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What would it take to get a message to another star?


How to calculate data rate of Voyager 1?Is it possible to extend high speed data transmission with lasers to the distance Earth to Mars?How to construct a message to send to the stars?If a MarCO-type CubeSat were in orbit around Bennu, what kind of power would it need to communicate with the Deep Space Network?What evidence would be needed to determine a signal was artificial in origin?What was the last message to Opportunity today (13 Feb '19)?What was the last message Opportunity sent?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








2












$begingroup$


Let's assume that we've somehow detected the existence of an extraterrestrial civilization through "passive" means, such as detecting a suspicious exoplanetary atmospheric signature. We're now hoping to send that civilization a "hello, neighbor!" message - more than just a flash, something that actually communicates information.



  1. Given our existing communication hardware, without having to build more, what's the furthest distance we could hope to send a signal before it gets lost in background noise?

  2. Assuming we could gather extraordinary resources (this could be first contact, after all...) what's the most powerful message that we could send, and how far could it reach?









share|improve this question









$endgroup$









  • 1




    $begingroup$
    Related: space.stackexchange.com/q/14660/58
    $endgroup$
    – called2voyage
    9 hours ago

















2












$begingroup$


Let's assume that we've somehow detected the existence of an extraterrestrial civilization through "passive" means, such as detecting a suspicious exoplanetary atmospheric signature. We're now hoping to send that civilization a "hello, neighbor!" message - more than just a flash, something that actually communicates information.



  1. Given our existing communication hardware, without having to build more, what's the furthest distance we could hope to send a signal before it gets lost in background noise?

  2. Assuming we could gather extraordinary resources (this could be first contact, after all...) what's the most powerful message that we could send, and how far could it reach?









share|improve this question









$endgroup$









  • 1




    $begingroup$
    Related: space.stackexchange.com/q/14660/58
    $endgroup$
    – called2voyage
    9 hours ago













2












2








2





$begingroup$


Let's assume that we've somehow detected the existence of an extraterrestrial civilization through "passive" means, such as detecting a suspicious exoplanetary atmospheric signature. We're now hoping to send that civilization a "hello, neighbor!" message - more than just a flash, something that actually communicates information.



  1. Given our existing communication hardware, without having to build more, what's the furthest distance we could hope to send a signal before it gets lost in background noise?

  2. Assuming we could gather extraordinary resources (this could be first contact, after all...) what's the most powerful message that we could send, and how far could it reach?









share|improve this question









$endgroup$




Let's assume that we've somehow detected the existence of an extraterrestrial civilization through "passive" means, such as detecting a suspicious exoplanetary atmospheric signature. We're now hoping to send that civilization a "hello, neighbor!" message - more than just a flash, something that actually communicates information.



  1. Given our existing communication hardware, without having to build more, what's the furthest distance we could hope to send a signal before it gets lost in background noise?

  2. Assuming we could gather extraordinary resources (this could be first contact, after all...) what's the most powerful message that we could send, and how far could it reach?






communication deep-space






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









kgutwinkgutwin

2942 silver badges7 bronze badges




2942 silver badges7 bronze badges










  • 1




    $begingroup$
    Related: space.stackexchange.com/q/14660/58
    $endgroup$
    – called2voyage
    9 hours ago












  • 1




    $begingroup$
    Related: space.stackexchange.com/q/14660/58
    $endgroup$
    – called2voyage
    9 hours ago







1




1




$begingroup$
Related: space.stackexchange.com/q/14660/58
$endgroup$
– called2voyage
9 hours ago




$begingroup$
Related: space.stackexchange.com/q/14660/58
$endgroup$
– called2voyage
9 hours ago










2 Answers
2






active

oldest

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5












$begingroup$

The Arecibo raqdio telescope has a 300m diameter mirror. Let's consider a radio wavelength of 3cm (10 GHz) for convenience of arithmetic. That gives a diffraction limited beam width of 100 $mu Rad$, so at 100 light years, the signal would be spread over an area $10^14$ meters across.



The Arecibo signal was transmitted at 450kW, so supposing the data rate was 1 bit per second, so that the bandwidth is just 1 Hz, the signal flux is the power per square meter, per steradian (of source width) per Hertz.



So that is 450kW divided by the beam area (roughly 10^28 m^2) divided by the receiving antenna beam solid angle ($10^-8$ steradians) divided by the bandwidth (1Hz). This comes to $4.5times 10^-15 W/m^2/steradian/Hz$ or about half a trillion Janskys. A decent radio telescope can detect a flux of 1 Jy over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $1kHz$ or more.






share|improve this answer











$endgroup$










  • 3




    $begingroup$
    I just spent 45 minutes confirming that this is the correct answer! ;-)
    $endgroup$
    – uhoh
    6 hours ago


















2












$begingroup$

@SteveLinton's answer is excellent and I'll just confirm below that is rationale and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon caries most of the total received power per second.



Radio



From this answer:




One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.



Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.



$$ P_RX = P_TX + G_TX - L_FS + G_RX $$




  • $P_RX$: Received Power


  • $P_TX$: Transmitted Power


  • $G_TX$: Gain of Transmitting antenna (compared to isotropic)


  • $L_FS$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / lambda^2$) because receive gain is relative to isotropic)


  • $G_RX$: Gain of Earth's Receiving antenna (compared to isotropic)

$$L_FS = 20 times log_10left( 4 pi fracRlambda right)$$



$$G_Dish sim 20 times log_10left( fracpi dlambda right)$$





Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $L_FS$ is 412 dB.



This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as



$$P_RX = 56.5 + 90 - 412 + 90 = -175.5 textdBW.$$



Assuming a bandwidth $Delta f$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $k_B T times Delta f$ (where $k_B$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!



You can read about the use of Shannon-Hartley](https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) in this context in this answer.



Optical Transmission



note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark, you use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.



This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?



You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then



$$P_RX = 10 + 156 - 507.6 + 156 = -185.6 textdBW.$$



That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.



Doing photon counting and using $E = hc/ lambda$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.



This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.



However there is this answer:




13 bits per photon has been demonstrated with laser communications.




and that's not a fundamental limit. You would use a pulsed laser with the same 10W average power and encode data in the time structure of the pulses, in this case at the millisecond or microsecond level.



Modulate the Sun



This answer links to the open access paper A cloaking device for transiting planets which mentions the use of masks or mirrors to modulate the power of the Sun in a specific direction. I think this is the best way, but it requires superstructures or megastructures and so won't be built any time soon!






share|improve this answer











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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    5












    $begingroup$

    The Arecibo raqdio telescope has a 300m diameter mirror. Let's consider a radio wavelength of 3cm (10 GHz) for convenience of arithmetic. That gives a diffraction limited beam width of 100 $mu Rad$, so at 100 light years, the signal would be spread over an area $10^14$ meters across.



    The Arecibo signal was transmitted at 450kW, so supposing the data rate was 1 bit per second, so that the bandwidth is just 1 Hz, the signal flux is the power per square meter, per steradian (of source width) per Hertz.



    So that is 450kW divided by the beam area (roughly 10^28 m^2) divided by the receiving antenna beam solid angle ($10^-8$ steradians) divided by the bandwidth (1Hz). This comes to $4.5times 10^-15 W/m^2/steradian/Hz$ or about half a trillion Janskys. A decent radio telescope can detect a flux of 1 Jy over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $1kHz$ or more.






    share|improve this answer











    $endgroup$










    • 3




      $begingroup$
      I just spent 45 minutes confirming that this is the correct answer! ;-)
      $endgroup$
      – uhoh
      6 hours ago















    5












    $begingroup$

    The Arecibo raqdio telescope has a 300m diameter mirror. Let's consider a radio wavelength of 3cm (10 GHz) for convenience of arithmetic. That gives a diffraction limited beam width of 100 $mu Rad$, so at 100 light years, the signal would be spread over an area $10^14$ meters across.



    The Arecibo signal was transmitted at 450kW, so supposing the data rate was 1 bit per second, so that the bandwidth is just 1 Hz, the signal flux is the power per square meter, per steradian (of source width) per Hertz.



    So that is 450kW divided by the beam area (roughly 10^28 m^2) divided by the receiving antenna beam solid angle ($10^-8$ steradians) divided by the bandwidth (1Hz). This comes to $4.5times 10^-15 W/m^2/steradian/Hz$ or about half a trillion Janskys. A decent radio telescope can detect a flux of 1 Jy over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $1kHz$ or more.






    share|improve this answer











    $endgroup$










    • 3




      $begingroup$
      I just spent 45 minutes confirming that this is the correct answer! ;-)
      $endgroup$
      – uhoh
      6 hours ago













    5












    5








    5





    $begingroup$

    The Arecibo raqdio telescope has a 300m diameter mirror. Let's consider a radio wavelength of 3cm (10 GHz) for convenience of arithmetic. That gives a diffraction limited beam width of 100 $mu Rad$, so at 100 light years, the signal would be spread over an area $10^14$ meters across.



    The Arecibo signal was transmitted at 450kW, so supposing the data rate was 1 bit per second, so that the bandwidth is just 1 Hz, the signal flux is the power per square meter, per steradian (of source width) per Hertz.



    So that is 450kW divided by the beam area (roughly 10^28 m^2) divided by the receiving antenna beam solid angle ($10^-8$ steradians) divided by the bandwidth (1Hz). This comes to $4.5times 10^-15 W/m^2/steradian/Hz$ or about half a trillion Janskys. A decent radio telescope can detect a flux of 1 Jy over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $1kHz$ or more.






    share|improve this answer











    $endgroup$



    The Arecibo raqdio telescope has a 300m diameter mirror. Let's consider a radio wavelength of 3cm (10 GHz) for convenience of arithmetic. That gives a diffraction limited beam width of 100 $mu Rad$, so at 100 light years, the signal would be spread over an area $10^14$ meters across.



    The Arecibo signal was transmitted at 450kW, so supposing the data rate was 1 bit per second, so that the bandwidth is just 1 Hz, the signal flux is the power per square meter, per steradian (of source width) per Hertz.



    So that is 450kW divided by the beam area (roughly 10^28 m^2) divided by the receiving antenna beam solid angle ($10^-8$ steradians) divided by the bandwidth (1Hz). This comes to $4.5times 10^-15 W/m^2/steradian/Hz$ or about half a trillion Janskys. A decent radio telescope can detect a flux of 1 Jy over a period of an hour or less, so this signal will stand out like a sore thumb, once the correct frequency has been detected. In fact, you could probably up the data rate to $1kHz$ or more.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 6 hours ago









    Mark Foskey

    2,70112 silver badges21 bronze badges




    2,70112 silver badges21 bronze badges










    answered 8 hours ago









    Steve LintonSteve Linton

    10.9k1 gold badge27 silver badges55 bronze badges




    10.9k1 gold badge27 silver badges55 bronze badges










    • 3




      $begingroup$
      I just spent 45 minutes confirming that this is the correct answer! ;-)
      $endgroup$
      – uhoh
      6 hours ago












    • 3




      $begingroup$
      I just spent 45 minutes confirming that this is the correct answer! ;-)
      $endgroup$
      – uhoh
      6 hours ago







    3




    3




    $begingroup$
    I just spent 45 minutes confirming that this is the correct answer! ;-)
    $endgroup$
    – uhoh
    6 hours ago




    $begingroup$
    I just spent 45 minutes confirming that this is the correct answer! ;-)
    $endgroup$
    – uhoh
    6 hours ago













    2












    $begingroup$

    @SteveLinton's answer is excellent and I'll just confirm below that is rationale and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon caries most of the total received power per second.



    Radio



    From this answer:




    One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.



    Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.



    $$ P_RX = P_TX + G_TX - L_FS + G_RX $$




    • $P_RX$: Received Power


    • $P_TX$: Transmitted Power


    • $G_TX$: Gain of Transmitting antenna (compared to isotropic)


    • $L_FS$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / lambda^2$) because receive gain is relative to isotropic)


    • $G_RX$: Gain of Earth's Receiving antenna (compared to isotropic)

    $$L_FS = 20 times log_10left( 4 pi fracRlambda right)$$



    $$G_Dish sim 20 times log_10left( fracpi dlambda right)$$





    Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $L_FS$ is 412 dB.



    This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as



    $$P_RX = 56.5 + 90 - 412 + 90 = -175.5 textdBW.$$



    Assuming a bandwidth $Delta f$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $k_B T times Delta f$ (where $k_B$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!



    You can read about the use of Shannon-Hartley](https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) in this context in this answer.



    Optical Transmission



    note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark, you use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.



    This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?



    You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then



    $$P_RX = 10 + 156 - 507.6 + 156 = -185.6 textdBW.$$



    That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.



    Doing photon counting and using $E = hc/ lambda$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.



    This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.



    However there is this answer:




    13 bits per photon has been demonstrated with laser communications.




    and that's not a fundamental limit. You would use a pulsed laser with the same 10W average power and encode data in the time structure of the pulses, in this case at the millisecond or microsecond level.



    Modulate the Sun



    This answer links to the open access paper A cloaking device for transiting planets which mentions the use of masks or mirrors to modulate the power of the Sun in a specific direction. I think this is the best way, but it requires superstructures or megastructures and so won't be built any time soon!






    share|improve this answer











    $endgroup$



















      2












      $begingroup$

      @SteveLinton's answer is excellent and I'll just confirm below that is rationale and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon caries most of the total received power per second.



      Radio



      From this answer:




      One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.



      Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.



      $$ P_RX = P_TX + G_TX - L_FS + G_RX $$




      • $P_RX$: Received Power


      • $P_TX$: Transmitted Power


      • $G_TX$: Gain of Transmitting antenna (compared to isotropic)


      • $L_FS$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / lambda^2$) because receive gain is relative to isotropic)


      • $G_RX$: Gain of Earth's Receiving antenna (compared to isotropic)

      $$L_FS = 20 times log_10left( 4 pi fracRlambda right)$$



      $$G_Dish sim 20 times log_10left( fracpi dlambda right)$$





      Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $L_FS$ is 412 dB.



      This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as



      $$P_RX = 56.5 + 90 - 412 + 90 = -175.5 textdBW.$$



      Assuming a bandwidth $Delta f$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $k_B T times Delta f$ (where $k_B$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!



      You can read about the use of Shannon-Hartley](https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) in this context in this answer.



      Optical Transmission



      note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark, you use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.



      This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?



      You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then



      $$P_RX = 10 + 156 - 507.6 + 156 = -185.6 textdBW.$$



      That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.



      Doing photon counting and using $E = hc/ lambda$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.



      This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.



      However there is this answer:




      13 bits per photon has been demonstrated with laser communications.




      and that's not a fundamental limit. You would use a pulsed laser with the same 10W average power and encode data in the time structure of the pulses, in this case at the millisecond or microsecond level.



      Modulate the Sun



      This answer links to the open access paper A cloaking device for transiting planets which mentions the use of masks or mirrors to modulate the power of the Sun in a specific direction. I think this is the best way, but it requires superstructures or megastructures and so won't be built any time soon!






      share|improve this answer











      $endgroup$

















        2












        2








        2





        $begingroup$

        @SteveLinton's answer is excellent and I'll just confirm below that is rationale and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon caries most of the total received power per second.



        Radio



        From this answer:




        One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.



        Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.



        $$ P_RX = P_TX + G_TX - L_FS + G_RX $$




        • $P_RX$: Received Power


        • $P_TX$: Transmitted Power


        • $G_TX$: Gain of Transmitting antenna (compared to isotropic)


        • $L_FS$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / lambda^2$) because receive gain is relative to isotropic)


        • $G_RX$: Gain of Earth's Receiving antenna (compared to isotropic)

        $$L_FS = 20 times log_10left( 4 pi fracRlambda right)$$



        $$G_Dish sim 20 times log_10left( fracpi dlambda right)$$





        Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $L_FS$ is 412 dB.



        This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as



        $$P_RX = 56.5 + 90 - 412 + 90 = -175.5 textdBW.$$



        Assuming a bandwidth $Delta f$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $k_B T times Delta f$ (where $k_B$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!



        You can read about the use of Shannon-Hartley](https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) in this context in this answer.



        Optical Transmission



        note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark, you use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.



        This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?



        You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then



        $$P_RX = 10 + 156 - 507.6 + 156 = -185.6 textdBW.$$



        That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.



        Doing photon counting and using $E = hc/ lambda$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.



        This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.



        However there is this answer:




        13 bits per photon has been demonstrated with laser communications.




        and that's not a fundamental limit. You would use a pulsed laser with the same 10W average power and encode data in the time structure of the pulses, in this case at the millisecond or microsecond level.



        Modulate the Sun



        This answer links to the open access paper A cloaking device for transiting planets which mentions the use of masks or mirrors to modulate the power of the Sun in a specific direction. I think this is the best way, but it requires superstructures or megastructures and so won't be built any time soon!






        share|improve this answer











        $endgroup$



        @SteveLinton's answer is excellent and I'll just confirm below that is rationale and numbers are correct. Then I'll show that you can do it optically as well, but with 10 meter telescopes instead of Arecibos you run into a challenge because each individual light photon caries most of the total received power per second.



        Radio



        From this answer:




        One standard way to estimate how well signals can be sent between points is to use a link budget calculation, where things are in a standardized format so Engineers can understand each part of the link separately, and to share the information with each other.



        Since the calculation is a series of multiplications and division, when you use dB, these become addition and subtraction of logarithms. I'm going to leave out the smaller corrections from the big equation shown here since this is an approximate calculation.



        $$ P_RX = P_TX + G_TX - L_FS + G_RX $$




        • $P_RX$: Received Power


        • $P_TX$: Transmitted Power


        • $G_TX$: Gain of Transmitting antenna (compared to isotropic)


        • $L_FS$: "Free space Loss", what we usually call $1/r^2$ (but also has $R^2 / lambda^2$) because receive gain is relative to isotropic)


        • $G_RX$: Gain of Earth's Receiving antenna (compared to isotropic)

        $$L_FS = 20 times log_10left( 4 pi fracRlambda right)$$



        $$G_Dish sim 20 times log_10left( fracpi dlambda right)$$





        Using 300 meters and 3 cm for an Arecibo antenna at each end as mentioned in the other answer, the gain (over an isotropic antenna) at each end is about 90 dB. The transmit power of 450 kW is 56.5 dBW. 100 light years is 9.5E+17 meters, so $L_FS$ is 412 dB.



        This gives the Arecibo to Arecibo at 100 Ly, 3 cm, 450 kW received power as



        $$P_RX = 56.5 + 90 - 412 + 90 = -175.5 textdBW.$$



        Assuming a bandwidth $Delta f$ of 1 Hz as in the other answer, and a receiver front-end temperature of 20 Kelvin (typical for practical Deep Space Network dishes) the NEP (Noise Equivalent Power) would be $k_B T times Delta f$ (where $k_B$ is the Boltzmann constant or 1.381E-23 J/K) is only -215.6 dBW, and would be -185.6 dBW for roughy 1 kHz, so @SteveLinton's answer is spot-on!



        You can read about the use of Shannon-Hartley](https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem) in this context in this answer.



        Optical Transmission



        note: After writing this section I realized that the Sun is going to drown out your signal unless you can find a narrow wavelength range where the Sun's emission is extremely dark, you use a very stable laser wavelength, and you hope that the people 100 light years away use a filter that isolates your laser wavelength taking into account the Doppler shift due to all of the motion between your planet and their planet.



        This is extremely unlikely to work, whereas the Sun is going to be much dimmer in a narrow radio band, giving you more room to work with. For more on that see answers to How far have individual stars been seen by radio telescopes?



        You can apply the same calculation to an optical link. Using a 10 meter telescope at each end, a 10W laser and a wavelength of 500 nm, you now get gains of 156 dB, a power of 10 dBW, and a path loss of 507.6 dB. The received power is then



        $$P_RX = 10 + 156 - 507.6 + 156 = -185.6 textdBW.$$



        That's surprisingly similar to the radio received power. If you used a temperature-based bolometer to measure the optical signal, you might think that you could do a similar comparison to NEP, but there's a problem because each visible photon carries so much energy.



        Doing photon counting and using $E = hc/ lambda$, the photon energy is about 4E-19 Joules means that -185.6 dBW (about 2.8E-19 Joules/sec) is going to be only about 1.3 photons per second.



        This means that if you were simply counting photons per 1 second bin, you wouldn't be able to do 1 kHz, and even 1 Hz would require a lot of statistical analysis.



        However there is this answer:




        13 bits per photon has been demonstrated with laser communications.




        and that's not a fundamental limit. You would use a pulsed laser with the same 10W average power and encode data in the time structure of the pulses, in this case at the millisecond or microsecond level.



        Modulate the Sun



        This answer links to the open access paper A cloaking device for transiting planets which mentions the use of masks or mirrors to modulate the power of the Sun in a specific direction. I think this is the best way, but it requires superstructures or megastructures and so won't be built any time soon!







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 6 hours ago

























        answered 6 hours ago









        uhohuhoh

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