, where the term moduli space is used for any space whose points correspond to some kind of geometric object. The projective space $\mathbb{P}(V)$, for example, is the moduli space of lines in the vector space $V$. ...

3

Well, there is the Kawai-Lewellen-Tye (KLT) relations, which says that a closed string amplitude is roughly speaking a product of two open string amplitudes. See e.g. Ref. 1. References: Z. Bern, Perturbative Quantum Gravity and its Relation to Gauge Theory, Living Rev. Relativity 5 (2002) 5; Section 3.1.

1

I suspect that Becker & Becker are referring to asymptotic safety, a theoretical programme that attempts to describe gravity with a quantum field theory. The philosophy is that a QFT is defined and sensible at all energy scales so long as all of its couplings are always finite. The simplest way for that to be the case is if all couplings flow to UV ...

2

It is not enough to be popular or enough to quantize gravity. A TOE has to be able to include special relativity ( Lorenz invariance) and to embed the whole standard model of particle physics. At the moment only string theories are able, have the group structures , to do this, but the specific string model is still a matter of research. this also mean ...

0

If I remember correctly this is discussed somewhere in the chapter "String interactions and Riemann surfaces" from Zwiebach's A first course in String Theory book.

1

1) First, looking at $(2.3.4)$, you see that $j^a$ is the coefficient of $\partial_a\rho$. An application of this $(2.3.12), (2.3.13)$. To make connection with this formalism, it is preferable to choose the variations : $X^\mu\rightarrow X^\mu-\epsilon \rho(\sigma) v^c\partial_c X^\mu$ From this, we get : $\partial_a X^\mu\rightarrow ... 2 To be clear what we're talking about (as I'm not totally sure this is what the question intended), I'll talk about the paradigmatic example of AdS/CFT, the equivalence between$\mathcal N=4$Yang-Mills on the one hand, and IIB string on (asymptotically)$AdS_5\times S^5$on the other (at general parameters: no t'Hooft limits etc). We are very much closer ... 2 In principle yes, but there are several conceptual and technical issues that make it unclear how this could be achieved. Even though the AdS/CFT correspondence is conjectured to be exact(with much evidence hinting at this), it is hard to prove this essentially because in order to do calculations, one still has to use approximations and perturbation theory on ... 1 The action is $$S = \frac{1}{2\pi \alpha'} \int d^2 \sigma \sqrt{\gamma} \gamma^{ab} \partial_a X^\mu \partial_b X_\mu$$ The definition of the stress tensor from GR is $$T_{ab} = \lambda\frac{4\pi}{\sqrt{\gamma}} \frac{ \delta S}{ \delta \gamma^{ab}}$$ Usually$\lambda = 1$, but different books use different conventions. I do not remember what convention ... 2 First of all, note that the radial operator ordering${\cal R}$is implicitly implied in many textbooks of CFT (e.g. Ref. 1). For instance, eq. (2.2.7) on p. 39 in Ref. 1 is discussing Wick's theorem between two operator ordering prescriptions. In this case between normal ordering$:~:$and radial ordering${\cal R}$. See also e.g. this Phys.SE post. The ... 4 When we say that they are unit vectors, we mean that the proper length is equal to one. The proper lengths of the two vectors are $$\gamma_{ab} t^a t^b=1,\quad \gamma_{ab}n^a n^b=1$$ and should be equal to one, i.e.$1\to 1$, at all times. (In the Minkowski signature, one of these squared lengths is minus one, but that won't change anything about the text ... 1 The different string vibration modes represent particles. There are infinitely many different excitations of increasing mass and experiments usually have a limited energy scale E, so we don't worry about particles with m>E. One can then only consider e.g. massless modes and then build an effective theory describing their interactions. One can do this by ... 1 Starting from the boundary condition $$\partial^{\sigma}X^{\mu}(\tau,0)=0$$ and lowering the index on the derivative using the metric gives $$\gamma^{\sigma\tau}\partial_{\tau}X^{\mu}(\tau,0)+\gamma^{\sigma\sigma}\partial_{\sigma}X^{\mu}(\tau,0) =0.$$ Apparently Polchinski wants to express this in terms of the metric$\gamma_{ab}$with its indices ... 1 You may think to a inner product$(T, T')$:$(T,T') = \int d^2\sigma \sqrt{g} \,T. T'$where$T$and$T'$are tensors of equal rank, and$T.T'$corresponds to a contraction on the tensor indices. Now, your expression is simply, that, for all the Diff-Weyl variations$\delta g_{DW}$, and for all the moduli variations$\delta g_M$, you have :$(\delta ...

1

To explicate what we mean by no significance, I would suggest that you understand the indexing as a convenient type of naming. We need some way to say which dimension we are referring to and using numbers lets us use convenient notation, but we could just as well have named the dimensions. The ones we experinece might be Tim, Alice, Bob, and Carol. Any new ...

3

There is no significance to the numbering of the dimensions. When we refer to a vector it's common to write is as $x^\alpha$, where $\alpha$ runs from zero to the number of spacetime dimensions minus one. $x^0$ is frequently used to refer to the timelike dimension, so $x^1$ to $x^n$ refer to the $n$ spatial dimensions. However there is no signficance as to ...

4

A soliton is a localized, non-dispersive solution of a nonlinear theory in Euclidean space. It certainly is a real object: you have a famous story about a certain John Russell who observed soliton-like waves made by a boat on a river (wikipedia knows everything about it!) The so-called morning glory clouds in Australia ...

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