# Perturbation

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1. ## Idea about black holes... 4 spacial dimensions?

Look up Kruskal-Sziekeres (not sure about the spelling) coordinates. Offers a more coherent depiction of the space-time geometry about a Schwarzschild black hole' date=' much more than the regular spherical polar coordiantes, which are rather elluding when it comes to Schwarzschild geometry. Yep, for Kerr black holes (or any rotating black hole [static]) the singularity is a ring. A singularity is not so called because the dimensions of it are singular, i.e. pointlike. A singularity is called this because it's the region where the curvature, which is essentially the gravitational field-strength of the hole, diverges to infinity. In the simple case of a non-rotating, spherically symmetric (stationary) black hole, that is the well known Schwarzschild black hole, the curvature diverges to infinity as one tends towards the centre of the black hole, so the singularity of this variety of hole is a point. For more general holes, however, the region where this divergence occurs needn't be a dimensionless point. To stress the point again: the singularity is where the field is infinite.

3. ## Is the Lorentz Transform the Complete Picture?

By saying "let's introduce some difference in phase" doesn't allow one to conclude that matter consists of out of phase magnetic and electric waves. If they were, it would be noticable. Any wave-form can be described by a complex number. And in that complex number you're saying that c, the speed of light, is a complex number with Re©=v. Furthermore, playing with the gamma factor doesn't change anything about the particle itself, the gamma factor is derived from the geometry of Minkowski space-time. The duality of matter is described well enough with QM.
4. ## Is the Lorentz Transform the Complete Picture?

Complex energies, momenta etc. of real particles don't make any sense physically and are just as undesirable as negative energies. Wave nature is well explained by quantum mechanics without having to resort to complex energies etc. That's an odd looking form of the Lorentz transformation for a boost. As for the Lorentz group [imath]SO^{\uparrow}_+(1, 3)[/imath], it's not the full picture, as it doesn't include translations, only rotations and boosts. The full symmetry group is the Poincare [imath]\cal{P}[/imath].
5. ## Graviton

It would operate in a similar manner to its wave counter part: gravitational radiation predicted by general relativity. Gravitational waves operate by causing the space between two points to change by altering the geometry (metric) of some region of space-time as they propogate through it. These perturbations in the geometry satisfy the following inhomgenous wave equation in linearised field theory $\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)h_{\alpha\beta}=16\pi \frac{G}{c^4}T_{\alpha\beta}$ Describing the propogation of linear perturbations of a background Minkowski space-time with phase velocity c, i.e. gravitational waves propogate at c, in the Lorentz(/harmonic/de Donder) gauge. The above equation does, however, only come about if we deal with a weak field and work in a specific gauge, though because the gauge transformations we apply have no physical relevance we're free to choose the gauge ourself. The graviton is the quantisation of this field, as the photon is of the electromagnetic. It's a tensor gauge boson, and thus has spin 2, and the potential it generates is always attractive for a positive mass. The current problem with gravitation on Planck scales comes from this quantisation: can we formulate a gauge invariant renormalisable field theory for gravity? Naive attempts in the 50's set about formulating gravity in a similar manner to the field theories around at the time, these effective field theories were however unrenormalisable (unable to remove divergences from loop-momentum integrals in ultraviolet/infrared limit) or suffered from some other physically undeseriable effect. It is often the case that for one to be able to formulate a quantum field theory of anything we have to have the geometry of the space-time presupposed; however in general relativity the geometry is a dynamic quantity, varying with position and time depending upon the local distribution of matter and its own behaviour. (Loop) Quantum gravity proceeds by constructing a gauge invariant field theory from the Diffeomorphic group (the group of isomorphic, differentiable maps between hyperplanes of the space-time) of the space-time. I've not studied it a great deal though, likewise with M-theory.
6. ## Help me choose a scientist.

Marconi, Hertz, Bell...?
7. ## String Theory

It pretty much consists of a set of procedures for computing bits and bobs in particle physics; last time I checked it was lacking in a firm conceptual structure that the more established branches of particle physics have, such as QFT.
8. ## Excuses for massiveness

Not in field theory we don't, if that's what you're saying. Could you actually give some justification to your theory, possibly show us some of your literature? What kind of response do you expect from us if all you're giving us are vague, unsubstantiated clauses?
9. ## polarized photons

Actually I just wrote a simplish explanation of virtual particles for someone. It's the post with the Feynman diagrams in it Virtual Particles
10. ## Excuses for massiveness

Vacuum radiation being what? Virtual particles pairs, vacuum polarisation etc.? This is perfectly well justified and verified physically. Look up some QFT.
11. ## polarized photons

Huh? Are you saying that the Coloumb potential is a result of the exchange of a scalar photon? In a Coloumb field we still have exchanges of vector bosons, the magnetic potential is simply zero. This isn't the same thing as exchanging a scalar. The "non-vacuum" supplying a charge sounds like vacuum polarisation, but I don't think this is what you're trying to say.
12. ## Graviton question

It's a result of quantum field theory, and there are many reasons why physics accepts the field theoretic approach to particle interactions.
13. ## non-linear ODE problem

The show part is easy enough (just remeber to use the original ODE for getting rid of g'). Well I guessed the solution g(t)=exp(t). Using the transformation we have $z'+(a+2bg)z=-b\Rightarrow z'+(1+2e^{-t}e^t)z=-e^{-t}$ Which has the solution [imath]z=-\frac{1}{2}e^{-t}[/imath] and transforming back will give you the general solution for y(t). If the show bit is what you're having trouble with, gimme a shout.
14. ## question about trig substitution

I completed the square in the denominator and used the substitution [imath]\tan\theta =x+\frac{3}{2}[/imath]. This gave me $\int \frac{x}{4x^2+12x+13}dx=-\frac{1}{4}\ln\left(\frac{2}{\sqrt{4x^2+12x+13}}\right)-\frac{3}{8}\tan^{-1}\left(x+\frac{3}{2}\right)+c$ $=\frac{1}{4}\ln\left(\frac{\sqrt{4x^2+12x+13}}{2}\right)-\frac{3}{8}\tan^{-1}\left(x+\frac{3}{2}\right)+c$ I differentiated this, and it gave me the correct original integrand. Maybe mathematica removed the power of a half or something from the log.
15. ## polarized photons

And that should say longitudinal.
16. ## polarized photons

[imath]j(j+1)\hbar^2[/imath] are the eigenvalues of the norm of the angular momentum operator [imath]J^2=J^2_x+J^2_y+J^2_z[/imath]. The squareroot is for the J.
17. ## Graviton question

Yes they do, but it's not much, and the luminosity is pretty small in any case. Gravitational radiation propogates at c and is very weak, so we have to sit around for a while and wait for a big enough quasar or whatever to shoot some gravitational radiation our way. This is the whole point of LIGO.
18. ## Graviton question

Gravitational radiation is always being radiated by a source, it doesn't matter if there's nothing for them to interact with, well as long as the source isn't spherically symmetric (or there's some other condition that means it won't radiate). You can have a free graviton, just as one could have a free photon. It'll have been created somewhere in practice, but in calculation it's acceptable to suppose that one can have an exchange particle not attached to two vertices (an external gauge particle).
19. ## polarized photons

Also, the transverse polarisations of a photon cannot emerge in practice due to the Ward identity of QED. $k_{\mu}\cal{M}^{\mu}$$=0$ Where M is the matrix element arising from a scattering process involving a photon with four-momentum k.
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