It was staying still but now because you are

running with the particle or say a person which was outside the bus,

if you're running with them, to you,

it seems like your bus is moving to the left side and you guys are standing still.

When that happens, when all of a sudden in your world,

bus starts to move to the left side,

then those that were standing still seems to move to left side,

and those that were running to the right side with the same velocity,

but with opposite direction,

seems to be standing still.

So, this is the plus charge,

so plus charge creates current to make a magnetic field, right?

Interestingly, because plus charges is going to the left,

the current seems to be going to the left as well.

As in the case when you are standing still,

when your electrons we're running to the right side because

the current is opposite to the flow of your electrons.

So, no matter you are standing still or running with your particle of interests,

in that wire, you will have current to the left side which will create magnetic field.

However, your particle of interests has no velocity, is standing still.

So, V cross B will be zero,

because V is zero.

So, that's why even though you have succeeded in creating a magnetic field,

even when running with the particle of interest,

we cannot exert a magnetic force on a particle.

So, now you may scratch your head, what happens?

How can I turn off the magnetic force,

and still see the same effect?

Still see the same force, right?

So, if there's any force on the particle,

it must come from electric field.

Why? [inaudible] is a multiple choice.

From Lawrence force, we hear either electric or magnetic or both.

If you lost magnetic,

somehow, you have to have electric, right?

So, it must come from an electric field,

but then you can scratch your head.

Why? It is impossible,

because you have the same number of charges,

plus and minus charge.

How can I suddenly create a non-electroneutral case, right?

So, it must be that the moving wire has produced an electric field.

Somehow, they have disturbed the local electron neutrality.

So, it must be that a neutral wire with a

current appears to be charged when set in motion.

So, let's see if this hypothesis makes sense.

So, we will discuss a little bit of a difficult concept

like contraction and dilation of space, and also time.

So, relativistic contraction of distances,

and its impact on charge density.

So, charges are always the same moving or not, period.

So, the charge Q on a particle is an invariant scalar quantity,

independent of the frame of reference, so that's constant.

We need only worry about the fact that the volume can change

because of the relativistic contraction of distances.

So, let me ask Melody.

In modern physics, what have you learned about the volume when you have a moving object?

Will the moving objects volume change or not?

I think it will decrease, right?

It will decrease, yes. It will decrease.

So, if we take a length L naught of

the wire in which there is a charge density rho naught of stationary charged,

it will contain the total charge Q is equal to rho naught,

L naught, A naught.

If the same charges are observed in a different frame to be moving with velocity v,

they will all be found in a piece of the material with the shorter lengths.

That is to say, the length L is equal to L naught,

times square root of one minus V squared, over V squared.

Although Vi is very small in comparison to C still,

it is subtracting your number that is multiplying L naught.

So, you can see lengths is being shorten,

but with the same area A naught,

since dimension transverse to the motion are unchanged.

Good. So, with this knowledge,

let's move on to the next slide.

So, if we call rho,

the density of charge in the frame in which they are moving,

the total charge Q will be,

rho L times A naught.

So, again, if we call rho the density of charge in the frame in which they are moving,

the total charge Q will be rho L, A naught.

Why is that so important?

Because, again when I was moving with the particle of interest,

which sign of charge were stationary and which sign of charge are mobile?

When you're moving, the plus sides are mobile.

Yes.

All the electrons are negative.

Exactly, so plus charge are mobile,

and electrons are stationary, okay?

So, that's a very important concept.

So, the charge density of moving

distribution varies in the same A as the relativistic mass of a particle.

So, you can see rho is equal to rho naught,

over square root of one minus V squared, over C squared.

So, in other words,

if the volume decreases,

your density increases, okay?

So, let's discuss rest density versus mobile density.

We now use this general result for the positive charge density,

rho sub plus of our wire.

These charges are addressed in frame S. In S prime,

where the wire moves with the speed v,

rho prime plus becomes rho prime sub plus,

is equal to rho sub plus over square root,

over one minus V squared over C squared.

Meaning in S prime,

the positive charge density increases, right?

The negative charges are at rest in S prime,

so that they have their rest density not in this frame,

right? In S prime frame.

So, rho sub minus will be the mobile density when we're in S prime.

So, minus rho will be equal to rho prime sub minus,

over square root one minus V squared, over C squared.

So, the rho prime sub minus will indeed decrease in S prime,

that means plus increases, minus decreases.

Then what happens? Will the wire be positively charged or negatively charged?

I think it will be positively charged.

Positively charged.

So, if I have positively charged wire and negative charge outside, what happens?

They'll have some attractive force.

Attractive force, there you go.

Now, let's think about the other case

where the negative charge were moving in this direction.

So, you have a current to the left,

and your wire has electrons moving to the same directions,

you have the same current.

If you have current in parallel,

what type of force do have?

Repulsive or attractive?

I think attractive.

Attractive. Now, you have the same result,

but through different tool.

For S, you have through magnetic field,

for S prime, through electric field.

So, now you understand why we have to think both of them.

They are only tools to understand the force interaction, okay?

So, now we know that we have electric force acting on the particle in S prime frame,

and the electric field at a distance r from the axis of a cylinder is E prime,

which is equal to rho prime A,

over two pi epsilon naught r,

which is equal to this equation.

Now, the force on the negatively charged particle is toward a wire as we know.

The electric force in S prime has the same direction as the magnetic force in S,

as Melody just mentioned.

The magnitude of the force in S prime is,

F prime is equal to,

q over two pi epsilon naught,

times rho sub plus A over R times V squared over C squared,

over square root of one minus V squared over C squared.

Now, comparing the results for F prime with our result for F,

we see F prime is slightly different,

is almost the same if V is close to 0.

Then, we can ignore this term,

so it's almost F prime is F. But you see still,

the finest velocity, F prime is slightly larger than F. So,

why is this the case?

Right? You may wonder how can force increase as you move around, right?