Why can't cables used for computer networking transfer data really fast, say at the speed of light?

I ask this because electricity travels at the speed of light. Take Ethernet cables for example, I looked them up on wikipedia.

Propagation speed   0.64        c

Why only 64% What does propagation speed mean? I know there are other variables affecting the latency and perceived speed of computer network connections, but surely this is a bottle neck.

In other words, I'm asking, what is it about a fiber-optics cable that makes it faster than an Ethernet cable?

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    $\begingroup$ "Ideal wires" that transferred information at c would only provide an incremental speedup (maybe 33%) over the wires we have! By contrast, improvements to the physical encoding of bits, higher-frequency carrier waves, frequency multiplexing, and so on offer order of magnitude speedups, and will continue doing so for some time (although, as Charles Stross once said, once you get up into X-ray frequencies your network card becomes indistinguishable from a death ray). $\endgroup$ – zwol Aug 6 '14 at 1:24
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    $\begingroup$ Having said that, c does put a hard floor on latency, and we're quite close to that floor in many cases. For instance, this is a significant part of why RAM access latency hasn't kept up with the CPU clock for many years. $\endgroup$ – zwol Aug 6 '14 at 1:31
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    $\begingroup$ @Zack Note that a 3 GHz clock signal (a common processor speed nowadays) sends out a rising edge every $c/f = 0.1\,\mathrm{meter} = 4\,\mathrm{inches}$. Throw in a factor of two for index-of-refraction effects like the first version of this question and you have the astounding result that for a processor more than a couple of inches across it is physically impossible for a 3 GHz clock signal to synchronize the entire chip. This is part of the reason why processor speeds topped out around 2 GHz for several years and you started to see multi-core processors instead of faster singletons. $\endgroup$ – rob Aug 6 '14 at 3:05
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    $\begingroup$ @rob: No, it doesn't mean that. Clock delays are very predictable, and CPU's don't change shape. If you're 3 mm from the clock input pin, you know that the clock is delayed by 10 ps. What it does mean is that you can't assume that the entire CPU settles in a given state near the end of each clock period. Different parts of the chip have different (and overlapping) clock periods. That's actually an advantage as large parts of a CPU are cache nowadays. There's a certain logic in running the cache at half a period offset from the CPU clock. $\endgroup$ – MSalters Aug 6 '14 at 11:03
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    $\begingroup$ @PhilFrost Consider not just one message, but a dialogue, in which each message is shorter than the bandwidth-delay product: this is a good model for random access to main memory, for instance, when the CPU has to wait for each read before it can issue the next one. Shorter cables really will make that go faster overall, as would "ideal wires". $\endgroup$ – zwol Aug 7 '14 at 12:45

As you've probably guessed the speed of light isn't the limitation. Photons in a vacuum travel at the speed of light ($c_o$). Photons in anything else travel slower, like in your cable ($0.64c_o$). The amount the speed is reduced by depends on the material by the permittivity.

Information itself is slower still. One photon doesn't carry much information. Information is typically encoded in the change of states of the energy. And these changes of states can only be propagated at lower rates than the fundamental transmission speed.

Detecting both the energy and the rates of change require physical materials to convert the photons into something more usable. This is because the channel used for transmission usually conducts energy at a maximum rate called bandwidth. The bandwidth of the channel is the first limit in network speeds. Fiber optics can transmit signals with high bandwidths with less loss than copper wires.

Secondly the encoded signals have a lot of overhead. There is a lot of extra data transmitted with error correction, routing information, encryption and other protocol data in addition to the raw data. This overhead also slows down data throughput.


Lastly the amount of traffic on a network can slow down the overall system speed as data gets dropped, collisions occur and data has to be resent.


EDIT: I see you've changed your question some....

In other words I'm asking, what is it about a fiber-optics cable that makes it faster than an Ethernet cable?

Fiber optics has the ability to conduct higher energy charges. Photons with higher energies, by definition are at higher frequencies.

$E_{photon}=hf$ where $h$ is the plank constant (h=6.63*10^-34 J.s) and $f$ is the frequency of the photon.

Why does frequency matter? Because of how communication systems work. Typically we setup a strong signal oscillating at the most efficient frequency for the transmission channel to conduct it. If the frequency is too low and we lose our signal's power and likewise too high and we lose power. This is due to how the medium responds to different levels of charge energy. So there's a $F_{max}$ and a $F_{min}$.

Then we add information to the oscillation by changing it at some rate. There are a many ways to add information but in general the amount of information you can add is proportional to the rate the channel can respond to or bandwidth of the system. Basically you have to stay in between $F_{max}$ and $F_{min}$.

It just so happens that the higher the operating frequency the easier it is to get wider and wider bandwidths. For example a radio at 1GHz with 10% channel width only allows for 100MHz max switching rates. But a fiber optic signal at 500THz a 10% channel width means a 50THz max switching rate. Big difference!

You might be wondering why channels have frequency limits and why 10%. I just picked 10% as a typical example. But transmission channels of all types have limits to what kind of energy levels they absorb, reflect, and conduct. For a rough example x-rays which are high frequency or high energy charges, they go right though a lot of materials, whereas heat which is a frequency lower than optical light doesn't transmit well through paper but it can through glass. So there are frequencies where photons can be used to carry energy and frequencies where they can't.

Yes they do all travel at $c_o$ in free space and slower in other media, but they can't carry information at that same rate or higher. You might be interested to read Shannon-Hartley's Theorem.

  • $\begingroup$ "This is because the channel used for transmission usually conducts energy at a maximum rate called bandwidth." could you explain what determines this maximum rate? First of all, the transfer of electrons is an example of the flow of energy, right? If yes, is the fact that different materials let electrons travel through them at different speeds what causes the maximum rate? $\endgroup$ – Celeritas Aug 6 '14 at 0:10
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    $\begingroup$ @Celeritas Not really. Electron movement is different than charge movement (see physics.stackexchange.com/questions/17741/…). Not every material responds equally to every frequency or energy level of charge. So the channel conducting the photon energy has physical efficiency limits based on frequency (or energy level) see en.wikipedia.org/wiki/Photon. For example fiber optics can efficiently conduct high energy photons (or high frequency) which allow for faster/bigger bandwidths. $\endgroup$ – user6972 Aug 6 '14 at 0:40
  • $\begingroup$ "higher energy charges"? $\endgroup$ – rob Aug 6 '14 at 2:57
  • $\begingroup$ @rob a photon with more energy $\endgroup$ – user6972 Aug 6 '14 at 4:33
  • $\begingroup$ The latency is also helped significantly in very long cables, because fiber optic cables need fewer repeaters - this is nicely seen in the trans-atlantic cable, which is devoid of any other infrastructure - just the cable and repeaters. $\endgroup$ – Luaan Aug 6 '14 at 10:41

Since cables carry electricity moving at the speed of light, why aren't computer networks much faster?

Perhaps I can address your confusion with a rhetorical question:

Since air carries sound moving at the speed of sound, why can't I talk to you much faster?

The speed of sound is much slower than light, but at 340 m/s in air, it's still pretty damn fast. However, this isn't the speed of the channel, it is its latency. That is, if you are 340 meters away, you will hear me 1s after I make a sound. That says nothing about how fast I can communicate with you, which is limited by how effectively I can speak, and how well you can hear me.

If we are in a quiet room, I can probably speak very quickly and you can still hear me. If we are far apart or the environment is noisy, I will have to speak more slowly and clearly.

With electrical communications the situation is much the same. The speed limit is not due to the latency, but rather how fast one end can transmit with the other end still being able to reliably receive. This is limited by noise picked up from the environment and distortions introduced by the cable.

As it turns out, especially for long distances, it is easier (and more economical) to manufacture a fiber optic cable that does not permit outside interference and introduces very little distortion, and that is why fiber optic cables are preferred for long distance, high speed networking.

The reasons for optical fiber's superior properties are many, but a significant development is single-mode fiber. These are fibers which, through carefully controlled geometry and research clever enough to earn a Nobel prize, support electromagnetic propagation in just one mode. This significantly reduces modal dispersion, which has the undesirable effect of "smearing" or "spreading" pulses which encode information. This is a kind of distortion that if excessive, renders the received signal unintelligible, thus limiting the maximum rate at which information can be transmitted.

A further advantage is that fiber optic communications operate at an extremely high frequency, which reduces chromatic dispersion, a distortion due to different frequencies propagating at different speeds. Typical wavelengths used in fiber are in the neighborhood of 1550 nm, or a frequency of around 193000 GHz. By comparison, category 6a cable is specified only up to 0.5 GHz. Now, in order to transmit information we must modulate some aspect of the signal. A very simple modulation would be turning the transmitter on and off. However, these transitions mean the signal can not consist of just one frequency of light (Fourier components), so the different frequency components of the pulse will be subject to chromatic dispersion. As we increase the carrier frequency but hold the bitrate the same, the fractional bandwidth decreases. That is, the transitions from the modulation become slower relative to the carrier frequency. Thus, chromatic dispersion is decreased, since the signal becomes more like just one frequency of light.

Modern single-mode fiber is so good that the information rate is usually limited by our technology to manufacture the receivers and transmitters at the ends, not by the cable. As an example, wavelength-division multiplexing was developed (and is constantly improved even today) to allow multiple channels to coexist on the same fiber. Several times, networks have been upgraded by upgrading the transceivers at the ends, leaving the cable unchanged. Considering the cost of upgrading a transcontinental cable, the economic advantage should be obvious.

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    $\begingroup$ This is a very pertinent answer. Now just add dispersion (limiting frequency of transmission) and a discussion of mono mode fiber to really be the best answer to the (modified) question being asked. $\endgroup$ – Floris Aug 6 '14 at 13:01
  • $\begingroup$ @Floris I've done my best to incorporate those concepts, though I must admit my understanding of the underlying physics in this area is a subset of what's on Wikipedia. Please let me know if you spot any errors. $\endgroup$ – Phil Frost Aug 6 '14 at 14:10
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    $\begingroup$ It's not a tech analogy without a car in it somewhere... It's not how fast the car can go at it's maximum. It's about how long it takes to get from A to B on the freeway... vs freeway at rush hours... vs the dirt road with a one lane bridge under construction. Fiber is a 6 lane super-highway. Copper is a 2 lane road with traffic lights. $\endgroup$ – WernerCD Aug 6 '14 at 19:58
  • $\begingroup$ Dispersion: different frequencies travel at slightly different velocities. Since any signal (think square wave) is really made up of many different frequencies (Fourier components), these different speeds mean that the waveform distorts, then becomes impossible to decode. You have to limit the carrier frequency (bit rate) to permit long distance transmission. In mono mode fiber the frequency of interest is the (monochromatic) wavelength of light. Thus no dispersion, thus (practically) fiber does not limit data rate (but the transceiver does). $\endgroup$ – Floris Aug 6 '14 at 20:39
  • $\begingroup$ @Floris can truly monochromatic light transmit information, though? If it's modulated at all, then there are Fourier components, thus different frequencies, thus some dispersion, right? My understanding of single-mode fiber is that it addresses just one kind of dispersion: that from multi-mode propagation. Which is a very significant limit, but not the only limit, to the maximum speed possible. $\endgroup$ – Phil Frost Aug 6 '14 at 20:44

A transmission line is made of a pair of conductors which have some resistance, inductance, capacitance, and leakage conductance. We can take all of these per unit length:

a cable

The wave equation for signals in this line, in the limit of a lossless cable with $R=0$, $G=0$, is $$ \frac{\partial^2 V(x)}{\partial x^2} + \omega^2 LC \cdot V(x) = 0 $$ You have to be a little careful with notation and dimensions here. In a typical circuit you use $L$ and $C$ for total inductance and capacitance, and $\sqrt{LC}$ is the characteristic frequency of the oscillator. Here, $L$ and $C$ are the inductance and capacitance per unit length, and so $1/\sqrt{LC}$ has units of speed.

In fact the derivation at wikipedia goes on to show that, in the limit of a lossless cable, the output is $$ V_\text{out}(x,t) \approx V_\text{in}(t-\sqrt{LC}x) $$ which is consistent with signals traveling down the cable with speed $v=1/\sqrt{LC}$.

Clearly the inductance and capacitance per unit length for a cable depends mostly on their geometry, and somewhat on the magnetic and dielectric properties of the space around and between the cables. It'd be interesting to come up with values of $L$ and $C$ that give $v=c$, or $v>c$; I haven't done this myself, but I suspect that geometric considerations alone will make this impossible for vacuum-separated parallel wires, coaxial wires, and other common geometries without introducing some magical meta-material.

  • $\begingroup$ You don't have to use traditional copper wire. Your transmission line could be radiation in free space, or laser or a fiber optic cable too and in special cases where $E_r$ is close to 1. $\endgroup$ – user6972 Aug 6 '14 at 1:02
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    $\begingroup$ Fiber optics don't work unless then have index of refraction large enough to maintain total internal reflections; typically $n\approx1.3–1.5$, corresponding to $v/c\approx0.6–0.8$. Of course you can use a waveguide for microwave frequencies, but I don't know that it makes sense to apply the $L,C$ formalism there. $\endgroup$ – rob Aug 6 '14 at 2:55
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    $\begingroup$ @rob waveguides have a characteristic impedance just like coaxial and twin-lead transmission lines. The same abstraction of inductance and capacitance per unit length still applies. $\endgroup$ – Phil Frost Aug 6 '14 at 12:08
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    $\begingroup$ @rob if you already have $E$ and $H$, then that's simpler for other transmission lines also. Usually the calculation of characteristic impedance from inductance and capacitance comes into play when the question is "what geometry do I need to manufacture a transmission line of [some given impedance]". Well-known engineering formulas can give impedance and capacitance per unit length for most common geometries, from which characteristic impedance can be calculated. $\endgroup$ – Phil Frost Aug 6 '14 at 12:40
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    $\begingroup$ It may be worth noting that once series resistance enters in, the useful speed of a cable, in the absence of repeaters, has a significant delay term which is proportional to the square of the length. It may also be worth noting that while it's possible to use repeaters that pass signals along as soon as they arrive, it's often better to have repeaters measure the timing of incoming signals, figure out what they should be, and send out new signals which precisely match what the received signals should have been. For example, if all pulses should be a multiple of 100ns, then... $\endgroup$ – supercat Aug 8 '14 at 15:29

"Surely this is a bottleneck" - No, it's really not. Any real-life network connection is not speed-limited by the propagation speed of the signal in the cable, but by the processing delays in the various routers, switches, and network interface processing at each end.

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    $\begingroup$ According to this paper "he two major variables that have the most effect on network latency are Distance Delay and Queue Delay" so distance delay is attributed to the speed of the cables. serviceassurancedaily.com/2008/06/latency-and-jitter $\endgroup$ – Celeritas Aug 6 '14 at 0:13
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    $\begingroup$ @Celeritas The figures quoted in your link seem to bear me out. Copper 5us/km = 40,000km to give the problematic delay of 200ms quoted. To put it another way, if you ping a remote IP address, the delay you see in the results is overwhelmingly due to processing in the routers etc en route. $\endgroup$ – peterG Aug 6 '14 at 0:43
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    $\begingroup$ @peterG it depends on how far are you pinging. If you are pinging your router on the desktop maybe, if you are pinging a server in India from USA cable here it is your 40,000km (you pay twice because your packet has to go and come back). $\endgroup$ – pqnet Aug 6 '14 at 19:35
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    $\begingroup$ @peterG my point is pretty straightforward: signal propagation is a significative component of the delay if you consider world wide networks. Right now from my computer I can ping the other side of the planet with about 300 ms roundtrip. More than half of that is due to signal propagation speed being finite, so I guess that I can say that in my case the delay is mostly determined by signal propagation. $\endgroup$ – pqnet Aug 6 '14 at 22:02
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    $\begingroup$ Don't know about 'mostly', and the OP was using the term 'bottleneck' which it certainly isn't; but I have to concede, having made a few more tests, the current internet is getting within a respectable distance of the limit set by the speed of light. My own measurements: Manchester (UK) to Magnitogorsk (RU) 3939km at c = 3.3us/km = 26ms round trip; ping is only 103ms, so it's closer than I expected! The internet's got faster since the last time I tried this! $\endgroup$ – peterG Aug 6 '14 at 23:14

How sure are you that electricity travels at the speed of light? Although electricity propagation moves at the speed of an E/M wave, and not electrons, its speed depends on the dielectric constant of the material. Only in a vacuum, I think, would it travel at the speed of light.

  • $\begingroup$ What is meant by "electricity"? Do we mean electrical energy? Electrons? Electric charge? The electric field? Changes in that electric field? $\endgroup$ – Phil Frost Aug 6 '14 at 14:54
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    $\begingroup$ I would say that light travels at the speed of light by definition. What you would like to say maybe is that c is the value for speed of light in the vacuum, while in other medium speed of light is lower $\endgroup$ – pqnet Aug 6 '14 at 22:05

Why only 64% What does propagation speed mean? I know there are other variables effecting the latency and perceived speed of computer network connections, but surely this is a bottle neck.

Speed of signal propagation is distance the signal (packet) travels in one second. It is usually lower than $c$ because EM waves that carry the information travel in metal or any material medium with lower velocity than $c$. See group velocity and theory of dispersion.

The speed of the signal determines minimum latency, but increasing it by changing medium or using different frequency band would have little effect on the maximum data rate of the transmission line (bits/s transferred). This is determined more by the electric power used for the transfer, intensity of noise and bandwidth used as well as capabilities of the electronics on the ends. Optical cables are used not for their faster signal propagation speed, but for their other benefits, such as much broader usable bandwidth.


Two reasons:

1) The speed of light in a "medium" is (almost*) always slower than the speed of light in a vacuum.

2) Electricity propagating in a wire is subject to inductive and capacitive effects which slow it's progress.

And even if wires were infinitely fast, integrated circuits are not. Again, inductive (a little) and capacitive (a lot) effects limit how rapidly an IC "gate" can "switch".

Interesting bit of trivia: The wires connecting points of the "backplane" of the (ca 1976) Cray 1 "supercomputer" were all the same length, whether the wire spanned an inch or 30 inches. This assured they had the same propagation delay.

(*) I vaguely recall that the lab guys have created weird scenarios where light propagates through certain special media "faster than light".

  • $\begingroup$ RE: *, no they didn't there was a mistake in the experiment when they thought they did. $\endgroup$ – Celeritas Oct 17 '14 at 8:10

The speed of electrons that flows in the cable, i.e. the current, is only a few m/s. The EM wave propagates much faster. Anyway, the speed of a computer no depends intrinsically of the speed of electrons, but the speed of energy transfers between electronics components.

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    $\begingroup$ Indeed, you can transmit information using an amplitude- or frequency-modulated AC signal, where the average charge drift velocity is zero. $\endgroup$ – rob Aug 5 '14 at 22:38
  • $\begingroup$ It's slower than that -- a few millimeters per hour (if Wikipedia is to be believed) $\endgroup$ – Bryan Boettcher Aug 5 '14 at 22:54
  • $\begingroup$ According to this answer:physics.stackexchange.com/a/13568/21817, they go very fast. $\endgroup$ – jinawee Aug 6 '14 at 9:30

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