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?
What is the difference between 王 and 皇?
Clarification on Integrability
Is this n-speak?
How should I write this passage to make it the most readable?
Is there a SQL/English like language that lets you define formulations given some data?
Are differences between uniformly distributed numbers uniformly distributed?
How can I communicate my issues with a potential date's pushy behavior?
Can the IPA represent all languages' tones?
Why are Tucker and Malcolm not dead?
Can a bald person be a Nazir?
How to "add" units to results of pgfmathsetmacro?
What sort of psychological changes could be made to a genetically engineered human
Is there any way to stop a user from creating executables and running them?
How would you translate this? バタコチーズライス
Heating Margarine in Pan = loss of calories?
A torrent of foreign terms
Can lodestones be used to magnetize crude iron weapons?
Why is tert-butoxide often used in elimination reactions when it is not necessary?
Website error: "Walmart can’t use this browser"
(A room / an office) where an artist works
Running code generated in realtime in JavaScript with eval()
If I animate and control a zombie, does it benefit from Undead Fortitude when it's reduced to 0 HP?
Dogfights in outer space
Do beef farmed pastures net remove carbon emissions?
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;
$begingroup$
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.
- 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?
- 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
$endgroup$
add a comment |
$begingroup$
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.
- 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?
- 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
$endgroup$
1
$begingroup$
Related: space.stackexchange.com/q/14660/58
$endgroup$
– called2voyage♦
9 hours ago
add a comment |
$begingroup$
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.
- 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?
- 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
$endgroup$
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.
- 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?
- 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
communication deep-space
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
I just spent 45 minutes confirming that this is the correct answer! ;-)
$endgroup$
– uhoh
6 hours ago
add a comment |
$begingroup$
@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!
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f38116%2fwhat-would-it-take-to-get-a-message-to-another-star%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
I just spent 45 minutes confirming that this is the correct answer! ;-)
$endgroup$
– uhoh
6 hours ago
add a comment |
$begingroup$
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.
$endgroup$
3
$begingroup$
I just spent 45 minutes confirming that this is the correct answer! ;-)
$endgroup$
– uhoh
6 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
@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!
$endgroup$
add a comment |
$begingroup$
@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!
$endgroup$
add a comment |
$begingroup$
@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!
$endgroup$
@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!
edited 6 hours ago
answered 6 hours ago
uhohuhoh
49.8k23 gold badges197 silver badges644 bronze badges
49.8k23 gold badges197 silver badges644 bronze badges
add a comment |
add a comment |
Thanks for contributing an answer to Space Exploration Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f38116%2fwhat-would-it-take-to-get-a-message-to-another-star%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Related: space.stackexchange.com/q/14660/58
$endgroup$
– called2voyage♦
9 hours ago