I 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}:
<br /> \def\uE{\stackrel{u}{E^a}}<br /> \def\vE{\stackrel{v}{E^a}}<br /> \def\uB{\stackrel{u}{B^a}}<br /> \def\vB{\stackrel{v}{B^a}}<br /> \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.

 

[aliens123]

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.

If this is the case, then why isn't the Lorentz transformation given in this link:
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Related by a "simple" Lorentz transformation? That is, $(E_G)_\mu \mapsto (E_G)'_\mu$
Is just
$$(E_G)_\mu = (E_G)'_\alpha \Lambda_{\mu}^{\ \ \alpha}$$
But this is not what the Lorentz transformation given in the link is. Or, in other words, what should we call
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
This is what I was calling the "Lorentz transformation." But this is equivalent to
$$(E_G)_\mu \mapsto (E_H)'_\mu$$

 

[aliens123]

• The Lorentz transformations for the electric field relates
  the components of \uE according to u^a
  with
  the components of \vE 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.

This doesn't seem to agree with PeterDonis. He is saying the Lorentz transformation relates
the components of \uE according to u^a
with
the components of \uE according to v^a.

 

[PeterDonis]

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?"

Let's unpack this question.

If you are at rest in frame G, your 4-velocity is $v_G$, and the electric field you measure is $E_G$.

If you are at rest in frame H, your 4-velocity is $v_H$, and the electric field you measure is $E_H$.

The abstract index notation for these two statements is:

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

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

In component notation in frame G, we have

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

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

If we want to transform these equations to frame H, we use the Lorentz transformation $\Lambda$. I'll use $\rho$ and $\sigma$ for the indexes in the primed frame, to keep them distinct from the indexes $\mu$ and $\nu$ in the unprimed frame:

$$(E_G)'_\rho = \Lambda^\mu{}_\rho (E_G)_\mu$$

$$(E_H)'_\rho = \Lambda^\mu{}_\rho (E_H)_\mu$$

What you are saying, however, is that we should have

$$(E_G)_\mu = (E_H)'_\rho$$

Which doesn't correspond to anything in the above.

If this is the case, then why isn't the Lorentz transformation given in this link:
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

Related by a "simple" Lorentz transformation? That is, $(E_G)_\mu \mapsto (E_G)'_\mu$
Is just
$$(E_G)_\mu = (E_G)'_\alpha \Lambda_{\mu}^{\ \ \alpha}$$
But this is not what the Lorentz transformation given in the link is

The Lorentz transformation given in the link is the translation into 3-vector notation of

$$F'_{\rho \sigma} = \Lambda^\mu{}_{\rho} \Lambda^\nu{}_\sigma F_{\mu \nu}$$

In other words, it is a description of how you transform the EM field tensor. It is not a description of how you transform either $E_G$ or $E_H$.

 

[PeterDonis]

• 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.

No, it doesn't. The Lorentz transformations for the electric and magnetic fields transform $F_{\mu \nu}$. They don't transform either \uE or \vE (or their magnetic counterparts). The latter objects are not the EM field tensor; they are 4-vectors describing particular observables, which transform as 4-vectors. The EM field tensor transforms as a 2nd-rank antisymmetric tensor, which, when translated into 3-vector notation, gives the Lorentz transformation equations on the page the OP referenced.

 

[aliens123]

What you are saying, however, is that we should have

$$(E_G)_\mu = (E_H)'_\rho$$

I am not saying that these should be equal. The use of the term "Lorentz transformation" perhaps should not have been used. But the formula in the link considers the following "transformation":
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
I should note that the author does not use the term "Lorentz transformation." Nonetheless, that is the transformation he is considering, and he does use the term "transformation."

The transformation
$$F_{\mu 0} \mapsto F'_{\mu 0}$$
is what I (and perhaps others) have been calling the "Lorentz transformation of the Electric field." (Which, is probably not a good idea considering that term already has a meaning).
https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node136.html

 

[etotheipi]

No, it doesn't. The Lorentz transformations for the electric and magnetic fields transform $F_{\mu \nu}$. They don't transform either \uE or \vE (or their magnetic counterparts). The latter objects are not the EM field tensor; they are 4-vectors describing particular observables, which transform as 4-vectors. The EM field tensor transforms as a 2nd-rank antisymmetric tensor, which, when translated into 3-vector notation, gives the Lorentz transformation equations on the page the OP referenced.

I'm not sure I entirely understand this; I don't see the issue. Given two observers of 4-velocities $u^a$ and $v^a$, the components of the EM tensor relative to their respective bases transform like
$$\tilde{F}^{\tilde{\mu} \tilde{\nu}} = \frac{\partial \tilde{x}^{\tilde{\mu}}}{\partial x^{\mu}} \frac{\partial \tilde{x}^{\tilde{\nu}}}{\partial x^{\nu}} F^{\mu \nu} = {\Lambda^{\tilde{\mu}}}_{\mu}{\Lambda^{\tilde{\nu}}}_{\nu} F^{\mu \nu}$$and you can find the transformation rules of the electric components by identifying $E^i = F^{i0}$ and $\tilde{E}^i = \tilde{F}^{i0}$.

Further, the $E^i$ are precisely the spatial components of the four-vector $\uE$ with respect to the basis of the $u^a$ observer, and the $\tilde{E}^i$ are precisely the spatial components of $\vE$ with respect to the basis of the $v^a$ observer. So what @robphy wrote is correct, it's just another way of looking at the same thing!

 

[PeterDonis]

I am not saying that these should be equal.

Ok.

the formula in the link considers the following "transformation":
$$F_{\mu 0} \mapsto F'_{\mu 0}$$

No, it does not. What you write here doesn't even make sense; $F_{\mu 0}$ is not even a valid geometric object to begin with, so talking about "transforming" it is nonsense.

The link you give talks about transforming the EM field tensor, just as I said. It then derives the 3-vector transformation equations, for the special case of a pure boost, from the 4-vector transformation equation.

You might be confused by the fact that, in the process of deriving the 3-vector equations, the article chooses the direction of the pure boost to be the "1" direction, and then explicitly shows what that implies for the transformed value of $F^{01}$, which is the E field in the "1" direction. That does not mean the article is deriving a transformation equation for $F_{\mu 0}$. It just means the article is leaving the details for the rest of the components of $E$ and $B$, and how they all combine to reach the final result given, for the reader to fill in.

 

[aliens123]

The link you give talks about transforming the EM field tensor, just as I said. It then derives the 3-vector transformation equations, for the special case of a pure boost, from the 4-vector transformation equation.

You might be confused by the fact that, in the process of deriving the 3-vector equations, the article chooses the direction of the pure boost to be the "1" direction, and then explicitly shows what that implies for the transformed value of $F^{01}$, which is the E field in the "1" direction. That does not mean the article is deriving a transformation equation for $F_{\mu 0}$. It just means the article is leaving the details for the rest of the components of $E$ and $B$, and how they all combine to reach the final result given, for the reader to fill in.

I believe that the following transformation (as in the article):

Is obtained by making the identifications
$$E' = F'_{i 0}, E = F_{j 0}, i, j \in \{1, 2, 3\}$$

 

[PeterDonis]

the $E^i$ are precisely the spatial components of the four-vector $\uE$, and the $\tilde{E}^i$ are precisely the spatial components of $\vE$.

Again, let's unpack this.

We have two frames, one in which an observer with 4-velocity $u$ is at rest, the other in which an observer with 4-velocity $v$ is at rest.

We have two "electric field as measured by" 4-vectors, $\uE$ and $\vE$. The equation defining these 4-vectors ensures that they are orthogonal to their respective 4-velocity vectors. (Instead of "orthogonal", @robphy called them "purely spatial"; same thing.)

Because they are purely spatial, we can "interpret" them as 3-vectors in the appropriate frames.

However, we have to be very, very careful in how we describe this "interpretation", because it's extremely easy, as pretty much all of the OP's posts in this thread show, to confuse this "interpretation" with the claim that transforming $\uE$ and $\vE$ from one frame to another is the same thing as "transforming the electric field". Which is wrong.

To transform "the electric field", you need to transform the EM field tensor. Which, in 3-vector notation, gives a pair of equations for $E$ and $B$ in the primed frame in terms of $E$ and $B$ in the unprimed frame. And the OP is looking at those equations and thinking they ought to be the right transformation equations for $\uE$ and $\vE$. Which of course they're not; $\uE$ and $\vE$ are 4-vectors and transform as 4-vectors. But "the electric field" is part of a 2nd-rank antisymmetric tensor, and doesn't transform as a 4-vector, and the 3-vector transformation law for "the electric field" is not a simple vector transformation law.

 

[PeterDonis]

I believe that the following transformation (as in the article):
View attachment 279551

Is obtained by making the identifications
$$E' = F'_{i 0}, E = F_{j 0}, i, j \in \{1, 2, 3\}$$

That, plus transforming the EM field tensor $F$ all as one thing, as a 2nd-rank antisymmetric tensor. Which is not the same as the transformation law for $E$ being "the transformation law for $F_{\mu 0}$".

 

[etotheipi]

To transform "the electric field", you need to transform the EM field tensor. Which, in 3-vector notation, gives a pair of equations for $E$ and $B$ in the primed frame in terms of $E$ and $B$ in the unprimed frame. And the OP is looking at those equations and thinking they ought to be the right transformation equations for $\uE$ and $\vE$. Which of course they're not; $\uE$ and $\vE$ are 4-vectors and transform as 4-vectors. But "the electric field" is part of a 2nd-rank antisymmetric tensor, and doesn't transform as a 4-vector, and the 3-vector transformation law for "the electric field" is not a simple vector transformation law.

Yes, I agree with all of that. The only meaningful components of the $\uE$ and $\vE$ are those with respect to the bases of the $u^a$ and $v^a$ observers respectively; these are simply $F^{i0}$ and $\tilde{F}^{i0}$. In that regard, whilst it's of course mathematically possible to write $\uE$ and $\vE$ in terms of any other bases, the results are not going to tell you anything meaningful physically. It's what makes this "electric-field-4-vector" a little hard to understand at first (or at least, it tripped me up when I first saw it!)

 

[pervect]

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
$$E'_a = F_{a0}v^{0} + F_{a1}v^{1} + F_{a2}v^{2} + F_{a3}v^{3}$$
$$\begin{bmatrix} E'_0 \\ E'_1 \\ E'_2 \\ E'_3 \end{bmatrix} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_x & -B_y & B_x & 0 \end{bmatrix} \begin{bmatrix} v_0 \\ v_1 \\ v_2 \\ v_3 \end{bmatrix}$$

Recall that the basis vectors for an observer in motion are not the same as the basis vectors for an observer at rest.

The E-field vector E, like any vector, is expressed as

$\vec{E} = E^0 e_0 + E^1 e_1 + E^2 e_2 + E^3 e_3$

where the $e_i$ are the basis vectors associated with a particular observer.

You have omitted the basis vectors $e_i$ from your formalism, writing only the components. This is common. But this has caused you to become confused, what you want to calculate is the electric field compnents in terms of the basis vectors of the moving observer, while what you've written is the electric field copmonents in terms of the basis vectors of the stationary observer.

But the basis vectors of the stationary observer and the moving observer are different, they're related by the Lorentz transform.

If you transform the basis vectors, you should find that Wald's statements are in agreement with your other statements - I have not done the work myself. Alternatively, you can just transform the components of the Faraday tensor from one set of basis vectors to the others, as other posters have suggested.

$$F'_{cd} = \Lambda^c{}_a \Lambda^d{}_b F_{ab}$$

where $\Lambda$ is the Lorentz transformation matrix.

 

[PeterDonis]

The E-field vector E, like any vector, is expressed as

$\vec{E} = E^0 e_0 + E^1 e_1 + E^2 e_2 + E^3 e_3$

where the $e_i$ are the basis vectors associated with a particular observer.

I'm not sure this way of putting it helps any, because it perpetuates the same confusion about what the term "the electric field" means that I have been trying to clarify in this thread.

Wald's 4-vector $E_a = F_{ab} v^b$ is a perfectly good 4-vector, yes, and in component form, if we include the basis vectors and raise an index, so we have $E^\mu = F^\mu{}_\nu v^\nu$, it looks like the RHS of what you wrote. But the LHS of what you wrote uses the notation $\vec{E}$, which customarily denotes a 3-vector, not a 4-vector.

If the 3-vector $\vec{E}$ is supposed to be "the spatial part" of Wald's 4-vector $E$, you can make it into a 4-vector by noting that "purely spatial" means that it is orthogonal to the basis vector $e_0$, which is also the observer's 4-velocity $v$, which means $E^0 = 0$. But there is still a confusion lurking here: if this 4-vector $E$ is supposed to be the same object as Wald's 4-vector $E$, it does not transform the way "the electric field" is supposed to transform. Wald's 4-vector $E$ transforms like an ordinary 4-vector. But "the electric field" is part of the EM field tensor, and there is no way to write a valid transformation law that just involves the "electric field" part of that tensor. You have to include the magnetic part as well, either by writing two 3-vector transformation laws, or by writing a single 2nd-rank antisymmetric tensor transformation law.

If, OTOH, you intend your $\vec{E}$ to be "the electric field" in the sense of a "piece" of the EM field tensor, then you cannot write it as a 4-vector the way you are doing. It simply is not that kind of object.

 

[Sagittarius A-Star]

Maybe it's easier to transform the electromagnetic field via the four-potential.

A four-current $\mathbf J = \begin{pmatrix} c \rho \\ \vec j \\ \end{pmatrix}$ creates a four-potential

$\mathbf A = \begin{pmatrix} \phi \\ \vec a \\ \end{pmatrix} =\frac {1}{c} \int \, \frac {\mathbf J}{r} dV$

The four-potential $\mathbf A$ can be Lorentz-transformed. In the new frame, the electric and magnetic fields are:

$\vec E' = - \nabla \phi ' - \frac {1}{c} \frac {\partial \vec a '}{\partial t}$
$\vec B' = - \nabla \times \vec a'$.

Source:
https://en.wikipedia.org/wiki/Electromagnetic_four-potential

Helmholtz worked already with the scalar- and vector-potential before SR was created:
https://physics.princeton.edu//~mcdonald/examples/helmholtz.pdf

 

[vanhees71]

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

I don't know, why this should be wrong. It's the standard Lorentz transformation of the electromagnetic field written in terms of $\vec{E}$ and $\vec{B}$. One should be aware that these are not the spatial components of four-vectors but the entries of the antisymmetric Faraday tensor.

See also Sect. 3.2 in

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

Just to make it clear again. In the Ricci calculus tensor components are understood to be defined wrt. a fixed basis (here a Minkowski-orthonormal basis). Now $F^{\mu \nu}$ are the components of the em. field-strength tensor field and $u^{\mu}$ are the components of the four-velocity of an observer. Then one can define the four-vectors
$$E_{\mu} = F_{\mu \nu} u^{\nu}, \quad B_{\mu}=^{\dagger} F_{\mu \nu} u^{\nu} \frac{1}{2} \epsilon_{\rho \sigma \mu \nu} F^{\rho \sigma} u^{\nu}.$$
For an observer at rest $u^{\mu}=1$, and you get the electromagnetic-field components in three-vector notation $\vec{E}$ and $\vec{B}$.

That's why you can say that for a general $u^{\mu}$ the four-vectors $E_{\mu}$ and $B_{\mu}$ are the electric and magnetic fields as observed by the observer moving with this four-velocity, but it's not the $\vec{E}'$ and $\vec{B}'$ components as measured in the restframe of the observer but the four-vector components $E_{\mu}$ and $B_{\mu}$ are the four-vector components in the "computational frame", i.e., to get $\vec{E}'$ and $\vec{B}'$ as measured in the rest frame of the observer, you have to perform the corresponding Lorentz boost to these four-vector components
$$E^{\prime \mu}={\Lambda^{\mu}}_{\nu} E^{\nu}, \quad B^{\prime \mu}={\Lambda^{\mu}}_{\nu} B^{\nu}.$$
Again: It's important to note that the three-vector components $\vec{E}$ and $\vec{B}$ in some fixed frame of reference are not spatial components of a four-vector but the components of the antisymmetric field-strength tensor (in the older literature for this reason also called a "6-vector" $(\vec{E},\vec{B})$, but I think that's even more confusing).