How can I get the solution of the wave equation for a membrane geometry, so that I can set the modes to 0, 1 and 2?

@FakeMod When you take a total derivative w.r.t. to time, you can no longer take a partial derivative w.r.t. $q$ or $\dot{q}$. You either have $L(q,\dot{q})$, no time dependence, or you have $L(t) = L(q(t),\dot{q}(t))$ for a path $t\mapsto (q(t), \dot{q}(t))$, no $q,\dot{q}$ dependence.

Can someone share an insight on this please physics.stackexchange.com/questions/571276/…

I can rotate a spin about an axis, to go from spin up state to spin down state. But I can't reverse the spin by some kind of spatial inverse transform to go from spin up to spin down. Why is that?

@B.Brekke Because a spatial inverse is not a rotation, so you don't get a representation of it on the state space just from the $S_i$ operators

So if I had an object in O(n) I would be able to inverse it, but a spin is in SU(2). It sounds silly to say "in a group", what is it really called?

Representations of a group act on a vector space

this should be tagged FAQ, no?

@EmilioPisanty It's only been viewed 767 times

do you have the impression that we get this question very often in other forms?

When you think about it, bosonic lightcone quantization is really just as simple as recognizing $[\alpha_m^{\mu},\alpha_n^{\nu}] = m \delta_{m+n,0} \eta^{\mu \nu}$ implies states of negative norm $<0|\alpha_{n}^0 \alpha_{-n}^0 |0> = - n$ due to the $\alpha_{n}^0$ oscillators, and saying wouldn't it be great if we could just set them to zero?

This would mean $X^{0} = x_0^0 + 2 \alpha' p^0 \tau + \sum_{n \neq 0} \frac{1}{n} \alpha_n^0 e^{-in \tau} \cos(n \sigma) = x_0^0 + 2 \alpha' p^0 \tau $. Well thanks to coordinate invariance of Polyakov and the fact that $x_0^0 + 2 \alpha' p^0 \tau$ satisfies the wave equation, we can indeed use this as one of the space-time coordinates.

The constraint $(\dot{X} + X')^2 = 0$ means we could then solve for the $ \alpha_n^1$ oscillators in terms of the other $\alpha_n^i$ oscillators, but the relation is quadratic, however going to lightcone coordinates, for which $X^2 = 2 \eta_{+-} X^+ X^- + ..$ we would find a linear expression for the $ \alpha_n^-$ oscillators if $X^+ = x_0^+ + 2 \alpha' p^+ \tau$

Should we raise a custom flag on a question if it has a bounty but we think it should be closed?

You really would be stuck with fermionic $b_0^{\mu}$ oscillators without local supersymmetry, this stupid thinking alone could be enough to discover susy in a sense

laziness = local susy

The first person to try supersymmetry must have been all out of idea, rly

it turns the scalar into spinor!

Mostly likely the thought came to them in the shower

Yeah I'd like to see the first occurrence of it in string theory

I don't rly know much about pre-string string theory

I should look it up someday

Is there any good book on Regge trajectories?

But it seems pretty clear that you could basically ignore susy in a model $S_b + S_f$ until you tried to quantize and were forced to face these negative norm states, having nothing but $X^{+} = x_0^+ + p^+ \tau$ and $\psi^+ = \text{oscillators}^+$ a symmetry like $\delta X^+ = \overline{\varepsilon} \psi^+$ would justify setting $ \psi^+ = 0$ thus getting rid of the oscillators responsible for negative norm states, which you know 'should' go away

This is the best/easiest book on Regge stuff I can find but I still barely understand Regge trajectories

@Slereah I (and a couple of others) actually had a long dinner conversation on this with Julius Wess - one of the nicest persons I've ever met.

actually there were hell-bend on cancelling vacuum contributions.

so they jerry-rigged their theory to have 2 fermions occur for every boson so they could make the 0-point energies cancel. They weren't thinking of supersymmetry at that stage but just use that as a starting point.

heh

It's the Dirac sea all over again

This cites the first paper to apparently do it, of course it's incomprehensible...

for all you guys here: 1.bp.blogspot.com/-RouDIOUQ_U4/XxLl8uzoSAI/AAAAAAACxHg/…

@BioPhysicist yes

@FakeMod What situation are you looking at here? Why do you think this equation holds?

I don't know what you mean by "derive the potential for an electromagnetic field"

If you're looking for an argument that the electromagnetic potential exists, then the Lorentz force tells you nothing about it.

@FakeMod No, it's not. The relation between the E-L potential and the force is not just $F = \nabla U$ if your potential depends on velocity!

yes

what stops you from writing it with gradients?

...write it as a gradient in $v$?

What's wrong with $\frac{\mathrm{d}}{\mathrm{d}t}\nabla_v U + \nabla_x U$?

if we accept that for a function $f(x,y)$ we write $\partial_x f$ for the derivative w.r.t. to the first slot and $\partial_y f$ for the derivative w.r.t. the second slot, then for a function of vectors $f(\vec v, \vec w)$ we can write $\nabla_v f$ for the gradient w.r.t. to the first slot and $\nabla_w f$ for the gradient w.r.t. the second slot, no?

David Zaslavsky was one of the first people to chat on this chatroom

can i web scrape this chatroom?

(the transcript)

according to chat.stackexchange.com/robots.txt I can