Recent content by cometzir

Sorry for unclearly description. This formula is used in a two dimensional system. v1 and v2 are velocity operators in x and y direction

Several papers (eg. Di Xiao, et. al, Berry phase effects on electronic properties, RevModPhys, 82,2010)mentioned a formula to calculate the Hall conductivity(See the picture).This formula is used in an two dimensional system, v1 and v2 are velocity operators in x and y direction, Phi0 and PhiN...

We say we can't predict everything exactly, but what has happened is not uncertain. Since what will happen is certain, why can't we know it before it happens?

Yes, when we measure some thing, the measurement itself will have a effcet on the thing, and then, as the Heisenburg uncertainty says, we will never get a absolutely accurate result, But I think the measurement error is because we separate the thing we measure from its external environment, and...

When we study something with our physics theory, we may always ignore some "unimportant" factors to simplify the culculation.And then, we get a approximation. But if we don't ingore any factors, we will get the absolutely accurate result. Is it possible? I think the physics theory isn't...

But how could the Linear Accelerator control the change of the E field? Does it take the same time for the particle travel from one plate to another?

Thank you, I think I have understand it.

Thank you. Another thing, think about the Linear Accelerator, it has many pair of plates. Are there any opposing E fields between the plates as the picture shows? And why is there no E field in the metal box of the cyclotron?

But a Metal Masks can also shield the E field, and there are many holes. The metal box of the cyclotron also has holes, but it can also shield the E field.

So the intensity of electric field in the box isn't zero even if the plate isn't in the box, and the electrostatic screen is just a approximation. Is it right?

It is true there can be an E field between the plate and the box, but in the second situation ,if the plate is not in contact with the box, the opposing E field will be perpendicular to the E field between the plate, even though the e field line isn't straight, it won't be "opposing".

Thank you for your detailed answers. I know it is impossible to let the equipment in the 1st and 3rd picture work. But in the 2nd pictrue, the pates can be smaller and aren't in contact with the box,as the following picture shows,if so ,where is the apposing E field? For approximation, I...

If the capacitor is out of the box, the opposing electric field can be like the red line in graph 1, but if the capacitor is on the border of the box like graph 2, or even in the box like graph 3, can you draw the opposing eletric field line?

So, there is another opposite electric field.But where is it? Is it in the metal box, or in the plate inside? If it is in the metal box, the electric field strength inside the box will not be zero. Is it possible?

Here is the picture