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2:02 AM
@Charlie It depends very much on what you think is a "proper" definition of Lorentz matrices :P
@NiharKarve I change it regularly - this one is Six from Heaven's Vault
 
 
1 hour later…
3:25 AM
@ACuriousMind ahh I didn't know - I had Google image searched that picture and I got only one hit on some Russian website
 
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4:09 AM
Hi Good Morning.. All
 
 
2 hours later…
5:53 AM
@Charlie of course you can. If $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu}$ you use the fact that vectors transform under the Minkowski metic to justify it, e.g. $A'^{\nu} = \eta^{\nu \sigma} A'_{\sigma}$ tells us $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu} = \Lambda^{\mu}_{\,\,\,\nu} \eta^{\nu \sigma} A'_{\sigma} = \Lambda^{\mu \sigma} A'_{\sigma}$ where I defined $\Lambda^{\mu \sigma} = \Lambda^{\mu}_{\,\,\,\nu} \eta^{\nu \sigma}$.
Similarly lowering the index on the LHS $A_{\rho} = \eta_{\rho \mu} A^{\mu} = \eta_{\rho \mu} \Lambda^{\mu \sigma} A'_{\sigma} = \Lambda_{\rho}^{\,\,\,\sigma} A'_{\sigma}$ lets you justify raising and lowering the other index.
For example, $A^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} A'^{\nu}$ in the t-x plane reads as
$$ \begin{bmatrix} A^0 \\ A^1 \end{bmatrix} = \begin{bmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{bmatrix} \begin{bmatrix} A'^0 \\ A'^1 \end{bmatrix} $$
so that
\begin{align}
\begin{bmatrix} A_0 \\ A_1 \end{bmatrix} &= \begin{bmatrix} A^0 \\ - A^1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix} \begin{bmatrix} A^0 \\ A^1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & - 1 \end{bmatrix} \begin{bmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{bmatrix} \begin{bmatrix} A'^0 \\ A'^1 \end
 
 
2 hours later…
7:53 AM
HELLO EVERYONE
Would it be possible to analyze exponential graphs using calculus, different series(such as Laplace transform etc) and in case you wanted to know, I am doing newton’s law of cooling.
he idea is that I deriving the equation of newton’s law of cooling but my high school teacher said that it will be a bit easy to derive an equation that could easily be found online so therefore I am thinking of how to make my report a bit more difficult by adding more analysis. Would it be possible? (Any help from you would be very useful
 
8:42 AM
@GeneralMO7 hi.
What sort of thing did you have in mind? How to graph the data? How to derive the equation for the temperature as a function of time?
 
8:54 AM
Hi there. Everyone .
Do anyone know how to make a chat room
 
@KrrishDhaneja go to this page and at the bottom of the page is a button to create a new room. Though it may required a minimum rep to create a room and I note your rep is quite low.
 
9:10 AM
@JohnRennie, the idea is to apply some analysis (a bit more complex than just rate of change, etc) to make my report more with my level for an HL math student.
Or maybe change a variable in the rate of cooling and compare
the main idea is to add into my report to make it a bit more harder
 
Thanks for help @JohnRennie
 
@GeneralMO7 Newton's law of cooling is basically very simple. It's not obvious to me what extra complexity you could add.
 
What would be the application of Laplace transform?
 
Laplace transform is commonly used to study electric circuits
 
The Laplace transform is useful for solving various types of differential equation, and that includes the heat equation.
But it's hard to see how you would use it for Newton's law of cooling.
 
9:32 AM
Can you please explain why it might be hard for Newton’s law of cooling?
 
Because Newton's law of cooling is such a simple differential equation that solving it is easy with needing a Laplace transform.
 
@GeneralMO7 is this for an IA or just a report?
 
10:03 AM
@NiharKarve, yes it is for my IA.
I have no clue how you figured it out.
And my supervisor said that it will be easy to just derive the equation and apply it to a real life situation.
I really need your help
 
@GeneralMO7 the equation for Newton's law of cooling?
 
Yes I derived the equation then applied it to a real life situation.
 
So you've done it? And presumably you've found it was easy?
 
Personally I found it easy and my teacher confirmed that but she said I can maybe further develop it to meet my level of mathematics.
 
10:38 AM
@GeneralMO7 If you want to add more detail that is not trivial, talk about how things like Stefan-Boltzmann and other things reduce to Newton's law of cooling in a linear approximation
7
Q: Are Newton's law of Cooling and Stefan's law related?

CheekuMany of Indian school textbooks claim a proof of Newton's law of cooling from Stefan's law of black-body radiation. As far as I am aware of, Newton's law is based on cooling from convection currents and Stefan's law on radiation. There is not supposed to be any relation between them. Question: ...

The graph stuff seems like a bad way to go
 
 
2 hours later…
12:28 PM
Hello
I have a question about compactification of $\Bbb R^{1,1}$
 
 
1 hour later…
1:47 PM
shoot
 
2:19 PM
3
Q: Editing a question invalidating answers

BuraianIn this question which the user @Ankit has posted, the question kept pushing for more or less the same idea till 21st November. At the time, the question then contained an explicit calculation where the author had tried to explain his work. Two days later in 22-Nov, @ACuriousmind does an edit rem...

 
2:53 PM
@NikeDattani sorry for the delay. Yes the question is very vague and it would seem easy to provide an explicit example. I can probably look it up but I presume people interested in the question would prefer something self-contained.
 
3:53 PM
@Slereah my question is about the construction of the conformal compactification of $\Bbb R^{1,1}$ (the Minkowski plane). So the definition of conformal compactification is: An embedding of a non-compact Lorentzian manifold into a compact Lorentzian manifold as a dense open subspace, such that the embedding is a conformal map. So what is the embedding? Is it $f: \Bbb R^{1,1} \to \Bbb R^{2,2}?$
 
The embedding is $\mathbb{R}^{1,1}$ into $\mathbb{R}^{1,1}$
you have probably seen the Penrose diagram
it's just a subset of $\mathbb{R}^{1,1}$
 
is the Penrose diagram a depiction of $\Bbb R^{1,1}$ embedded into $\Bbb R^{1,1}$?
 
Yes
Although... IIRC it's actually $\mathbb{R} \times S$
at least the most common compactification
 
@bolbteppa I can see how it's done in principle, my problems was whether it's actually a valid thing to do since it has two indices in two different coordinate systems :/ So expressions like $$\eta_{\mu'\nu'}{\Lambda^{\mu'}}_\sigma=\Lambda_{\mu'\sigma}$$ and $$\eta^{\mu'\nu'}{\Lambda^\sigma}_{\mu'}=\Lambda^{\sigma\nu'}$$ necessarily contain summations in different coordinate systems. Also the Lorentz transformation matrices aren't tensorial objects right?
 
But the mapping has the same topology, of course
The subset still has $\mathbb{R}^{1,1}$ as a topology
 
4:00 PM
ok I messed up the first equation and I don't want to ping you over and over correcting it sorry, you can see what it should be though
well, I guess the issue of having indices in different coordinate systems isn't so bad, the problem is mostly that the $\Lambda$ aren't tensors
and so using the metric which essentially provides the mapping between $TM$ and $T^*M$ doesn't make sense there
 
@Slereah okay I understand this a little bit better now. I would like to go through the exercise of constructing a conformal compactification of $\Bbb R^{1,1}$ using different coordinates. My question is, if you use different coordinates, will the Penrose diagram look different, after compactifying?
 
A tensor in Minkowski space is a quantity which transforms under a Lorentz transformation, why wouldn't a Lorentz transformation transform under a Lorentz transformation
 
@geocalc33 Depends what you mean by "different", but overall yes
 
You map a vector to a new frame under a Lorentz transformation, one index of the Lorentz transformation is uncontracted, then you perform another Lorentz transformation, of course that index is going to have to transform under the Lorentz transformation, it's just like a vector, so it can be raised and lowered like a vector index
 
@Slereah like for example, if Instead of using the metric $g=dxdy$ for $\Bbb R^{1,1}$, you use the conformal metric $g=\frac{dxdy}{xy}.$ I just want to see how the diagram changes when you switch coordinates
 
4:07 PM
hmm that's a good point
 
If you look at $x^{\mu} = \Lambda^{\mu}_{\,\,\,\nu} x'^{\nu}$, just focus on $\Lambda^{\mu}_{\,\,\,\nu} x'^{\nu}$, only the $\mu$ is free and it has to behave like a vector index just like the LHS does
 
I don't know exactly what embedding would work using the metric $g=\frac{dxdy}{xy}$
Do you have any hints that I can take with me?
 
you wouldn't get the entire Minkowski space with such a conformal metric
since it's not define along zeros
also 1/x doesn't compactify $\mathbb{R}$
so it wouldn't be a compactification of Minkowski space
 
But x,y are only positive. I think you do get the entire Minkowski structure, it's just transported to the open first quadrant
I've shown this part. The structure in the first quadrant is diffeomorphic to Minkowski. I got the metric $g$ from pulling back the metric $f=dxdy$
 
5:11 PM
Hello again, I have made some research and found that if we assume that the temperature of the room is ± a certain temperature we can use Laplace transform for newton’s law of cooling so what other variables that could possibly include Laplace transform or a bit more difficult math more than just newton’s law equation.
 
1
A: How to integrate Newton's law of cooling?

Jan Eerland First way, use Laplace transform: $$\text{T}'\left(t\right)=\text{K}\cdot\left(\text{T}\left(t\right)-\text{A}\right)\to\text{s}\cdot\text{T}\left(\text{s}\right)-\text{T}\left(0\right)=\text{K}\cdot\left(\text{T}\left(\text{s}\right)-\text{A}\cdot\frac{1}{\text{s}}\right)$$ Solving $\text{T}\...

It's just a linear first order ordinary differential equation with constant coefficients, it can be solved in multiple ways, the Laplace transform method is just one way to do it which is more complicated than other methods and involves using the inverse Laplace transform also
 
5:39 PM
0
Q: Why is the ångström not a metric unit? And why is the ångström spelt with the Swedish/Finnish letters “å” and “ö”?

Arunabh BhattacharyaOne of my questions from October 2019, Why is the ångström not a metric unit? And why is the ångström spelt with the Swedish/Finnish letters "å" and "ö"? has got useful answers. However, my question is downvoted to -4. I cannot understand why my question has got downvotes.

 
 
4 hours later…
9:49 PM
Hello there
I have been researching for a while looking for what to include more than just deriving the equation of Newton’s law of cooling and a website gave me inspiration by stating what if the temperature of the environment is changing (± C). In conclusion, I wasn’t sure if that person carried out the mathematical process in the right way and the only place I could ask for accurate information is here so please let me know what you think about the mathematical process he went for and if it is right or wrong.
 

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