Nobody complains about physicists' math?

[jostpuur]

One might think that you can find anything on the internet, but I haven't found any site where somebody would be complaining about physicists' way of using mathematics. I wonder why. Wouldn't physicists math be an easy thing to make fun of?

 

[Mororvia]

If it describes reality, why would it be made fun of?

 

[zoobyshoe]

Wouldn't physicists math be an easy thing to make fun of?

http://www.aip.org/history/einstein/images/ae17.jpg

"What do you mean, funny? Let me understand this cause, I don't know maybe it's me, I'm a little ****ed up maybe, but I'm funny how? I mean, funny like I'm a clown, I amuse you? I make you laugh... I'm here to ****in' amuse you? What do you mean funny, funny how? How am I funny?"

 

[Jimmy Snyder]

You mean like 'Quantum physics is classical physics in the limit as zero approaches h?'

 

[mgb_phys]

In physics 0/0 or infinity/infinity cancels - it's obvious!

 

[jostpuur]

This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

$$D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)$$

but of course he wouldn't use already proven simple rigorous theorem. Instead he does this with with "differentials". First the lecturer assumes that this is clear

$$d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2$$

and then divides by du, and gets

$$\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}$$

which is the desired result.

Q: What precisely is the dx and du?

A: Well they are some kind of infinitesimal quantities, but there's no need to be rigorous here, because this is physics.

Q: Why was the result derived like that?

A: Well mathematicians would have probably done this more complicatedly with epsilons and deltas, but a simpler proof such as this is sufficient for us, because this is physics.

 

[Gilligan08]

http://www.aip.org/history/einstein/images/ae17.jpg

"What do you mean, funny? Let me understand this cause, I don't know maybe it's me, I'm a little ****ed up maybe, but I'm funny how? I mean, funny like I'm a clown, I amuse you? I make you laugh... I'm here to ****in' amuse you? What do you mean funny, funny how? How am I funny?"

Goodfellas? What movie is that from?

 

[jostpuur]

The conclusion

$$f(x)=1+Ax+O(x^2)\quad\implies\quad f(x)=e^{Ax}$$

is among the most unbelievable ones.

For example when you solve a quadratic equation, you can write the expression

$$x^2 + Ax$$

in a form

$$(x+A/2)^2 - A^2/4.$$

Here you write the old expression in a different new form. Similarly, once a particle physicist is given a function

$$f(x)=1 + Ax+O(x^2),$$

he can write it in a form

$$f(x)=e^{Ax}.$$

 

[ZapperZ]

This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

$$D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)$$

but of course he wouldn't use already proven simple rigorous theorem. Instead he does this with with "differentials". First the lecturer assumes that this is clear

$$d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2$$

and then divides by du, and gets

$$\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}$$

which is the desired result.

Q: What precisely is the dx and du?

A: Well they are some kind of infinitesimal quantities, but there's no need to be rigorous here, because this is physics.

Q: Why was the result derived like that?

A: Well mathematicians would have probably done this more complicatedly with epsilons and deltas, but a simpler proof such as this is sufficient for us, because this is physics.

1. Does this violate any part of mathematics, within the confines of what it is being used for?

2. Does this give a consistently correct description of the system it is describing?

Zz.

 

[jim mcnamara]

I think somebody's cage got rattled, hmm, Zz?

 

[ZapperZ]

Someone is in a cage?

Zz.

 

[Integral]

AFAIK nearly all mathematicians laugh at the way Physicists do math. Of course, not nearly as hard as they laugh at the way engineers do math.
But this is so unsurprising that it is not discussed outside of Math dept coffee rooms.

 

[Mk]

AFAIK nearly all mathematicians laugh at the way Physicists do math.

Hey, physicists are the ones that always have to draw pictures before they can understand a problem

 

[zoobyshoe]

Goodfellas? What movie is that from?

Yeah, That's Newton doing the Joe Pesci line from "Goodfellas".

 

[robphy]

AFAIK nearly all mathematicians laugh at the way Physicists do math.

Maybe that's why some of us are physicists.
Somehow... maybe with luck or maybe with physical intuition... in spite of the physicist's sloppy and impatient mathematics, the physicist often gets the answer correct. (Certainly there are cases when the mathematics is just plain wrong and leads to nonsense... for [almost] everybody.)

I have encountered [overly-]mathematical colleagues who are too wrapped up in the math that they miss the physics under discussion.

I certainly appreciate the mathematical physicist who can use enough mathematics to show the physics [and its abstract structure]... but who is able [when pressed] to deliver the mathematical details.

 

[FredGarvin]

AFAIK nearly all mathematicians laugh at the way Physicists do math. Of course, not nearly as hard as they laugh at the way engineers do math.
But this is so unsurprising that it is not discussed outside of Math dept coffee rooms.

Yeah. And we laugh at mathematicians at the way they try to talk to girls.

 

[Chi Meson]

Yeah, That's Newton doing the Joe Pesci line from "Goodfellas".

Wow, Newton's pretty good. Has he been in any movies himself? Didn't he host SNL way back?

 

[zoobyshoe]

Wow, Newton's pretty good. Has he been in any movies himself? Didn't he host SNL way back?

He used to work the Catskills every summer, and had an off-Broadway one man show one season that Johnathan Demme was interested in translating to film, but the deal fell apart in negotiation due to Newton's temper.

He does a great Brando, too:

http://www.corrosion-doctors.org/Biographies/images/Newton.jpg

"Bonasera. Bonasera. What have I ever done to make you treat me so disrespectfully? If you had come to me in friendship then this scum that ruined your daughter would be suffering this very day. And if by chance a man like yourself should makes enemies then they would become my enemies. And then they would fear you."

 

[G01]

In physics 0/0 or infinity/infinity cancels - it's obvious!

I think I actually canceled infinity over infinity once on a optics test, and got full credit!

 

[NeoDevin]

to find a flux per unit area through an infinite plane we can do

$$\frac{1}{A}\int_S\Phi da$$

Where $S$ is the entire infinite plane, and $A$ is it's area.

 

[Pythagorean]

so is there a mathematical proof for why 0! = 1?

 

[animalcroc]

One might think that you can find anything on the internet, but I haven't found any site where somebody would be complaining about physicists' way of using mathematics. I wonder why. Wouldn't physicists math be an easy thing to make fun of?

I think that's way overrated.
"rigor" in math came after the giants who developed modern math in the first place, and they're "giants".

 

[Dragonfall]

Of course, not nearly as hard as they laugh at the way engineers do math.

I don't laugh at the way they do math. I'm terrified at it. Think about it the next time you're crossing a bridge or ride an elevator.

 

[Thrice]

Ah physics, wherein every function equals the first term of its taylor series.

 

[ice109]

so is there a mathematical proof for why 0! = 1?

http://www.adonald.btinternet.co.uk/Factor/Zero.html

Ah physics, wherein every function equals the first term of its taylor series.

most annoying thing ever

 

[Hurkyl]

This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

$$D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)$$

but of course he wouldn't use already proven simple rigorous theorem. Instead he does this with with "differentials". First the lecturer assumes that this is clear

$$d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2$$

and then divides by du, and gets

$$\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}$$

which is the desired result.

Q: What precisely is the dx and du?

A: Well they are some kind of infinitesimal quantities, but there's no need to be rigorous here, because this is physics.

Q: Why was the result derived like that?

A: Well mathematicians would have probably done this more complicatedly with epsilons and deltas, but a simpler proof such as this is sufficient for us, because this is physics.

If one of my mathematical colleagues wrote that derivation, I would have considered it perfectly rigorous -- it's an elementary manipulation of differential forms, coupled with a typical but harmless¹ ambiguity in notation. Alas, it's a seems that the scientific community has a strong disdain for mathematican sophistication, and you're stuck with rubbish explanations like this.


1: At least, it's harmless if you know what's going on...

 

[Hurkyl]

Similarly, once a particle physicist is given a function

$$f(x)=1 + Ax+O(x^2),$$

he can write it in a form

$$f(x)=e^{Ax}.$$

If $f(x)=1 + Ax+O(x^2)$, then it is exactly right that $f(x)=e^{Ax}+O(x^2)$.

 

[jostpuur]

If one of my mathematical colleagues wrote that derivation, I would have considered it perfectly rigorous -- it's an elementary manipulation of differential forms, coupled with a typical but harmless¹ ambiguity in notation. Alas, it's a seems that the scientific community has a strong disdain for mathematican sophistication, and you're stuck with rubbish explanations like this.1: At least, it's harmless if you know what's going on...

This trickery is irrational, because the equation

$$df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2$$

comes out from nowhere. If you want to use mathematics as a tool, why not just take the chain rule as it is, and then use it? The physicists could also merely write

$$\frac{df}{du} = \frac{\partial f}{\partial x_1} \frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}$$

and say "this is a known result, and we can use it".

Why start with something else, and then do some kind of pseudo proof for the chain rule? And why insist, that this pseudo proof was the easier way? That's basically deriving the derivative out of Taylor series.

 

[jostpuur]

If $f(x)=1 + Ax+O(x^2)$, then it is exactly right that $f(x)=e^{Ax}+O(x^2)$.

The physicists can also get the precise

$$f(x)=e^{Ax}$$

from this. The thing is, that if some expression involves exponential function, there's no way preventing physicist somehow messing with it.

Suppose a physicist wants to postulate, that some spinor transforms, in rotations, according to action given by matrices

$$e^{-i\theta\cdot\sigma/2}$$

The physicist will not postulate this directly (because then everything would still be rigor). Instead he postulates that the transformation matrices are

$$1 - \frac{i}{2}\theta\cdot\sigma + O(|\theta|^2)$$

and then derives the full transformation, and does the derivation somehow wrong, but gets the desired result.

I don't understand what is achieved by this.

 

[Jimmy Snyder]

so is there a mathematical proof for why 0! = 1?

It's defined that way for convenience, not as the result of any proof. Surely a mathematician would be justified in laughing at someone who tried to prove definitions.