# Is there a physical limit to data transfer rate?

Is there a physical limit to data transfer rate (e.g. for USB $3.0$, this rate can be a few Gbit per second)? I am wondering if there is a physical law giving a fundamental limit to data transfer rate, similar to how the second law of thermodynamics tells us perpetual motion cannot happen and relativity tells us going faster than light is impossible.

• Well the speed of light is the maximum anything can go so I would say that.
– Naz
Apr 30 '18 at 12:26
• This is not a meaningful question, because it brings exotic answers like a pipe with black holes moving near the speed of light. A meaningful question would be, for example, the maximum transfer rate at a given transmission power or some other reasonable physical limitation that would relate to practical applications. Apr 30 '18 at 14:29
• @Naz that's how fast you can receive the first bit, has nothing to do with bandwidth :) Apr 30 '18 at 19:38
• This is more of a practical concern than a theoretical one, but Charlie Stross once quipped that "once you get up into X-ray frequencies your network card becomes indistinguishable from a death ray."
– zwol
Apr 30 '18 at 23:19
• @safesphere: Somebody once told me that the data rate achieved by a pipe containing black holes moving at the speed of light is equal to the data rate you get for signal-to-noise ratio when you plug Planck-temperature thermal radiation for signal, and vacuum fluctuations for noise, into Shannon's formula. So your speculation indeed seems to be the correct answer to the question, and the maximum rate can be calculated exactly—not that it's anywhere close to achievable. May 1 '18 at 15:44

tl;dr- The maximum data rate you're looking for would be called the maximum entropy flux. Realistically speaking, we don't know nearly enough about physics yet to meaningfully predict such a thing.

But since it's fun to talk about a data transfer cord that's basically a $1\mathrm{mm}$-tube containing a stream of black holes being fired near the speed of light, the below answer shows an estimate of $1.3{\cdot}{10}^{75}\frac{\mathrm{bit}}{\mathrm{s}}$, which is about $6.5{\cdot}{10}^{64}$ faster than the current upper specification for USB, $20\frac{\mathrm{Gbit}}{\mathrm{s}}=2{\cdot}{10}^{10}\frac{\mathrm{bit}}{\mathrm{s}}$.

# Intro

You're basically looking for an upper bound on entropy flux:

• entropy: the number of potential states which could, in theory, codify information;

• flux: rate at which something moves through a given area.

So,$$\left[\text{entropy flux}\right]~=~\frac{\left[\text{information}\right]}{\left[\text{area}\right]{\times}\left[\text{time}\right]}\,.$$ Note: If you search for this some more, watch out for "maximum entropy thermodynamics"; "maximum" means something else in that context.

In principle, we can't put an upper bound on stuff like entropy flux because we can't claim to know how physics really works. But, we can speculate at the limits allowed by our current models.

# Speculative physical limitations

Wikipedia has a partial list of computational limits that might be estimated given our current models.

In this case, we can consider the limit on maximum data density, e.g. as discussed in this answer. Then, naively, let's assume that we basically have a pipeline shipping data at maximum density arbitrarily close to the speed of light.

The maximum data density was limited by the Bekenstein bound:

In physics, the Bekenstein bound is an upper limit on the entropy $S$, or information $I$, that can be contained within a given finite region of space which has a finite amount of energy—or conversely, the maximum amount of information required to perfectly describe a given physical system down to the quantum level.

"Bekenstein bound", Wikipedia [references omitted]

Wikipedia lists it has allowing up to$$I ~ \leq ~ {\frac {2\pi cRm}{\hbar \ln 2}} ~ \approx ~ 2.5769082\times {10}^{43}mR \,,$$where $R$ is the radius of the system containing the information and $m$ is the mass.

Then for a black hole, apparently this reduces to$$I ~ \leq ~ \frac{A_{\text{horizon}}}{4\ln{\left(2\right)}\,{{\ell}_{\text{Planck}}^2}} \,,$$where

• ${\ell}_{\text{Planck}}$ is the Planck length;

• $A_{\text{horizon}}$ is the area of the black hole's event horizon.

This is inconvenient, because we wanted to calculate $\left[\text{entropy flux}\right]$ in terms of how fast information could be passed through something like a wire or pipe, i.e. in terms of $\frac{\left[\text{information}\right]}{\left[\text{area}\right]{\times}\left[\text{time}\right]}.$ But, the units here are messed up because this line of reasoning leads to the holographic principle which basically asserts that we can't look at maximum information of space in terms of per-unit-of-volume, but rather per-unit-of-area.

So, instead of having a continuous stream of information, let's go with a stream of discrete black holes inside of a data pipe of radius $r_{\text{pipe}}$. The black holes' event horizons have the same radius as the pipe, and they travel at $v_{\text{pipe}} \, {\approx} \, c$ back-to-back.

So, information flux might be bound by$$\frac{\mathrm{d}I}{\mathrm{d}t} ~ \leq ~ \frac{A_{\text{horizon}}}{4\ln{\left(2\right)}\,{{\ell}_{\text{Planck}}^2}} {\times} \frac{v_{\text{pipe}}}{2r_{\text{horizon}}} ~{\approx}~ \frac{\pi \, c }{2\ln{\left(2\right)}\,{\ell}_{\text{Planck}}^2} r_{\text{pipe}} \,,$$where the observation that $\frac{\mathrm{d}I}{\mathrm{d}t}~{\propto}~r_{\text{pipe}}$ is basically what the holographic principle refers to.

Relatively thick wires are about $1\,\mathrm{mm}$ in diameter, so let's go with $r_{\text{pipe}}=5{\cdot}{10}^{-4}\mathrm{m}$ to mirror that to estimate (WolframAlpha):$$\frac{\mathrm{d}I}{\mathrm{d}t} ~ \lesssim ~ 1.3{\cdot}{10}^{75}\frac{\mathrm{bit}}{\mathrm{s}} \,.$$

Wikipedia claims that the maximum USB bitrate is currently $20\frac{\mathrm{Gbit}}{\mathrm{s}}=2{\cdot}{10}^{10}\frac{\mathrm{bit}}{\mathrm{s}}$, so this'd be about $6.5{\cdot}{10}^{64}$ times faster than USB's maximum rate.

However, to be very clear, the above was a quick back-of-the-envelope calculation based on the Bekenstein bound and a hypothetical tube that fires black holes near the speed of light back-to-back; it's not a fundamental limitation to regard too seriously yet.

• @JohnDvorak Probably about 280 times the mass of Earth before factoring in that they're all moving arbitrarily close to the speed of light; but, gotta make some allowances to have the latest tech, right?
– Nat
Apr 30 '18 at 12:55
• Now we just need an SI prefix so we can properly advertise cables instead of saying the transfer rate is 1.3 ridicubits Apr 30 '18 at 14:15
• This is a great derivation, but my semitrailer full of flash drives can beat the pants off your average data rate. (and if it doesn't I'll just add a second trailer section) Apr 30 '18 at 15:06
• @rob Yeah, technically I was referring to the limit at which storage media would collapse into a black hole if any more dense. Wikipedia included "It happens that the Bekenstein-Hawking Boundary Entropy of three-dimensional black holes exactly saturates the bound [...]" -"Bekenstein bound", so I used the black hole equation as the limit, as seemed to be the heuristic argument others had put forth. Was it misapplied here?
– Nat
Apr 30 '18 at 18:53
• @DavidStarkey Given that we're using quantum phenomena, wouldn't that be ridiqubits instead?
– user
May 1 '18 at 15:35

The Shannon-Hartley theorem tells you what the maximum data rate of a communications channel is, given the bandwidth.

$$C = B \log_2\left(1+\frac{S}{N}\right)$$

Where $C$ is the data rate in bits per second, $S$ is the signal power and $N$ is the noise power.

Pure thermal noise power in a given bandwidth at temperature $T$ is given by:

$$N = k_BTB$$

So for example, if we take the bandwidth of WiFi (40MHz) at room temperature (298K) using 1W the theoretical maximum data rate for a single channel is:

$$40 \times 10^6 \times \log_2\left(1 + \frac{1}{1.38\times 10^{-23} \times 298 \times 40 \times 10^6}\right) = 1.7 \times 10^9 = 1.7 \mathrm{\;Gbs^{-1}}$$

In a practical system, the bandwidth is limited by the cable or antenna and the speed of the electronics at each end. Cables tend to filter out high frequencies, which limits the bandwidth. Antennas will normally only work efficiently across a narrow bandwidth. There will be significantly larger sources of noise from the electronics, and interference from other electronic devices which increases $N$. Signal power is limited by the desire to save power and to prevent causing interference to other devices, and is also affected by the loss from the transmitter to the receiver.

A system like USB uses simple on-off electronic signal operating at one frequency, because that's easy to detect and process. This does not fill the bandwidth of the cable, so USB is operating a long way from the Shannon-Hartley limit (The limiting factors are more to do with the transceivers, i.e. semiconductors). On the other hand, 4G (and soon 5G) mobile phone technology does fill its bandwidth efficiently, because everyone has to share the airwaves and they want to pack as many people in as possible, and those systems are rapidly approaching the limit.

• Comments are not for extended discussion; this conversation has been moved to chat.
– rob
May 2 '18 at 3:26

No, there is no fundamental limit on overall transfer rate. Any process that can transfer data at a given rate can be done twice in parallel to transfer data at twice that given rate.

• Tell that to my DSL provider! :D
– AnoE
Apr 30 '18 at 13:01
• Could the downvoters explain their votes? This seems a reasonable answer to me, but I could be missing something. Thanks. Apr 30 '18 at 14:02
• The answer is (a) deliberately not answering the spirit of the question and (b) wrong anyway. Even if you assumed that adding extra cables was an option you would eventually hit a limit due to the physical space they would require and the need to combine the data from each cable. If you assume the end-points are points then only one cable can have a minimum distance and the other cables must take a longer route leading eventually to timing issues. Anyway, who wants a multi-strand cable that is thicker than it is wide? In the end, you must answer the evaded question - "what is the serial rate?" Apr 30 '18 at 14:13
• I really don't see why this answer should be downvoted. Sure it's incomplete, but it brings a legitimate and interesting point. Apr 30 '18 at 14:22
• @rghome: No, if you want to suggest improvements to the question or request clarification, you add a comment. If you just want to make a point, then you invite the other party to a chat room. :) This is an answer, albeit a wrong one for the reason you have laid out (already a +1 from me on that :P) Apr 30 '18 at 14:44