GREPhysics.NET: GR9677 #49

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GR9677 #49

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Electromagnetism $\Rightarrow$ }Relativistic Fields

For motion in the x direction, one has the following equations for the E and B fields,

$\begin{eqnarray} E_x &=& E_x'\\ <br /> B_x &=& B_x'\\ E_y &=& \gamma (E_y' - v B_z')\\ B_y &=& \gamma (B_y' - v/c^2 E_z')\\ E_z &=& \gamma (E_z' - v B_y')\\ B_z &=& \gamma (B_z' - v/c^2 E_y')<br /> \end{eqnarray}$

Since $E_z' = \sigma/(2\epsilon_0)$ (with all other primed components 0), the transformed field is just $E_z=\gamma E_z'$ , as in choice (C).

(Recall that $\gamma = 1/\sqrt{1-\beta^2}$ , where $\beta = v/c$ )

If one forgets the Lorentz-transformed fields, one can also quickly derive the answer for this case. Since the transformed charge density is Lorentz contracted in one of its area dimensions, one has $\sigma = \gamma \sigma '$ . One can tell by symmetry of the surface that the other field components cancel, and one again arrives at the result for $E_z$ as above.

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Comments

syreen
2013-09-11 11:48:17

Is it just me, or are your primes & unprimes swapped? Primed should be the moving frame, shouldn't it?
See GR8677 #22 for the same transformations, but with primes and unprimes switched.

Or does it not matter for this E&M transformations?

syreen
2013-09-11 11:50:43

Oops, just saw that the two are different.
Still, though, shouldn't we be solving for primed?

Jovensky
2013-09-23 22:18:49

syreen, we are looking from the rest frame of second observer, so the first observer is moving relative to the second observer, thus we are solving for the rest frame variables (the unprimed one).

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NEC

mpdude8
2012-04-20 15:09:34

Hmm... I just said that as you increase your speed toward c, the electric field that you measure should increase.

As you increase v towards c, the square root term gets very small. Thus, it should be on the denominator. C is the only choice that has this feature -- and A should be eliminated right off the bat because it has no dependence on v in the first place.

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NEC

wittensdog
2009-11-03 17:47:18

I'm pretty confident that ETS is getting at the issue of length contraction here. I would say that the field transformation formulas are a nice thing to know, if you can remember them, but in my opinion understanding how length contraction and other relativistic effects come into play has more to do with the real physics going on. If you remember the formula sigma / 2*epsilon, and you realize that the charge density should go up by a factor of gamma from length contraction, you have your answer in about two seconds.

The fact that the plane is infinite means that regardless of your velocity with respect to it, you still see it as infinite, so we know we can treat the problem as an infinite plane with a different charge density.

WoolfianOperator
2009-11-05 18:27:55

i agree. the other options have either no effect (A) on the field or the field getting smaller because of the $\frac{1}{\gamma$ terms (B,D,E). remember gamma is always greater than or equal to 1

flyboy621
2010-10-24 16:51:33

good solution!

blairwitch
2011-03-12 09:32:24

could you expound on how you get (derive) sigma/2* epsilon

asdfuogh
2011-10-08 18:13:42

Great point, looking at it as if the charge density became greater due to length contraction.

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NEC

tin2019
2007-10-16 07:54:20

I find it simpler to deal with a potential 4-vector $(\phi,\mathbf{A})$ which has to be invariant under Lorentz transformations. The field in question has a potential $\phi=Ez$ because, for using $\Delta \phi=\partial^2\phi/\partial x^2 +\partial^2\phi/\partial y^2+\partial^2\phi/\partial z^2=0$ and the fact that potential must be independent of both x and y we get $d^2\phi/dz^2=0$ or, because $d\phi/dz=E_z=E$ $\phi=Ez$ . Now applying Lorentz transformations to $\phi$ we getrnrn $\phi'=\frac{\phi+\frac{v}{c}A_x}{\sqrt{1-v^2/c^2}}$ rnrnand since $\mathbf{B}=0$ we can take the vector potential to be $\mathbf{A}=0$ , and thus we havernrn $\phi'=\frac{\phi}{\sqrt{1-v^2/c^2}}$ rnrnNow $\mathbf{E'}=-\nabla\phi'-\partial\mathbf{B}/\partial t$ , for there is magnetic field now. $x$ component of the vector potential isrnrn $A_x=\frac{\frac{v}{c}\phi}{\sqrt{1-v^2/c^2}}$ rnrnHowever this is time independent, and so our equation for electric field reduces tornrn $\nabla\phi'=-E$ . rnrnTaking the gradient of relation $\phi'=\phi/\sqrt{1-v^2/c^2}$ we see that the electric field in primed system must have only a $z$ component, as $\phi$ is independent of $x$ and $y$ . This yieldsrnrn $E'=E/\sqrt{1-v^2/c^2}$ rnrnI realize that this is not the simplest solution, but it is easier for me to remember that the 4-vector is Lorentz invariant. It could be that I'm using a cannon to kill a mosquito.

dumbguy
2007-10-16 19:10:00

yeah but you killed that mosquito

kronotsky
2018-10-22 02:53:30

I think this is a lot easier than remembering how the faraday tensor transforms (thanks for the tip!), but lorentz contraction is the way to go.

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NEC

hungrychemist
2007-10-05 03:19:52

E = sigma/2*epsilon

Consider sigma (charge density) = Q/Area = Q / X * Y but respect to second observer length in X must be shorten.

find the shorten length by X' = sqrt[1-(v/c)^2] * X

Then the new sigma is Q/(sqrt[1-(v/c)^2] * X*Y)

plug new sigma above to E equation, to get C

greenfruit
2008-11-01 10:06:03

awesome.

mike
2009-11-06 13:07:05

This is the best way to do it, and is a nice fact to remember

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NEC

Post A Comment!

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I find it simpler to deal with a potential 4-vector $(\phi,\mathbf{A})$ which has to be invariant under Lorentz transformations. The field in question has a potential $\phi=Ez$ because, for using $\Delta \phi=\partial^2\phi/\partial x^2 +\partial^2\phi/\partial y^2+\partial^2\phi/\partial z^2=0$ and the fact that potential must be independent of both x and y we get $d^2\phi/dz^2=0$ or, because $d\phi/dz=E_z=E$ $\phi=Ez$ . Now applying Lorentz transformations to $\phi$ we getrnrn $\phi'=\frac{\phi+\frac{v}{c}A_x}{\sqrt{1-v^2/c^2}}$ rnrnand since $\mathbf{B}=0$ we can take the vector potential to be $\mathbf{A}=0$ , and thus we havernrn $\phi'=\frac{\phi}{\sqrt{1-v^2/c^2}}$ rnrnNow $\mathbf{E'}=-\nabla\phi'-\partial\mathbf{B}/\partial t$ , for there is magnetic field now. $x$ component of the vector potential isrnrn $A_x=\frac{\frac{v}{c}\phi}{\sqrt{1-v^2/c^2}}$ rnrnHowever this is time independent, and so our equation for electric field reduces tornrn $\nabla\phi'=-E$ . rnrnTaking the gradient of relation $\phi'=\phi/\sqrt{1-v^2/c^2}$ we see that the electric field in primed system must have only a $z$ component, as $\phi$ is independent of $x$ and $y$ . This yieldsrnrn $E'=E/\sqrt{1-v^2/c^2}$ rnrnI realize that this is not the simplest solution, but it is easier for me to remember that the 4-vector is Lorentz invariant. It could be that I'm using a cannon to kill a mosquito.

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