What is the relationship between relativity and the Doppler effect?Relativistic Doppler effect derivationHow realistic is the game “A slower speed of light”?Doppler effect and lightDoppler effect and acceleration's impactSpecial relativity radar doppler shiftDoppler Effect and BeatDoppler effect and different framesWhat is the difference between these formulas? (Doppler effect)Doppler Effect and RelativityWhat is this connection between clocks and the Doppler effect?Doppler Effect and Speed relativity
Jem'Hadar, something strange about their life expectancy
Are hand made posters acceptable in Academia?
When did hardware antialiasing start being available?
What happens when the centripetal force is equal and opposite to the centrifugal force?
Parts of mini page are not placed properly
Why does Surtur say that Thor is Asgard's doom?
Not hide and seek
Why didn’t Eve recognize the little cockroach as a living organism?
The English Debate
If I cast the Enlarge/Reduce spell on an arrow, what weapon could it count as?
Is it possible to deploy Apex code which uses Person Accounts to an org which doesn't have Person Accounts enabled?
Exit shell with shortcut (not typing exit) that closes session properly
Why doesn't the fusion process of the sun speed up?
How do you justify more code being written by following clean code practices?
Why is this tree refusing to shed its dead leaves?
How to test the sharpness of a knife?
Creating points with attributes from coordinates in ArcPy
Why didn't Voldemort know what Grindelwald looked like?
Should I be concerned about student access to a test bank?
God... independent
Box half filled color
Output visual diagram of picture
Why is "la Gestapo" feminine?
What is this high flying aircraft over Pennsylvania?
What is the relationship between relativity and the Doppler effect?
Relativistic Doppler effect derivationHow realistic is the game “A slower speed of light”?Doppler effect and lightDoppler effect and acceleration's impactSpecial relativity radar doppler shiftDoppler Effect and BeatDoppler effect and different framesWhat is the difference between these formulas? (Doppler effect)Doppler Effect and RelativityWhat is this connection between clocks and the Doppler effect?Doppler Effect and Speed relativity
$begingroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
$endgroup$
add a comment |
$begingroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
$endgroup$
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago
add a comment |
$begingroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
New contributor
$endgroup$
My sister just watched this video about space contraction (Spanish), and asked me if this is related to doppler effect.
In the clip they also introduce the idea that a bat would be affected by similar effects when measuring an object's length, due to the time it takes for sound to propagate.
I told her that:
Doppler effect is about alteration of the perceived frequency of a
signal produced by the relative movement between transmitter and
receiver. The quoted video is about relativity, which is a "deeper" effect.
Maybe doppler effect can be understood as the effect of relativity on
a wave phenomena.
Now I'm wondering about her intuition. If she's right, should I be able to take a sin function, apply a Lorentz transform to it, and arrive to same results as with the doppler formula? Unfortunately, the maths are beyond my skills.
Can someone shed some light about the relation between doppler and relativity, if any? Can be doppler effect explained by relativity/Lorentz alone?
special-relativity waves doppler-effect
special-relativity waves doppler-effect
New contributor
New contributor
edited 2 days ago
Chair
4,40072241
4,40072241
New contributor
asked 2 days ago
jjmontesjjmontes
1715
1715
New contributor
New contributor
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago
add a comment |
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = fracv - v_ov - v_s f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = fracv - v_ov - v_s gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = fracc - v_rc f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = fracc - v_rc gamma_s f_s = sqrtfrac1 - v_r/c1 + v_r/c f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
add a comment |
$begingroup$
As pointed out in earlier answers, Doppler effect and relativistic effects are independent.
There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.
However, it's interesting to note that mathematically there are parallels in the descriptions of the two.
In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.
General wave equation in one dimension:
$$
fracpartial^2 phi partial x^2 = frac1u^2 fracpartial^2 phipartial t^2
$$
(u = propagation speed of the wave)
As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.
Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.
Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.
That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.
Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)
The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.
$endgroup$
add a comment |
$begingroup$
Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis.
If $Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$Delta t+fracbeta Delta t1-beta=fracDelta t1-beta=Delta T$$.
From the invariance of spacetime interval $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$
$endgroup$
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
);
);
jjmontes 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%2f466848%2fwhat-is-the-relationship-between-relativity-and-the-doppler-effect%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = fracv - v_ov - v_s f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = fracv - v_ov - v_s gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = fracc - v_rc f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = fracc - v_rc gamma_s f_s = sqrtfrac1 - v_r/c1 + v_r/c f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
add a comment |
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = fracv - v_ov - v_s f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = fracv - v_ov - v_s gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = fracc - v_rc f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = fracc - v_rc gamma_s f_s = sqrtfrac1 - v_r/c1 + v_r/c f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
add a comment |
$begingroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = fracv - v_ov - v_s f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = fracv - v_ov - v_s gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = fracc - v_rc f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = fracc - v_rc gamma_s f_s = sqrtfrac1 - v_r/c1 + v_r/c f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
$endgroup$
The ordinary Doppler effect is independent of relativity; it's basically just a fact of kinematics. It's not even really a wave phenomenon; it also applies to particles. For example, the Doppler effect explains why your car windshield gets wetter faster when you're driving than when you're parked.
The formula for the Doppler effect is
$$f_o = fracv - v_ov - v_s f_s$$
where $f_o$ is the observed frequency, $f_s$ is the source's emitted frequency, and $v_0$ and $v_s$ are the velocities of the observer and source. These are absolute velocities; they have to be defined with respect to the medium, e.g. the air for a sound wave. Relativity adds a correction to this formula because both the source and the observer will experience time dilation, so we should really have
$$gamma_0 f_0 = fracv - v_ov - v_s gamma_s f_s.$$
This is a very small correction assuming the speeds are small.
When people talk about the relativistic Doppler effect, they usually mean the Doppler effect for light waves specifically, with full relativistic corrections. Light waves are exceptional because they have no medium, so we aren't tied to a specific frame. It's instead more convenient to go to the observer's frame, where we naively have
$$f_o = fracc - v_rc f_s$$
where $v_r$ is the relative velocity. Relativity corrects this formula in two ways. First, velocities don't quite add linearly, so $v_r neq v_o - v_s$ in general. Second, we have to remember the time dilation factor for the source,
$$f_o = fracc - v_rc gamma_s f_s = sqrtfrac1 - v_r/c1 + v_r/c f_s.$$
There is no time dilation factor for the observer, because we're in the observer's frame, where they are at rest. This last formula is what people usually call "the relativistic Doppler effect", but again it's pretty close to the nonrelativistic result as long as $v_r ll c$.
edited 2 days ago
answered 2 days ago
knzhouknzhou
45.1k11122219
45.1k11122219
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
add a comment |
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
$begingroup$
the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
$endgroup$
– craq
yesterday
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
add a comment |
$begingroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
$endgroup$
There is a Doppler effect even without Special or General Relativity, just arising from Galilean relative motion. For example, neither of these theories is necessary to explain the fact that the pitch of an ambulance siren changes as it passes by.
However, relativity does have to be taken into account when calculating the Doppler effect for a fast-moving object or one in a strong gravitational field. In other words, there are relativistic corrections to the Doppler effect.
If you use a Lorentz transformation to derive the Doppler effect, you will get the right answer for any velocity, but you won’t get the Doppler effect for a gravitational field.
edited 2 days ago
answered 2 days ago
G. SmithG. Smith
9,10911427
9,10911427
add a comment |
add a comment |
$begingroup$
As pointed out in earlier answers, Doppler effect and relativistic effects are independent.
There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.
However, it's interesting to note that mathematically there are parallels in the descriptions of the two.
In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.
General wave equation in one dimension:
$$
fracpartial^2 phi partial x^2 = frac1u^2 fracpartial^2 phipartial t^2
$$
(u = propagation speed of the wave)
As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.
Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.
Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.
That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.
Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)
The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.
$endgroup$
add a comment |
$begingroup$
As pointed out in earlier answers, Doppler effect and relativistic effects are independent.
There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.
However, it's interesting to note that mathematically there are parallels in the descriptions of the two.
In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.
General wave equation in one dimension:
$$
fracpartial^2 phi partial x^2 = frac1u^2 fracpartial^2 phipartial t^2
$$
(u = propagation speed of the wave)
As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.
Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.
Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.
That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.
Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)
The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.
$endgroup$
add a comment |
$begingroup$
As pointed out in earlier answers, Doppler effect and relativistic effects are independent.
There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.
However, it's interesting to note that mathematically there are parallels in the descriptions of the two.
In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.
General wave equation in one dimension:
$$
fracpartial^2 phi partial x^2 = frac1u^2 fracpartial^2 phipartial t^2
$$
(u = propagation speed of the wave)
As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.
Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.
Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.
That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.
Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)
The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.
$endgroup$
As pointed out in earlier answers, Doppler effect and relativistic effects are independent.
There are several ways to show that, one way is to consider transverse velocity. For instance the case of particles moving at relativistic speed, for instance in a ring-shaped particle accelerator. Relative to the center of the ring that is a transverse velocity. The relativistic velocity gives rise to a measurable time dilation effect for the particles. Due to the time dilation effect there is a frequency shift of any radiation that is emitted or absorbed. That is relativistic effect, not Doppler effect.
However, it's interesting to note that mathematically there are parallels in the descriptions of the two.
In the 1880's a german called 'Waldemar Voigt' published an article in which he presented some work on how to represent Doppler effect mathematically. Specifically, Voigt discussed how solutions to general general wave equation transform from one frame of reference to another.
General wave equation in one dimension:
$$
fracpartial^2 phi partial x^2 = frac1u^2 fracpartial^2 phipartial t^2
$$
(u = propagation speed of the wave)
As part of his 'Reflections on Relativity' Kevin Brown writes about these explorations by Voigt in an article titled The relativity of light.
Voigt arrived at transformations very close to the Lorentz transformations, and Kevin Brown discusses that if Voigt would have aimed for full symmetry he would readily have arrived at the Lorentz transformations. Again, this was in the course of a general exploration of Doppler effect.
Years later, in the 1890's, Lorentz arrived at the Lorentz transformations in the course of exploring properties of Maxwell's theory of electricity and magnetism.
As we know, the Lorentz transformations are at the heart of special relativity.
That makes one wonder:
how did it come about that Lorentz arrived at the Lorentz transformations in the course of working on deeper understanding of how Maxwell's equations describe the physical world? Maxwell's theory of electricity and magnetism was formulated decades before the introduction of Einstein's Special Relativity.
Back when Maxwell had introduced his theory of electricity and magnetism Maxwell had noticed that his equations implied that the electromagnetic field would have the following capability: propagating undulations of the electromagnetic field. That is: propagating waves. (Maxwell's theory is so powerful that Maxwell had a way of calculating the speed of the electromagnetic waves from first principles, finding that this calculated speed was the same as the know speed of light, to within the measuring accuracy available. That is: Maxwell's equations imply that light is electromagnetic waves.)
The Lorentz transformations from Maxwell's equations:
If you put two and two together it is apparent that the wave propagation is the crucial element. The fact that Maxwell's equations give rise to the Lorentz transformations relates to the fact that there are solutions to Maxwell's equations that describe propagating waves.
answered 2 days ago
CleonisCleonis
2,085714
2,085714
add a comment |
add a comment |
$begingroup$
Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis.
If $Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$Delta t+fracbeta Delta t1-beta=fracDelta t1-beta=Delta T$$.
From the invariance of spacetime interval $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$
$endgroup$
add a comment |
$begingroup$
Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis.
If $Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$Delta t+fracbeta Delta t1-beta=fracDelta t1-beta=Delta T$$.
From the invariance of spacetime interval $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$
$endgroup$
add a comment |
$begingroup$
Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis.
If $Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$Delta t+fracbeta Delta t1-beta=fracDelta t1-beta=Delta T$$.
From the invariance of spacetime interval $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$
$endgroup$
Imagine that the observer(S') is receeding away from the source(S) with velocity v along their common $x$ axis.
If $Delta t$ is the time interval between two emmision of photons in the S frame,then the interval of receiving these two photons in the S' frame as seen from S frame will be $$Delta t+fracbeta Delta t1-beta=fracDelta t1-beta=Delta T$$.
From the invariance of spacetime interval $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$
answered 2 days ago
Apashanka DasApashanka Das
177
177
add a comment |
add a comment |
jjmontes is a new contributor. Be nice, and check out our Code of Conduct.
jjmontes is a new contributor. Be nice, and check out our Code of Conduct.
jjmontes is a new contributor. Be nice, and check out our Code of Conduct.
jjmontes 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%2f466848%2fwhat-is-the-relationship-between-relativity-and-the-doppler-effect%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
$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago