Mixing
Could be possible to build a 4-vector in special relativity whose spatial component was the electric field E?
$begingroup$
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
special-relativity tensor-calculus classical-electrodynamics lorentz-symmetry
asked Jan 13 at 18:21
DaniDani
183
$endgroup$
add a comment |
$begingroup$
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
special-relativity tensor-calculus classical-electrodynamics lorentz-symmetry
asked Jan 13 at 18:21
DaniDani
183
$endgroup$
add a comment |
$begingroup$
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
special-relativity tensor-calculus classical-electrodynamics lorentz-symmetry
asked Jan 13 at 18:21
DaniDani
183
$endgroup$
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
special-relativity tensor-calculus classical-electrodynamics lorentz-symmetry
special-relativity tensor-calculus classical-electrodynamics lorentz-symmetry
asked Jan 13 at 18:21
DaniDani
183
asked Jan 13 at 18:21
DaniDani
183
asked Jan 13 at 18:21
DaniDani
183
asked Jan 13 at 18:21
DaniDani
183
asked Jan 13 at 18:21
DaniDani
183
183
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
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.
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
$endgroup$
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
add a comment |
$begingroup$
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.
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
$endgroup$
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
add a comment |
$begingroup$
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?
answered Jan 13 at 19:09
DaniDani
183
$endgroup$
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
add a comment |
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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.
answered Jan 13 at 19:08
robphyrobphy
1,982238
$endgroup$
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
add a comment |
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5 Answers
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active
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votes
5 Answers
5
active
oldest
votes
active
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votes
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$begingroup$
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.
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
$endgroup$
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.
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
answered Jan 13 at 19:08
Ben CrowellBen Crowell
50.9k6155299
50.9k6155299
add a comment |
add a comment |
$begingroup$
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.
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
$endgroup$
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
add a comment |
$begingroup$
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.
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
$endgroup$
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
add a comment |
$begingroup$
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.
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
$endgroup$
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.
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
edited Jan 13 at 18:32
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
answered Jan 13 at 18:26
G. SmithG. Smith
7,32611425
7,32611425
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
add a comment |
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
1
1
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
Here is a nice article about the unification of $vec{E}$ and $vec{B}$. arxiv.org/abs/1111.7126
$endgroup$
– N. Steinle
Jan 13 at 18:50
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
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? Thanks!
$endgroup$
– Dani
Jan 13 at 18:51
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You don’t have a deduction to criticize! You failed to find a four-vector that satisied the equations you wrote down. Since your second equation involves the magnetic field, how could you possibly expect to satisfy it, when your first equation requires $E^0$ to be a combination of the components of the electric field? This seems like proof to me that what you want is impossible. And, of course, if what you were trying to do were possible, it would have been done 100 years ago.
$endgroup$
– G. Smith
Jan 13 at 20:16
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
$begingroup$
You may “know about the electromagnetic tensor and its properties” but you didn’t grasp its relevance. If electric and magnetic fields mix together under a Lorentz transformation, then they cannot also stay unmixed under a Lorentz transformation, as in your failed attempt. Things either mix or they don’t. They can’t do both.
$endgroup$
– G. Smith
Jan 13 at 20:22
add a comment |
$begingroup$
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.
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
$endgroup$
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
add a comment |
$begingroup$
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.
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
$endgroup$
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
add a comment |
$begingroup$
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.
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
$endgroup$
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.
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
answered Jan 14 at 0:03
Jahan ClaesJahan Claes
4,9661033
4,9661033
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
add a comment |
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
1
1
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
$begingroup$
(I'll add that you also can't solve your second equation at all if you allow arbitrary magnetic fields. You can easily see this by taking the derivative with respect to any components of the magnetic field on both sides of that equation. One side with have the derivative be zero, the other will not. So it simply can't be solved)
$endgroup$
– Jahan Claes
Jan 14 at 0:16
add a comment |
$begingroup$
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?
answered Jan 13 at 19:09
DaniDani
183
$endgroup$
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
add a comment |
$begingroup$
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?
answered Jan 13 at 19:09
DaniDani
183
$endgroup$
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
add a comment |
$begingroup$
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?
answered Jan 13 at 19:09
DaniDani
183
$endgroup$
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?
answered Jan 13 at 19:09
DaniDani
183
answered Jan 13 at 19:09
DaniDani
183
answered Jan 13 at 19:09
DaniDani
183
answered Jan 13 at 19:09
DaniDani
183
183
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
add a comment |
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
1
1
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
$begingroup$
The problem is your time coordinate depends on $beta$. It's perfectly fine to have it depend on spacial stuff.
$endgroup$
– Jahan Claes
Jan 14 at 0:13
add a comment |
$begingroup$
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.
answered Jan 13 at 19:08
robphyrobphy
1,982238
$endgroup$
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
add a comment |
$begingroup$
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.
answered Jan 13 at 19:08
robphyrobphy
1,982238
$endgroup$
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
add a comment |
$begingroup$
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.
answered Jan 13 at 19:08
robphyrobphy
1,982238
$endgroup$
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.
answered Jan 13 at 19:08
robphyrobphy
1,982238
edited Jan 13 at 20:09
answered Jan 13 at 19:08
robphyrobphy
1,982238
answered Jan 13 at 19:08
robphyrobphy
1,982238
answered Jan 13 at 19:08
robphyrobphy
1,982238
1,982238
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
add a comment |
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
1
1
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This reasoning seems a little circular. Of course if you assume that the putative electric field four-vector is equal to the first row of the electromagnetic field tensor, then it's trivially true that has a zero timelike component and has the wrong transformation properties.
$endgroup$
– Ben Crowell
Jan 13 at 23:23
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
$begingroup$
This formulation is based on a tensorial development of the field tensor and Maxwell Equations, as found in Misner-Thorne-Wheeler [Ch 3.1] and in Wald [Ch 4.2], which is more geometrical and elegant compared to matrix representations and clumsy 3-vector formulations. The magnetic field is defined analogously with the Hodge-dual *F. (In other words, is there a more elegant way to describe the clumsier coordinate-based calculations and transformation formulas to demonstrate Lorentz invariance? Yes, use tensors through out.)
$endgroup$
– robphy
Jan 14 at 0:34
add a comment |
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