What is the relationship between relativity and the Doppler effect?












4












$begingroup$


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?










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
















4












$begingroup$


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
    18 mins ago














4












4








4





$begingroup$


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 19 mins ago









Chair

4,39072241




4,39072241






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asked 7 hours ago









jjmontesjjmontes

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1214




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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
    18 mins ago


















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
















$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
18 mins ago




$begingroup$
Related: physics.stackexchange.com/q/61946
$endgroup$
– Chair
18 mins ago










2 Answers
2






active

oldest

votes


















7












$begingroup$

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
$$f_o = frac{v - v_o}{v - v_s} f_s$$
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
$$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
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
$$f_o = frac{c - v_r}{c} f_s$$
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,
$$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
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











$endgroup$





















    2












    $begingroup$

    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











    $endgroup$













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      2 Answers
      2






      active

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      2 Answers
      2






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      active

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      7












      $begingroup$

      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
      $$f_o = frac{v - v_o}{v - v_s} f_s$$
      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
      $$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
      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
      $$f_o = frac{c - v_r}{c} f_s$$
      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,
      $$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
      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











      $endgroup$


















        7












        $begingroup$

        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
        $$f_o = frac{v - v_o}{v - v_s} f_s$$
        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
        $$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
        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
        $$f_o = frac{c - v_r}{c} f_s$$
        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,
        $$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
        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











        $endgroup$
















          7












          7








          7





          $begingroup$

          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
          $$f_o = frac{v - v_o}{v - v_s} f_s$$
          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
          $$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
          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
          $$f_o = frac{c - v_r}{c} f_s$$
          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,
          $$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
          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
          $$f_o = frac{v - v_o}{v - v_s} f_s$$
          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
          $$gamma_0 f_0 = frac{v - v_o}{v - v_s} gamma_s f_s.$$
          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
          $$f_o = frac{c - v_r}{c} f_s$$
          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,
          $$f_o = frac{c - v_r}{c} gamma_s f_s = sqrt{frac{1 - v_r/c}{1 + v_r/c}} f_s.$$
          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 6 hours ago

























          answered 7 hours ago









          knzhouknzhou

          45k11122218




          45k11122218























              2












              $begingroup$

              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












                $begingroup$

                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












                  2








                  2





                  $begingroup$

                  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 7 hours ago

























                  answered 7 hours ago









                  G. SmithG. Smith

                  8,98611427




                  8,98611427






















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