Nobody complains about physicists' math?

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In summary, physicists often use mathematical shortcuts and approximations in their work, which may seem sloppy or incorrect to mathematicians. However, these shortcuts are often sufficient for physicists to accurately describe and understand the physical systems they are studying. Both physicists and mathematicians have their own strengths and weaknesses, and ultimately, it is the combination of their skills that leads to a complete understanding of complex phenomena.
  • #1
jostpuur
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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?
 
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  • #2
If it describes reality, why would it be made fun of?
 
  • #3
jostpuur said:
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?"
 
  • #4
You mean like 'Quantum physics is classical physics in the limit as zero approaches h?'
 
  • #5
In physics 0/0 or infinity/infinity cancels - it's obvious!
 
  • #6
This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

[tex]
D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)
[/tex]

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

[tex]
d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2
[/tex]

and then divides by du, and gets

[tex]
\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}
[/tex]

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.
 
  • #7
zoobyshoe said:
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?
 
  • #8
The conclusion

[tex]
f(x)=1+Ax+O(x^2)\quad\implies\quad f(x)=e^{Ax}
[/tex]

is among the most unbelievable ones.

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

[tex]
x^2 + Ax
[/tex]

in a form

[tex]
(x+A/2)^2 - A^2/4.
[/tex]

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

[tex]
f(x)=1 + Ax+O(x^2),
[/tex]

he can write it in a form

[tex]
f(x)=e^{Ax}.
[/tex]
 
  • #9
jostpuur said:
This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

[tex]
D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)
[/tex]

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

[tex]
d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2
[/tex]

and then divides by du, and gets

[tex]
\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}
[/tex]

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.
 
  • #10
I think somebody's cage got rattled, hmm, Zz?
 
  • #11
Someone is in a cage?

Zz.
 
  • #12
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. :smile:
 
  • #13
Integral said:
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
 
  • #14
Gilligan08 said:
Goodfellas? What movie is that from?
Yeah, That's Newton doing the Joe Pesci line from "Goodfellas".
 
  • #15
Integral said:
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.
 
  • #16
Integral said:
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. :smile:
Yeah. And we laugh at mathematicians at the way they try to talk to girls.
 
  • #17
zoobyshoe said:
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?
 
  • #18
Chi Meson said:
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."
 
  • #19
mgb_phys said:
In physics 0/0 or infinity/infinity cancels - it's obvious!

:smile: I think I actually canceled infinity over infinity once on a optics test, and got full credit! :smile:
 
  • #20
to find a flux per unit area through an infinite plane we can do

[tex] \frac{1}{A}\int_S\Phi da[/tex]

Where [itex]S[/itex] is the entire infinite plane, and [itex]A[/itex] is it's area.
 
  • #21
so is there a mathematical proof for why 0! = 1?
 
  • #22
jostpuur said:
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".
 
  • #23
Integral said:
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.
 
  • #24
Ah physics, wherein every function equals the first term of its taylor series.
 
  • #25
Last edited by a moderator:
  • #26
jostpuur said:
This (or things like this) happened often in physics lectures. A physicists wants to use the derivative rule of composite functions

[tex]
D_u f(x_1(u), x_2(u)) = (\partial_1 f) x'_1(u) + (\partial_2 f) x'_2(u)
[/tex]

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

[tex]
d f = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2
[/tex]

and then divides by du, and gets

[tex]
\frac{df}{du} = \frac{\partial f}{\partial x_1}\frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}
[/tex]

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...
 
Last edited:
  • #27
jostpuur said:
Similarly, once a particle physicist is given a function

[tex]
f(x)=1 + Ax+O(x^2),
[/tex]

he can write it in a form

[tex]
f(x)=e^{Ax}.
[/tex]
If [itex]f(x)=1 + Ax+O(x^2)[/itex], then it is exactly right that [itex]f(x)=e^{Ax}+O(x^2)[/itex].
 
  • #28
Hurkyl said:
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

[tex]
df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2
[/tex]

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

[tex]
\frac{df}{du} = \frac{\partial f}{\partial x_1} \frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}
[/tex]

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.
 
Last edited:
  • #29
Hurkyl said:
If [itex]f(x)=1 + Ax+O(x^2)[/itex], then it is exactly right that [itex]f(x)=e^{Ax}+O(x^2)[/itex].

The physicists can also get the precise

[tex]
f(x)=e^{Ax}
[/tex]

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

[tex]
e^{-i\theta\cdot\sigma/2}
[/tex]

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

[tex]
1 - \frac{i}{2}\theta\cdot\sigma + O(|\theta|^2)
[/tex]

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.
 
  • #30
Pythagorean said:
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.
 
  • #31
jostpuur said:
This trickery is irrational, because the equation

[tex]
df = \frac{\partial f}{\partial x_1} dx_1 + \frac{\partial f}{\partial x_2} dx_2
[/tex]

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

[tex]
\frac{df}{du} = \frac{\partial f}{\partial x_1} \frac{dx_1}{du} + \frac{\partial f}{\partial x_2} \frac{dx_2}{du}
[/tex]

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?

But you still haven't told me what is WRONG with it. All you have done is argue it based on a matter of TASTES.

That's like arguing that one shouldn't use a screw driver to open the lid of a can of paint, because the screw driver was designed to be used in a certain way. You seem to forget the important point here : it works!

That phrase "it works" has always been severely undermined. Yet, it is THE most powerful argument there is. As long as the usage of the "tool" does not break any "laws" (I didn't use the screw driver to murder someone to get to that paint can to paint my house), then the claim that it works validates its usage. That is why I asked you what is mathematically wrong with it. I'm not talking about "canceling" 0/0, which would be breaking mathematical "laws", and in fact, results in something that doesn't work (look at the Fraunhoffer diffraction pattern). I'm talking about shortcut in notations that you are highlighting here.

Most physicists use mathematics as a tool, whether you like it or not. We need to know what the tools are, and how to use it correctly within its limits of validity. Once we know that, HOW we use it really shouldn't be the sore point of mathematicians. Physicists really do not have the patience nor the inclinations to focus on the "tools". If we do, what's the use of mathematicians?

In the experimental facility that I work at, we have this stainless steel plate that's mounted on a low ceiling by 4 bolts. The ends of the bolts stick down from the plate. While it is high enough for most of us not to hit it, someone around 6' tall or taller could hit his/her head on it. The safety regulation requires us to do something about it, and this could include shaving the bolt to a shorter length, putting in a cover over the whole contraption, etc... But we came out with something easier. Puncture 4 tennis balls, and stick the protruding ends of the bolt into the tennis balls. The bright, fluorescent color provides the advance warning to anyone approaching that area, and even if someone hits his/her head on it, it would not hurt. It was a quick, easy, and CHEAP solution to a problem. Yet, we had used something for what it wasn't meant to be used. Even our safety inspector was impressed. Why? Because IT WORKS!

The ability of physicists to adapt the mathematics to suit their needs, without really violating any mathematical laws, shows their creativity and imagination to solve the problem at hand. I'm sure this is done in many other fields as well, especially in engineering. I really don't understand why this would be a subject of ridicule or to be laughed at. In fact, I would think that the ability to take something and use it in a different manner while still maintaining its validity, is something that should be admired.

Zz.
 
  • #32
Thrice said:
Ah physics, wherein every function equals the first term of its taylor series.
The sum of the first and second terms no?
 
  • #33
jimmysnyder said:
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.

Are you sure that this is not derived via the Gamma function?

Zz.
 
  • #34
My favorite story about such laughings comes from Feynman (who else?). I'm mangling his story, but here goes:

A small group of math majors was looking at a french curve and pondering its shape. A french curve is a piece of plastic (or other material) with an edge that looks somewhat spirally. It's used by draftsmen in drawing curved lines. Because of the spirally edge, it approximates arcs of any radius in a range. The mathematicians knew this, but pondered why it had the particular shape that it had. Was it optimal in some way. Feynman told them that this particular shape had the following property. In no matter what orientation you hold it, the tangent at the bottom was parallel to the ground. The math majors satisfied themselves that this was true and thanked him for his help.

What he told them would be true of any curve. That's the fundamental theorem of calculus, that the slope of a curve at an extremum is zero.
 
  • #35
ZapperZ said:
Are you sure that this is not derived via the Gamma function?

Zz.
Yes. 0! is defined to be 1. The Gamma function is a late comer.
 
Last edited:

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