User talk:Light current/My Sandbox

Freds Adventures in electromagnetic fields.
Fred is an experimental scientist of yesteryear. Like many of todays physicists, Fred knows nothing about electrical engineering other than EM radiation. He doesnt know what it is, but he know it travels at a constant rate 'c' depending on the medium because he has an accurate clock.

Unlike today's physicists, he has not yet discovered the electron, charge, current, magnetism or anything similar, yet he is far more informed about the fundamental nature of EM radiation and knows that it always travels at a constant rate. He is thus far more informed than the real scientists pre Michelson-Morely etc.

Freds experiments.
Fred starts building his knowledge from what little he knows (as do we all). Fred found he could send em down a pair of wires and he could detect this em when it arrived. What could Fred deduce about things purely by oberving em radiation? Fred surmises that this phenomenon could be energy and sets out to test this theory. Fred is familar with the heat version of energy as he uses this to cook his meals stay warm etc.

If Fred sent some of this stuff down his long line, then left it, what did he get? By taking a resistor, and connecting it across the end, Fred could determine that, by the heating effect, the stored substance did appear to have energy.

What other properties of em could Fred discover?
Fred found that he could only store a finite amount of energy in his line. By experiment, he found that the energy stored was proportional to the length of the line, the voltage to which it was charged (or the amount of energy inserted). He also found that the time it takes to remove all the energy depended on the value of the resistor that he put on the end. If Fred put just the right value on, all the energy came out in one quick dollop, otherwise it seemed to dribble out more slowly. What would the Fred, intelligent scientist of yesteryear make of that.?

By doing energy measurements the Fred would quickly find that the energy stored was prop to the length of the line and propl to the square of the voltage across the line, it did not matter how fast the energy was extracted, it was constant. THe square factor might disturb him as he is not used to this sort of thing.

At this point Fred, decides to try to look at the problem a different way. He has a flash of inspiration one night in his cave and decides that this energy might just be due to two factors and not one. Next morning he therefore proceeds to investigate fundamentals. When Fred attached his resistor once more to the end of his line, he noticed a voltage across it but only as long as the energy was coming out. When he changed the value of the resistor, he noticed the voltage change as well.

Conveyor belt analogy of Transmission line

Assume a conveyor belt as used in coal mines, coal iis transported on top an the belt returns underneath. Now imagine buckets attached to the belt. When the buckets get to the far end, they are tipped up and drop their load. If not tipped, they return their load to the start point.

-->> UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU --U load(receiving) end <<--                                                         U UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU --

Diagram rep of conveyor

Freds argument
(Note: In the following, subscript notation is used for 1st and 2nd partial derivatives.)

For an ideal lossless two conductor transmission line (TL), the voltage across the conductors and the current through the conductors must satisfy the following equations:


 * $$V_x + L I_t = 0 \,$$


 * $$I_x + C V_t = 0 \,$$

where L and C are the inductance and capacitance per unit length of the TL.

These equations lead to the following wave equations for the voltage and current:


 * $$V_{xx} - LC V_{tt} = 0 \,$$


 * $$I_{xx} - LC I_{tt} = 0 \,$$

Consider such a TL of unit length that is unterminated at both ends. The current at the ends of the TL is constrained to be zero. '''No! for each individual travelling wave, I <> 0. I_t = I(+) + I(-)'''

With these boundary conditions, the solution to the current wave equation is:


 * $$I(x,t) = \sum_{n=1}^\infty a_n(t) \sin(n \pi x) \,$$

where


 * $$a_n(t) = 2 \left [ \left ( \cos(\frac{n \pi t}{\sqrt{LC}}) \int_{0}^{1} I(x,0) \sin(n \pi \tau)\, d\tau \right ) + \left ( \sin(\frac{n \pi t}{\sqrt{LC}}) \int_{0}^{1} I_t(x,0) \sin(n \pi \tau)\, d\tau \right ) \right ]$$

For


 * $$I(x,t) = 0 \,$$

the only solution is the trivial solution:


 * $$a_n(t) = 0 \,$$

which can only be true when the initial conditions are:


 * $$I(x,0) = I_t(x,0) = 0 \,$$

Thus:


 * $$V_x = 0 \Rightarrow \ V(x,t) = constant \,$$

That is, for a finite length ideal lossless TL that is unterminated at both ends, there is no possible superposition of waves that give the solution V(x,t) = constant, I(x,t) = 0.

My comments
How has voltage reflection boundary condtion at the ends been accounted for? Is it hidden in I=0 at the ends?

This argument is totally fallacious in that it presuposes no current at the end of the line even when being fed by a current. If this assumption is made, its obvious there will be no voltage on the line

My argument
Alfred only seems to be considering waves travelling one way! THe two way complete equations are:
 * $$I(+)(x,t) = Vo/Z\sum_{n=1}^\infty a_n(t) \sin n(t \omega  -kx) \,$$
 * $$I(-)(x,t) = -Vo/Z\sum_{n=1}^\infty a_n(t) \sin n(t \omega +kx) \,$$


 * $$V(+)(x,t) = Vo\sum_{n=1}^\infty a_n(t) \sin n(t \omega -kx) \,$$
 * $$V(-)(x,t) = Vo\sum_{n=1}^\infty a_n(t) \sin n(t \omega + kx) \,$$

So Itot = I(+) + I(-) and Vtot = V(+) + V(-) Now
 * $$A_n(t) = 2/T \left [ \cos(\frac{n \pi t}{\sqrt{LC}}) \int_{0}^{1} I(x,0) \cos(n \pi \tau)\, dt \right ]$$


 * $$B_n(t) = 2/T \left [ \sin(\frac{n \pi t}{\sqrt{LC}}) \int_{0}^{1} I(x,0) \sin(n \pi \tau)\, dt \right ]$$

Total current in line is:


 * $$I_t(x,t) = Vo/Z[\sum_{n=1}^\infty A_n(t) \sin n(t \omega -kx) \ -\sum_{n=1}^\infty A_n(t) \sin n(t \omega +kx)] = 0$$

Assuming the summation can be performed later,


 * $$I_t(x,t) = Vo/Z\sum_{n=1}^\infty [A_n(t) \sin n(t \omega -kx) \ -A_n(t) \sin n(t \omega +kx)] = 0$$


 * $$I_t(x,t) = Vo/Z\sum_{n=1}^\infty [A_n(t) 2\cos n \omega t\sin nkx] = 0$$

Cant get the sine wave to approach a dc level without squaring (rectifying) it. When the sinusiondal functions of time and space are added,(sin A + sin B) the voltage seems to be a sine wave amplitude modualted by a cosine wave. Will this give rectification??

The line is of finite length so this puts limits on the fundamental and harmonic frequencies supported. THe lowest freq poss for a voltage wave is 1/(2T) wher T is the line delay. THe lowest freq poss for a current wave is 1/2T (ie 1/2 wavelength in line)

Similarly,
 * $$V_t(x,t) = Vo=\sum_{n=1}^\infty [A_n(t) 2\sin n \omega t\cos nkx] $$

Quantitative Treatment of Radiation
In the original edition of his book, Purcell went no further than drawing these pictures and interpreting them qualitatively. But in his second edition he added an appendix in which he computes the radiation field quantitatively.

Labeled diagram for computing strength of transverse field

Derivation of Larmor formula
I've never understood why the Larmor formula isn't even mentioned in most introductory textbooks. In fact I never saw it in all of my undergraduate education. Yet it's simple to interpret, and it has all sorts of interesting uses. I've already mentioned antennas. You can also understand why the sky is blue and polarized, and explain why a classical Rutherford atom, with electrons orbiting around the nucleus, would be unstable, radiating away all its energy in a fraction of a nanosecond. More generally, the Larmor formula ties electromagnetic waves to their source, making it clear that radiation requires not just motion, but also acceleration of charged particles. Once you have the Larmor formula, why bother to cover Maxwell's equations?