- #1

etotheipi

Hola amigos, I was doing some stuff and got a bit stuck. To transform components of the em tensor between different bases in the minkowski space you can do, just like any other tensor, $$\overline{F}^{\bar{\mu} \bar{\nu}} = \frac{\partial \bar{x}^{\bar{\mu}}}{\partial x^{\mu}} \frac{\partial \bar{x}^{\bar{\nu}}}{\partial x^{\nu}} F^{\mu \nu} = {\Lambda^{\bar{\mu}}}_{\mu}{\Lambda^{\bar{\nu}}}_{\nu} F^{\mu \nu}$$e.g. to transform the ##x^1## component of the electric field, for the traditional case of uniform motion along the ##x^1## direction at ##c\beta \mathbf{e}_1##, then$$\begin{align*}

\overline{E}^{1} = c\overline{F}^{10} = c{\Lambda^{1}}_{\mu} {\Lambda^{0}}_{\nu} F^{\mu \nu} = c{\Lambda^{1}}_{0}{\Lambda^{0}}_{1}F^{01} + c{\Lambda^{1}}_{1}{\Lambda^{0}}_{0}F^{10} + 0 + 0 &= c\beta^2 \gamma^2 \left( -\frac{E^1}{c} \right) + c \gamma^2 \left( \frac{E^1}{c} \right) \\

&= \gamma^2 E^1 (1-\beta^2) = E^1

\end{align*}$$i.e. that ##\overline{E}^1 = E^1##. But then I tried a different method, knowing that the electric field measured by an observer with 4-velocity ##U = \gamma c (\mathbf{e}_0 + \beta \mathbf{e}_1) = c\overline{\mathbf{e}}_0## should be the contraction of the em tensor and the 4-velocity of the observer, i.e. the resulting rank 1 tensor (vector) with one empty slot has spatial components ##{E_U}^i = F^{i\nu} U_{\nu}##. However, when I tried to work this out [for the same scenario as before, with the ##\{ \overline{\mathbf{e}}_{i} \}## coordinate system moving at ##c\beta \mathbf{e}_1## w.r.t. the ##\{ \mathbf{e}_{i} \}## coordinate system], I get$$(E_U)^1 = F^{10} U_0 + F^{11}U_1 = \left( \frac{E^1}{c} \right) \gamma c + 0 = \gamma E^1$$which is different to what I got before. But I don't see why it shouldn't work, because when I evaluate the same contraction in the other coordinate system, I get$$(\overline{E_U})^1 = \overline{F}^{1\nu} \overline{U}_{\nu} = \overline{F}^{1 0} \overline{U}_{0} + 0+ 0 + 0 = \left( \frac{\overline{E}^1}{c} \right) c = \overline{E}^1$$which is fine. So I'm wondering, where I made my mistake in evaluating the contraction of those two tensors in the first coordinate system, or maybe something else is wrong. Thank you

\overline{E}^{1} = c\overline{F}^{10} = c{\Lambda^{1}}_{\mu} {\Lambda^{0}}_{\nu} F^{\mu \nu} = c{\Lambda^{1}}_{0}{\Lambda^{0}}_{1}F^{01} + c{\Lambda^{1}}_{1}{\Lambda^{0}}_{0}F^{10} + 0 + 0 &= c\beta^2 \gamma^2 \left( -\frac{E^1}{c} \right) + c \gamma^2 \left( \frac{E^1}{c} \right) \\

&= \gamma^2 E^1 (1-\beta^2) = E^1

\end{align*}$$i.e. that ##\overline{E}^1 = E^1##. But then I tried a different method, knowing that the electric field measured by an observer with 4-velocity ##U = \gamma c (\mathbf{e}_0 + \beta \mathbf{e}_1) = c\overline{\mathbf{e}}_0## should be the contraction of the em tensor and the 4-velocity of the observer, i.e. the resulting rank 1 tensor (vector) with one empty slot has spatial components ##{E_U}^i = F^{i\nu} U_{\nu}##. However, when I tried to work this out [for the same scenario as before, with the ##\{ \overline{\mathbf{e}}_{i} \}## coordinate system moving at ##c\beta \mathbf{e}_1## w.r.t. the ##\{ \mathbf{e}_{i} \}## coordinate system], I get$$(E_U)^1 = F^{10} U_0 + F^{11}U_1 = \left( \frac{E^1}{c} \right) \gamma c + 0 = \gamma E^1$$which is different to what I got before. But I don't see why it shouldn't work, because when I evaluate the same contraction in the other coordinate system, I get$$(\overline{E_U})^1 = \overline{F}^{1\nu} \overline{U}_{\nu} = \overline{F}^{1 0} \overline{U}_{0} + 0+ 0 + 0 = \left( \frac{\overline{E}^1}{c} \right) c = \overline{E}^1$$which is fine. So I'm wondering, where I made my mistake in evaluating the contraction of those two tensors in the first coordinate system, or maybe something else is wrong. Thank you

**Edit**: Actually, maybe writing this out helped to pin down the problem. The ##E_U## is still a 4-vector, so it's components will be different in the two coordinate systems. I guess, in that case, we need to transform the components of ##E_U## to the ##\{ \overline{\mathbf{e}}_{\mu} \}## coordinate system in order to get the components ##\mathbf{E}## measured by the guy with four velocity ##U##...
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