Ndtyhij

Hi everyone and sorry for my English.

I would like to know if I can build a legitimate 4-vector as $E^alpha=(E^0,mathbf{E})$.

I'd like you to check if my way is correct.

1- We already know that $mathbf{E}$ transforms under Lorentz boost as:
begin{equation}label{sdf}
begin{aligned}
mathbf{E}'&=gammaleft(mathbf{E}+vec{beta}timesmathbf{B} right)-dfrac{gamma^2}{gamma+1}vec{beta}left(vec{beta}cdotmathbf{E}right)\[5mm]
&text{So:}\[5mm]
E'_parallel&=E_parallel\
mathbf{E}_perp'&=gammaleft(mathbf{E}_perp+vec{beta}timesmathbf{B} right)
end{aligned}
end{equation}
2- While the spatial component of any 4-vector must obey the following rule:
begin{equation}
begin{aligned}
E_parallel'&=gamma(E_{parallel}-beta E^0)\
mathbf{E}_perp'&=mathbf{E}_perp
end{aligned}
end{equation}
So both expressions must to be equal:
begin{equation}
left{
begin{aligned}
gamma E_parallel-gammabeta E^0&=E_parallel\
gammamathbf{E}_perp+gammavec{beta}timesmathbf{B}&=mathbf{E}_perp
end{aligned}
right.
end{equation}
From the first one we can conclude that time component of 4-vector must be $E^0=dfrac{gamma-1}{gammabeta}E_parallel$ or $E^0=dfrac{gamma-1}{gammabeta^2}vec{beta}cdotmathbf{E}$
But what can we conclude for the second one? Is therefore possible to build that 4-vector $E^alpha$?

Thank you very much

I thought G. Smith's answer was fine in terms of explaining the physics involved, but the OP says:

Thank you G.Smith. I know about the electromagnetic tensor and its properties, but i was wondering if there is a formal proof about the imposibility of building that $E^alpha$ tensor, based on the allowed transformations, as I've tried. What's wrong with my deduction?

Your deduction is of the form X => Y, where X seems to be the proposition that one can make a four-vector of the form $(E^0,textbf{E})$. What is not totally clear to me about your X is what other data you think should be allowed to be encoded in $E^0$, but anyway I think it's possible to give a nonexistence proof without needing to clarify that point.

You've proved some equations involving $E^0$ which vanish when the electric field is zero. Therefore when the field is zero, your 4-vector vanishes. But a Lorentz transformation on a zero vector always gives a zero vector, so you've proved that if an electric field is zero in one frame of reference, it's zero in all other frames. This is false, so we have a proof by contradiction that X is false.

No, it is not possible to make a four-vector from the electric field. But from the electric field and the magnetic field together you can make a four-tensor, $F_{munu}$.

https://en.m.wikipedia.org/wiki/Electromagnetic_tensor

This is because electric and magnetic fields transform into each other under Lorentz transformations. The transformed electric field is a linear combination of the untransformed electric field and the untransformed magnetic field. Amd similarly for the transformed magnetic field.

The lesson is that electric and magnetic fields are just two aspects of one unified thing, the electromagnetic field.

Your formula for $E_0$ depends on $beta$. If there WERE a legal four-vector for the electric field, it's components can't depend on the Lorentz transformation you do. Your formula for $E_0$ should be independent of $beta$. But as you show with your algebra above, this is not possible.

Maybe I could say that my derivation is not possible because the time coordinate $E^0$ depends on another coordinate ($E_parallel$) and it is not allowed because coordinates in a 4-vector must be independent? Is this a factible answer that proofs what I want to?

As observed by an inertial observer,
the Electric Field is a spatial vector,
which means that its time-component in that frame is always zero.

In addition, the Magnetic Field is also a spatial vector... and thus has zero time-component.

As @G. Smith notes, the electric and magnetic fields transform by mixing components (because the electric and magnetic fields are components of a two-index tensor).. and remain spatial,

which are not like 4-vectors (since the time-component of a 4-vector won't generally stay zero after transformation).

update:

Up to sign conventions, $$E_b=F_{ab}u^a$$
is the electric-field according to the observer with 4-velocity $u^a$.
(It is an observer-dependent four-vector.)

But since $F_{ab}=F_{[ab]}$, it follows that
$$E_bu^b=F_{ab}u^a u^b=0,$$
that is, the observer with 4-velocity $u^b$ measures the time-component of $E_b$ to be zero. Thus, $E_b$ has only spatial-components for that observer.