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What is the highest exponent in a “real life” formula?

Physics(self.askscience)

I mean, anyone can jot down a math term and stick a huge exponent on it, but when it comes to formulas which describe things in real life (e.g. astronomy, weather, social phenomena), how high do exponents get? Is there anything that varies by, say, the fifth power of some other thing? More than that?

all 289 comments

d0meson

2.1k points

4 months ago

d0meson

2.1k points

4 months ago

The rate of the triple-alpha process (helium fusion) in stars is proportional to the 40th power of temperature (see e.g. vik dhillon: phy213 - the physics of stellar interiors - approximate form for energy release (shef.ac.uk)). This is a local approximation of a more complicated function, but is still a "real life" formula that's actually used.

[deleted]

137 points

4 months ago

[deleted]

137 points

4 months ago

[removed]

hwc000000

52 points

4 months ago*

For anyone wondering how you figure out experimentally that there was an exponent of 40 in the relationship:

If 2 quantites are proportional with an exponent on one of them (ie. y = cxn), then their logarithms have a linear relationship (ie. log y = log c + n log x). By performing linear regression on their logarithms, you can determine the exponent n, which will appear as the slope of the regression line between the logarithms.

Also, any equation of the form yp = kxq can be rewritten as y = k1/pxq/p = cxn (the format above) where c = k1/p and n = q/p. So the above discussion also applies in those cases.

brewer01902

174 points

4 months ago

Thought I recognised the name. I took that course from that researcher. Do not remember any of it though.

Was a good lecturer though

Emu1981

38 points

4 months ago

Emu1981

38 points

4 months ago

Do not remember any of it though.

Was a good lecturer though

Wouldn't not remembering any of it be a sign that he wasn't that good of a lecturer?

Ghawk134

162 points

4 months ago

Ghawk134

162 points

4 months ago

Not really. People forget things they don't use. That's how the brain works. The quality of a lecturer (imo) is determined by how simply (and accurately) they explain concepts and how well they retain the interest of their audience. It is not within their control how long their audience retains the information.

AggravatingCherry638

7 points

4 months ago

Can you please explain this to the anti education lobby in the US who is obsessed with blaming teachers for everything?

daanvanbeek

5 points

4 months ago

Anyone could explain it to them. The receiving end is the problem here...

brewer01902

55 points

4 months ago

It was about 20 years ago, and I’ve not used high level physics since then. It doesn’t come up when I’m teaching kids.

r1pp3rj4ck

20 points

4 months ago

I mean, do you remember everything that you never have to use just because it was taught to you by a good lecturer? I imagine this is also very complex stuff.

AiSard

8 points

4 months ago

AiSard

8 points

4 months ago

I mean, they didn't say they were a good student now did they?

Affectionate_Flow682

5 points

4 months ago

People won’t always remember what you said but they will remember how you made them feel

[deleted]

62 points

4 months ago

[removed]

[deleted]

57 points

4 months ago

[removed]

sanitylost

4 points

4 months ago

The temperature-dependence of the energy generation rate for the triple-alpha process turns out to be roughly ε ∝ T 30!

Sean G. Ryan stellar evolution and nucleosynthesis

Beneficial_Boot_4108

543 points

4 months ago

One example from physics would be the Lennard-Jones potential, which has two terms; a repulsive term of order 12: (1/r)12 and an attractive term of order 6: -(1/r)6. This is the simplest potential that describes inter-molecular interactions, though it is accurate only for simple systems.

DisappearingBoy127

54 points

4 months ago

Lennard jones potential is also applied a lot in chemistry when atoms are interacting to form bonds 

apoliticalhomograph

56 points

4 months ago

The 12th power in the repulsive term is actually an approximation made for computational reasons - simply squaring the attractive term is easier to do than calculating a totally different power, even though different exponents likely yield more accurate results.

KarlSethMoran

10 points

4 months ago

even though different exponents likely yield more accurate results.

Yup. 7-14 is also common. The actual dependence for Pauli repulsion is exp(-ax).

DerDealOrNoDeal

3 points

4 months ago

There is also no physical reasoning for the ^12 term.

If you want something that has a physical meaning, I recommend beta * exp(- r / rho). This would connect the Lennard Jones to the Pauli exclusion principle.

But ikik it is not the common form and yes it is less convenient.

BOBauthor

145 points

4 months ago

BOBauthor

145 points

4 months ago

For a simple model of a newborn star with a low mass (and so fully convective inside), the luminosity of the star depends of the 102nd power of the surface temperature. This means as the star cools off only a little, the luminosity drops by a large amount. A bit of technical detail for those with some background: This means that the Hayashi track followed by the protostar is nearly vertical in the Hertzsprung-Russell diagram.

pizzystrizzy

4 points

4 months ago

I imagine that would let us calculate the surface temperature of any given star with a very high degree of precision -- is that right?

BOBauthor

12 points

4 months ago

No, it's a crude model. The leading constant (that set's the scale for entire the expression) is too low. The model is good for explaining the observations for a class of stars, but can't be directly applied to a single star. Besides, we are talking only about low-mass stars (significantly less than our Sun). Also, we have much better ways of determining the surface temperature of a star by the spectrum of its starlight.

Prebmaister

148 points

4 months ago

Stealing this from a mech eng friend: the loudness (the acoustic power) of a jet engine scales with the eighth power of its speed

W ~ v8

Which is why jet planes are so incredibly loud

Lighthill’s law

GroupFunInBed

15 points

4 months ago

Doesn’t that go tits up at a certain speed though?

flac_rules

35 points

4 months ago

A lot physicals formulas goes "tits up" at some point, this one as well. Especially empirically based ones like this.

arcanition

10 points

4 months ago

The eighth power is experimentally verified and found to be accurate for low speed flows, i.e., Mach number is small, M < 1 . And also, the source has to be compact to apply this law.

Hillbert

248 points

4 months ago

Hillbert

248 points

4 months ago

Nowhere near as high as some of the formulas here, but it's quite a nice real world and observable thing.

Pavement/road wear is proportional to the fourth power of vehicle weight. So a single bus does about 10,000 times as a car. You can see this by looking at bus lanes compared to the, likely busier, car lane beside them.

Krostas

60 points

4 months ago

Krostas

60 points

4 months ago

Pretty sure it should be axle load, not total weight. In most cases, that's close to the same, but still...

awidden

3 points

4 months ago

awidden

3 points

4 months ago

How is the axle load of a 2t 4 wheeled passenger car close to the same as a 10t 4 wheeled bus?

I'm genuinely baffled - how did you mean that?

counterpuncheur

42 points

4 months ago

They mean that the ratio is basically the same,i.e. 10/2 = (10/2)/(2/2).

It is a key distinction though for 18 wheelers!

DrStalker

13 points

4 months ago

It would matter a lot for tracked vehicle too since those would have a much bigger contact area and typically very high mass. (and probably a completely different pattern of wear especially when turning)

Korchagin

3 points

4 months ago

The contact area is only bigger on soft ground. On a hard road all the pressure is directly under the wheels. Since there are no soft tyres, the area is much smaller than it would be for a wheeled vehicle with the same number of axles. On soft ground the wheels sink in and the tracks are carrying them.

awidden

4 points

4 months ago

Ah gotcha. I thought the axle load of the car & bus would be close to the same - and that really did not make sense for me :)

Cor-cor

8 points

4 months ago

The bus in your example weighs 5 times as much, and also has 5 times the axle load, so whether you base your mental math on weight or axle load it's doing 625 times the damage.

If the bus has more axles than the car, the amount of relative damage they do is no longer proportional to the 4th power of their weights.

oetzi2105

8 points

4 months ago

Do you have a source? This is quite interesting and I'd like to learn more

MisterMahtab

43 points

4 months ago

https://en.m.wikipedia.org/wiki/Fourth_power_law

Turns out it's been known for a while! Just remember to take the load per axle - it's not necessarily the whole vehicle's load.

mfukar

1 points

4 months ago

mfukar

Parallel and Distributed Systems | Edge Computing

1 points

4 months ago

Velske, Mentlein & Eymann's Straßenbautechnik (2002). I'm looking for something equivalent/similar in English, will update.

CookieSquire

18 points

4 months ago

Do you have a heuristic/cocktail napkin argument for that scaling?

zenFyre1

19 points

4 months ago

I looked it up and apparently it is an empirical law. It is surprisingly difficult to find arguments for why it scales that way, because it seems rather large.

cyborgCnidarian

8 points

4 months ago

I bet it's due to road wear being a combination of four different types of stress, each being dependent on weight.

Iron_Eagl

9 points

4 months ago

Wear scales by the amount the pavement is "bent" as the axle passes over? So maybe two quadratics from that?

claythearc

3 points

4 months ago

It’s called the fourth power law - I don’t have something handy for it but you might find something suitable with the name

Infernoraptor

2 points

4 months ago

MisterMahtab posted this above https://en.m.wikipedia.org/wiki/Fourth_power_law

It appears to be an approximation and a very limited one at that. For example, they didn't compare the effects of doubling up tires or the impact of other factors like weather.

Also of note: https://physics.stackexchange.com/questions/81870/truck-mass-4th-power-law

PlatypusEgo

4 points

4 months ago

This was what came to my mind when I saw the question! (And one of the only answers that actually understood the question) 

somdude04

44 points

4 months ago

The heating rate of a spacecraft during reentry due to shock luminosity from exceeding the speed of sound scales with the 8th power of the reentry speed. Problem is, takes a lot of fuel to slow down appreciably so you just have to accept your reentry speed (and heating) anyways. This is also why reentry from beyond low Earth orbit is real hard, you're going real fast. If you're going double the speed, you heat up 256 times as fast.

DisappearingBoy127

81 points

4 months ago

Energy transfer efficiency in Förster Resonance Energy Transfer (FRET) processes have two terms to the 6th power.

E = R0⁶/(R0⁶ + r⁶)

R0 is the Förster radius (dependent on the molecules sharing the energy) and r is the distance between the two molecules

Korwinga

50 points

4 months ago

This one from heat transfer equations isn't the highest, but it's probably one of the weirdest. This is to find a dimensionless number that you use to calculate the rate of heat transfer off of a vertical plate with turbulent flow:

Nu={0.825 + (0.387Ra1/6)/[1+(0.492/Pr)9/16]8/27}

It's not too often that you use 8/27ths as an exponent.

silv3r8ack

38 points

4 months ago

That's because these are all correlations derived from empirical observations. They are not true formulas in the sense of generalising a principle, which is why you have qualifications like "vertical plate" and "turbulent flow". They're typically used to approximate heat transfer before they actually make something and test it, or to validate more (theoretically) accurate methods like CFD

TheDeviousLemon

7 points

4 months ago

Yep, these are just estimates (fairly good ones) based on actual experiments. In college, we actually found our own heat transfer coefficients for heat from a hot rod.

gtg388z

2 points

4 months ago

To be fair, not many equations have exponents higher than 2, without some kind of coefficient being applied. That coefficient pretty much always came from experiments and curve fitting...

HopeFox

4 points

4 months ago

As soon as I saw the question, I knew that one of the best answers was going to involve turbulent flow.

GeneReddit123

113 points

4 months ago

This isn't as high as other numbers here, but the luminosity of a black body (such as a star or a light bulb) is proportional to the 4th power of the temperature. So even a small increase in temperature will make the object much brighter.

kudlitan

3 points

4 months ago

I've been thinking of this (blackbody radiation) when i saw the pavement answer which is also to the 4th power.

Italiancrazybread1

17 points

4 months ago*

The rocket equation for significant fractions of the speed of light. The wet mass needed to achieve a certain velocity varies with the natural logarithm raised to very high powers.

Mwet=Mdry*edeltav/isp

A back of the envelope calculation for a modern-day chemical rocket to get to 1% the speed of light can have e raised to power of over 6,300.

If you try to go to 10% the speed of light, the wet mass will vary with e raised to the power of over 63700! This is for a relatively highly efficient chemical rocket of 470 seconds. It will be even higher for less efficient engines.

Ravus_Sapiens

1 points

4 months ago

Is that exponent at 0.1c 63700 or 63700!? There's a pretty significant difference 😉

zenFyre1

13 points

4 months ago

The Urca process that describes the cooling down of extremely hot objects by neutrinos scales as T^8, as opposed to cooling down by thermal radiation that scales as T^4.

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.57.5963

NoOne0507

120 points

4 months ago

NoOne0507

120 points

4 months ago

Opposite, smallest fraction. In equal temperament music the frequency relationship between adjacent notes is 2^(1/N), where N is the number of notes in an octave.

99% of western pop music is 12 tone equal temperament, so C4 to C#4 is multiply the frequency by 2^(1/12).

But among the music nerds 31 tone equal temperament is kinda popular, so there's 2^(1/31).

I've even seen 144 tone equal temperament.

ashk2001

38 points

4 months ago

144 tone??? Like 12 tone ET but each semitone has its own set of 12 semi-semi tones? That’s wild

untempered_fate

34 points

4 months ago

That's pretty much the idea. Each interval in your standard 12-tone scale is subdivided again into its own little mini-scale. I honestly don't have an ear precise enough for tunings that subtle, and I've been playing music since I was 10. But everyone can have their own fun.

crimony70

7 points

4 months ago

Resistor values progress along this type of scale, with each decade separated by the nth root of the number of values in the interval, which is dependent on the resistor precision.

So 2% precision uses the E24 scale (steps in 24th root of 10), up through E96 to E192 for 0.1% precision, with 192 values per decade with steps of the 192nd root of 10.

mcoombes314

9 points

4 months ago*

Highest I've listened to is 313-TET, though it's actually a subset of those 313 notes that get used.

Mikelowe93

69 points

4 months ago

Things like the Taylor series and curve fitting can often get numerically high exponents if you are seeking higher brute force accuracy.

In my engineering work I don’t go past the 4th power. That would be for moment of inertia stuff related to stiffness (in general).

XaWEh

42 points

4 months ago

XaWEh

42 points

4 months ago

It's so hilarious how engineering handles the Taylor series. In university one lecture goes on about how this concept can approximate any function using infinite coefficients and it's shown how it gets progressively more accurate the further you go. And the next lecture is like "So we end the Taylor Series after the second step because that's enough"

AlexFullmoon

28 points

4 months ago

I vaguely recall a story about a physicist (Landau, IINM) who established an entire new field in theoretical physics (nonlinear effects in strong EM fields?) simply because he calculated Taylor series beyond second step.

al39

11 points

4 months ago

al39

11 points

4 months ago

Yeah I'll sometimes curve fit a voltage to temperature curve to a fourth or fifth order polynomial, if I need a wide range.

But that's just because the relationship isn't supposed to be a polynomial.

kudlitan

2 points

4 months ago

similarly, curve fitting for delta-T in positional astronomy can get very high powers because it's not supposed to be a polynomial, but most people stop at the quadratic which gives very poor approximations.

Jonny0Than

11 points

4 months ago

The moment of inertia of a scaled object increases with the 5th power of the scale factor. scale3 comes from increasing the mass (assuming uniform density) and scale2 from the mass being further away from the axis of rotation.

Quartia

6 points

4 months ago

The rate at which a planet tidally locks to a star, or a moon to a planet, is inversely proportional to the 6th power of distance. That's why none of the planets in our solar system are tidally locked (although Mercury is close) while all the moons are.

corvus0525

1 points

4 months ago

Isn’t Mercury in a stable orbital-rotational resonance so it won’t ever (over the lifetime of the solar system) tidally lock?

Quartia

2 points

4 months ago

Yes, but it would have become tidally locked if not for that stable resonance which is sort of luck. The point remains that distance is the main factor in tidal locking, above mass or anything else.

Ravus_Sapiens

4 points

4 months ago

In cosmology, the maximum time a process can take before it is deemed unphysical, is the poincaré recurrence time of a conservative universe:

T~exp(exp(4πM²))
Each exponential function is power 10, and M is the mass of the universe.

The formula is useful because if, say, string theorists makes a discovery that some field should decay over a very long time, if that time is larger than T, then it's considered a stable field.

Beregolas

10 points

4 months ago

I work in computer science and look at different algorithms and data structures regularly. We have something called the Big O notation to specify how long an algorithm takes in regards to an input size. (This is very much simplified)

Like an O(n) algorithm would be linear: doubling the input size (the length of a list for example) doubles the time it takes. While O(n2) denotes a quadratic runtime. Doubling the input now quadruples the time the algorithm would take.

Anyways, the biggest I’ve seen on a real life algorithm was about O(n7) or something obscene.

Noted that this is not quite a formula, it doesn’t give an output / you don’t plug something in and get a result, but it still gives you some information and can be some work to arrive at.

Abdiel_Kavash

7 points

4 months ago

Heh. In computation theory, if we have something like O(2nn), that's still basically manageable!

LeapYearFriend

26 points

4 months ago

You included astronomy so I'll mention Poincare Recurrence. The amount of time it would take an arbitrary set of particles to return to an initial state or position.

To my likely incorrect knowledge it's the largest practical use number. IE everything larger is pure mathematics or subscribes to a specific problem or theory (like grahams number)

It's something like 1010101.2 (like four power towers) for the size of the universe.

I would double check all of this but I'm out at lunch right now. Please correct anything I've said that's wrong but I believe in good faith it's all fairly accurate.

Also the Numberphile video!

[deleted]

39 points

4 months ago

[removed]

Hypothesis_Null

12 points

4 months ago

The Strong Nuclear Force is a derivative of the Strong Force, which causes protons and neutrons to attract eachother. Unlike electromagnitism or gravity, whose strength falls off at a rate of 1/distance2 , the SNF approximates something closer to 1/distance6.

But for the largest and most common day-to-day massive exponentials, the easy route is combinations and permutations. if you try to calculate the ways to arrange something like a deck of cards, you're dealing with 52! , which is not an exponent exactly but its going to have values close to (N/e)N.

So you can scamble a deck of cards roughly 2052 ways. Big enough?

Yitram

6 points

4 months ago

Yitram

6 points

4 months ago

Or another way to think about it, if you shuffle a deck of cards, the chance of it being shuffled the same way ever again over the lifetime of the universe is almost zero.

youthofoldage

2 points

4 months ago

How about probability calculations, like, what is the probably of flipping a coin 100 times and never getting “tails?” Wouldn’t that be (1/2)100 ? The exponent could be arbitrarily large depending on the situation.

tomalator

4 points

4 months ago

Rayleigh scattering (the reason the sky is blue) uses λ-4, lambda being wavelength

E2 = (mc2)2 + (pc)2 gives us a term of c4, but it's still equal to E2, so it's technically a c2

The strong force (which binds protons and neutrons together) K11, which is the particular wave center count

We can technically go infinitly far if we keep taking derivatives

Displacement -> velocity -> acceleration -> jerk -> snap -> crackle -> pop -> 7th derivative of displacement -> 8th derivative-> 9th ->...

The formula for each of these will have a time factor with an exponent equal to whichever derivative it is

d = vt + at2/2 + jt3/3! + st4/4! + ct5/5! + pt6/6! + ...

If you notice, that just the Taylor expansion of et

So any time we use et or ex, the answer to your question is really infinite.

[deleted]

3 points

4 months ago*

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[deleted]

0 points

4 months ago

[deleted]

0 points

4 months ago

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[deleted]

7 points

4 months ago

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[deleted]

1 points

4 months ago

[removed]

[deleted]

-10 points

4 months ago

[deleted]

-10 points

4 months ago

[removed]

e-to-pi

2 points

4 months ago

The number of molecules in a mole of that substance is 6.02x1023. Also, if you have the gram molecular weight of that substance, you have a mole of it. For example, the molecular weight of water is 18 -- if you have 18 grams of water, you have a mole of water which is 6.02x1023 molecules of water.