Does light lose energy when it reflects from a surface?

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When light reflects from a surface, at least the direction of its momentum changes. Since the total momentum must be conserved, there has to be something going on within the atoms of the surface.



So my question is that does reflected light increase the internal energy of the surface even if it is a really really tiny amount?



P.S. I am not talking about the fraction of light that gets absorbed by the surface. I know the energy of that fraction contributes to the internal energy of the surface. My concern is only about the photons being reflected.






energy photons momentum reflection







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    up vote
    3
    down vote

    favorite
    1












    When light reflects from a surface, at least the direction of its momentum changes. Since the total momentum must be conserved, there has to be something going on within the atoms of the surface.



    So my question is that does reflected light increase the internal energy of the surface even if it is a really really tiny amount?



    P.S. I am not talking about the fraction of light that gets absorbed by the surface. I know the energy of that fraction contributes to the internal energy of the surface. My concern is only about the photons being reflected.






    energy photons momentum reflection







    share|cite|improve this question

























      up vote
      3
      down vote

      favorite
      1









      up vote
      3
      down vote

      favorite
      1






      1





      When light reflects from a surface, at least the direction of its momentum changes. Since the total momentum must be conserved, there has to be something going on within the atoms of the surface.



      So my question is that does reflected light increase the internal energy of the surface even if it is a really really tiny amount?



      P.S. I am not talking about the fraction of light that gets absorbed by the surface. I know the energy of that fraction contributes to the internal energy of the surface. My concern is only about the photons being reflected.






      energy photons momentum reflection







      share|cite|improve this question















      When light reflects from a surface, at least the direction of its momentum changes. Since the total momentum must be conserved, there has to be something going on within the atoms of the surface.



      So my question is that does reflected light increase the internal energy of the surface even if it is a really really tiny amount?



      P.S. I am not talking about the fraction of light that gets absorbed by the surface. I know the energy of that fraction contributes to the internal energy of the surface. My concern is only about the photons being reflected.






      energy photons momentum reflection




      energy photons momentum reflection






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      share|cite|improve this question













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      edited 4 hours ago

























      asked 4 hours ago









      physicsguy19

      583115




      583115




















          2 Answers
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          Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.



          The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.






          share|cite|improve this answer






















          • So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
            – physicsguy19
            4 hours ago







          • 1




            If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
            – Dale
            3 hours ago










          • To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
            – Gilbert
            23 mins ago

















          up vote
          0
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          When a photon is reflected there is no energy lost. There is a brief time when the photon interacts with the electron, momentum is deposited and then transferred right back to the photon. Only when there is absorption is momentum transferred to the material. Interestingly when photons are emitted there is a momentum change as well.






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          • I think that explanation clearly contradicts the conservation of momentum.
            – physicsguy19
            3 hours ago










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          2 Answers
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          active

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          2 Answers
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          active

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          active

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          active

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          up vote
          2
          down vote













          Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.



          The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.






          share|cite|improve this answer






















          • So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
            – physicsguy19
            4 hours ago







          • 1




            If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
            – Dale
            3 hours ago










          • To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
            – Gilbert
            23 mins ago














          up vote
          2
          down vote













          Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.



          The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.






          share|cite|improve this answer






















          • So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
            – physicsguy19
            4 hours ago







          • 1




            If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
            – Dale
            3 hours ago










          • To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
            – Gilbert
            23 mins ago












          up vote
          2
          down vote










          up vote
          2
          down vote









          Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.



          The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.






          share|cite|improve this answer














          Yes, this is the principle behind Doppler radar. The frequency/energy increases if the surface is moving towards the source. The frequency/energy decreases if the reflecting surface is moving away from the source.



          The only time the frequency/energy will be unchanged is if the surface initially has the opposite momentum of the light.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 4 hours ago









          Dale

          2,178415




          2,178415











          • So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
            – physicsguy19
            4 hours ago







          • 1




            If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
            – Dale
            3 hours ago










          • To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
            – Gilbert
            23 mins ago
















          • So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
            – physicsguy19
            4 hours ago







          • 1




            If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
            – Dale
            3 hours ago










          • To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
            – Gilbert
            23 mins ago















          So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
          – physicsguy19
          4 hours ago





          So the frequency of reflected light is less than the incoming light evem if the surface is stationary?
          – physicsguy19
          4 hours ago





          1




          1




          If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
          – Dale
          3 hours ago




          If the surface is stationary then there will be a slight decrease. If the surface has the opposite momentum of the photon then the energy/frequency will be unchanged. I will add this to the answer
          – Dale
          3 hours ago












          To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
          – Gilbert
          23 mins ago




          To clarify, upon reflection from a stationary object, the object recoils, giving it kinetic energy. The photon’s energy then needs to decrease (redshift) in order to account for the kinetic energy. The amount by which the frequency of the photon shifts is the Doppler factor. Thus both energy and momentum are conserved.
          – Gilbert
          23 mins ago










          up vote
          0
          down vote













          When a photon is reflected there is no energy lost. There is a brief time when the photon interacts with the electron, momentum is deposited and then transferred right back to the photon. Only when there is absorption is momentum transferred to the material. Interestingly when photons are emitted there is a momentum change as well.






          share|cite|improve this answer




















          • I think that explanation clearly contradicts the conservation of momentum.
            – physicsguy19
            3 hours ago














          up vote
          0
          down vote













          When a photon is reflected there is no energy lost. There is a brief time when the photon interacts with the electron, momentum is deposited and then transferred right back to the photon. Only when there is absorption is momentum transferred to the material. Interestingly when photons are emitted there is a momentum change as well.






          share|cite|improve this answer




















          • I think that explanation clearly contradicts the conservation of momentum.
            – physicsguy19
            3 hours ago












          up vote
          0
          down vote










          up vote
          0
          down vote









          When a photon is reflected there is no energy lost. There is a brief time when the photon interacts with the electron, momentum is deposited and then transferred right back to the photon. Only when there is absorption is momentum transferred to the material. Interestingly when photons are emitted there is a momentum change as well.






          share|cite|improve this answer












          When a photon is reflected there is no energy lost. There is a brief time when the photon interacts with the electron, momentum is deposited and then transferred right back to the photon. Only when there is absorption is momentum transferred to the material. Interestingly when photons are emitted there is a momentum change as well.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 4 hours ago









          PhysicsDave

          61726




          61726











          • I think that explanation clearly contradicts the conservation of momentum.
            – physicsguy19
            3 hours ago
















          • I think that explanation clearly contradicts the conservation of momentum.
            – physicsguy19
            3 hours ago















          I think that explanation clearly contradicts the conservation of momentum.
          – physicsguy19
          3 hours ago




          I think that explanation clearly contradicts the conservation of momentum.
          – physicsguy19
          3 hours ago

















           

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