Any chance anyone can help me with this question? It's gotten some comments, but no one so far was able to help. Is there a chance this discussion in Hamill's book has an error? physics.stackexchange.com/questions/496852/…

Groot Mourn in

Final schematic. What do you think?

@TKT 1. It would help if you gave your (or rather the book's) definition of "canonical" for a transformation 2. It is non-standard to have a transformation on phase space that involves complex numbers, since the phase space is not required to carry a complex structure. There is a meaningful way to use complex variables also in classical mechanics, but this is emphatically not basic Hamiltonian mechanics.

@JohnRennie Are you sure the voltage required to move TiO2 particles in a $1cm^3$ 2mm plexiglass container right? Would a lower voltage work?

@NovaliumCompany I don't know what voltage is needed. But from what people have posted here the commercial screens use a voltage of 15 volts and a cell size of 100 microns. That gives a field strength of 1500V/cm.

The force on a charged particle is equal to the charge times the field strength, so if you have a 1cm cell then to get the same force you need a voltage 100 times as big.

@JohnRennie Got it, thanks

@JohnRennie Another really stupid question. What's the main difference between a conventional transformer and an audio transformer except that one uses separate windings and the other is self-induced? Also, I'm looking at a 100V audio transformer, what's that voltage?

Regular transformers are optimised to work at 50Hz while audio transformers are optimised for higher frequencies.

I haven't looked in detail at your circuit but it looks as if you're using a signal generator at audible frequencies so hundreds of hertz. In that case an audio transformer will have lower losses than a mains transformer.

@JohnRennie Yep, @PM2Ring advised me to use an audio transformer. I'm finding out the frequency of the circuit now, one sec.

But bear in mind that I know next to nothing about electronics so this is an opinion rather than a statement of fact!

@JohnRennie If you know next to nothing, I know -nothing :D so I'll take my chances

Re the voltage, the high field gradient is required to make the cell switch fast i.e. to make the screen refresh fast. If you're just experimenting you could use a lower voltage. It would make your cell switch more slowly but that wouldn't matter.

@JohnRennie All I want for now, is to make a single cell switch between black and white, that's it :D

So yeah, I'll probably start with voltage around 880

I was just thinking that if you kept the voltage down to say 400V that would be easier to generate.

You'd get slow switching but that wouldn't matter for your experiment.

Well, I have the circuit working, all I'm struggling is to find a proper transformer that's available to buy in a local store, but I don't know what to look for :P

@NovaliumCompany something like this possibly

But e-reader displays have a lot of very small cells, which would result in a one giant cell,which would require quite a lot of voltage, and I don't see how they fit a giant transformer in there

@NovaliumCompany the cells are all in parallel so they all have just 15V across them.

@JohnRennie bit confused. If in the future I decide to add more 1cm^3 cells and make a display, I could somehow split that 1500V into many 1500V?

Let me draw a quick diagram:

I just used a speaker and a phone app to find out that the frequency is 2160Hz (maybe I'm mistaken, I'll check again)

Suppose you want your cells to show this simple image. You need the voltages to be as I've marked them on the image.

The way you'd write the cells is:

So you just need a single voltage $+V$

I've shown the cells wired together with actual wires. In an eInk display transistors are used to direct the voltage to the individual cells.

Of course... so simple... stupid me

I actually knew this, but I was being a stupid homunculus as always

:-)

But if you have many LEDs in parallel, and a 9V source, they wouldn't all get 9V between them? They would all get the same amount of voltage but lower, no?

When you wire LEDs in parallel they are all connected to the same battery so they all get the same voltage.

I mean, I've hooked an LED to a 9V and it shine bright. When I hooked 5 LEDs in parallel, they shine not so bright.

If they draw too much current then it would reduce the battery voltage due to the internal resistance of the battery.

Got it

To avoid this you use a voltage source with a low internal resistance.

Thank you. About the transformer, how do you determine that this is the right one and why is it so expensive :((

@NovaliumCompany I don't know that is the right transformer, but it can be configured as different step up ratios. Have a look at the diagram of the windings:

Yeah, got it. I guess I'll look for something similar in a local store.

If you apply the input signal across C and C the primary is 21 turns. So if you take the ouput between 1 and 2 the secondary is 4000 turns so the step up ratio is 4000/31 = 129.

That would step up a 9V signal to about 1200V.

Do I have to pay attention to any other specification except the ratios? (when I browse stores)

@NovaliumCompany I don't know.

@JohnRennie Oh ok. No problems. Thank you so much for the help! Now, lunch time.

@PM2Ring I have baby oil and I'll buy different kinds of black paint to see which one works. I'll put NaOH in the baby oil and then TiO2 to charge them. Then I'll try different kinds of black paint to see which one will work. Is that a good plan?

I have a project in which instead of one delta function potential in am infinite 1d potential I have two delta function potentials. The whole well will have 3 regions. I can solve for two regions but not the one in the middle which is bordered by the delta function potentials. I have read the two research papers regarding the one with only one delta function potential but I cannot get any ideas what to do.

I have also asked this multiple times and is probably annoying but some nice hints would be great

@Korra What have you tried to do with this problem so far? If all you have are delta functions, all you need to do is solve in each region and then match the wavefunctions at the boundaries with the appropriate boundary conditions given by the delta functions

I have tried factorization methods

a=(p+ifn(x))/√2m

a(dagger)=(p-ifn(x))/√2m

My fn comes out to be √2mE cot ( (√2mE/hbar)(x-b))

Where b is constant of integration

Applying singularity conditions at x=0 and x=a give me value of b as (√2mE/hbar)(a-b)=π

All this comes straightforward from the problem of one delta potential.

Nevertheless for the three "regions" formed by the delta functions, V=0 is the same as in the case of one delta potential. Therefore the above procedure applies to both cases of one delta potential and two delta potential

Even the singularity conditions but the problem is coming in the middle region which is bounded by two delta potentials

The wavefunction is not zero over there not can we apply the singularity conditions over there as our fn will not be finite of we do this

Sorry I meant f1 not fn

Also E1 and not just E

@NovaliumCompany I guess so. As I said before, finding a suitable black material might be tricky because most black paints will contain a suspension of solid particles, which will probably interfere with the TiO2 particles. You really want a black oil-soluble dye, as Wikipedia says. Dyes that are used to colour timber are called stains.

I suggest that you try to get your cell working first without the black stuff. That way it'll be easier to see how the TiO2 interacts with the field. Once you've got it working with clear oil, then try it with stained oil.

@PM2Ring Good idea. Thanks!

@JohnRennie LEP technology. Similar to LED tech but LEP standing for Light Emitting Phosphor. How does it work? I know a blue laser is used to make the intensity of the beam. And the blue light hits a piece of phosphor creating white light. But what does what at the smaller level and what creates what?

@ACuriousMind Thank you! The author simply describes them as transformations that preserve Hamilton's equations, although with a different Hamiltonian, K. He adds that H = K + the partial of F with respect to t) where F is the generating function of the transformation. I think this book might just be too advanced?

I rather think the book might be too vague :P

@ALL There are many people asking questions that have to do with both electrical engineering and physics. Most of the time they ask where to ask the questions they have. And I tell them to ask at both sites. Here and the EE.SE site. I think it would be good to open a chat which has the combination of the great people at EE.SE and The H Bar. Do you agree?

@ScientistSmithYT Please do not tell people to ask the same question on 2 or more sites. That's called cross-posting, and it is discouraged on Stack Exchange.

OTOH, if you are talking about questions in chat rooms, that's a bit different. But even in that case, it's better to not have 2 isolated parallel conversations going on simultaneously.

hello =)

@heather hey, long time no see :) How's it going?

@ACuriousMind pretty well. about to start junior year. how are you?

Very fine, just got back from vacations (went to a metal festival + a good friend's wedding) and enjoying a quiet time at work (because everyone else is also on vacation). I also have no idea what junior year is :P

3rd out of 4 years of american highschool.