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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













14












$begingroup$


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?










share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    Related: physics.stackexchange.com/q/61946
    $endgroup$
    – Chair
    2 days ago















14












$begingroup$


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?










share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    Related: physics.stackexchange.com/q/61946
    $endgroup$
    – Chair
    2 days ago













14












14








14


1



$begingroup$


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?










share|cite|improve this question









New contributor




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







$endgroup$




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






share|cite|improve this question









New contributor




jjmontes 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




jjmontes 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 days ago









Chair

4,40072241




4,40072241






New contributor




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









asked 2 days ago









jjmontesjjmontes

1715




1715




New contributor




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





New contributor





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






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











  • $begingroup$
    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




$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
2 days ago










4 Answers
4






active

oldest

votes


















15












$begingroup$

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 $v_r ll c$.






share|cite|improve this answer











$endgroup$












  • $begingroup$
    the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
    $endgroup$
    – craq
    yesterday


















5












$begingroup$

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.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      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 $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$






      share|cite|improve this answer









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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

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        active

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        active

        oldest

        votes









        15












        $begingroup$

        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 $v_r ll c$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
          $endgroup$
          – craq
          yesterday















        15












        $begingroup$

        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 $v_r ll c$.






        share|cite|improve this answer











        $endgroup$












        • $begingroup$
          the front windshield gets wetter faster. The opposite applies to the back windshield. Unless you're driving backwards.
          $endgroup$
          – craq
          yesterday













        15












        15








        15





        $begingroup$

        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 $v_r ll c$.






        share|cite|improve this answer











        $endgroup$



        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 $v_r ll c$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        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
















        • $begingroup$
          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











        5












        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$

















          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$















            5












            5








            5





            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$



            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            G. SmithG. Smith

            9,10911427




            9,10911427





















                2












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    CleonisCleonis

                    2,085714




                    2,085714





















                        0












                        $begingroup$

                        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 $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          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 $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            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 $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$






                            share|cite|improve this answer









                            $endgroup$



                            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 $$c^2Delta tau^2=c^2(Delta T)^2(1-beta^2)$$$$Delta tau=Delta Tsqrt1-beta^2$$$$Delta tau=Delta t sqrtfrac1+beta1-beta$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 days ago









                            Apashanka DasApashanka Das

                            177




                            177




















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