Mixing
Is the direction of the electric field relative to the magnetic field in an electromagnetic wave a...
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In an electromagnetic wave.
Could the magnetic field be mirrored around the xy-plane? Is there a specific reason that the 2 fields are oriented this way? Is it just a convention?
electromagnetism waves electromagnetic-radiation polarization
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
$endgroup$
add a comment |
$begingroup$
In an electromagnetic wave.
Could the magnetic field be mirrored around the xy-plane? Is there a specific reason that the 2 fields are oriented this way? Is it just a convention?
electromagnetism waves electromagnetic-radiation polarization
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
$endgroup$
5
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
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I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
$endgroup$
– takintoolong
Jan 3 at 3:46
add a comment |
$begingroup$
In an electromagnetic wave.
Could the magnetic field be mirrored around the xy-plane? Is there a specific reason that the 2 fields are oriented this way? Is it just a convention?
electromagnetism waves electromagnetic-radiation polarization
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
$endgroup$
In an electromagnetic wave.
Could the magnetic field be mirrored around the xy-plane? Is there a specific reason that the 2 fields are oriented this way? Is it just a convention?
electromagnetism waves electromagnetic-radiation polarization
electromagnetism waves electromagnetic-radiation polarization
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
edited Jan 2 at 20:07
Qmechanic♦
102k121831169
102k121831169
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
asked Jan 2 at 19:04
Alexander AmeyeAlexander Ameye
636
636
5
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
$begingroup$
I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
$endgroup$
– takintoolong
Jan 3 at 3:46
add a comment |
5
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
$begingroup$
I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
$endgroup$
– takintoolong
Jan 3 at 3:46
5
5
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
$begingroup$
I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
$endgroup$
– takintoolong
Jan 3 at 3:46
$begingroup$
I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
$endgroup$
– takintoolong
Jan 3 at 3:46
add a comment |
5 Answers
5
active
oldest
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If you were to flip the direction of the magnetic field in the above picture, it would describe light propagating in the opposite direction—so no, in this way, its direction is not a convention.
However, it's important to keep in mind that the direction of the magnetic field is a convention at a more fundamental level. It represents an oriented plane perpendicular to its direction via an orientation rule. If we all decided to use the "left hand rule" instead of the right hand rule for the cross product, then the magnetic field would point in the opposite direction. It is a bivector, or a pseudovector, field.
The electric field is a true vector field. At least in the three-dimensional description, which is itself a (very decent) convention, as long as we acknowledge that this picture depends on our reference frame.
In the four-dimensional picture, electric and magnetic fields can be understood as a single bivector field, such that a choice of reference frame determines a splitting of this bivector into independent "timelike" (electric) and "spacelike" (magnetic) planes.
That may be more than you were originally looking for, but I think this question makes for a good entry point to many of the other choices of representation that we make in physics.
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
$endgroup$
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
add a comment |
$begingroup$
The direction of the fields of an electromagnetic wave is not conventional but it's determined by Maxwell's equations. For the particular case of a plane electromagnetic wave of angular frequency $omega$ in vacuum, which can be represented by the picture you posted, Maxwell's equations require that the wave vector $boldsymbol{k}$, the electric field $boldsymbol{E}$ and the magnetic flux density $boldsymbol{B}$ obey the relationships (the dot represent the scalar product and the cross the vector product)
$$begin{align}&boldsymbol{k}cdot boldsymbol{E} = 0, \ &boldsymbol{k}cdot boldsymbol{B} = 0, \ & boldsymbol{B}=frac{1}{omega}boldsymbol{k}times boldsymbol{E}.end{align}$$
This means that $boldsymbol{k}$, $boldsymbol{E}$ and $boldsymbol{B}$ are three orthogonal vectors and that the direction of any one of them is determined by the other two. Therefore, no, you cannot mirror the magnetic field in the picture.
Note that if it were possible to mirror just the magnetic field, then, by superposition, you would be able to construct a wave having nonzero electric field but zero magnetic field.
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
$endgroup$
1
$begingroup$
You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
$endgroup$
– Bruce Greetham
Jan 2 at 20:17
3
$begingroup$
@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
add a comment |
$begingroup$
The arrows are pointing in the "positive" direction of the fields. Which direction is "positive" is arbitrary (you can say that the "positive" direction in an electrical field in the direction in which the electrical potential energy for a positively charged particle is increasing, but what charge is "positive" is itself a convention). Moreover, those conventions are chosen independently. However, once those conventions are chosen, the directions of the fields for an electromagnetic wave are fixed. So if you're asking whether the conventions could have been chosen such that the magnetic field would be flipped, the answer is "yes". If you're asking whether we can flip the magnetic field, without affecting any of the other conventions in physics, the answer is "no".
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
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B is defined by the expression of the Lorentz force. In covariant notation, this is written as $f^nu = j_mu F^{munu}$, where
${displaystyle F^{mu nu }={begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\E_{x}/c&0&-B_{z}&B_{y}\E_{y}/c&B_{z}&0&-B_{x}\E_{z}/c&-B_{y}&B_{x}&0end{bmatrix}}.} $
One can see this as defining the Lorentz force as $vec f = q vec E + q vec v times vec B$, as well as $vec E$ and $vec B$.
One can replace B by -B, or E by -E. This would be useless as the measurable physical quantities such as forces and energies would not change. Students would have to learn cumbersome definitions like like "$vec E$ is the opposite of the force per charge".
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
$endgroup$
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Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
add a comment |
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Electrons in an antenna rod, accelerated in one direction, all emit photons with the same spin of their electric and magnetic field components. This is obvious, since the receiving antennas could be made of a rod to receive the changing electric field or could be made of a ring (available in the German wiki only) to receive the magnetic field. If accelerated in the same direction electrons would have spin clockwise and anti clockwise one would not be able to use the magnetic component (because both variations will cancel out each over).
I’m not sure but think that protons and positrons will have the opposite spin to electrons and antiprotons. Clarification from specialists will be nice.
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
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4
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This question is not related to spin.
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– my2cts
Jan 2 at 21:17
2
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To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
add a comment |
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5 Answers
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5 Answers
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$begingroup$
If you were to flip the direction of the magnetic field in the above picture, it would describe light propagating in the opposite direction—so no, in this way, its direction is not a convention.
However, it's important to keep in mind that the direction of the magnetic field is a convention at a more fundamental level. It represents an oriented plane perpendicular to its direction via an orientation rule. If we all decided to use the "left hand rule" instead of the right hand rule for the cross product, then the magnetic field would point in the opposite direction. It is a bivector, or a pseudovector, field.
The electric field is a true vector field. At least in the three-dimensional description, which is itself a (very decent) convention, as long as we acknowledge that this picture depends on our reference frame.
In the four-dimensional picture, electric and magnetic fields can be understood as a single bivector field, such that a choice of reference frame determines a splitting of this bivector into independent "timelike" (electric) and "spacelike" (magnetic) planes.
That may be more than you were originally looking for, but I think this question makes for a good entry point to many of the other choices of representation that we make in physics.
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
$endgroup$
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
add a comment |
$begingroup$
If you were to flip the direction of the magnetic field in the above picture, it would describe light propagating in the opposite direction—so no, in this way, its direction is not a convention.
However, it's important to keep in mind that the direction of the magnetic field is a convention at a more fundamental level. It represents an oriented plane perpendicular to its direction via an orientation rule. If we all decided to use the "left hand rule" instead of the right hand rule for the cross product, then the magnetic field would point in the opposite direction. It is a bivector, or a pseudovector, field.
The electric field is a true vector field. At least in the three-dimensional description, which is itself a (very decent) convention, as long as we acknowledge that this picture depends on our reference frame.
In the four-dimensional picture, electric and magnetic fields can be understood as a single bivector field, such that a choice of reference frame determines a splitting of this bivector into independent "timelike" (electric) and "spacelike" (magnetic) planes.
That may be more than you were originally looking for, but I think this question makes for a good entry point to many of the other choices of representation that we make in physics.
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
$endgroup$
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
add a comment |
$begingroup$
If you were to flip the direction of the magnetic field in the above picture, it would describe light propagating in the opposite direction—so no, in this way, its direction is not a convention.
However, it's important to keep in mind that the direction of the magnetic field is a convention at a more fundamental level. It represents an oriented plane perpendicular to its direction via an orientation rule. If we all decided to use the "left hand rule" instead of the right hand rule for the cross product, then the magnetic field would point in the opposite direction. It is a bivector, or a pseudovector, field.
The electric field is a true vector field. At least in the three-dimensional description, which is itself a (very decent) convention, as long as we acknowledge that this picture depends on our reference frame.
In the four-dimensional picture, electric and magnetic fields can be understood as a single bivector field, such that a choice of reference frame determines a splitting of this bivector into independent "timelike" (electric) and "spacelike" (magnetic) planes.
That may be more than you were originally looking for, but I think this question makes for a good entry point to many of the other choices of representation that we make in physics.
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
$endgroup$
If you were to flip the direction of the magnetic field in the above picture, it would describe light propagating in the opposite direction—so no, in this way, its direction is not a convention.
However, it's important to keep in mind that the direction of the magnetic field is a convention at a more fundamental level. It represents an oriented plane perpendicular to its direction via an orientation rule. If we all decided to use the "left hand rule" instead of the right hand rule for the cross product, then the magnetic field would point in the opposite direction. It is a bivector, or a pseudovector, field.
The electric field is a true vector field. At least in the three-dimensional description, which is itself a (very decent) convention, as long as we acknowledge that this picture depends on our reference frame.
In the four-dimensional picture, electric and magnetic fields can be understood as a single bivector field, such that a choice of reference frame determines a splitting of this bivector into independent "timelike" (electric) and "spacelike" (magnetic) planes.
That may be more than you were originally looking for, but I think this question makes for a good entry point to many of the other choices of representation that we make in physics.
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
edited Jan 2 at 21:29
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
answered Jan 2 at 20:14
Luke BurnsLuke Burns
459616
459616
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
add a comment |
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
2
2
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
$begingroup$
Thanks for pointing out that the direction of the magnetic field is an arbitrary convention, but at a deeper level than the question is asking about.
$endgroup$
– Buzz
Jan 3 at 2:56
4
4
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
$begingroup$
The direction of electric field also a convention, just at a deeper level (i.e. signs assigned to kinds of charges).
$endgroup$
– Džuris
Jan 3 at 11:48
add a comment |
$begingroup$
The direction of the fields of an electromagnetic wave is not conventional but it's determined by Maxwell's equations. For the particular case of a plane electromagnetic wave of angular frequency $omega$ in vacuum, which can be represented by the picture you posted, Maxwell's equations require that the wave vector $boldsymbol{k}$, the electric field $boldsymbol{E}$ and the magnetic flux density $boldsymbol{B}$ obey the relationships (the dot represent the scalar product and the cross the vector product)
$$begin{align}&boldsymbol{k}cdot boldsymbol{E} = 0, \ &boldsymbol{k}cdot boldsymbol{B} = 0, \ & boldsymbol{B}=frac{1}{omega}boldsymbol{k}times boldsymbol{E}.end{align}$$
This means that $boldsymbol{k}$, $boldsymbol{E}$ and $boldsymbol{B}$ are three orthogonal vectors and that the direction of any one of them is determined by the other two. Therefore, no, you cannot mirror the magnetic field in the picture.
Note that if it were possible to mirror just the magnetic field, then, by superposition, you would be able to construct a wave having nonzero electric field but zero magnetic field.
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
$endgroup$
1
$begingroup$
You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
$endgroup$
– Bruce Greetham
Jan 2 at 20:17
3
$begingroup$
@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
add a comment |
$begingroup$
The direction of the fields of an electromagnetic wave is not conventional but it's determined by Maxwell's equations. For the particular case of a plane electromagnetic wave of angular frequency $omega$ in vacuum, which can be represented by the picture you posted, Maxwell's equations require that the wave vector $boldsymbol{k}$, the electric field $boldsymbol{E}$ and the magnetic flux density $boldsymbol{B}$ obey the relationships (the dot represent the scalar product and the cross the vector product)
$$begin{align}&boldsymbol{k}cdot boldsymbol{E} = 0, \ &boldsymbol{k}cdot boldsymbol{B} = 0, \ & boldsymbol{B}=frac{1}{omega}boldsymbol{k}times boldsymbol{E}.end{align}$$
This means that $boldsymbol{k}$, $boldsymbol{E}$ and $boldsymbol{B}$ are three orthogonal vectors and that the direction of any one of them is determined by the other two. Therefore, no, you cannot mirror the magnetic field in the picture.
Note that if it were possible to mirror just the magnetic field, then, by superposition, you would be able to construct a wave having nonzero electric field but zero magnetic field.
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
$endgroup$
1
$begingroup$
You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
$endgroup$
– Bruce Greetham
Jan 2 at 20:17
3
$begingroup$
@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
add a comment |
$begingroup$
The direction of the fields of an electromagnetic wave is not conventional but it's determined by Maxwell's equations. For the particular case of a plane electromagnetic wave of angular frequency $omega$ in vacuum, which can be represented by the picture you posted, Maxwell's equations require that the wave vector $boldsymbol{k}$, the electric field $boldsymbol{E}$ and the magnetic flux density $boldsymbol{B}$ obey the relationships (the dot represent the scalar product and the cross the vector product)
$$begin{align}&boldsymbol{k}cdot boldsymbol{E} = 0, \ &boldsymbol{k}cdot boldsymbol{B} = 0, \ & boldsymbol{B}=frac{1}{omega}boldsymbol{k}times boldsymbol{E}.end{align}$$
This means that $boldsymbol{k}$, $boldsymbol{E}$ and $boldsymbol{B}$ are three orthogonal vectors and that the direction of any one of them is determined by the other two. Therefore, no, you cannot mirror the magnetic field in the picture.
Note that if it were possible to mirror just the magnetic field, then, by superposition, you would be able to construct a wave having nonzero electric field but zero magnetic field.
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
$endgroup$
The direction of the fields of an electromagnetic wave is not conventional but it's determined by Maxwell's equations. For the particular case of a plane electromagnetic wave of angular frequency $omega$ in vacuum, which can be represented by the picture you posted, Maxwell's equations require that the wave vector $boldsymbol{k}$, the electric field $boldsymbol{E}$ and the magnetic flux density $boldsymbol{B}$ obey the relationships (the dot represent the scalar product and the cross the vector product)
$$begin{align}&boldsymbol{k}cdot boldsymbol{E} = 0, \ &boldsymbol{k}cdot boldsymbol{B} = 0, \ & boldsymbol{B}=frac{1}{omega}boldsymbol{k}times boldsymbol{E}.end{align}$$
This means that $boldsymbol{k}$, $boldsymbol{E}$ and $boldsymbol{B}$ are three orthogonal vectors and that the direction of any one of them is determined by the other two. Therefore, no, you cannot mirror the magnetic field in the picture.
Note that if it were possible to mirror just the magnetic field, then, by superposition, you would be able to construct a wave having nonzero electric field but zero magnetic field.
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
edited Jan 2 at 23:03
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
answered Jan 2 at 20:01
Massimo OrtolanoMassimo Ortolano
1,9451615
1,9451615
1
$begingroup$
You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
$endgroup$
– Bruce Greetham
Jan 2 at 20:17
3
$begingroup$
@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
add a comment |
1
$begingroup$
You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
$endgroup$
– Bruce Greetham
Jan 2 at 20:17
3
$begingroup$
@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
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– Acccumulation
Jan 2 at 23:06
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@Acccumulation I know: see my comment above to Bruce.
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– Massimo Ortolano
Jan 2 at 23:10
1
1
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You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
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– Bruce Greetham
Jan 2 at 20:17
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You may want to start with "Having defined the direction of B by the Lorentz rule ... " (BTW -1 was not me).
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– Bruce Greetham
Jan 2 at 20:17
3
3
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@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
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– Massimo Ortolano
Jan 2 at 20:21
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@BruceGreetham The point is that the magnetic field is defined in that way: once you have a certain definition, you cannot mirror the magnetic field (which is, I think, what the OP is asking).
$endgroup$
– Massimo Ortolano
Jan 2 at 20:21
2
2
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
The third equation you give is written in terms of the cross product, and the direction of the cross product (right hand rule) is due to convention.
$endgroup$
– Acccumulation
Jan 2 at 23:06
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
$begingroup$
@Acccumulation I know: see my comment above to Bruce.
$endgroup$
– Massimo Ortolano
Jan 2 at 23:10
add a comment |
$begingroup$
The arrows are pointing in the "positive" direction of the fields. Which direction is "positive" is arbitrary (you can say that the "positive" direction in an electrical field in the direction in which the electrical potential energy for a positively charged particle is increasing, but what charge is "positive" is itself a convention). Moreover, those conventions are chosen independently. However, once those conventions are chosen, the directions of the fields for an electromagnetic wave are fixed. So if you're asking whether the conventions could have been chosen such that the magnetic field would be flipped, the answer is "yes". If you're asking whether we can flip the magnetic field, without affecting any of the other conventions in physics, the answer is "no".
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
$endgroup$
add a comment |
$begingroup$
The arrows are pointing in the "positive" direction of the fields. Which direction is "positive" is arbitrary (you can say that the "positive" direction in an electrical field in the direction in which the electrical potential energy for a positively charged particle is increasing, but what charge is "positive" is itself a convention). Moreover, those conventions are chosen independently. However, once those conventions are chosen, the directions of the fields for an electromagnetic wave are fixed. So if you're asking whether the conventions could have been chosen such that the magnetic field would be flipped, the answer is "yes". If you're asking whether we can flip the magnetic field, without affecting any of the other conventions in physics, the answer is "no".
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
$endgroup$
add a comment |
$begingroup$
The arrows are pointing in the "positive" direction of the fields. Which direction is "positive" is arbitrary (you can say that the "positive" direction in an electrical field in the direction in which the electrical potential energy for a positively charged particle is increasing, but what charge is "positive" is itself a convention). Moreover, those conventions are chosen independently. However, once those conventions are chosen, the directions of the fields for an electromagnetic wave are fixed. So if you're asking whether the conventions could have been chosen such that the magnetic field would be flipped, the answer is "yes". If you're asking whether we can flip the magnetic field, without affecting any of the other conventions in physics, the answer is "no".
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
$endgroup$
The arrows are pointing in the "positive" direction of the fields. Which direction is "positive" is arbitrary (you can say that the "positive" direction in an electrical field in the direction in which the electrical potential energy for a positively charged particle is increasing, but what charge is "positive" is itself a convention). Moreover, those conventions are chosen independently. However, once those conventions are chosen, the directions of the fields for an electromagnetic wave are fixed. So if you're asking whether the conventions could have been chosen such that the magnetic field would be flipped, the answer is "yes". If you're asking whether we can flip the magnetic field, without affecting any of the other conventions in physics, the answer is "no".
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
answered Jan 2 at 23:13
AcccumulationAcccumulation
1,886210
1,886210
add a comment |
add a comment |
$begingroup$
B is defined by the expression of the Lorentz force. In covariant notation, this is written as $f^nu = j_mu F^{munu}$, where
${displaystyle F^{mu nu }={begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\E_{x}/c&0&-B_{z}&B_{y}\E_{y}/c&B_{z}&0&-B_{x}\E_{z}/c&-B_{y}&B_{x}&0end{bmatrix}}.} $
One can see this as defining the Lorentz force as $vec f = q vec E + q vec v times vec B$, as well as $vec E$ and $vec B$.
One can replace B by -B, or E by -E. This would be useless as the measurable physical quantities such as forces and energies would not change. Students would have to learn cumbersome definitions like like "$vec E$ is the opposite of the force per charge".
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
$endgroup$
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
add a comment |
$begingroup$
B is defined by the expression of the Lorentz force. In covariant notation, this is written as $f^nu = j_mu F^{munu}$, where
${displaystyle F^{mu nu }={begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\E_{x}/c&0&-B_{z}&B_{y}\E_{y}/c&B_{z}&0&-B_{x}\E_{z}/c&-B_{y}&B_{x}&0end{bmatrix}}.} $
One can see this as defining the Lorentz force as $vec f = q vec E + q vec v times vec B$, as well as $vec E$ and $vec B$.
One can replace B by -B, or E by -E. This would be useless as the measurable physical quantities such as forces and energies would not change. Students would have to learn cumbersome definitions like like "$vec E$ is the opposite of the force per charge".
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
$endgroup$
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
add a comment |
$begingroup$
B is defined by the expression of the Lorentz force. In covariant notation, this is written as $f^nu = j_mu F^{munu}$, where
${displaystyle F^{mu nu }={begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\E_{x}/c&0&-B_{z}&B_{y}\E_{y}/c&B_{z}&0&-B_{x}\E_{z}/c&-B_{y}&B_{x}&0end{bmatrix}}.} $
One can see this as defining the Lorentz force as $vec f = q vec E + q vec v times vec B$, as well as $vec E$ and $vec B$.
One can replace B by -B, or E by -E. This would be useless as the measurable physical quantities such as forces and energies would not change. Students would have to learn cumbersome definitions like like "$vec E$ is the opposite of the force per charge".
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
$endgroup$
B is defined by the expression of the Lorentz force. In covariant notation, this is written as $f^nu = j_mu F^{munu}$, where
${displaystyle F^{mu nu }={begin{bmatrix}0&-E_{x}/c&-E_{y}/c&-E_{z}/c\E_{x}/c&0&-B_{z}&B_{y}\E_{y}/c&B_{z}&0&-B_{x}\E_{z}/c&-B_{y}&B_{x}&0end{bmatrix}}.} $
One can see this as defining the Lorentz force as $vec f = q vec E + q vec v times vec B$, as well as $vec E$ and $vec B$.
One can replace B by -B, or E by -E. This would be useless as the measurable physical quantities such as forces and energies would not change. Students would have to learn cumbersome definitions like like "$vec E$ is the opposite of the force per charge".
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
edited Jan 3 at 0:13
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
answered Jan 3 at 0:06
my2ctsmy2cts
4,7802618
4,7802618
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
add a comment |
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
$begingroup$
Well, one could define charge with an opposite sign as well, and so on.
$endgroup$
– rubenvb
Jan 3 at 11:44
add a comment |
$begingroup$
Electrons in an antenna rod, accelerated in one direction, all emit photons with the same spin of their electric and magnetic field components. This is obvious, since the receiving antennas could be made of a rod to receive the changing electric field or could be made of a ring (available in the German wiki only) to receive the magnetic field. If accelerated in the same direction electrons would have spin clockwise and anti clockwise one would not be able to use the magnetic component (because both variations will cancel out each over).
I’m not sure but think that protons and positrons will have the opposite spin to electrons and antiprotons. Clarification from specialists will be nice.
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
$endgroup$
4
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
2
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
add a comment |
$begingroup$
Electrons in an antenna rod, accelerated in one direction, all emit photons with the same spin of their electric and magnetic field components. This is obvious, since the receiving antennas could be made of a rod to receive the changing electric field or could be made of a ring (available in the German wiki only) to receive the magnetic field. If accelerated in the same direction electrons would have spin clockwise and anti clockwise one would not be able to use the magnetic component (because both variations will cancel out each over).
I’m not sure but think that protons and positrons will have the opposite spin to electrons and antiprotons. Clarification from specialists will be nice.
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
$endgroup$
4
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
2
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
add a comment |
$begingroup$
Electrons in an antenna rod, accelerated in one direction, all emit photons with the same spin of their electric and magnetic field components. This is obvious, since the receiving antennas could be made of a rod to receive the changing electric field or could be made of a ring (available in the German wiki only) to receive the magnetic field. If accelerated in the same direction electrons would have spin clockwise and anti clockwise one would not be able to use the magnetic component (because both variations will cancel out each over).
I’m not sure but think that protons and positrons will have the opposite spin to electrons and antiprotons. Clarification from specialists will be nice.
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
$endgroup$
Electrons in an antenna rod, accelerated in one direction, all emit photons with the same spin of their electric and magnetic field components. This is obvious, since the receiving antennas could be made of a rod to receive the changing electric field or could be made of a ring (available in the German wiki only) to receive the magnetic field. If accelerated in the same direction electrons would have spin clockwise and anti clockwise one would not be able to use the magnetic component (because both variations will cancel out each over).
I’m not sure but think that protons and positrons will have the opposite spin to electrons and antiprotons. Clarification from specialists will be nice.
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
answered Jan 2 at 19:27
HolgerFiedlerHolgerFiedler
4,15031134
4,15031134
4
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
2
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
add a comment |
4
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
2
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
4
4
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
$begingroup$
This question is not related to spin.
$endgroup$
– my2cts
Jan 2 at 21:17
2
2
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
$begingroup$
To the extent that all questions about electromagnetism involve spin, this answer is still quite wrong. The radiation from a rod antenna is linearly polarized. The photons which make up linearly-polarized radiation are in a superposition of left- and right-handed spin states. The sign of the more mobile charges in the antenna does not contribute.
$endgroup$
– rob♦
Jan 3 at 0:59
add a comment |
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5
$begingroup$
Please, report the source of the picture.
$endgroup$
– Massimo Ortolano
Jan 2 at 20:12
$begingroup$
I am not sure if you are implying physical space, but the graph is not meant to imply the physical 3D x,y,z. The graph is a portrayal of vectors, mirroring would change values of vectors by negative or positive.
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– takintoolong
Jan 3 at 3:46