There's sooo much going on today. I'm back in the lab again, but I'm also studying for the last little bit of my EM II class. Here are the EM notes for today. Hopefully, I'll get a lab book up again in the morning.

We'll have a particel mofin along hte path $\vec{r} = \vec{r_o}\left(t\right)$. There is a quite lengthy explanation of IRFs, but I'll skip that for now and keep careful track of whether or not this comes back to bite me in the butt. We define $\vec{R}\left(t^\prime\right) = \vec{r} - \vec{r_0}\left(t\right)$ which is the vector from the point charge at time $t^\prime$ to the observatin poitn $\left(\vec{r}, t\right)$. This gives us a retarded time, $t^\prime$ determined by $t - t^\prime = R\left(t^\prime\right)$, where $R\left(t^\prime\right) = |\vec{R}\left(t^\prime\right)|$. This makes far more sense if you translate one of the ever present ever invisible $1$s to a c to get $c\left(t - t^\prime\right) = R\left(t^\prime\right)$

The potentials in the IRf can be written as

$\phi = \dfrac{e}{R\left(t^\prime\right)}$, $\vec{A} = 0$.

A charge at rest will have 4-veclocity $U^\mu = \left(1, 0, 0, 0\right)$.

We can noew define the 4 potential to be $A^\mu = f U^\mu$. We can also form a four vector version of $R^\mu$ as $R^\mu = \left(t - t^\prime, r - r_0\left(t^\prime\right)\right) = \left(t - t^\prime, \vec{R}\left(t^\prime\right)\right)$. Looking at this, you should see a four space distance without the time axis turned negative. In a sense, this fits because it isnt' squared yet. In a sense it doesnt' because if it isn't squared, then the time componet shoudl have an $i$ out in front. This is somewhat Wick rotated, to coin a somewhat fancy phrase.

In the special case where the charge truly isn't moving, then $f$ above shoudl be $e/R\left(t^\prime\right)$. For the more general case where the charge is moving with four velocity $U^\mu$, we get

$f = \dfrac{e}{\left(-U^\nu R_\nu\right)}$, so $A^\mu = -\dfrac{eU^\mu}{U^\nu R^\nu}$

Here, we have rather mysteriously gotten our negative sign back in front of the time coordinate. Ask the professor about this tomorrow. The pertinent point is near equation 7.29.

Now, on to problem number 2

The strategy is just to bludgeon through the curl of $\vec{A}$ equation and get the final result. Then, bludgeon through the cross product of $\vec{E}$ and show that the results are equivalient.

Looking at the Leinard-Wiechert Potentials.

We'll have a particel mofin along hte path $\vec{r} = \vec{r_o}\left(t\right)$. There is a quite lengthy explanation of IRFs, but I'll skip that for now and keep careful track of whether or not this comes back to bite me in the butt. We define $\vec{R}\left(t^\prime\right) = \vec{r} - \vec{r_0}\left(t\right)$ which is the vector from the point charge at time $t^\prime$ to the observatin poitn $\left(\vec{r}, t\right)$. This gives us a retarded time, $t^\prime$ determined by $t - t^\prime = R\left(t^\prime\right)$, where $R\left(t^\prime\right) = |\vec{R}\left(t^\prime\right)|$. This makes far more sense if you translate one of the ever present ever invisible $1$s to a c to get $c\left(t - t^\prime\right) = R\left(t^\prime\right)$

The potentials in the IRf can be written as

$\phi = \dfrac{e}{R\left(t^\prime\right)}$, $\vec{A} = 0$.

A charge at rest will have 4-veclocity $U^\mu = \left(1, 0, 0, 0\right)$.

We can noew define the 4 potential to be $A^\mu = f U^\mu$. We can also form a four vector version of $R^\mu$ as $R^\mu = \left(t - t^\prime, r - r_0\left(t^\prime\right)\right) = \left(t - t^\prime, \vec{R}\left(t^\prime\right)\right)$. Looking at this, you should see a four space distance without the time axis turned negative. In a sense, this fits because it isnt' squared yet. In a sense it doesnt' because if it isn't squared, then the time componet shoudl have an $i$ out in front. This is somewhat Wick rotated, to coin a somewhat fancy phrase.

In the special case where the charge truly isn't moving, then $f$ above shoudl be $e/R\left(t^\prime\right)$. For the more general case where the charge is moving with four velocity $U^\mu$, we get

$f = \dfrac{e}{\left(-U^\nu R_\nu\right)}$, so $A^\mu = -\dfrac{eU^\mu}{U^\nu R^\nu}$

Here, we have rather mysteriously gotten our negative sign back in front of the time coordinate. Ask the professor about this tomorrow. The pertinent point is near equation 7.29.

Now, on to problem number 2

For 2.a., and b, see the Wake Forest notes:

\url{http://users.wfu.edu/natalie/s13phy712/lecturenote/lecture27/lecture27latexslides.pdf}

expression 17 and up give the appropriate curl. If time allows take a look at Dr. Nevels article on graphically protraying E and B fields. There's no reason everything shouldn't apply here since the L\&W potentials were derived classically.

The strategy is just to bludgeon through the curl of $\vec{A}$ equation and get the final result. Then, bludgeon through the cross product of $\vec{E}$ and show that the results are equivalient.

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