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Given a four-potential (coordinates are written in Einstein's notation) $A=(x^1, 2x^1, x^3 - 2x^2, 2x^3)$ which electric and magnetic field are generated?

  1. $E=(-c,0,2c), B=(0,1,-2)$
  2. $E=(c,0,2c), B=(1,0,-2)$
  3. $E=(-c,-2c,2c), B=(1,0,1)$
  4. $E=(-c,0,0), B=(1,0,0)$
  5. $E=(c,0,0), B=(0,-2,-2)$

I compute

$\dfrac{E^1}c = \dfrac{\partial A^0}{\partial x^1} - \dfrac{\partial A^1}{\partial x^0} = 1$ and so $E^1=c$, $E^2=0 , E^3=0$.

So intuitively the answer is $5$. but components of the magnetic field are not the same I found, can someone help me?

Andrews
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Ant2198
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  • 5

2 Answers2

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The magnetic field is not zero, as $A^2$ depends on $x^3$, yielding $B_1=1$.

You didn't say whether the components you specified for $A$ are $A^\mu$ or $A_\mu$. The notation $E^1$ doesn't make sense, since $E$ is a three-vector with normal indices, not a four-vector with covariant and contravariant indices. Also,

$$ \frac{\partial A^0}{\partial x^1} - \frac{\partial A^1}{\partial x^0} $$

doesn't make sense, since this is $\partial_1A^0-\partial_0A^1$, that is, a mixture of terms from $F_1^0$ and $F_0^1$. You need

$$ \partial^1A^0-\partial^0A^1=\frac{\partial A^0}{\partial x_1} - \frac{\partial A^1}{\partial x_0}\;. $$

I suspect if you sort all these things out, you'll get a different sign for $E_1$, in agreement with answer $4)$.

joriki
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-1

Using $$ F^{\mu \nu} = \partial^\mu A^\nu - \partial^\nu A^\mu = \begin{pmatrix} 0 & -E_1/c & -E_2/c & -E_3/c \\ E_1/c & 0 & -B_3 & B_2 \\ E_2/c & B_3 & 0 & -B_1 \\ E_3/c & -B_2 & B_1 & 0 \end{pmatrix} $$ I find that all three components of the magnetic field are zero, so either we both made a mistake, or the question is wrong. Is that the same you are getting?

Wizact
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