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
New contributor
$endgroup$
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
New contributor
$endgroup$
$begingroup$
"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
1 min ago
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
New contributor
$endgroup$
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
New contributor
New contributor
edited 41 mins ago
David Z♦
63.5k23136252
63.5k23136252
New contributor
asked 6 hours ago
Bálint TataiBálint Tatai
291
291
New contributor
New contributor
$begingroup$
"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
1 min ago
add a comment |
$begingroup$
"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
1 min 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
1 min 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
1 min ago
add a comment |
1 Answer
1
active
oldest
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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Bálint Tatai is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f461393%2fwhy-do-electromagnetic-waves-have-the-magnetic-and-electric-field-intensities-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 6 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.
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.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f461393%2fwhy-do-electromagnetic-waves-have-the-magnetic-and-electric-field-intensities-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
"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
1 min ago