Should pi be replaced with tau in mathematics?

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In summary, some mathematicians are suggesting that instead of using pi, it would be better to use 'tau', which is equal to 2*pi. This is because pi often occurs with a 2 nearby, and using tau would make certain calculations and concepts more intuitive. However, this idea has been met with criticism and it is unlikely that this change will happen due to the already established use of tau in other fields of science and the fact that pi is a constant and not a convention.
  • #36
Any arc length can subtend an angle of 1 radian. Just make the radius equal to the arc length.

If the arc length is e, and the radius is e, then the arc length subtends an angle of 1 radian. Diameter is then 2e, circumference is 2(pi)e, area is (pi)e^2.

This is fun! Insert your favorite constant please.:smile:
 
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  • #37
I just had a thought about how all this developed. In ancient times the diameter of a circle was more important than the radius, since they were measuring land. In modern times the radius is more important, because mathematicians care about the unit circle and its circumference. If aliens had math but not agriculture, they might take tau as being more fundamental than pi.

I hope SETI knows about this. It would be a shame if the aliens are transmitting 6.28... and we're missing it :smile:
 
  • #38
SteveL27 said:
I hope SETI knows about this. It would be a shame if the aliens are transmitting 6.28... and we're missing it :smile:
As long as they transmit in binary we're fine.
 
  • #39
Thanks, guys. That last bit about SETI and binary was hilarious. I've been laughing through the next several several questions.
 
  • #40
I think there's a lot more annoying conventions than pi. I never even thought about pi being annoying until reading this thread. However, when we learned about how current is opposite to the flow of electrons, because it's the flow of imaginary positive charge carriers (or the negative flow of negative charges) I immediately thought that was an irritating convention.

To be honest, I don't think scaling pi by a half will really help much of anything. I'm taking an electrical engineering signals course right now, so I'm doing a lot with periodic signals, which is one place people say that tau would make more sense. I do personally get confused when working with fundamental periods and frequencies and formulas like T=1/f = 1/(2*pi*omega). However, my confusion almost always comes from where the 2pi is supposed to go. That would be just as much of an issue if there was a switch to tau or pi bar or something representing 2 pi. I have no trouble remembering that 2pi is the period of a sinusoid.
 
  • #41
thegreenlaser: see my earlier thread from here:
https://www.physicsforums.com/showthread.php?t=426341

We can cut out not just the "2", but also the "pi" completely by often measuring in revolutions (or cycles if you prefer). Since I'm making my own calculator, I'll be adding the new functions rsin, rcos, rtan etc. to represent this seemingly obvious function.

E.g.:
rsin(1/8) = 0.7071... (1/8th of a turn or circle)
ratan(1÷1) = 1/8... (1/8th of a turn or circle)
 
  • #42
Since I'm making my own calculator

In which case you may be interested in a paper I wrote some years back published in the
Empire Survey Review and entitled

The use of the 5th quadrant.

go well
 
  • #43
Studiot: What was it about? Any link to the paper?
 
  • #44
Twas not really a full blown paper, more of a simple article and appeared in June 1986.

A simple computation that causes much confusion in navigation, surveying and engineering is the calculation of partial coordinates by distance and bearing.

Calculators and computers are programmed to provide angular functions according to the mathematical definition.
That is the defining rotating arm rotates anticlockwise from the horizontal or x axis.

Distance and bearing coordinates are defined by an arm that rotates clockwise from the y or vertical or north axis, resulting in cumbersome 'sign rules' to be applied or up to three computer tests if programmed.

It is possible to remedy both these issues by using the fact that calculators will accept input angles greater than 360 degrees and modifying the traditional formulae to suit.

A suitable input angle may be obtained by noting that as a result of the 90o shift the whole circle bearing (wcb) and the mathematical angle(ma) may be related by the equation

ma + wcb = 450o

This takes us into the fifth quadrant.

In order to cope with the differing directions of rotation the usual projections onto the x and y axes must be inverted so

[itex]\Delta[/itex]Y = Rsin(450-wcb) = Rsin(ma)

[itex]\Delta[/itex]X = Rcos(450-wcb) = Rcos(ma)


These formulae are automatically correct for sign rules, unlike the traditional
ones.

go well
 
  • #45
He makes use of Euler's identity to strengthen his claim but that doesn't make any sense.

It works with pi too.

[tex]{e}^{i\pi}=0-1[/tex]

To change it to the form of Euler's identity, it would be

[tex]{e}^{i\tau}-1=0[/tex]

Tau is also very annoying as writings involving tau would be very confusing. He is not even taking trigonometry and area into consideration.
 
  • #46
dimension10 said:
He makes use of Euler's identity to strengthen his claim but that doesn't make any sense.

It works with pi too.

[tex]{e}^{i\pi}=0-1[/tex]

To change it to the form of Euler's identity, it would be

[tex]{e}^{i\tau}-1=0[/tex]

Tau is also very annoying as writings involving tau would be very confusing. He is not even taking trigonometry and area into consideration.

Not to mention, the latter equation tells us LESS. Try taking the square root of e^{i \tau} and 1. You'll get e^{i \pi} and... ±1. So which one is e^{i \pi}? The latter equation doesn't tell us.
 
  • #47
Char. Limit said:
Not to mention, the latter equation tells us LESS. Try taking the square root of e^{i \tau} and 1. You'll get e^{i \pi} and... ±1. So which one is e^{i \pi}? The latter equation doesn't tell us.

Not too sure if you could say that. Even with pi there is a problem. It doesn't tell us [tex]\exp(i \frac{pi}{2})[/tex]

[tex]\sqrt{\exp(i \pi)}= \pm i [/tex] .

But the correct one is i sin (pi/2)+cos(pi/2)=i+0=i not -i.

So the best is just [tex] \exp(ix)=i \sin x +\cos x [/tex].
 
  • #48
dimension10 said:
Not too sure if you could say that. Even with pi there is a problem. It doesn't tell us [tex]\exp(i \frac{pi}{2})[/tex]

The point is that the equation corresponding to tau can be algebraically derived from the equation correponding to pi. So Char.limit is correct, essentially the equation for tau gives less information, because it cannot derive the equation for pi algebraically.
 
  • #49
disregardthat said:
The point is that the equation corresponding to tau can be algebraically derived from the equation correponding to pi. So Char.limit is correct, essentially the equation for tau gives less information, because it cannot derive the equation for pi algebraically.

And that for pi cannot derive that for pi/2 algebraically so there is not best equation whatsoever.
 
  • #50
dimension10 said:
And that for pi cannot derive that for pi/2 algebraically so there is not best equation whatsoever.

You can't derive Fermat's last theorem either, so what?
 
  • #51
dimension10 said:
And that for pi cannot derive that for pi/2 algebraically so there is not best equation whatsoever.

I don't think you get the meaning of the word "better". "Better" does not mean "best". "Less" does not mean "the least". I never said that e^(i pi) = -1 was the "best" equation. I said it was "better" and gave "more" information than e^(i tau) = 1, something that's undeniably true.
 
  • #52
Char. Limit said:
I don't think you get the meaning of the word "better". "Better" does not mean "best". "Less" does not mean "the least". I never said that e^(i pi) = -1 was the "best" equation. I said it was "better" and gave "more" information than e^(i tau) = 1, something that's undeniably true.

Yup, but there are an infinite number of equations better than that.

[tex]\exp(i \frac{\pi}{2})=i[/tex]

[tex]\exp(i \frac{\pi}{4})=\frac{i+1}{\sqrt{2}}[/tex]

[tex]\exp(i \frac{\pi}{8})=\sqrt{\frac{i+1}{\sqrt{2}}}[/tex]
 
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  • #53
disregardthat said:
You can't derive Fermat's last theorem either, so what?

There does exist a proof for Fermat's last theorem. And here, I am saying that an infinite number of identities you cannot derive using [tex]{\exp}^{i}(\pi)=-1[/tex]
 
  • #54
dimension10 said:
Yup, but there are an infinite number of equations better than that.

[tex]\exp(i \frac{\pi}{2})=i[/tex]

[tex]\exp(i \frac{\pi}{4})=\frac{i+1}{\sqrt{2}}[/tex]

[tex]\exp(i \frac{\pi}{8})=\sqrt{\frac{i+1}{\sqrt{2}}}[/tex]

Okay, now I can see you're not even trying to dispute my argument. You're just making other arguments.
 
  • #56
I have to say I'm for using this; that said I wouldn't stop using pi, I'd use them together. In some cases it would be useful to define a constant that represents the ratio of radius to circumference, rather than diameter to circumference. In practice, it seem to me that radius is more fundamental than diameter: while we measure diameter, we use radius. I personally might start using that convention in my own work, as it keeps the constants from getting in the way. And it isn't like pi goes away, you just write that [itex]\tau[/itex] = 2 [itex]\pi[/itex] on your paper and everyone is on the same page as you. Also, I think 4[itex]\pi[/itex]2 occurs a lot in formulas I've seen, so that just becomes [itex]\tau2[/itex]. Also, if you integrated to get something squared, the constant should be 1/2. I'd rather use circumference = tau r and area = 1/2 tau r2 because I can tie that into my knowledge of integration and remember it better. Also, the existence of tau would make using radians even easier than it is now, because you could convert from a fraction of a circle to radians very intuitively by simply memorizing a few places of tau; using 6.3 is fairly accurate, comparable to 3.14. Trigonometric functions are the same, as the input they take is in radians anyways. This would make trigonometry so radically simple.
 
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