The h Bar

1:37 AM

Are the vectors in SR a distinct mathematical object from the vectors in GR since SR vectors need to transform under Lorentz transformations and GR vectors have more general transformations?

Actually I guess they're all the same, just dependent on the metric being used

dmckee

2:25 AM

I would describe them as a special case. Special relativity is general relativity in the limit of weak gravitational effects (i.e. flat space-time), so the four-vectors of special relativity are a sub-set of those in general relativity.

On the other hand it is a subset characterized by enough additional guarantees that it might be seen to warrant separate study.

vzn

2:56 AM

@danielunderwood for an answer try looking at the vectors in here :) arxiv.org/abs/1603.07655

danielunderwood

3:07 AM

Thanks I'll check out that paper. I've actually been wondering if there's anything like a modern view on ether theory and that looks like it may be in that direction

I think they may be the same as far as the question that was in my head. I struggled for a while to figure out what vectors and tensors are, only to realize that the mathematical notion is a bit different than the physical notion and was confused by jumping between the two without knowing. What I was wondering here is if a similar thing may be happening where physical vectors were different based on context. I think they may b…

vzn

3:46 AM

@danielunderwood think your questions cut to the heart of the problem. there are ways of thinking of the GR vectors that are considered nonphysical under conventional wisdom but candidly, suspect conventional wisdom may be wrong. we've had several debates in the room on this subj, search transcript if you have some spare time. heres a recent related question to ponder, maybe somewhat inspired by similar considerations...

18

Q: Does light really travel more slowly near a massive body?

It is a routine problem for beginners in general relativity to calculate the coordinate velocity of light for the Schwarzschild metric. Starting from the metric: $$ ds^2 = -\left(1-\frac{r_s}{r}\right)c^2dt^2 + \frac{dr^2}{1-\frac{r_s}{r}} + d\Omega^2 $$ We use the fact that light travels on a ...

...

just had to share this with room, another tragic-yet-riveting story about unrestrained young testosterone? o_O Thailand cave search: Divers close in on missing soccer team / cnn

Secret

4:46 AM

downvote for lack of detail — Secret 14 secs ago

Probably one of the most honest driven downvotes I have ever casted

John Rennie

5:44 AM

Someone upvoted this? Really?

1

Q: What is a good vacuum that can go from carper to floor and can pick up pet hair w/o a problem?

Right now I have a Eureka light weight and it doesnt do me a bit of good. The fliter becomes clogged with pet hair before the canister is even 1/16 full.

vacuum home-experiment

Cows

7:31 AM

Been coding all day yesterday . . on and off. Thinking about attempting some very very very very trivial baby in diapers math proof that I can't seem to do. Will spend some time thinking about it in the morning. For now, I will retire for the day . .

Anonymous

@ACuriousMind Thanks. josh's answer was helpful!

Anonymous

@JohnRennie Morning. I'm confused about a very basic QM thing. Is it possible to experimentally distinguish between two states $\Psi$ and $e^{i\theta}\Psi$?

Anonymous

Is it possible to detect the global phase?

John Rennie

You're asking two different questions.

The global phase is just a gauge because you can choose any origin for $\theta$ that you want.

But if you're asking whether you can detect a phase difference then of course you can.

Anonymous

@JohnRennie Yes, that's what I meant

Anonymous

7:45 AM

What would be the experimental method for that?

Anonymous

Actually take something simple like $|0\rangle$ and $e^{i\theta}|0\rangle$

John Rennie

@Blue Young's slits?

Anonymous

Umm, how exactly? Say I have a qubit (using electron spin) in state $|0\rangle$ and another electron spin in state $e^{i\theta}|0\rangle$

Anonymous

Or you could use the differently polarized lights

Anonymous

Actually, I should explain the context of the question first maybe:

Anonymous

7:51 AM

1

A: Why should state of the "control register" change in the Phase estimation algorithm?

Imagine you have an eigenvector $|u\rangle$ of $U$. If you have a state such as $|1\rangle|u\rangle$ and you apply controlled-$U$ to it, you get out $e^{i\phi}|1\rangle|u\rangle$. The phase isn't attached to a specific register, it's just an overall multiplicative factor. Now let's use a superpo...

Anonymous

DaftWullie says in the comment that there is no way to distinguish between a composite state $(|0\rangle)_A \otimes (e^{i\theta}|0\rangle)_B$ and $(e^{i\theta}|0\rangle)_A \otimes (|0\rangle)_B$

Anonymous

There is no physical difference. Think about it this way: what experiment would you do to distinguish the two? If there's a physical difference, there must be a way to distinguish them. — DaftWullie 58 mins ago

Anonymous

in The Classical Channel, 48 mins ago, by Blue

@DaftWullie Interesting. So, I can't think of any experiment to distinguish the "phase". But one thing which is still a bit of unsatisfactory to me: Why is it just that our failure to create an experiment which can distinguish the phase, dictate on which qubit that phase actually should be in? Why should physics depend on our restrictions at all?

Anonymous

in The Classical Channel, 47 mins ago, by DaftWullie

physics doesn't. quantum theory is our description of physics (as best we know it), and it doesn't contain the possibility to distinguish it. Perhaps there's a different theory that does permit it.

Anonymous

"Think about it this way: what experiment would you do to distinguish the two?" i.e. $(|0\rangle)_A \otimes (e^{i\theta}|0\rangle)_B$ & $(e^{i\theta}|0\rangle)_A \otimes (|0\rangle)_B$ ----- is that true, in the first place? That's why I asked the question about whether we can distinguish the presence of a phase difference

Anonymous

7:59 AM

@JohnRennie I mean what we need to detect is not just the presence of the phase difference, but in fact the relative phase difference. That is, does qubit A have a phase of $e^{i\theta}$ w.r.t qubit B or does qubit B have a phase of $e^{i\theta}$ w.r.t qubit A

Anonymous

Such an experiment would help to distinguish between $(|0\rangle)_A \otimes (e^{i\theta}|0\rangle)_B$ and $(e^{i\theta}|0\rangle)_A \otimes (|0\rangle)_B$

Novalium Company

8:11 AM

Guys, I was thinking about a voltage divider with a load (= resistor) and what we can do is sum the second resistor with the load resistor (which are in parallel) and we get an equivalent one, after that we can view the system as if there was no load and no current going out of the output?

Anonymous

@NovaliumCompany Why would there be no current and no load?

Anonymous

Can you draw the situation?

Novalium Company

Ok, let me simplify my question. If I wanted to build a voltage divider, where I have initial 9V and I want my load to receive 0.1A and 5V how can I do that? How can I calculate what resistors I would need?

(I made the numbers up (0.1A, 5V...), I don't know if they'll work)

Anonymous

$9 V = (1.5 A)R$

Anonymous

$R=?$

Anonymous

8:15 AM

Trying to see if they can be put in series

Novalium Company

So first I find the resistance of the load?

ups

nvm.

Wait, my book explains that, let me read again and I'll come back after 5 mins.

Anonymous

If net resistance of load is less than 6 ohms it can be put in series with another resistance such that the net is 6

Anonymous

If it is more than 6, we'll have to device some other branching

Novalium Company

My book gives an example where load is taken into consideration but it talks about some 10 percent rule, $Vout = Vin\frac{R2}{R1 + R2}$ won't work anymore.. and it just gets complicated. The exercise is: You have a 10V supply, create a voltage divider to supply a load that is rated at 3V and draws 9.1mA.

Anonymous

It's rated 3V and draws 9.1 mA. Find its resistance first

Anonymous

8:25 AM

Simple V=IR

Novalium Company

R = 27.3mOhms.

Anonymous

Fair, now solve for $r$ from $10 V=(27.3 m\Omega + r)(9.1 mA)$ I guess?

Novalium Company

What is that, why?

I now know the resistance of the load and the desired voltage.

And I need to find resistor values?

Anonymous

Sid

@NovaliumCompany are you sure about that calculation?

Novalium Company

8:32 AM

No, my book says different...

My book says 330ohms.

for the load resistance.

Anonymous

If your load is $R_2$, then you need to find $R_1$

Sid

I ask because if the current is in mA, the Resistance is usually in kilo-ohms..

Anonymous

In the above diagram

Anonymous

@Sid Yeah, that calculation @NovaliumCompany did looks wrong

Novalium Company

@Blue we don't have R2 yet.

Anonymous

8:35 AM

R2 is your load resistance

Anonymous

Which you calculated wrongly

Anonymous

8 mins ago, by Novalium Company

R = 27.3mOhms.

Novalium Company

no, I need to find R1 and R2. We know Rl.

Anonymous

@NovaliumCompany Can you show your circuit diagram? I don't know what $R_1$ and $R_2$ is supposed to mean in your context.

Anonymous

Novalium Company

8:36 AM

taleri.files.wordpress.com/2014/02/… go to page 81.

Anonymous

Here there is no $R_L$ separately

Novalium Company

That's a wrong diagram then xD

The book is using some sort of 10 percent rule...

which is confusing even more.

Anonymous

So, basically their constraint is $I_2=0.1 I_L$

Anonymous

Now it's just solving a couple of simultaneous linear equations

Anonymous

8:42 AM

Write down the KVL equations

Novalium Company

$I_2=0.1 I_L$ confuses me.

Why would you want your I2 to be 10% of IL?

Anonymous

Apparently it minimizes some power loss

Novalium Company

Can we first do an example without that complication?

(on the same diagram)

Anonymous

We'll have to do the math, to check if their claim is actually true

Novalium Company

So lets first calculated the resistance of the load?

Anonymous

8:46 AM

@NovaliumCompany You could try, Unfortunately, I need to leave now. Anyhow, as far as minimizing power loss is concerned that rule doesn't seem accurate to me (you could ask on the main site though)

Novalium Company

Oh, ok. See you then :)

I'll have to leave as well.

Novalium Company

11:41 AM

youtube.com/watch?v=JGXdi7XcQi8 At 17:36. What I don't understand is why if we want to keep the voltage at 3V across R2 (and Rload), why the resistance of R2 has to be 100 times less than the resistance of the load?

Secret

12:27 PM

@Cows, you have became famous

Secret

12:58 PM

newscientist.com/article/…

If reality is a pond, and particle are the ripples from the splashes, what exactly is water?

and moreover, how meaningful we can go down this rabbit hole before it ceased to be relevant

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