Electric Field seen by an observer in motion

[aliens123]

In Robert Wald's General Relativity textbook page 64 reads:
__________________________
In prerelativity physics, the electric field $\vec{E}$ and magnetic field $\vec{B}$ each are spatial vectors. In special relativity these fields are combined into a single spacetime tensor field $F_{ab}$ which is antisymmetric in its indices, $F_{ab} = - F_{ba}$. Thus $F_{ab}$ has six independent components. For an observer moving with 4-velocity $v^{a}$, the quantity
$$E_a = F_{ab}v^{b}$$
is interpreted as the electric field measured by that observer, while
$$B_a = -\frac{1}{2} \epsilon_{ab}^{\ \ \ cd}F_{cd}v^b$$
is interpreted as the magnetic field, where $\epsilon_{abcd}$ is the totally antisymmetric tensor of positive orientation with norm $\epsilon_{abcd} \epsilon^{abcd}=-24, \epsilon_{0123}=1.$
__________________________
I am confused by this bit. Of course, in the rest frame we have $v^{b} = (1,0,0,0)$ and so it is not too hard to see that $E_a = F_{a0}$. For an observer not at rest; however, this says that electric field for them (which I'll denote $E'_a$) is given by
$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$
$$\begin{bmatrix} E'_0 \\ E'_1 \\ E'_2 \\ E'_3 \end{bmatrix} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_x & -B_y & B_x & 0 \end{bmatrix} \begin{bmatrix} v_0 \\ v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

On the other hand, if we Lorentz transform to the rest frame of the observer, then the electric field transforms like:
$$\begin{align}
\mathbf{E}'& \!=\!\gamma\mathbf{E}\!-\!(\gamma\!-\!1)(\mathbf{n}\cdot\mathbf{E})\mathbf{n}+\:\gamma\left(\mathbf{v}\times \mathbf{B}\right)
\tag{ft-02a}\\
\mathbf{B}'& \!=\!\gamma\mathbf{B}\!-\!(\gamma\!-\!1)(\mathbf{n}\cdot \mathbf{B})\mathbf{n}\!-\!\gamma\left(\mathbf{v}\times\mathbf{E}\right)
\tag{ft-02b}
\end{align}$$

See https://physics.stackexchange.com/q...ctromagnetic-tensors-by-matrix-multiplication
(Note that $v$ appearing in this equation is the velocity of the Lorentz boost transformation, whereas the $v^a$ in Wald's text is the 4-velocity. The two are related by a factor of $\gamma$.)

These two equations are not in agreement. Specifically, they differ by the term containing $\mathbf{n}$. So what is Wald trying to say?

 

[Dale]

Wald is trying to say exactly what he said. There is no confusion. If a random internet post contradicts a statement in Wald which is both covariant and clearly physically motivated then said random internet post is wrong.

For an observer moving with 4-velocity $v^{a}$, the quantity
$$E_a = F_{ab}v^{b}$$
is interpreted as the electric field measured by that observer, ... in the rest frame we have $v^{b} = (1,0,0,0)$ and so it is not too hard to see that $E_a = F_{a0}$.

This really is the end of it. It is a clearly correct interpretation, as you recognize.

 

[PeterDonis]

For an observer moving with 4-velocity $v^{a}$, the quantity
$$E_a = F_{ab}v^{b}$$
is interpreted as the electric field measured by that observer

Yes, but this only applies if you take the components of $F_{ab}$, $v^b$, and $E_a$ all in the same frame. You can't use this equation to transform between frames.

For an observer not at rest; however, this says that electric field for them (which I'll denote $E'_a$) is given by
$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$

$$\begin{bmatrix} E'_0 \\ E'_1 \\ E'_2 \\ E'_3 \end{bmatrix} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_x & -B_y & B_x & 0 \end{bmatrix} \begin{bmatrix} v_0 \\ v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

No, that's not how you Lorentz transform the EM field tensor, which is what you need to do if you want to derive transformation equations for the $E$ and $B$ fields using the 4-tensor formalism. To Lorentz transform the EM field tensor, $F_{ab}$, you need to apply the Lorentz transformation matrix twice, one for each index on the tensor.

 

[robphy]

There are two formalisms under discussion here:

• 4-dimensional vectors and tensors (possibly decomposed with respect to an observer's 4-velocity "spacetime-splitting")
• 3-dimensional vectors and tensors in the spatial subspace appearing the observer-splitting

The 4-dimensional tensorial one is cleaner because statements are generally independent of an observer,
and are arguably a clearer presentation of the physics.

The 3-dimensional vectors may feel more familiar... however, the corresponding expressions (e.g. involving cross-products) are observer-dependent and are rather complicated.

Some non-trivial effort is needed to reconcile the two formalisms...
it's essentially translating
"tensor algebra and calculus" into
"observer-dependent vector algebra and calculus".

(Over the years, I have been trying to be comfortable with both and to find ways to more easily reconcile them. In my experience, this is not an easy task... one needs patience and one needs practice with each formalism... there are no quick answers. (Sign conventions complicate the translation.) )For example:

http://home.uchicago.edu/~geroch/Course Notes
See "7. The Geometry of World-Lines" and "13. Electromagnetic Fields: Decomposition by an Observer" from Geroch's General Relativity (1972)
[ https://www.amazon.com/dp/0987987178/?tag=pfamazon01-20 - final edit]
(This may be a useful supplement to Wald's text. Both Bobs are at U. Chicago.)

http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/1202.1.K.pdf
See "2.6 Particle Kinetics in Index Notation and in a Lorentz Frame"
and "2.11 Nature of Electric and Magnetic Fields; Maxwell’s Equations"
in APPLICATIONS OF CLASSICAL PHYSICS (2012-2013 Version)
by Roger D. Blandford and Kip S. Thorne

https://arxiv.org/abs/gr-qc/0105096
Gravitoelectromagnetism: Just a Big Word?
Robert T. Jantzen, Paolo Carini, Donato Bini

https://www.semanticscholar.org/paper/Understanding-Spacetime-Splittings-and-Their-or-:-Jantzen-Carini/b270307a513b29825104cc53835fe849096ec1c4
Understanding Spacetime Splittings and Their Relationships or Gravitoelectromagnetism : the User Manual
R. Jantzen, P. Carini, D. Bini

https://luth.obspm.fr/~luthier/gourgoulhon/fr/present_rec/imcce_syrte10.pdf
Special relativity from an accelerated observer perspective
Eric Gourgoulhon
(there are some formulas in the inertial case)

 

[vanhees71]

Yes, but this only applies if you take the components of $F_{ab}$, $v^b$, and $E_a$ all in the same frame. You can't use this equation to transform between frames.
No, that's not how you Lorentz transform the EM field tensor, which is what you need to do if you want to derive transformation equations for the $E$ and $B$ fields using the 4-tensor formalism. To Lorentz transform the EM field tensor, $F_{ab}$, you need to apply the Lorentz transformation matrix twice, one for each index on the tensor.

The reason that this is not the Lorentz transform of the electromagnetic field components in terms of $\vec{E}$ and $\vec{B}$ is that you still get this components in the computational frame and not in the rest frame of the observer. So you have to do a Lorentz transformation to the rest frame of the observer to get the complete transformation rules of $\vec{E}$ and $\vec{B}$. Equivalently you can of course just do the transformation directly to the Faraday-tensor components. For details, see Sect. 3.2 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The issue with the covariant vectors $E^{\mu}$ and $B^{\mu}$ is discussed starting with (3.2.33). Note that I have a different convention than Wald (west-coast signature (+---) and my four-velocity $u^{\mu}$ is dimensionless, i.e., normalized such that $u_{\mu} u^{\mu}=1$).

 

[aliens123]

Wald is trying to say exactly what he said. There is no confusion. If a random internet post contradicts a statement in Wald which is both covariant and clearly physically motivated then said random internet post is wrong. This really is the end of it. It is a clearly correct interpretation, as you recognize.

Said "random internet post" is correct.
See https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

 

[aliens123]

No, that's not how you Lorentz transform the EM field tensor, which is what you need to do if you want to derive transformation equations for the $E$ and $B$ fields using the 4-tensor formalism. To Lorentz transform the EM field tensor, $F_{ab}$, you need to apply the Lorentz transformation matrix twice, one for each index on the tensor.

Yes, I know this. That is what the second part of my post was about.

 

[PeterDonis]

Said "random internet post" is correct.

No, it isn't. See below.

I know this. That is what the second part of my post was about

The second part of your post describes transforming 3-vectors. That part is correct. What is not correct is the first part of your post, where you incorrectly state how to transform the EM field tensor. Since you have incorrectly stated that in the first part of your post, of course what you come up with is not going to match up with the correct statement of the 3-vector transformation law in the second part of your post. But a correct statement of the transformation law for the EM field tensor will match up with the 3-vector transformation laws for the electric and magnetic field 3-vectors.

What you also do not appear to understand is that Wald's equations for $E$ and $B$ in terms of the EM field tensor have nothing to do with Lorentz transformations. They only work, as I said before, when all of the quantities appearing in the equations are evaluated in the same frame. You cannot use them to derive anything about transforming quantities between frames.

 

[Dale]

Said "random internet post" is correct.
See https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

If that is true (I have not checked), then it doesn't conflict with Wald.

 

[aliens123]

No, it isn't. See below. The second part of your post describes transforming 3-vectors. That part is correct.

The second part of my post is the random internet post. How can that be both correct and not correct?

What is not correct is the first part of your post, where you incorrectly state how to transform the EM field tensor.

In the first part of my post I do not try to transform the EM field tensor.

 

[aliens123]

If that is true (I have not checked), then it doesn't conflict with Wald.

Well, yes; I'm aware that there is no conflict. As I interpret Wald's statement there is a conflict. I know that there is no conflict. So I know that I am misinterpreting Wald's statement. Hence my question.

 

[PeterDonis]

The second part of my post is the random internet post.

No, your entire post is the random internet post.

In the first part of my post I do not try to transform the EM field tensor.

If that way of phrasing it confuses you, try this way: the first part of your post does not give a valid equation relating the electric field vector in one frame (the primed frame) to quantities in another frame (the unprimed frame).

I know that I am misinterpreting Wald's statement.

Yes. You are interpreting Wald's statement as saying something about some kind of relationship between quantities in different frames. It doesn't.

 

[Dale]

Well, yes; I'm aware that there is no conflict. As I interpret Wald's statement there is a conflict. I know that there is no conflict. So I know that I am misinterpreting Wald's statement. Hence my question.

Ah, got it. I actually thought you were saying that Wald was wrong because he disagreed with a SE answer.

I think there are two issues. One is transforming the tensor, as mentioned by @PeterDonis

The other is a little more subtle. When you do Wald’s covariant operation you wind up with a covariant quantity. That covariant quantity is a weird one.

It is the E-field measured by the observer. The E-field is not covariant but the E-field measured by a specific observer is covariant because everyone agrees what they will measure. However, the components of this covariant quantity are not the same as the components of the E-field in the observer’s frame. So you have a covariant quantity that is the E-field measured by an observer expressed in coordinates that may not be the observer’s coordinates.

That is a tricky quantity, so my guess is that is where the misunderstanding arises.

 

[etotheipi]

I think I asked basically the exact same question back in December. Have a look at @Ibix's nice (as always) reply here: https://www.physicsforums.com/threads/transforming-some-es-and-fs.997414/post-6431784

 

[aliens123]

Ah, got it. I actually thought you were saying that Wald was wrong because he disagreed with a SE answer.

I think there are two issues. One is transforming the tensor, as mentioned by @PeterDonis

The other is a little more subtle. When you do Wald’s covariant operation you wind up with a covariant quantity. That covariant quantity is a weird one.

It is the E-field measured by the observer. The E-field is not covariant but the E-field measured by a specific observer is covariant because everyone agrees what they will measure. However, the components of this covariant quantity are not the same as the components of the E-field in the observer’s frame. So you have a covariant quantity that is the E-field measured by an observer expressed in coordinates that may not be the observer’s coordinates.

That is a tricky quantity, so my guess is that is where the misunderstanding arises.

Oh okay I think I see. So would it be correct to say that

$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$

Is the electric field seen by an observer with 4-velocity $v^a$. And

$$\mathbf{E}'& \!=\!\gamma\mathbf{E}\!-\!(\gamma\!-\!1)(\mathbf{n}\cdot\mathbf{E})\mathbf{n}+\:\gamma\left(\mathbf{v}\times \mathbf{B}\right)
\tag{ft-02a}\\
\end{align}$$

Is also the electric field seen by an observer with 4-velocity $v^a$, but they appear to be different because they have different components because they have a different set of basis vectors?

I think I asked basically the exact same question back in December. Have a look at @Ibix's nice (as always) reply here: https://www.physicsforums.com/threads/transforming-some-es-and-fs.997414/post-6431784

Yep this is the exact same question thank you

 

[aliens123]

The reason that this is not the Lorentz transform of the electromagnetic field components in terms of $\vec{E}$ and $\vec{B}$ is that you still get this components in the computational frame and not in the rest frame of the observer. So you have to do a Lorentz transformation to the rest frame of the observer to get the complete transformation rules of $\vec{E}$ and $\vec{B}$. Equivalently you can of course just do the transformation directly to the Faraday-tensor components. For details, see Sect. 3.2 in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The issue with the covariant vectors $E^{\mu}$ and $B^{\mu}$ is discussed starting with (3.2.33). Note that I have a different convention than Wald (west-coast signature (+---) and my four-velocity $u^{\mu}$ is dimensionless, i.e., normalized such that $u_{\mu} u^{\mu}=1$).

When I first read this I didn't understand what you were trying to see but it appears crystal clear in hindsight

 

[PeterDonis]

So would it be correct to say that

$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$

Is the electric field seen by an observer with $4-velocity$v^a.##

No. You are mixing up quantities in two different frames. That's not correct.

And

\mathbf{E}'& \!=\!\gamma\mathbf{E}\!-\!(\gamma\!-\!1)(\mathbf{n}\cdot\mathbf{E})\mathbf{n}+\:\gamma\left(\mathbf{v}\times \mathbf{B}\right)
\tag{ft-02a}\\
\end{align}$$

Is also the electric field seen by an observer with 4-velocity $v^a,$ but they appear to be different because they have different components because they have a different set of basis vectors?

No. Lorentz transforming 3-vectors from one frame to another is a different thing from computing a direct observable.

 

[aliens123]

No. You are mixing up quantities in two different frames. That's not correct.

The statement which you are saying is not correct is literally what Wald said.

 

[PeterDonis]

The statement which you are saying is not correct is literally what Wald said.

No, it isn't. Here is what you said:

$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$

Here is what Wald said, expanded out the way you did in the above:

$$E_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$

If those two statements look identical to you, you haven't looked hard enough. Sure, the difference is only one symbol, but that one symbol is causing you a lot of confusion.

 

[etotheipi]

You just need to be careful. Note that here I'll use Wald's abstract index notation (Roman letters identify the underlying abstract tensors, whilst Greek letters refer to the components with respect to a basis). It's basically just repeating what @vanhees71 said earlier...

As I understand it, let's say you want the electric field measured by someone with four-velocity $v^a$. You can build an "electric-field-4-vector measured by the observer with 4-velocity $v^a$", i.e. $(E_v)^a$, like$$(E_v)^a = F^{ab} v_b$$here $(E_v)^a$ is the abstract 4-vector itself.

You can now ask what the components $(E_v)^{\mu}$ of this 4-vector with respect to the basis $\{ \mathbf{e}_{\mu} \}$ carried by the observer with 4-velocity $v^a$ are; this gives you the electric field components measured by that observer. Each component is simply obtained by contracting this 4-vector with the corresponding basis vector of his, in turn.

Of course you can also find the components of this same 4-vector with respect to the basis carried by any other observer, but I believe those are not meaningful. If you do this, you'll need to do a further poincare transformation to once again arrive at the meaningful components mentioned in the previous paragraph.

 

[Dale]

...
Is the electric field seen by an observer with 4-velocity va. And
...
Is also the electric field seen by an observer with 4-velocity va, but they appear to be different because they have different components because they have a different set of basis vectors?

I hesitate to say that simply because I have not gone through the math to verify that second expression. However, in principle sort of.

Again, the first quantity is a covariant quantity, the components of which are expressed in whatever frame $F_{\mu \nu}$ was originally expressed in. The second expression (assuming it is correct) is not a covariant quantity, it is the components of the E field in the observer's frame.

 

[PeterDonis]

The second expression (assuming it is correct) is not a covariant quantity, it is the components of the E field in the observer's frame.

I don't think that's what the second expression is. I think the second expression is the Lorentz transformation equation for the E field in the 3-vector formalism. That's not the same as the components of the E field in the observer's frame.

 

[Dale]

I don't think that's what the second expression is. I think the second expression is the Lorentz transformation equation for the E field in the 3-vector formalism. That's not the same as the components of the E field in the observer's frame.

I am not certain what that expression is and as I said I have not verified it myself, so "buyer beware" for all my comments about it.

For what it's worth, I probably am not ever going to sit down and actually verify that thing. Covariant quantities are preferable IMO. Non-covariant quantities are just more effort than they are worth in relativity.

 

[aliens123]

I don't think that's what the second expression is. I think the second expression is the Lorentz transformation equation for the E field in the 3-vector formalism. That's not the same as the components of the E field in the observer's frame.

What does the Lorentz transformation equation for the E field tell you, if not the components of the E field in the observer's frame?

 

[PeterDonis]

What does the Lorentz transformation equation for the E field tell you, if not the components of the E field in the observer's frame?

It tells you the relationship between the 3-vector E and B fields in one frame, and the 3-vector E and B fields in another frame.

That is not the same as the relationship between the EM field tensor, the observer's 4-velocity, and the E field measured by that observer. The latter is what Wald is describing in the formula you gave in the OP. That relationship does not require doing any Lorentz transformation at all. It is not a relationship between quantities in different frames. It is a relationship between invariant geometric objects. Note that the "E field" in Wald's formula is not a 3-vector; it's a 4-vector. So it isn't even the same kind of thing as what appears in the 3-vector Lorentz transformation equations you give.

 

[robphy]

Each $E^a$ is a 4-vector that is spatial according that corresponding observer $v^a$ . The resulting spatial components form the 3-vectors in each frame.

The Lorentz transformations for the electric field relate those 3-vectors...
the v-components of E=vF with
the v’-components of E’=v’F.

In the spirit of @PeterDonis ‘s comment,
this is not the same as finding
the v’-components of E=vF or
the v-components of E’=v’F.

 

[aliens123]

Let me try to be more clear. Suppose we have two observers G and H. Let observer G be at rest in the unprimed frame, and let H be at rest in the primed frame.

There are two different covariant quantities: the electric field measured by observer G, and the electric field measured by observer H.

For each covariant quantity we can express it in terms of components in two ways: the components in the rest frame of observer G, and the components in the rest frame of observer H.

As an abstract vector, the electric field measured by observer G is given by
$$ (E_G)_a = F_{ab}(v_G)^b $$
where $(v_G)$ is the four velocity of observer G.
The components in the rest frame of observer G are given by
$$(E_G)_\mu = F_{\mu \nu}(v_G)^\nu = F_{\mu 0}.$$
The components in the rest frame of observer H are given by
$$(E_G)'_\mu = F'_{\mu \nu}(v_G)'^\nu$$

As an abstract vector, the electric field measured by observer H is given by
$$(E_H)_a = F_{ab}(v_H)^b$$
where $(v_H)$ is the four velocity of observer H.
The components in the rest frame of observer H are given by
$$(E_H)'_\mu = F'_{\mu \nu}(v_H)'^\nu = F'_{\mu 0}$$
The components in the rest frame of observer G are given by
$$(E_H)_\mu = F_{\mu \nu}(v_H)^\nu $$

Question 1:
And it is not the case that $(E_G)'_\mu = (E_H)'_\mu$ ?
EDIT:
Followup:
The 3-vector Lorentz transformation of E (which was given in OP) is essentially the transformation
$$(E_G)_\mu \mapsto (E_H)'_\mu$$
?

 

[PeterDonis]

The Lorentz transformations for the electric field relate those 3-vectors...
the v-components of E=vF with
the v’-components of E’=v’F.

The second of these should be

the v' components of E'=v'F'

You have to transform all of the objects to the primed frame, including the EM field tensor.

 

[robphy]

The second of these should be

the v' components of E'=v'F'

You have to transform all of the objects to the primed frame, including the EM field tensor.

My F is the field tensor (in abstract indices), not the components of F in a basis. (I’m on my phone and am too lazy to write the abstract indices).

 

[PeterDonis]

Suppose we have two observers G and H. Let observer G be the unprimed frame, and let H be the primed frame.

More precisely, I assume you mean that observer G is at rest in the unprimed frame, and H is at rest in the primed frame.

There are two different covariant quantities: the electric field measured by observer G, and the electric field measured by observer H.

Yes. Note that both of these are 4-vectors, not 3-vectors, since you are using the term "covariant quantities".

For each covariant quantity we can express it in terms of components in two ways: the components in the rest frame of observer G, and the components in the rest frame of observer H.

Yes.

As an abstract vector, the electric field measured by observer G is given by
$$(E_G)_a = F_{ab}(v_G)^b$$
where $(v_G)$ is the four velocity of observer A.

I assume you mean the four velocity of observer G. Yes, this is correct. Note that it is a covariant equation, valid in any frame.

The components in the rest frame of observer G are given by
$$(E_G)_\mu = F_{\mu \nu}(v_G)^\nu = F_{\mu 0}.$$

Yes.

The components in the rest frame of observer H are given by
$$(E_G)'_\mu = F'_{\mu \nu}(v_G)'^\nu$$

Yes.

As an abstract vector, the electric field measured by observer H is given by

$$(E_H)_a = F_{ab}(v_H)^b$$

where $(v_H)$ is the four velocity of observer H.

The components in the rest frame of observer H are given by

$$(E_H)'_\mu = F'_{\mu \nu}(v_H)'^\nu = F'_{\mu 0}$$

The components in the rest frame of observer G are given by

$$(E_H)_\mu = F_{\mu \nu}(v_H)^\nu $$

All good.

And it is not the case that $(E_G)'_\mu = (E_H)'_\mu$ ?

Correct. This should be obvious from your two equations:

$$(E_H)'_\mu = F'_{\mu \nu}(v_H)'^\nu = F'_{\mu 0}$$

$$(E_G)'_\mu = F'_{\mu \nu}(v_G)'^\nu$$

Obviously the second is different from the first, since $(v_G)'^\nu \neq (v_H)'^\nu$, and therefore we will not have $(E_G)'_\mu = F'_{\mu 0}$.

 

[PeterDonis]

My F is the field tensor (in abstract indices), not the components of F in a basis. (I’m on my phone and am too lazy to write the abstract indices).

If you're doing that for F, you should also do it for E and v, which means there is no such thing as "primed" versions of anything, you just have E = v F, as a relationship between covariant objects with abstract indices.

 

[etotheipi]

If you're doing that for F, you should also do it for E and v, which means there is no such thing as "primed" versions of anything, you just have E = v F, as a relationship between covariant objects with abstract indices.

I interpreted @robphy's notation to be, that E' and E are two different electric-field-4-vectors corresponding to the electric fields measured by two different observers with 4-velocities v' and v.

In other words, the primes in this case don't represent primed components, they represent different underlying quantities.

 

[PeterDonis]

Question 1:
And it is not the case that $(E_G)'_\mu = (E_H)'_\mu$ ?

I have already responded to this in my previous post just now, but I will add that you should not expect this equality to be true, because $E_G$ and $E_H$ are different geometric objects. This is obvious from the Wald-style formulas that you give for each of them:

$$(E_G)_a = F_{ab}(v_G)^b$$

$$(E_H)_a = F_{ab}(v_H)^b$$

Obviously these are two different things--one contracts the EM field tensor $F$ with the 4-velocity of observer G; the other contracts the same EM field tensor $F$ with a different 4-velocity, that of observer H. Why would you expect these two different things to have the same components in any frame?

Followup:
The 3-vector Lorentz transformation of E (which was given in OP) is essentially the transformation
$$(E_G)_\mu \mapsto (E_H)'_\mu$$
?

No, it isn't. That is the point I have been trying to get through to you all along.

The correct statements of what the Lorentz transformation does are:

$$(E_G)_\mu \mapsto (E_G)'_\mu$$

$$(E_H)_\mu \mapsto (E_H)'_\mu$$

And, for completeness:

$$F_{\mu \nu} \mapsto F'_{\mu \nu}$$

$$(v_G)^\nu \mapsto (v_G)'^\nu$$

$$(v_H)^\nu \mapsto (v_H)'^\nu$$

In other words: a Lorentz transformation doesn't change which object you are dealing with. It just transforms the components of that object from one frame to another. But going from $E_G$ to $E_H$ changes which object you are dealing with. That is simply a different thing from a Lorentz transformation.

 

[aliens123]

The correct statements of what the Lorentz transformation does are

(EG)μ↦(EG)μ′

Okay, so the transformation given in this link: https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Does not answer the question: "If I am at rest in frame G, then boost to frame H, what will be the new electric field I measure?"

 

[robphy]

Each $E^a$ is a 4-vector that is spatial according that corresponding observer $v^a$ . The resulting spatial components form the 3-vectors in each frame.

The Lorentz transformations for the electric field relate those 3-vectors...
the v-components of E=vF with
the v’-components of E’=v’F.

In the spirit of @PeterDonis ‘s comment,
this is not the same as finding
the v’-components of E=vF or
the v-components of E’=v’F.

Now that I am back on my laptop, let me rewrite this in a clearer notation.

I actually dislike '-notation because of its ambiguity... but on a phone, it's easy.
(I prefer Alice and Bob... to unprimed and primed.)

All indices are abstract... no indexed-components.

Let $u^a$ and $v^a$ be 4-velocities for two observers
They each determine an electric field (an observer-dependent spatial 4-vector) from the field tensor $F_{ab}$:
$\def\uE{\stackrel{u}{E^a}} \def\vE{\stackrel{v}{E^a}} \def\uB{\stackrel{u}{B^a}} \def\vB{\stackrel{v}{B^a}} \uE = F_{ab} u^b$ and $\vE= F_{ab} v^b$.

Note: $0=u_a\uE = u_a g^{ac} F_{cb} u^b$ (so $\uE$ is purely spatial to $u^a$), and similarly for $0=v_a\vE$.

• The Lorentz transformations for the electric field relates
  the components of $\uE$ (and $\uB$) according to $u^a$
  with
  the components of $\vE$ (and $\vB$) according to $v^a$.
• This is different from finding
  the components of $\uE$ according to $v^a$
  and
  the components of $\vE$ according to $u^a$.