Transforming the Electric Field Measured by an Observer with 4-velocity U

[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

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##...

 

[Ibix]

I think you've got it with the edit.

A more coordinate-free way to put it is that the components of vectors are meaningless. What you measure as the "x component of the electric field" is actually the inner product of the electric field with a unit vector in the x direction. So what your observer with four velocity ##U## measures (the quantity you called ##\overline{ E}^1##) is ##\overline{e}_{(1)}^\mu U^\nu F_{\mu\nu}##, where ##\overline e_{(1)}## is the observer's first spacelike basis vector. Since ##\overline e_{(1)}## isn't (0,1,0,0) in your unbarred coordinates, ##\overline e_{(1)}^\mu U^\nu F_{\mu\nu}\neq U^\nu F_{1\nu}##.

 

[etotheipi]

@Ibix thanks, that's a really great way of explaining it, it's clear to me now. I do like the geometrical point of view a lot. Maybe I'll check the transformation later, but I'm pretty sure it'll work as expected.