# Don't understand something in proof

1. Aug 27, 2009

### WiFO215

Alfven's theorem:

The change of flux through any closed loop moving through a magnetic field of conducting material is zero.

Imagine a loop S moving with speed v and a time later it is a little ahead. It is now loop S', it's dimensions have somehow changed. The "band" which connects S to S' is R. All this happens in a time dt. Initially, the field was B(t). After some time, it is B(t+dt).

(a) J = $$\sigma$$(E + v X B); J finite, $$\sigma$$ --> $$\infty$$, E + (v XB) = O.Take the curl: $$\nabla$$xE + $$\nabla$$x(v XB) = O. But
8B 8B
Faraday's law says $$\nabla$$xE = -dB/dt. So dB/dt = $$\nabla$$x(v XB). qed
(b) $$\nabla$$.B = 0 --> $$\oint$$ B. da = a for any closed surface. Apply this at time (t + dt) to the surface consisting of
S, S', and R:

$$\int$$S' B(t + dt) .da + $$\int$$R B(t + dt) .da - $$\int$$S B(t + dt) .da =a

(the sign change in the third term comes from switching outward da to inward da).

d$$\phi$$=$$\int$$S' B(t + dt) .da - $$\int$$S B(t) .da = $$\int$$ [B(t + dt) - B(t)] .da - $$\int$$R B(t + at) .da ~

d$$\phi$$= $$\int$$dB/dt . da} dt -$$\int$$ B(t + dt) . [(dl X v)

Since the second term is already first order in dt, we can replace B(t + dt) by B(t) (the distinction would be
second order)
:
d$$\phi$$=dt $$\int$$ (dB/dt.da) - $$\int$$B.(dlxv) [this term is the same as (vxB).dl] =dt{$$\int$$dB/dt .da - $$\nabla$$x(vXB).da}.

So,
d$$\phi$$/dt = 0

Now, I don't understand the part in bold. What does first order and second order refer to?

Last edited: Aug 27, 2009
2. Aug 27, 2009

### CompuChip

Nice how you copied the theorem and the proof. But you mind telling us what you don't understand, or should we just go through it step by step for you?

3. Aug 27, 2009

### WiFO215

Oi! When I edited the damn thing didn't update. I noticed that as soon as I typed it in. Stupid thing. I've updated now.

4. Aug 27, 2009

### nirax

this is a non rigorous proof, where dt is ignored in comparison to t. just think of dt as a small change in t. so wherever a small number is added to a large number, the smaller one is ignored. however if you divide one small number by another small number the result may not be small and should not be ignored.

5. Aug 27, 2009

### WiFO215

I'm sorry. I don't get you.

6. Aug 27, 2009

### Born2bwire

The magnetic field is continuous and slowly varying in time in comparison to dt. dt is some very small infinitesimal time increment, so a first order approximation would be that the magnetic field at time t and time t+dt is the same.

7. Aug 28, 2009

### WiFO215

Oh. What does "first order" mean? Does it mean making a good approximation for B as it varies slowly with respect to t?

8. Aug 28, 2009

### nirax

first order means that the small number is present as dt. if it is present as a multiple of two small numbers as in dt.ds (or any other thing) it is called second order. second order approximations are even finer and when you are solving problems only crudely, you should ignore all second order terms.

9. Aug 28, 2009

### Born2bwire

To more explicitly show this you would want to do an expansion of f(t+dt).

$$f(t+\Delta t) = f(t) + \Delta t f'(t) + \Delta t^2 \frac{1}{2}f''(t) + O(\Delta t^3)$$

So the first order approximation would drop any terms with \Delta t and above. Keeping the first two terms would be the second order approximation and so on.

10. Aug 28, 2009

### nirax

let me put it in new words but same in spirit to born2bwire

suppose you want to check the value of $$(t+\Delta t)(s+ \Delta s)$$. you expand this as $$(t.s + s.\Delta t + t.\Delta s + \Delta t .\Delta s)$$. now what you should take depends upon how closely you want to approximate.

1. suppose you want a very crude answer, you just ignore all the small terms (ie the ones involving $$\Delta$$'s). you get $$t.s$$. this is zeroth order approximation.

2. if you want to approximate more closely, you keep all terms involving single $$\Delta$$'s. that gives you $$(t.s + s.\Delta t + t.\Delta s )$$. this wud be termed as first order approximation.

3. if you are still not satisfied you can keep the terms involving two (or less) $$\Delta$$'s. that gives you full $$(t.s + s.\Delta t + t.\Delta s + \Delta t .\Delta s)$$. we call this second order approximation. in this case it turns out that this is exact. but there may be cases where third or more orders of approximation is useful.

i think you get the general idea.

11. Aug 28, 2009

### WiFO215

Right. That cleared it up. Thanks guys!