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

• B
• etotheipi
In summary, the conversation discusses transforming components of the electromagnetic tensor between different bases in Minkowski space using the transformation rule for tensors. The conversation also explores a different method of transforming the electric field measured by an observer with 4-velocity, which leads to a discrepancy in the results. However, the issue is resolved by understanding that the components of vectors are meaningless and the observer's measurement of the electric field is actually the inner product of the electric field with a unit vector in the observer's direction of motion.
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##...

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aliens123 and Dale
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}##.

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aliens123, PeterDonis and 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.

## 1. What is the purpose of transforming E's and F's in science?

The purpose of transforming E's and F's in science is to convert one form of energy or matter into another form that is more useful for a particular application or experiment. This transformation allows scientists to manipulate and control the properties of energy and matter to better understand their behavior and interactions.

## 2. How do scientists transform E's and F's?

Scientists use a variety of methods to transform E's and F's, depending on the specific substances and conditions involved. Some common techniques include chemical reactions, nuclear reactions, and electromagnetic radiation. These methods can be used to convert matter into energy or vice versa.

## 3. Can transforming E's and F's have negative consequences?

Yes, transforming E's and F's can have negative consequences if not done carefully and responsibly. For example, some transformations may produce harmful byproducts or have unintended environmental impacts. It is important for scientists to consider the potential risks and benefits of any transformation before conducting experiments.

## 4. What are some real-world applications of transforming E's and F's?

Transforming E's and F's has many practical applications in fields such as energy production, medicine, and materials science. For example, converting sunlight into electricity through solar panels, using radiation therapy to treat cancer, and synthesizing new materials with specific properties are all examples of transforming E's and F's for practical use.

## 5. How does the study of transforming E's and F's contribute to our understanding of the natural world?

Studying the transformation of E's and F's allows scientists to gain a deeper understanding of the fundamental laws and principles that govern the behavior of energy and matter. By manipulating and transforming E's and F's, scientists can uncover new insights and make discoveries that contribute to our understanding of the natural world and how it functions.

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