Is electric current relative?
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Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?
electromagnetism electric-current relative-motion
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up vote
2
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Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?
electromagnetism electric-current relative-motion
Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?
electromagnetism electric-current relative-motion
Motion as we know is relative. According to this current which is the flow of charges should be also relative . That means that if someone is moving with the same velocity with repsect to the velocity of electrons (current) would he see no current at all ?
electromagnetism electric-current relative-motion
electromagnetism electric-current relative-motion
asked 3 hours ago
ado sar
1488
1488
Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago
add a comment |
Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago
Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago
Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago
add a comment |
2 Answers
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Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$
where $gamma=1/sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
add a comment |
up vote
3
down vote
Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$
where $gamma=1/sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
add a comment |
up vote
5
down vote
Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$
where $gamma=1/sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
add a comment |
up vote
5
down vote
up vote
5
down vote
Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$
where $gamma=1/sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
Charge density $rho$ and current density $vec{J}$ form a Lorentz four-vector $(crho, vec{J})$ that transforms under a Lorentz transformation just like $(ct, vec{r})$ does.
For example, if two frames are in relative motion in the $z$-direction, with the primed frame having velocity $vhat{mathbf{z}}$ relative to the unprimed frame, then the transformation is
$$begin{align}
crho^prime &= gammaleft(crho - frac{v}{c}J_zright)\
J_x^prime &= J_x \
J_y^prime &= J_y \
J_z^prime &= gammaleft(J_z - frac{v}{c}(crho)right)
end{align}$$
where $gamma=1/sqrt{1-v^2/c^2}$.
If you had a beam of electrons in space, and you observed it from a reference frame moving with the same velocity as the beam, there would be no current.
Since charge density and current density are frame-dependent, so are electric and magnetic fields. (For example, in the frame moving along with the beam, there is no magnetic field.) However, the transformation rules for the fields are a bit more complicated; they transform as components of a two-index four-tensor rather than a one-index four-vector.
edited 2 hours ago
answered 2 hours ago
G. Smith
3,673917
3,673917
add a comment |
add a comment |
up vote
3
down vote
Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
add a comment |
up vote
3
down vote
Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
add a comment |
up vote
3
down vote
up vote
3
down vote
Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
Typically, we observe current in a solid, such as copper wire. So if you move with the same velocity as the average velocity of electrons in a current-carrying copper wire you will observe current of nuclei and bound electrons.
answered 2 hours ago
akhmeteli
17.6k21740
17.6k21740
add a comment |
add a comment |
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Related: Is magnetic field due to an electric current a relativistic effect? and Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?.
– The Photon
3 hours ago