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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 3 months |
| seen | 17 hours ago | |
| stats | profile views | 8 |
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Mar 31 |
comment |
Is there a mechanism for time symmetry breaking? I'll let someone else answer but I believe magnetic fields are key here. |
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Mar 9 |
comment |
I don't understand what we really mean by voltage drop (Part III) This steady state of flow is a current. And the only thing now that defines the Voltage (or voltage drop - look at my answer) is the resistance between your two probe points. Probe a wire (pretty much zero resistance), no voltage drop. Probe across a resistor you will get a finite voltage drop. I hope this is clear. |
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Mar 9 |
comment |
I don't understand what we really mean by voltage drop (Part II) Energetically we could talk about an effective potential which now takes into account this new resistive element. It mightn't be a bad idea to think about voltage like this. I can't get around it - but the basis of this concept is $V=IR$ which is why my answer focused on it. Note it is the relationship between a flow, impedance and potential difference. One needs to think about the entire circuit at once, macroscopically - I don't know what happens individual electrons. Once the battery is connected, practically instantly the circuit will reach a steady state of flow. |
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Mar 9 |
awarded | Commentator |
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Mar 9 |
comment |
I don't understand what we really mean by voltage drop I might need to think about this more but I've read Edit 2 a few times. Certainly a battery works by a separation of charge, but when in a circuit I think one must be careful about thinking about things in terms of "difference in number of electrons". Its all about flow (current). The Voltage is the potential difference which allows flow to occur (think gravitational potential difference allowing an object to fall from high to low). What if something impedes this motion? In our classical mechanics example we could introduce a viscous liquid or pinball maze to divert flow. |
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Mar 9 |
awarded | Enthusiast |
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Mar 5 |
awarded | Editor |
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Mar 5 |
revised |
I don't understand what we really mean by voltage drop Further example |
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Mar 5 |
answered | I don't understand what we really mean by voltage drop |
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Mar 4 |
comment |
Greens function in EM with boundary conditions confusion I said coincidence since this was the confusing point for me - when it is physical, and when it is not. It is very intuitive for me to say: Lets build up a charge distribution from dirac deltas. Then we jump from the greens to the convolution of G and $\rho$. Thanks for the clarification though! |
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Feb 28 |
comment |
Greens function in EM with boundary conditions confusion If this comes up I recommend mentioning that in an unbounded region the Greens function by coincidence (upto a normalization factor) is the potential of a point charge. So here it kind of has a physical meaning. When using it to solve problems, specifically when we consider boundaries or finite regions of space it is purely a tool. We use our conditions, jackson 1.42 and our freedom in $G$, Namely $G(x,x')=1/|x-x'|+F(x,x')$ where $F$ must satisfy the laplace equation to find the solution. (That is briefly what I got from your answers and my research). Graduate TA's would be nice to have. |
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Feb 28 |
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Greens function in EM with boundary conditions confusion It sure took a while, but I got it! I'm glad I am an experimentalist because this stuff flies way above my head - not to mention my simple angular momentum/torque blunder a few days ago. You are an asset to this website, Thank you. |
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Feb 28 |
accepted | Greens function in EM with boundary conditions confusion |
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Feb 28 |
comment |
Greens function in EM with boundary conditions confusion (hopefully I am being clear where my issue is, I don't want to be repeating the same thing over and over) |
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Feb 28 |
comment |
Greens function in EM with boundary conditions confusion Well from the mathematics its clear but for the problem I don't understand why we are adding charges that aren't there so to speak. $1/|x-x'|$ is the greens function of a point charge, and sure adding two together symmetrically will construct the desired boundary condition in this case but doesn't this have ramifications elsewhere? For example, is the greens function for the same problem with a REAL charge at $z=+x$ the same? If not, I don't understand the difference. Abstractly, mathematically I am fine. Thinking physically as above is where I am confused. |
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Feb 28 |
answered | Cable TV version of infinite ladder network |
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Feb 28 |
comment |
Greens function in EM with boundary conditions confusion I think what you said is perfectly clear - I am just taking some time to digest it. I am understanding more about how the boundary conditions come into play. Take E.q. 1.42 for example. If we specify dirichlet conditions such that 1.43 must be satisfied, why don't both surface integrals go to zero? (i.e. if $G_D=0$ then why doesn't $\frac{dG_n}{dn'}=0$ also?) I have no problem accepting your intuition statement except for the fact I don't see how charges get involved at all. Is this where I am applying physical meaning to the greens function where there is none? |
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Feb 28 |
awarded | Scholar |
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Feb 28 |
accepted | Is angular momentum always conserved in the absence of an external torque? |
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Feb 28 |
asked | Greens function in EM with boundary conditions confusion |
