The h Bar

12:58 AM

@ACuriousMind If $\mathrm{O}(n,m)$ denotes the subset of $\mathrm{Hom}(\Bbb R^n,\Bbb R^m)$ of injective orthogonal maps, is it compact?

rob

I love it when I realize in the middle of writing an answer that my answer is wrong. That's when I learn the most things around here.

dmckee

@rob Or that I've got most of it right, but I've been blithely papering over a bit of ignorance for years, and I won't be done until that hole is filled up.

David Z

3:24 AM

@WrichikBasu so, I see it's not an issue anymore in this case, but for the record, asking not to have your question put on hold doesn't stop us from putting it on hold. Of course it's not a reason to put a question on hold either, but we have noticed a pretty strong correlation between questions where the asker requests it not be put on hold and questions that deserve to be put on hold.

In the end, asking not to have your question put on hold does nothing except invite some extra-close scrutiny from the mods.

The Raiders of Las Vegas

(removed)

Secret

4:33 AM

(Demover)

The Raiders of Las Vegas

(unremoved)

Secret

Breit–Wheeler process

Breit–Wheeler process or Breit–Wheeler pair production is a physical process in which a pair of positron–electron is created in the collision of two photons. It is the simplest mechanism by which pure light can be potentially transformed into matter. The process can take the form of γ γ′ → e+ e− where γ and γ′ are two light quanta . The multiphoton Breit-Wheeler process, also refers to as nonlinear Breit-Wheeler or strong field Breit-Wheeler in the literature is the extension of the pure photon-photon Breit-Wheeler process when a high-energy probe photon decays into pairs propagating through an...

> As of January 2016, Breit Wheeler pair production hasn't yet been achieved in the laboratory.

What? I thought it is already experimentally confirmed

> The multiphoton Breit-Wheeler has however already been observed and studied experimentally

Now this memory fragment is finally making sense:

Sep 20 '15 at 18:22, by John Duffield

@GlenTheUdderboat : the Breit Wheeler process. That's where photons interact with photons. Directly. Most people are taught that they don't, but they do. See this report re research at Imperial.

That is a experiment proposal paper, not an actual experiment detection paper

cannot believe I had that idea wrong for 2.5 years

Rumplestillskin

Does anyone have any idea what it means for a solution of the einstein field equations to involve arbitrary functions as the metric components?

Secret

1

Q: Can electromagnetic fields interact with each other?

For example - if one were to generate an electromagnetic field using super-cooled magnets, and the magnets were shut off, is it possible for another field to interact with the collapsing one and essentially fire the remains out of a theoretical barrel like a shotgun? This would require each field...

electromagnetism

This question triggers that memory and brought up the previous discussion

This is why memories can be very scary

However, it does not matter anymore. With that inconsistency resolved, it will not pop up anymore

John Rennie

4:49 AM

@Rumplestillskin The FLRW metric is an example. The solution contains an arbitrary scale factor $a(t)$ that is determined by the initial conditions.

Rumplestillskin

@JohnRennie this is so difficult for me to understand. What if the scale factor was a(t) = 1/t? Isn't it singular at t=0? Also couldn't that mean that you have an infinite family of solutions?

John Rennie

Can you give an example or a citation to where the claim is made?

Rumplestillskin

@JohnRennie I just made that claim there.

What I mean is I don't have a citation.

John Rennie

So you've just made this up? Or is it something someone has told you?

Rumplestillskin

Wait - You just told me that the FLRW has an arbitrary function on time a(t) - which of course means it satisfies the EFE for any a(t) right?

Or am I missing something?

DanielSank

4:56 AM

What's going on in here?

John Rennie

In the FLRW metric $a(t)$ is determined by the types and densities of matter present. So many different functions are possible.

Rumplestillskin

I see - But are the field equations satisfied for an arbitrary function is what I am getting at? If so, then there are infinite solutions and just as many singularities - - possibly? correct?

John Rennie

I now have no idea what you are asking.

You can take any arbitrary metric and solve the field equations in reverse to work out the corresponding stress-energy tensor, if that's what you mean.

The resulting stress-energy tensor is generally unphysical i.e. involves exotic matter.

Rumplestillskin

Sorry, I am not being clear - - - Do you know any vacuum solutions to the Einstein field equations that contain arbitrary functions that satisfy the EFE?

Wrichik Basu

Anyone interested please support:

area51.stackexchange.com/proposals/113083/…

2

Rumplestillskin

5:01 AM

I should have said vacuum solutions

John Rennie

@Rumplestillskin no

Rumplestillskin

Hmmm

Secret

Hmm, let's see if I understood this discussion correctly: e.g. Given a metric tensor $g=a(x,y,z)dt^2+b(t)d\Sigma^2$ can a nontrivial solution T be always found when g is plugged into the EFE for arbitrary $a(x,y,z),b(t)$?

John Rennie

Yes, though the resulting stress-energy tensor may be unphysical

Rumplestillskin

Yes given a line element (ds)^2 = a(r,t) dt^2 + b(r,t) d\Sigma^2 such that the EFE are all satisfied for any a and b. Or put another way have you ever heard of a metric tensor that always satisfies the field equations for an arbitrary metric component? And also, we are considering vacuum solutions so the stress energy tensor \equiv 0.

John Rennie

5:12 AM

If T = 0 then I don't know of any metrics that contain arbitrary functions

Secret

$$G_{\mu\nu}+\Lambda g_{\mu\nu}=0$$ sounds too restricted to have arbitrary functions unless it is an overall scalar multiplication (but then the physics will not be changed by multiplying and overall scalar)

Rumplestillskin

Very peculiar.

Secret

Actually:

$$G_{\mu\nu}=-\Lambda g_{\mu\nu}$$

The required $G_{\mu\nu}$ must then be a scalar multiple of the metric

not sure if that rules out arbitrary functions that are not overall scalar multiplications

Rumplestillskin

Hmmm im not sure just something that came to mind! In what situation in GR do you use \Lambda g_{\mu \nu}? I have only ever worked with G_{\mu \nu} = 0 - - very rarely having a non-zero RHS. The only time I've ever had a non-zero RHS is calculating the post newtonian approximation.

Secret

Actually, what does EFE look like expressed entirely in terms of the metric (I can imagine the riemannian curvature tensor part will be highly nonlinear...)

Rumplestillskin

5:25 AM

Im not sure.. probably quite the mess!

Secret

I suspect it will be at least second order in terms of the metric since you need to raise index to get from the riemanian curvature tensor to the ricci tensor and then take the trace to get the ricci scalar

ah, almost forgot, there are derivatives of the metric as well....

It's a nonlinear PDE system in terms of the metric basically...

0celóñe7

5:38 AM

I smell my kind of math

@Rumplestillskin What? In any kind of cosmological theory it will be nonzero.

Secret

Quick sanity check: $g^{\alpha\beta}g_{\beta\gamma}=g^{\beta\alpha}g_{\beta\gamma}=\delta^{\alpha}_{‌\gamma}$?

(g is metric)

0celóñe7

yes

Secret

ok good

Rumplestillskin

5:55 AM

@0celóñe7 Ah I thought it was zero for the FLRW metric?

Secret

Sanity check again: $\delta^{\alpha}_{\beta}g_{\gamma\delta,\alpha}=g_{\gamma\delta,\beta}$?

Rumplestillskin

That is correct!

Secret

cool

Riemanian tensor entirely in terms of the metric:
$$R_{\alpha\beta\gamma\delta}=\frac{1}{2}(g_{\alpha\beta,\gamma\delta}+g_{\gamma\delta,\alpha\beta}-g_{\alpha\gamma,\beta\delta}-g_{\beta\delta,\alpha\gamma})+\frac{1}{4}\left((g_{2\beta,\gamma}+g_{2\gamma,\beta}-g_{\beta\gamma,2})g^{25}(g_{5\alpha,\delta}+g_{5\delta,\alpha}-g_{\alpha\delta,5})-(g_{2\beta,\delta}+g_{2\delta,\beta}-g_{\beta\delta,2})g^{27}(g_{7\alpha,\gamma}+g_{7\gamma,\alpha}-g_{\alpha\gamma,7})\right)$$

EFE coming shortly...

(NB I will fix those number indices later, they are just rough work to make it easier to track indices)

0celóñe7

6:15 AM

@Rumplestillskin no, it's a perfect fluid.

Secret

6:30 AM

Now..., good luck solving this:

$$R_{\beta\delta}-\frac{1}{2}Rg_{\beta\delta}+\Lambda g_{\beta\delta} = 0$$
becomes:

...
$$=2(g_{\alpha\beta,\gamma\delta}+g_{\gamma\delta,\alpha\beta}-g_{\alpha\gamma,\beta\delta}-g_{\beta\delta,\alpha\gamma})+\left((g_{\epsilon\beta,\gamma}+g_{\epsilon\gamma,\beta}-g_{\beta\gamma,\epsilon})g^{\epsilon\zeta}(g_{\zeta\alpha,\delta}+g_{\zeta\delta,\alpha}-g_{\alpha\delta,\zeta})-(g_{\epsilon\beta,\delta}+g_{\epsilon\delta,\beta}-g_{\beta\delta,\epsilon})g^{\epsilon\eta}(g_{\eta\alpha,\gamma}+g_{\eta\gamma,\alpha}-g_{\alpha\gamma,\eta})\right)+4\Lambda g_{\alpha\gamma}g_{\beta\delta}=0$$

So, the homogeneous EFE is a 2nd order 2nd degree nonlinear PDE (which is also the case even if $T_{\beta\delta} \neq 0$)

Rumplestillskin

Wow!

Secret

Reason of those sanity check is because I tried to simplify it by hand

So for a $ds^2 = a(x,y,z)dt^2 + b(t) d\Sigma^2$ the good news is that all our gs are diagonal, thus it might be possible to simplify it further to find if there are constraints on $a, b$

Secret

7:03 AM

$$g_{\alpha\beta,\gamma\delta}=\begin{pmatrix}\begin{pmatrix}\partial_{tt}a & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{pmatrix} & 0 & 0 & 0\\ 0 & \begin{pmatrix}0 & 0 & 0 & 0\\0 & \partial_{xx}b & \partial_{xy}b & \partial_{xz}b\\0 & \partial_{yx}b & \partial_{yy}b & \partial_{yz}b\\0 & \partial_{zx}b & \partial_{zy}b & \partial_{zz}b\end{pmatrix} & 0 & 0\\
0 & 0 & \begin{pmatrix}0 & 0 & 0 & 0\\0 & \partial_{xx}b & \partial_{xy}b & \partial_{xz}b\\0 & \partial_{yx}b & \partial_{yy}b & \partial_{yz}b\\0 & \partial_{zx}b & \partial_{zy}b & \partial_{zz}b\end{pmatrix} & 0\\0 …

but then, I have no idea how to distinguish between $g_{\alpha\beta,\gamma\delta}$ vs $g_{\gamma\delta,\alpha\beta}$

Anyway, this particular g and its derivatives have a simple block matrix representation:

$$g_{\alpha\beta,\gamma\delta}=(\partial_{tt}a\oplus 0) \oplus (0 \oplus H(b)) \oplus (0 \oplus H(b)) \oplus (0 \oplus H(b))$$
$$g_{\epsilon\beta,\gamma} = (\partial_{tt}a\oplus 0) \oplus (0\oplus \nabla b) \oplus (0\oplus \nabla b) \oplus (0\oplus \nabla b)$$

where $H$ is the hessian and $\nabla$ is the gradient

I am afraid I have to leave it here for now. Unless I can figure out how to distinguish between $g_{αβ,γδ}$ vs $g_{γδ,αβ}$ I cannot continue

Physics Meta

7:19 AM

1

Q: Dismissal of Non-Homework Question

Recently, one of my non homework questions for which I'd provided my attempt at the solution and conclusions, was dismissed. The question was clearly not a homework question, and even some of my instructors haven't been able to solve it yet. I was hoping for discussion of various methods to app...

discussion support feature-request homework specific-question

The Raiders of Las Vegas

7:41 AM

How about a Homework Site:

cs50.stackexchange.com

The Raiders of Las Vegas

7:58 AM

What's next, a Math 55 site?

Math 55

Math 55 is a two-semester long first-year undergraduate mathematics course at Harvard University, founded by Lynn Loomis and Shlomo Sternberg. The official titles of the course are Honors Abstract Algebra (Math 55a) and Honors Real and Complex Analysis (Math 55b). Previously, the official title was Honors Advanced Calculus and Linear Algebra. Some claim it is the most difficult undergraduate math course in the United States. == Description == The Harvard University Department of Mathematics claims that "This [Math 55] is probably the most difficult undergraduate math class in the country". Formerly...

Secret

8:11 AM

ok, so...: