5 ms·

I've often wondered about the feasbility of a private firm doing something similar to dominate the HFT scene by getting a massive lead on information from the o

I've often wondered about the feasbility of a private firm doing something similar to dominate the HFT scene by getting a massive lead on information from the opposite end of the globe.

SpaceRaccoon 6y ago
They would dominate indeed, but potentially at the detriment of Neutrino research detector sensitivity world-wide.
schoen 6y ago
Avoid the pesky curvature of the Earth's surface by going directly along a chord through the mantle rather than along a great circle? Using https://planetcalc.com/7725/ https://planetcalc.com/7725/ and https://planetcalc.com/73/ https://planetcalc.com/73/ I just concluded that the straight-line chord distance between Tokyo and San Francisco is 1552 km whereas the great-circle distance is 8270 km. (Neither of these is very precise because it's unclear where in each city you should measure from, and unclear whether either calculator uses data about the irregularity of the Earth's curvature.) It does make a noticeable difference for HFT applications, I guess: 1552 km/c is about 5 ms while 8270 km/c is about 28 ms. (The neutrinos might do better in another way because I guess light in a fiber doesn't directly follow the curve of the fiber itself, since it's getting repeatedly reflected off of the inside surface of the glass.)
- hinkley 6y ago
  Speed of light in a fiber is slower than the speed of light in a vacuum, so it's the straight line speed of light versus the great circle speed of light / index of refraction. So you're looking at somewhere north of 35 milliseconds (because the fiber is NOT great circle, and repeaters and routers will add little snips of time every time they touch the signal). But can you decode a neutrino signal in real time?
  - DaiPlusPlus 6y ago
    Pardon my ignorance, but what would a neutrino signal look like? Can’t it be pulsed on/off like an electron beam - does it have a wavelength/frequency like a photon?
    - sbierwagen 6y ago
      Pulses, yes. Making a neutrino beam is remarkably convoluted. https://en.wikipedia.org/wiki/Accelerator_neutrino https://en.wikipedia.org/wiki/Accelerator_neutrino >The process of the muon neutrino or muon antineutrino beam production consists of the following steps[1][2]: >Acceleration of a primary proton beam in a particle accelerator. >Proton beam collision with a fixed target. In such a collision secondary particles, mainly pions and kaons, are produced. >Focusing, by a set of magnetic horns, the secondary particles with a selected charge: positive to produce the muon neutrino beam, negative to produce the muon anti-neutrino beam. >Decay of the secondary particles in flight in a long (of the order of hundreds meters) decay tunnel. Charged pions decay[3] in more than 99.98% into a muon and the corresponding neutrino according to the principle of preserving electric charge and lepton number... There's several different places in this chain where you could modulate the beam by turning various magnets on or off. Probably not in the proton accelerator. Unfortunately, "long string of expensive experimental equipment" means "not efficient". From the paper: >A neutrino source delivering muons at a rate of 10^14 s^−1 with an energy of 150 GeV would require about 4 MW in proton beam power and 2.4 MW acceleration power, which for a 10% electrical efficiency translates into a total power consumption of roughly 65 MW. Yow. Something like ten times more power than the NuMI neutrino beam. He concludes this transmitter could do something like 100 bits/s to a stationary detector string anchored at the ocean bed. Deeper the better, for shielding against cosmic rays and solar radiation. Would be tough to put a neutrino detector close enough to a financial hub and still get useful bandwidth.
      - londons_explore 6y ago
        65 MW is only $6M/year at the cheapest available electricity prices. Even less if you turn it off at night or when markets aren't as volatile. 100 bits/sec is plenty to make lots of money... If you consider 100 bits/sec is 1 bit per 10 milliseconds, with say a 5% bit error rate (can't do error correction without introducing delay). Each end of the connection can make two candidate investment strategies based on globally available data. The 'bit' decides which strategy to go for. Easy money for anyone who can turn the science into reality... Easy money that will dry up as soon as a few other people start doing the same...
- femto 6y ago
  Are those numbers correct? In the worst case, the chord would be the diameter and the great circle would be half the circumference, so the ratio would be (pi.d/2)/d = pi/2 = 1.57. For fun the cos rule can be used to derive a general formula for the ratio of great circle to chord, given the angle (theta) in radians subtended at the centre of the Earth: theta / sqrt( 2-2.cos(theta) ) where: theta = (great circle distance) / radius (Assuming I haven't mucked up my algebra.) -- Not sure why my comment is rendering in italics. No emphasis is meant. Figured it out, it was the asterisk I used for a multiplication symbol.
  - schoen 6y ago
    Thank you for the sanity check. It looks like the web site has a bug!
- deleted 6y ago
  [deleted]
- ianhorn 6y ago
  The app you linked has some issues with data entry. If you re-enter the minutes field for SF, it will give a straight line distance of about 7800km, instead of the 1500km you listed. As another commenter pointed out the worse case is ratio is about 1.5, so I'm guessing either it's totally buggy or 7800km is the correct one.
- rocqua 6y ago
  Light in fiber travels at about 2/3 s of the speed of light in a vacuum. Not because of bouncing (the fiber creates a waveguide) but just because it is inside glass. This different speed of light is what gives glass its refractive index, which makes it optically useful.
  - schoen 6y ago
    Thanks, I feel like I've just experienced Feynman's complaint about his Brazilian students not applying their knowledge of the concept of "index of refraction" to an actual physical object. That is, I didn't think about how the whole idea of the index of refraction of a material is directly derived from the different speed of light in that material. I'm still confused because I thought I had learned that fiber optical cables work by total internal reflection, in which the light inside is repeatedly reflected by the surface of the fiber. Is this an overly simplistic view for thinking about the path that the light will follow in the fiber?
    - rocqua 6y ago
      It is a somewhat simplistic view because light isn't just a particle bouncing off walls, its a wave. What happens is that the fiber creates a 'waveguide'. This will gently curve the wavefront around the fiber. I barely understand this myself. But the way I imagine this is essentially as a water wave running through a trough. Lets consider a simple sinusoidal wave originating at one end of the trough. Lets call a wavefront any line across the trough that follows the peak of a wave (really it can be any fixed phase, but peaks are nicer to visualize). What happens to these wavefronts as time goes on? They bend around the corners. Essentially the water just in front of the wavefront is going to be pushed up next, and when this happens has a lot more to do with distance from the current wavefront than whether the actual water particles are bouncing off the wall or not. My explanation of wavefronts moving forward doesn't quite explain why the wavefront 'rotates' in a curved waveguide. I would guess something about path interference being different on the inside of a corner than the outside. Someone who actually studied this stuff probably knows a lot better.
JimTheMan 6y ago
Things that improve HFT sadden me because I don't really see the net benefit to mankind. I see someone else's share of the pie getting smaller, someone's gets bigger and the rest of us remain starving.
- rocqua 6y ago
  There are marginal benefits outside of HFT. Mostly, the spread in markets (difference between sell price and buy price) shrinks because of HFT. Hence people who want to make financial transactions have less friction. In order to interpret this as a global positive, you need to see a more efficient financial system as better. The story there is usually that a more efficient allocation of capital allows for the most growth, pulling more of the world out of poverty. (I fully think HFT is good for financial world, less certain on the financial world being better for the wider world)
- cbzoiav 6y ago
  If one firm develops this their competitors look to do it. You end up with huge investment in the science behind it with everyone trying to do it with higher throughput, lower latency and lower build / run costs than their competitors. Maybe two decades in the future this becomes a defacto communication technology.
  - JimTheMan 6y ago
    I like your optimism
jankotek 6y ago
Neutrino detection is very slow, it is averaging signal over several hours to transfer a few bytes. Useless for HFT.
hchz 6y ago
I think you might lose for two reasons: - Conventional RF systems can function with very low latency. The signal processing chain for radiation detection systems is comparatively large, and can require correlation (or anti-correlation) and significant noise reduction or signal separation to construct a useful signal. - It requires an unbelievable capital investment for low bandwidth. You need a large time advantage to make your PnL on the few names this would support compared to microwaves and other conventional means.
- cbzoiav 6y ago
  As somebody else points out the bandwidth isn't a major issue. You can have two opposing strategies and use a single bit to toggle between them. Put enough behind those strategies and you're making millions a day. Until your competitors notice and refuse to trade with you or do the same.
  - hchz 6y ago
    Sorry, this doesn't sound realistic (I work for a major HFT). Do you have evidence that someone net a billion in a year making a market on a single name? This seems to ignore that the effective bandwidth could be so low that the latency is greater than conventional transmission systems like RF. Since scintillation radiation detectors are highly stochastic compared to semiconductor RF receivers, I think it's probable it would take too long to receive the bit with sufficient confidence. Scintillation detectors don't work like a solid state detector and have worse time characteristics.
    - cbzoiav 6y ago
      It doesn't need to be a single name. You can select the best target based on expected volatility/volume - for example around announcements. As long as both ends are synchronised ahead of time (which can be done over traditional networks). The other chain (https://news.ycombinator.com/item?id=23903796 https://news.ycombinator.com/item?id=23903796) suggested 100 bits per second with a 5% error rate. With the (major) assumptions on that error rate and that you can modify / detect the beam in realtime that gives you plenty to work with. Of course there is a massive difference between technically possible vs actually implementable.

I've often wondered about the feasbility of a private firm doing something similar to dominate the HFT scene by getting a massive lead on information from the opposite end of the globe.

SpaceRaccoon 6y ago

They would dominate indeed, but potentially at the detriment of Neutrino research detector sensitivity world-wide.

schoen 6y ago

Avoid the pesky curvature of the Earth's surface by going directly along a chord through the mantle rather than along a great circle? Using https://planetcalc.com/7725/ https://planetcalc.com/7725/ and https://planetcalc.com/73/ https://planetcalc.com/73/ I just concluded that the straight-line chord distance between Tokyo and San Francisco is 1552 km whereas the great-circle distance is 8270 km. (Neither of these is very precise because it's unclear where in each city you should measure from, and unclear whether either calculator uses data about the irregularity of the Earth's curvature.) It does make a noticeable difference for HFT applications, I guess: 1552 km/c is about 5 ms while 8270 km/c is about 28 ms. (The neutrinos might do better in another way because I guess light in a fiber doesn't directly follow the curve of the fiber itself, since it's getting repeatedly reflected off of the inside surface of the glass.)

hinkley 6y ago
Speed of light in a fiber is slower than the speed of light in a vacuum, so it's the straight line speed of light versus the great circle speed of light / index of refraction. So you're looking at somewhere north of 35 milliseconds (because the fiber is NOT great circle, and repeaters and routers will add little snips of time every time they touch the signal). But can you decode a neutrino signal in real time?
- DaiPlusPlus 6y ago
  Pardon my ignorance, but what would a neutrino signal look like? Can’t it be pulsed on/off like an electron beam - does it have a wavelength/frequency like a photon?
  - sbierwagen 6y ago
    Pulses, yes. Making a neutrino beam is remarkably convoluted. https://en.wikipedia.org/wiki/Accelerator_neutrino https://en.wikipedia.org/wiki/Accelerator_neutrino >The process of the muon neutrino or muon antineutrino beam production consists of the following steps[1][2]: >Acceleration of a primary proton beam in a particle accelerator. >Proton beam collision with a fixed target. In such a collision secondary particles, mainly pions and kaons, are produced. >Focusing, by a set of magnetic horns, the secondary particles with a selected charge: positive to produce the muon neutrino beam, negative to produce the muon anti-neutrino beam. >Decay of the secondary particles in flight in a long (of the order of hundreds meters) decay tunnel. Charged pions decay[3] in more than 99.98% into a muon and the corresponding neutrino according to the principle of preserving electric charge and lepton number... There's several different places in this chain where you could modulate the beam by turning various magnets on or off. Probably not in the proton accelerator. Unfortunately, "long string of expensive experimental equipment" means "not efficient". From the paper: >A neutrino source delivering muons at a rate of 10^14 s^−1 with an energy of 150 GeV would require about 4 MW in proton beam power and 2.4 MW acceleration power, which for a 10% electrical efficiency translates into a total power consumption of roughly 65 MW. Yow. Something like ten times more power than the NuMI neutrino beam. He concludes this transmitter could do something like 100 bits/s to a stationary detector string anchored at the ocean bed. Deeper the better, for shielding against cosmic rays and solar radiation. Would be tough to put a neutrino detector close enough to a financial hub and still get useful bandwidth.
    - londons_explore 6y ago
      65 MW is only $6M/year at the cheapest available electricity prices. Even less if you turn it off at night or when markets aren't as volatile. 100 bits/sec is plenty to make lots of money... If you consider 100 bits/sec is 1 bit per 10 milliseconds, with say a 5% bit error rate (can't do error correction without introducing delay). Each end of the connection can make two candidate investment strategies based on globally available data. The 'bit' decides which strategy to go for. Easy money for anyone who can turn the science into reality... Easy money that will dry up as soon as a few other people start doing the same...
femto 6y ago
Are those numbers correct? In the worst case, the chord would be the diameter and the great circle would be half the circumference, so the ratio would be (pi.d/2)/d = pi/2 = 1.57. For fun the cos rule can be used to derive a general formula for the ratio of great circle to chord, given the angle (theta) in radians subtended at the centre of the Earth: theta / sqrt( 2-2.cos(theta) ) where: theta = (great circle distance) / radius (Assuming I haven't mucked up my algebra.) -- Not sure why my comment is rendering in italics. No emphasis is meant. Figured it out, it was the asterisk I used for a multiplication symbol.
- schoen 6y ago
  Thank you for the sanity check. It looks like the web site has a bug!
deleted 6y ago
[deleted]
ianhorn 6y ago
The app you linked has some issues with data entry. If you re-enter the minutes field for SF, it will give a straight line distance of about 7800km, instead of the 1500km you listed. As another commenter pointed out the worse case is ratio is about 1.5, so I'm guessing either it's totally buggy or 7800km is the correct one.
rocqua 6y ago
Light in fiber travels at about 2/3 s of the speed of light in a vacuum. Not because of bouncing (the fiber creates a waveguide) but just because it is inside glass. This different speed of light is what gives glass its refractive index, which makes it optically useful.
- schoen 6y ago
  Thanks, I feel like I've just experienced Feynman's complaint about his Brazilian students not applying their knowledge of the concept of "index of refraction" to an actual physical object. That is, I didn't think about how the whole idea of the index of refraction of a material is directly derived from the different speed of light in that material. I'm still confused because I thought I had learned that fiber optical cables work by total internal reflection, in which the light inside is repeatedly reflected by the surface of the fiber. Is this an overly simplistic view for thinking about the path that the light will follow in the fiber?
  - rocqua 6y ago
    It is a somewhat simplistic view because light isn't just a particle bouncing off walls, its a wave. What happens is that the fiber creates a 'waveguide'. This will gently curve the wavefront around the fiber. I barely understand this myself. But the way I imagine this is essentially as a water wave running through a trough. Lets consider a simple sinusoidal wave originating at one end of the trough. Lets call a wavefront any line across the trough that follows the peak of a wave (really it can be any fixed phase, but peaks are nicer to visualize). What happens to these wavefronts as time goes on? They bend around the corners. Essentially the water just in front of the wavefront is going to be pushed up next, and when this happens has a lot more to do with distance from the current wavefront than whether the actual water particles are bouncing off the wall or not. My explanation of wavefronts moving forward doesn't quite explain why the wavefront 'rotates' in a curved waveguide. I would guess something about path interference being different on the inside of a corner than the outside. Someone who actually studied this stuff probably knows a lot better.

JimTheMan 6y ago

Things that improve HFT sadden me because I don't really see the net benefit to mankind. I see someone else's share of the pie getting smaller, someone's gets bigger and the rest of us remain starving.

rocqua 6y ago
There are marginal benefits outside of HFT. Mostly, the spread in markets (difference between sell price and buy price) shrinks because of HFT. Hence people who want to make financial transactions have less friction. In order to interpret this as a global positive, you need to see a more efficient financial system as better. The story there is usually that a more efficient allocation of capital allows for the most growth, pulling more of the world out of poverty. (I fully think HFT is good for financial world, less certain on the financial world being better for the wider world)
cbzoiav 6y ago
If one firm develops this their competitors look to do it. You end up with huge investment in the science behind it with everyone trying to do it with higher throughput, lower latency and lower build / run costs than their competitors. Maybe two decades in the future this becomes a defacto communication technology.
- JimTheMan 6y ago
  I like your optimism

jankotek 6y ago

Neutrino detection is very slow, it is averaging signal over several hours to transfer a few bytes. Useless for HFT.

hchz 6y ago

I think you might lose for two reasons: - Conventional RF systems can function with very low latency. The signal processing chain for radiation detection systems is comparatively large, and can require correlation (or anti-correlation) and significant noise reduction or signal separation to construct a useful signal. - It requires an unbelievable capital investment for low bandwidth. You need a large time advantage to make your PnL on the few names this would support compared to microwaves and other conventional means.

cbzoiav 6y ago
As somebody else points out the bandwidth isn't a major issue. You can have two opposing strategies and use a single bit to toggle between them. Put enough behind those strategies and you're making millions a day. Until your competitors notice and refuse to trade with you or do the same.
- hchz 6y ago
  Sorry, this doesn't sound realistic (I work for a major HFT). Do you have evidence that someone net a billion in a year making a market on a single name? This seems to ignore that the effective bandwidth could be so low that the latency is greater than conventional transmission systems like RF. Since scintillation radiation detectors are highly stochastic compared to semiconductor RF receivers, I think it's probable it would take too long to receive the bit with sufficient confidence. Scintillation detectors don't work like a solid state detector and have worse time characteristics.
  - cbzoiav 6y ago
    It doesn't need to be a single name. You can select the best target based on expected volatility/volume - for example around announcements. As long as both ends are synchronised ahead of time (which can be done over traditional networks). The other chain (https://news.ycombinator.com/item?id=23903796 https://news.ycombinator.com/item?id=23903796) suggested 100 bits per second with a 5% error rate. With the (major) assumptions on that error rate and that you can modify / detect the beam in realtime that gives you plenty to work with. Of course there is a massive difference between technically possible vs actually implementable.