6 I finally figured it out. edited Nov 10 '17 at 8:38 J Thomas 44533 silver badges1111 bronze badges Liénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of $$E(r,t)$$ from Wikipedia. That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at $$v = 0.5$$. This is a picture of the magnitude and direction of the force, $$v = 0.5$$. SO: The Griffiths derivation and the Liénard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming thatI eventually got a copy of Griffiths and found out what he's talking about. L-W uses retarded time. Griffiths does not. His distance $$n\cdot\beta = \beta\sin\theta$$ we get$$|r-r_s|$$ is the distance between the charges in present time. His angle $$\theta$$ is the angle between the constant velocity and the location line in current time. $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulasBasicly he's saying "This is wrongwhat the force would be if it acted instantaneously between charges right now. Someone made a mistake calculating" That works for constant velocity because the E fieldretarded-time position can be computed easily and directly from the potentialcurrent time position (and constant velocity), and no one has caught it because no one actually uses these formulas for the E fieldvice versa.  Here I'm not clear what this result is someone who says Lorentzgood for, but this is wrong. I don't have an opinion yetwhat it's about it. Liénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of $$E(r,t)$$ from Wikipedia. That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at $$v = 0.5$$. This is a picture of the magnitude and direction of the force, $$v = 0.5$$. SO: The Griffiths derivation and the Liénard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. Liénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of $$E(r,t)$$ from Wikipedia. That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at $$v = 0.5$$. This is a picture of the magnitude and direction of the force, $$v = 0.5$$. SO: The Griffiths derivation and the Liénard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ I eventually got a copy of Griffiths and found out what he's talking about. L-W uses retarded time. Griffiths does not. His distance $$|r-r_s|$$ is the distance between the charges in present time. His angle $$\theta$$ is the angle between the constant velocity and the location line in current time. Basicly he's saying "This is what the force would be if it acted instantaneously between charges right now." That works for constant velocity because the retarded-time position can be computed easily and directly from the current time position (and constant velocity), and vice versa.  I'm not clear what this result is good for, but this is what it's about. 5 added 23 characters in body edited Oct 31 '17 at 10:11 Emilio Pisanty 88.9k2323 gold badges226226 silver badges462462 bronze badges LienardLiénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of E(r,t)$$E(r,t)$$ from wikipedia. That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at v = 0.5. $$v = 0.5$$. This is a picture of the magnitude and direction of the force. v = 0.5, $$v = 0.5$$. SO: The Griffiths derivation and the LienardLiénard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. HereHere is someone who says Lorentz is wrong. I don't have an opinion yet about it. https://sites.google.com/site/testsofphysicaltheories/English/lorentz-einstein?pli=1 Lienard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of E(r,t) from wikipedia. https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at v = 0.5. This is a picture of the magnitude and direction of the force. v = 0.5 SO: The Griffiths derivation and the Lienard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. https://sites.google.com/site/testsofphysicaltheories/English/lorentz-einstein?pli=1 Liénard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of $$E(r,t)$$ from Wikipedia. That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at $$v = 0.5$$. This is a picture of the magnitude and direction of the force, $$v = 0.5$$. SO: The Griffiths derivation and the Liénard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. 4 changed a misleading name edited Oct 31 '17 at 9:51 J Thomas 44533 silver badges1111 bronze badges Lienard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of E(r,t) from wikipedia. https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at v = 0.5. This is a picture of the magnitude and direction of the force. v = 0.5 SO: The "Biot-Savart"Griffiths derivation and the Lienard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. https://sites.google.com/site/testsofphysicaltheories/English/lorentz-einstein?pli=1 Lienard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of E(r,t) from wikipedia. https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at v = 0.5. This is a picture of the magnitude and direction of the force. v = 0.5 SO: The "Biot-Savart" derivation and the Lienard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. https://sites.google.com/site/testsofphysicaltheories/English/lorentz-einstein?pli=1 Lienard-Wiechert: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ where $$n =\overline{r-r_s}$$ $$\beta = v/c$$ This is the first term of E(r,t) from wikipedia. https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential#Corresponding_values_of_electric_and_magnetic_fields That thing in the denominator $$\frac{1}{1-n \centerdot\beta}$$ is not symmetric in the direction of motion. the dot product is positive when n and beta are the same direction, and negative when they are opposite directions. We divide by a number that's less than one in one direction and bigger than one in the other. It will give a graph that looks a lot like the first picture in the original question (flipped around if the signs of the source and target charges are different) and not like the second picture at all. This is a picture of the magnitude of the force at the red dot from particles moving to the right at v = 0.5. This is a picture of the magnitude and direction of the force. v = 0.5 SO: The Griffiths derivation and the Lienard-Wiechert derivations are different as follows: $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-n\cdot\beta)^3}\frac{n-\beta}{|r-r_s|}$$ $$E = \frac{q}{4\pi\epsilon_0} \frac{(1-\beta^2)}{(1-\beta^2\sin^2\theta)^\frac{3}{2}}\frac{n-\beta}{|r-r_s|}$$ Assuming that $$n\cdot\beta = \beta\sin\theta$$ we get $$\sqrt{1-\beta^2\sin^2\theta} = \sqrt{(1-\beta\sin\theta)(1+\beta\sin\theta)}$$ in one case, and $$(1-\beta\sin\theta) = \sqrt{(1-\beta\sin\theta)(1-\beta\sin\theta)}$$ in the other case. Presumably someone has made a sign error. One or the other of these formulas is wrong. Someone made a mistake calculating the E field from the potential, and no one has caught it because no one actually uses these formulas for the E field. Here is someone who says Lorentz is wrong. I don't have an opinion yet about it. https://sites.google.com/site/testsofphysicaltheories/English/lorentz-einstein?pli=1 3 added explanation which shows the specific difference between the two formulas. One has a sign error. edited Oct 31 '17 at 9:01 J Thomas 44533 silver badges1111 bronze badges 2 Improved answer. Made formula more understandable, added images edited Oct 31 '17 at 7:54 J Thomas 44533 silver badges1111 bronze badges 1 answered Oct 25 '17 at 4:02 J Thomas 44533 silver badges1111 bronze badges