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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  • $begingroup$
    "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
    11 mins 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
    11 mins ago














5












5








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.







$endgroup$




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









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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.









share|cite|improve this question




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edited 50 mins ago









David Z♦

63.5k23136252




63.5k23136252






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asked 6 hours ago









Bálint TataiBálint Tatai

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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
    11 mins 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
    11 mins 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
11 mins 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
11 mins ago










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.






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    1 Answer
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    1 Answer
    1






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    active

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    active

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    8












    $begingroup$

    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









    $endgroup$


















      8












      $begingroup$

      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









      $endgroup$
















        8












        8








        8





        $begingroup$

        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 6 hours ago









        Emilio PisantyEmilio Pisanty

        83.4k22203417




        83.4k22203417






















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