# Faraday's law and complex number

1. Oct 9, 2008

### spidey

Faraday's law is ∫ E dl = - ∂Ф/∂t
if we take sqaure root on both sides,

√∫ E dl = √- ∂Ф/∂t
√∫ E dl = i √ ∂Ф/∂t

Now the r.h.s has "i" in it. Does this mean anything? Having "i" in a equation means anything?
I have seen "i" in schrodinger equation and dirac equation. As like those equations, does the above equation also has any meaning?

2. Oct 9, 2008

### Redbelly98

Staff Emeritus
We don't really have an i, because one of the following will be true:

∫ E dl is negative, so √∫ E dl also gets a factor of i. Divide both sides of the equation by i, and there will be no more i's.

or

∂Ф/∂t is negative, in which case -∂Ф/∂t is positive, and
√ -∂Ф/∂t = √ (a positive value) → no factor of i

or

∫ E dl and ∂Ф/∂t are both zero → no factor of i

Of course, you could save a lot of trouble by not taking the square root in the first place!

3. Oct 9, 2008

### Antenna Guy

1) Under what conditions would ∫ E dl be "negative"?

2) Under what conditions would ∫ E dl be "positive"?

3) What is the sign of dl/dt in both cases?

Regards,

Bill

4. Oct 9, 2008

### Redbelly98

Staff Emeritus
Bill,

Are you familiar with induced emf's and Lenz' Law? That's the type of situation where ∫ E dl is nonzero.

Mark

5. Oct 9, 2008

### Antenna Guy

Hi Mark,

If E is propagating, dl/dt is not zero. The questions have to do with what your sign convention is.

Regards,

Bill

6. Oct 9, 2008

### spidey

supposing for this case ∫ E dl and ∂Ф/∂t are both positive, then -∂Ф/∂t is negative and so it will have an i. Under this condition, does this have any meaning?

My basic question is, not only for this equation, for any equation in physics,if it has an "i",then what it tells us?

7. Oct 9, 2008

### Antenna Guy

I would assume that dl/dt is negative in that case. I think that is the standard physics/optics convention.

Regards,

Bill

8. Oct 9, 2008

### Redbelly98

Staff Emeritus
They can't both be positive! Because

∫ E dl = -∂Ф/∂t

so if one is positive, the other must be negative.

9. Oct 9, 2008

### Antenna Guy

Not true if dl/dt is positive. Care to show otherwise?

Regards,

Bill

10. Oct 9, 2008

### Redbelly98

Staff Emeritus
Since dl is a vector (a length element with a direction), how does it even make sense to talk about dl/dt being positive?

But the problem with the OP's question is: given any equation in physics with a "-" sign on one side of the equation. Take the square root of both sides of the equation, and now you have a mysterious factor of "i", and he is trying to find meaning in that. For example, F=-kx for a spring.

11. Oct 9, 2008

### Antenna Guy

Because the notion of "propagation" has a direction too.

Regards,

Bill

12. Oct 10, 2008

### Redbelly98

Staff Emeritus
Bill,

I'm having trouble following your argument, beginning with this statement:

I am not even sure what dl/dt means at this point. I thought you were describing a situation where the closed path of the integral is moving and changing shape.

If so, I think this unnecessarily complicates the OP's question because dl/dt does not appear in the equation in question,

$$\int \vec{E} \cdot d\vec{l} = - \ \frac{d\Phi_B}{dt}$$

Anyway, could you clarify what dl/dt is for me? Let's make sure we're talking about the same thing.

13. Oct 10, 2008

### ibc

the minus sign is only to get the directions right, it's not that important, even less important when you take the square root of it =o
it's usualy more convenient to use avsolute values in such equations, and find the direction later.

about your question, I know that in RCL circuits' equations "imaginary" solutions are taken in cosideration aswell, imaginary currents, imaginary resistance, and stuff like that.
I think what it means is that these solutions (currents, resistances) are sinusoidal , but I'm not sure what's the physical meaning of these "imaginary values" if there is one at all, or maybe at the end only the real part of the value is taken, I'm not sure.

14. Oct 10, 2008

### marcusl

In many 2 dimensional problems, real and imaginary parts are used to indicate direction. For instance, in exp(i*phi)=cos(phi)+i*sin(phi), you can plot real and imag parts on the x and y axes, respectively. This is used to solve problems in potential theory (through conformal mapping, for instance). Elsewhere, real and imaginary indicate in-phase and quadrature components of a signal or a field. There are many other examples, some of them more subtle.

In other cases, real and imag parts indicate allowed and non-allowed modes. A light wave hitting a polished metal surface will reflect. The incident and reflected waves have real propagation constants; the evanescent wave that penetrates a nearly infinitesimal distance into the metal cannot propagate and is described by an imaginary propagation constant.

ibc mentions correctly that i is sometimes used as a bookkeeping tool and that the real part of an expression is taken at the end to find the physically significant quantity.

In short, i definitely has physical meaning in many cases.
??
You might profit from review of your E&M texts. First of all, see Redbelly's responses. Second, there's no propagation here--Faraday's law deals with the line integral of the field around a fixed path L that may be physical (a wire) or virtual.

15. Oct 10, 2008

### ArjenDijksman

As indicated by marcusl, complex numbers give directions in planes. When you have a factor "i" in a differential equation for a vector or vector derived function, it means that the vector differential is at an angle 90° to the vector itself. I've tried to make that clear in my video clip on the "www.youtube.com/watch?v=JmEMVJYbTu8"[/URL]. Excerpt: "There is the operation where the arrow is rotated by an angle alpha. We multiply ket |A> by a so called complex number to describe this rotation: exp(i alpha). So when you see a complex number in quantum-mechanical expressions, it is real physics. It simply means that ket |A> has undergone a rotation in its spinning surface."

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