Physics Help Forum Help with derivation of potential involve moving point charge.

 May 1st 2011, 07:03 PM #1 Member   Join Date: Nov 2010 Posts: 40 Help with derivation of potential involve moving point charge. I mainly have problem is in the final step of the derivation, but I am going to list the steps leading to that: The question is: Find the potentials of a point charge moving with constant velocity. Relevant equation: $\vec w(t) = \;\hbox { is position of q at time t. and}\; |\vec r - \vec w (t_r) |=c(t-t_r) \;\hbox { and }\; \eta =\vec r - \vec w(t_r) \;\hbox { and }\; V_{(\vec r, t)} = \frac 1 {4\pi \epsilon _0} \frac {qc}{(\eta c -\vec {\eta} \cdot \vec v)} \;\hbox { and }\; \vec A_{(\vec r,t)} = \frac {\vec v}{c^2} V_{(\vec r, t)}$ Solution from the book is: Let the particle passes through the origin at t=0 so that $\vec w(t) = \vec v t\Rightarrow \; |\vec r -\vec v_t_r|=c(t-t_r) \Rightarrow \; t_r=t ^+_-\frac r c$ I skip a few step above to arrive to the solution: $\eta = |\vec r - \vec v t_r| \;\hbox { and } \hat { \eta} =\frac {\vec r -\vec v t_r}{c(t-t_r)}$ $\Rightarrow\;\eta\left (1-\hat {\eta} \cdot \frac{\vec v(t_r)}{c}\right )= \frac 1 c [(c^2t -\vec r \cdot \vec v)-(c^2-v^2)t_r] = \frac 1 c \sqrt {(c^2t-\vec r \cdot \vec v)^2 + (c^2-v^2)(r^2 – c^2 t^2)}$ I don’t get how to get the last step. It is really algebra, but I put in the question just in case. I just cannot get the final solution. Please help. Last edited by yungman; May 1st 2011 at 07:20 PM.
 May 2nd 2011, 02:35 AM #2 Forum Admin     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,684 Again I have had to delete all of your images from before. There's a lot of algebra to do here. For starters, the last step follows from squaring the term, then taking the square root. That is to say: $\sqrt{[(c^2t - \vec{r} \cdot \vec{v}) - (c^2 - v^2)t_r]^2} = \sqrt{(c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2)(r^2 - c^2t^2)}$ I can't put too many images in here, but the key is to expand out the LHS under the square root. We get (ignoring the square root: $\begin{matrix} (c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2)^2t_r^2 - 2(c^2t - \vec{r} \cdot \vec{v})(c^2 - v^2)t_r \\ = (c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2) \left [ (c^2 - v^2)t_r^2 - 2(c^2t - \vec{r} \cdot \vec{v})t_r \right ] \end{matrix}$ We need to get rid of the r (dot) v inside of the brackets and we can do that from $| \vec{r}- \vec{v} t_r | = c(t - t_r)$ (This equation is mistyped in your post.) Square both sides and solve for r (dot) v. Then put that into your term inside the brackets. Multiply the heck out of everything in those brackets and you will find that practically everything cancels out except for that r^2 - c^2t^2. If you need more steps let me know. Maybe a can try a step by step over a number of posts. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here. Last edited by topsquark; May 2nd 2011 at 02:38 AM.
 May 2nd 2011, 09:24 AM #3 Member   Join Date: Nov 2010 Posts: 40 Thanks for the reply. I did the squaring and square rooting before. Because the path pass through origin at t=0 given, w(tr)=vtr. I just skip a step due to limited number of image. Again due to limited number of image, I just put down the solved equation of |r-vt|=c(t-tr) as tr=t-(+/-)(r/c) And due to t>0, tr=t-r/c. I tried to put it in and still cannot solve the equation. I don't know how to solve r dot v because that involve knowing the cosine angle between them and there is no information. that's where I really got stuck on the first place. What image you deleted? I see just as many equation from my original post.
May 2nd 2011, 12:07 PM   #4

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 Originally Posted by yungman Again due to limited number of image, I just put down the solved equation of |r-vt|=c(t-tr) as tr=t-(+/-)(r/c) And due to t>0, tr=t-r/c. I tried to put it in and still cannot solve the equation. I don't know how to solve r dot v because that involve knowing the cosine angle between them and there is no information. that's where I really got stuck on the first place. What image you deleted? I see just as many equation from my original post.
I usually quote the previous post. I was saying I couldn't do that.

To get r (dot) v:

$|\vec{r} - \vec{v}t |^2 = c^2(t - t_r)^2$

$r^2 - 2 \vec{r} \cdot \vec{v} t + v^2t^2 = c^2(t - t_r)^2$

Now solve for r (dot) v.

You don't need the tr = t - r/c. The tr terms all cancel out on their own.

-Dan
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 May 2nd 2011, 06:46 PM #5 Member   Join Date: Nov 2010 Posts: 40 I have been working on this for a few hours and I got close except I cannot solve the +/- sign. The way I have is even a little different from the book, please help me also to check whether I did it correctly. I have to split into a few posts to put in all the steps: $\eta =|\vec r -\vec v t_r|= \sqrt{ r^2-2t_r \vec r \cdot \vec v +v^2t_r} = c(t-t_r) \;\hbox { and } \vec {\eta} \cdot \vec v = (\vec r - t_r\vec v) \cdot \vec v = \vec r \cdot \vec v -v^2 t_r$ $v_{(\vec r,t)} = \frac {cq}{4\pi \epsilon_0} \frac 1 {\eta c - \vec{\eta} \cdot \vec v} = \frac {cq}{4\pi \epsilon_0} \frac 1 {\eta c - \vec r \cdot \vec v + v^2t_r}$ Last edited by yungman; May 2nd 2011 at 07:09 PM.
 May 2nd 2011, 07:04 PM #6 Member   Join Date: Nov 2010 Posts: 40 $t_r=t-\frac {\eta}{c} = t-\frac{\sqrt{r^2-2t_r\vec r\cdot\vec v +v^2t_r}} c \;\Rightarrow \; (c^2-v^2)t_r^2 -2(c^2t-\vec r \cdot \vec v)t_r -(r^2-v^2)=0 \;\Rightarrow \; t_r=\frac {[c^2t- (\vec r \cdot \vec v ) ]\;^+_-\sqrt{(c^2t - \vec r \cdot \vec v)^2 + (c^2-v^2)(r^2-c^2t^2)}}{(c^2-v^2)}$ (3) $V_{(\vec r,t)}= \frac {cq}{4\pi \epsilon_0} \left [\frac 1 {\eta c - \vec{\eta} \cdot \vec v}\right ] = \frac {cq}{4\pi \epsilon_0} \left [ \frac 1 {\eta c - \vec r \cdot \vec v + v^2t_r}\right ] = \frac {cq}{4\pi \epsilon_0} \left [ \frac 1 {c^2t +(v^2-c^2)t_r - \vec r \cdot \vec v}\right ]$ $V_{(\vec r,t)} = \frac {cq}{4\pi \epsilon_0} \left [\frac 1 { c^2t -(v^2-c^2)\left [ \frac {(c^2t-\vec r \cdot \vec v) \;^+_- \sqrt{(c^2t-\vec r \cdot \vec v)^2 + (c^2-v^2)(r^2-c^2t^2)}}{(c^2-v^2)}\right ]\right ] -\vec r \cdot \vec v} = \frac {cq}{4\pi \epsilon_0} \left [ \frac 1 {^-_+ \sqrt{(c^2t-\vec r \cdot \vec v)^2 + (c^2-v^2)(r^2-c^2t^2)}}\right ]$ I don't know how to get rid of the +/- sign. Please help with the final step. The only thing I can think of is if you look at (3) above, if you use the -ve, tr can be negative and we start out t=0 and both t and tr has to be either 0 or +ve, -ve is not allow. Thanks Alan Last edited by yungman; May 2nd 2011 at 08:58 PM.
May 3rd 2011, 07:06 AM   #7

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 Originally Posted by yungman I don't know how to get rid of the +/- sign. Please help with the final step. The only thing I can think of is if you look at (3) above, if you use the -ve, tr can be negative and we start out t=0 and both t and tr has to be either 0 or +ve, -ve is not allow. Thanks Alan
I'm a little unclear about a few things. First there is an easier way to get to the formula for the potential, which doesn't involve tr, and there is a square root missing in your answer.

Note:
$\eta c - \vec{\eta} \cdot \vec{v} = c \eta \left ( 1 - \hat{\eta} \cdot \frac{\vec{v}}{c} \right ) = \sqrt{(c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2)(r^2 - c^2t^2)}$

This is from the last line of your original post. (You eventually got something similar, but for some reason it still contained tr for some reason.) You can plug this in for the denominator of V. If your answer is allowed to depend on v, and I can see no reason why not, then this is a much easier derivation.

-Dan
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May 3rd 2011, 07:45 AM   #8
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 Originally Posted by topsquark I'm a little unclear about a few things. First there is an easier way to get to the formula for the potential, which doesn't involve tr, and there is a square root missing in your answer. Note: $\eta c - \vec{\eta} \cdot \vec{v} = c \eta \left ( 1 - \hat{\eta} \cdot \frac{\vec{v}}{c} \right ) = \sqrt{(c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2)(r^2 - c^2t^2)}$ This is from the last line of your original post. (You eventually got something similar, but for some reason it still contained tr for some reason.) You can plug this in for the denominator of V. If your answer is allowed to depend on v, and I can see no reason why not, then this is a much easier derivation. -Dan

I don't see when I miss the square root, I re-check my work already. I got $\eta c - \vec{\eta} \cdot \vec{v} = c \eta \left ( 1 - \hat{\eta} \cdot \frac{\vec{v}}{c} \right ) = \sqrt{(c^2t - \vec{r} \cdot \vec{v})^2 + (c^2 - v^2)(r^2 - c^2t^2)}$ as shown in my last post.

I don't see an easier way. I did try your suggestion and solve for r dot v and don't have much luck. Besides, usually you don't get tr, more likely you get the w(t) where v(t) = w'(t). Can you show me?

Thanks

Alan

May 3rd 2011, 02:06 PM   #9

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 Originally Posted by yungman I don't see an easier way. I did try your suggestion and solve for r dot v and don't have much luck. Besides, usually you don't get tr, more likely you get the w(t) where v(t) = w'(t). Can you show me? Thanks Alan
Sorry about the square root. I looked back over your post and don't know how I missed it.
$\displaystyle \begin{matrix}|\vec{r} - \vec{v}t_r|^2 =r^2 - 2 \vec{r} \cdot \vec{v}t_r + v^2t_r^2 = c^2(t - t_r)^2 \\ . \\ \vec{r} \cdot \vec{v} = \frac{1}{2t_r} ( r^2 + v^2t_r^2 - c^2(t - t_r)^2 ) \end{matrix}$
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Last edited by topsquark; May 3rd 2011 at 02:09 PM.

May 3rd 2011, 02:11 PM   #10

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 Originally Posted by yungman I don't see an easier way. I did try your suggestion and solve for r dot v and don't have much luck. Besides, usually you don't get tr, more likely you get the w(t) where v(t) = w'(t). Can you show me? Thanks Alan
Sorry about the square root. I looked back over your post and don't know how I missed it.
$\begin{matrix}|\vec{r} - \vec{v}t_r|^2 =r^2 - 2 \vec{r} \cdot \vec{v}t_r + v^2t_r^2 = c^2(t - t_r)^2 \\ . \\ \vec{r} \cdot \vec{v} = \frac{1}{2t_r} ( r^2 + v^2 - c^2(t - t_r)^2 ) \end{matrix}$

-Dan
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 Tags charge, derivation, involve, moving, point, potential

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