Why do electromagnetic waves have the magnetic and electric field intensities in the same phase?
$begingroup$
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
edited 19 mins ago
David Z♦
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
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$begingroup$
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
edited 19 mins ago
David Z♦
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
241
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.
$endgroup$
add a comment |
$begingroup$
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
edited 19 mins ago
David Z♦
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
241
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
edited 19 mins ago
David Z♦
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
241
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.
edited 19 mins ago
David Z♦
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
241
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.
edited 19 mins ago
David Z♦
63.5k23136252
edited 19 mins ago
David Z♦
63.5k23136252
edited 19 mins ago
David Z♦
63.5k23136252
63.5k23136252
asked 6 hours ago
Bálint TataiBálint Tatai
241
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.
asked 6 hours ago
Bálint TataiBálint Tatai
241
asked 6 hours ago
Bálint TataiBálint Tatai
241
241
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.
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
votes
active
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votes
$begingroup$
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.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
add a comment |
$begingroup$
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.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
$endgroup$
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.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.4k22203417
83.4k22203417
add a comment |
add a comment |
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
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