Why do electromagnetic waves have the magnetic and electric field intensities in the same phase?












5












$begingroup$


My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.






electromagnetic-radiation







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Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    "changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
    $endgroup$
    – andars
    2 hours ago










  • $begingroup$
    It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
    $endgroup$
    – dmckee♦
    1 hour ago


















5












$begingroup$


My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.






electromagnetic-radiation







share|cite|improve this question









New contributor




Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    "changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
    $endgroup$
    – andars
    2 hours ago










  • $begingroup$
    It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
    $endgroup$
    – dmckee♦
    1 hour ago
















5












5








5


1



$begingroup$


My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.






electromagnetic-radiation







share|cite|improve this question









New contributor




Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




My question is: in electromagnetic waves, if we consider the electric field as a sine function, the magnetic field will be also a sine function, but I am confused why that is this way.



If I look at Maxwell's equation, the changing magnetic field generates the electric field and the changing electric field generates the magnetic field, so according to my opinion if the accelerating electron generates a sine electric field change, then its magnetic field should be a cosine function because $frac{d(sin x)}{dx}=cos x$.






electromagnetic-radiation




electromagnetic-radiation






share|cite|improve this question









New contributor




Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









David Z♦

63.5k23136252




63.5k23136252






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Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 8 hours ago









Bálint TataiBálint Tatai

341




341




New contributor




Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Bálint Tatai is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    "changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
    $endgroup$
    – andars
    2 hours ago










  • $begingroup$
    It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
    $endgroup$
    – dmckee♦
    1 hour ago




















  • $begingroup$
    "changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
    $endgroup$
    – andars
    2 hours ago










  • $begingroup$
    It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
    $endgroup$
    – dmckee♦
    1 hour ago


















$begingroup$
"changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
$endgroup$
– andars
2 hours ago




$begingroup$
"changing magnetic field generates the electric field and the changing electric field generates the magnetic field" - I think this is misleading. Maxwell's equations aren't statements of cause and effect. Although we talk about one field changing inducing another, they happen at the same time. An increasing magnetic field doesn't really cause a curl to exist in the electric field, they are physically the same - an increasing magnetic field cannot exist without the curl in the electric field.
$endgroup$
– andars
2 hours ago












$begingroup$
It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
$endgroup$
– dmckee♦
1 hour ago






$begingroup$
It's worth stating clearly that the in-phase nature of the waves is true in the far field (i.e. when the waves are examined much farther from the source than the size of the source), but that this is not the case in the near field (i.e. when you are close to the source).
$endgroup$
– dmckee♦
1 hour ago












1 Answer
1






active

oldest

votes


















9












$begingroup$

The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
begin{align}
nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
end{align}
where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
    $endgroup$
    – dmckee♦
    1 hour ago











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









9












$begingroup$

The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
begin{align}
nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
end{align}
where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
    $endgroup$
    – dmckee♦
    1 hour ago
















9












$begingroup$

The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
begin{align}
nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
end{align}
where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
    $endgroup$
    – dmckee♦
    1 hour ago














9












9








9





$begingroup$

The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
begin{align}
nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
end{align}
where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.






share|cite|improve this answer









$endgroup$



The Maxwell equations that relate electric and magnetic fields to each other read (in vacuum, in SI units) as
begin{align}
nabla times mathbf E & = -frac{partialmathbf B}{partial t} \
nabla times mathbf B & = frac{1}{c^2} frac{partialmathbf E}{partial t},
end{align}
where the notation $nabla times{cdot}$ is a spatial derivative (the curl). This means that both sides have derivatives, and if you're applying them to a function like $cos(kx-omega t)$, then they will both change the cosine into a sine. This is what locks the phase of both waves to equal values.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 8 hours ago









Emilio PisantyEmilio Pisanty

83.5k22203417




83.5k22203417












  • $begingroup$
    Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
    $endgroup$
    – dmckee♦
    1 hour ago


















  • $begingroup$
    Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
    $endgroup$
    – dmckee♦
    1 hour ago
















$begingroup$
Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
$endgroup$
– dmckee♦
1 hour ago




$begingroup$
Just to comment with my comment on the questions, this explanation works nicely if spatial arrangement of the fields is uniform enough (as in a plane wave, which is to say in he far-field), but misses important details if the wave has a no-planar structure (as in the near-field).
$endgroup$
– dmckee♦
1 hour ago










Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.










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