Nobody complains about physicists' math?

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The discussion centers on the perception that physicists' use of mathematics often lacks the rigor found in pure mathematics, yet it still yields correct results in practical applications. Participants note that while mathematicians may find humor in physicists' approaches, this is rarely discussed openly. The conversation highlights how physicists sometimes rely on informal methods, such as using differentials without rigorous justification, to simplify complex concepts. Despite these perceived shortcomings, many physicists successfully derive accurate results, suggesting a balance between intuition and mathematical formality. Ultimately, the thread raises questions about the validity and acceptance of physicists' mathematical practices within the broader scientific community.
  • #101
What's the opposite color of blue?
Well the opposite of the oppisite in true-infinity wouldn't be blue but it would be blue. Both but both in two difrent way's. this might not make sences or relate to this thread, but this would be more funny than math. which them stating that would make it relate to this thread.. but hey wait it wasnt but then it was? chaos? or control of perception of chaos through means of understanding what makese it chaos so its no longer chaos. :D
 
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  • #102
Poop-Loops said:
The way I see it, physicists knowingly use "bad math", so they can tell you "it works here because we're making an approximation" or some such. For example, when instead of an sum (big sigma, forget what it's called), you use an integral, because the number of particles you are integrating over is so huge it roughly works out.

They're not doing it because they don't know any better, they are doing it because it makes life a lot easier, and they acknowledge the limitations.

Make a gauge, For example it couldn't be any lower than 2 in any way shape or form under any reason, then say it couldn't be any higher than 8 for the same reason's. you wouldn't need to known the exact number unless you making somthing that's why using A,B,C and D for subs are easy don't need to be right just know what's wrong
 
  • #103
Mathematicians make fun of physicist, physicist make fun of engineers, engineer make fun of computer scientist,

see the connection?
 
  • #104
Who make fun of mathematicians?
 
  • #105
People with girlfriends.
 
  • #106
True. :smile:
 
  • #107
I'm writing this up quick so forgive the lack of quotes when I address certain claims. I believe I've vaguely heard of explain's first example, which is a very good one. The last two are very disturbing indeed. Which brings me to the critique on renormalisation practices. The messy asymptotics can be nausiating I agree, however certain regularization techniques used are mathematically quite legitimate. For example zeta regularization and Pauli-Villars regularization among others. The calculation of #96 is valid under a few summation senses. We may also take it as the value of the Dirichlet eta at the origin from zeta regularization.
What must be eradicated is the continued ignorance practising physicists have on these procedures. By blundering on the commutativity of a regularization, the physicist ends up with wrong Casimir energies for one instance.
Furthermore, this approximation warfare going on in this thread is not what I am concerned of. As a true mathematician one is bound by powers of definition and truth. If in their often necessary approximations, were physicists to use O-notation at the end or even a humble ~ instead of strict equality(=), the subtleties would surely add to rigour more than most while appeasing Poincare for his remarkable work.
Yet things are not to stop there, explain's latter two examples speak volumes of the troubles with physicist mathematics today. Swallow your lack of subtlety and compute, with rigour and understanding.
 
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  • #108
As a physics student, I've always wanted to have mathematical rigor myself.

I don't think, however, that it's required for most jobs in the field (well not the degree we find ideal). I'm assuming this is something more important to theoretical physicists... who had better have mathematical rigor or else don't mind selling out to the entertainment industry. I'm also assuming that there are far less theoretical physicists than experimentalists. This makes mathematical rigor less significant to physics in academics. Also, it would be arrogant for a mathematician to assume that they know more about physics because of their rigor. All that time we don't waste on math is spent understanding physics, which is not math. If the programs (math and physics) are to be approximately equivalent, then you can't teach all the math (proofs, advanced calc, and abstract algebra namely) as well as the physics.
 
  • #109
Well Pythagorean, you're modern physics acts in the framework it does thanks to the efforts of a mathematician, namely Hilbert; who claimed what was then an inconvenient truth(what with mathematics and physics taking on different paths as compared with the times of Euler, Gauss, Newton, etc.): clearly physics is too difficult for physicists!

Rigour allows us to explore phenomena of physics at their most extremal conditions. Further quite contrary to your notion of 'wasting time on the math', today's calculation savvy experimentalist is best off getting his math right the first time rather than 'wasting' time in revisiting and correcting ill formed equations.
Our understanding of a subject is reflected solely by what kind of mathematical characterization we can give it. The more mathematical the better understood.
That is why if a grand unified theory of reality is to be formed it will be a purely mathematical description.
In fact this holds true as we go into the facets of prevailing physics. Quantum theory and relativity are more mathematically inclined than Newtonian physics, whereas string theories are even more so and so on.
In that sense physics is simply a particular and admittedly subjectively interesting case of the very many considerable in mathematics. Since such generalization is uncomfortable to most we have a separate field that goes by 'physics'.
 
  • #110
yasiru89 said:
Since such generalization is uncomfortable to most we have a separate field that goes by 'physics'.

HA! you're trolling me with this one, I mean... seriously... did you spell separate incorrectly on purpose too? Clever.

I shouldn't have said "wasted time" but the point I'm making is still valid. Physics is not mathematics.

I plan to double major in math and physics because I don't like not understanding steps that are hand-waved in derivations. What you don't seem to accept is that we wouldn't get through the physics material if we studied mathematical rigor. We'd be mathematicians clueless about its practical applications to physical phenomena (which is what we're interested in).

Somebody put a list here earlier..

paraphrasing: mathees make fun of physees make fun of engees, etc...

I see the hierarchy the other way around. I envy the engineer who can fix his own fridge and car, and I have no clue what draws people to abstract mathematical concepts that have nothing to do with our physical reality; just not my thing... I want to learn mathematical rigor to be able to make better physical predictions.

Good on Hilbert, I'm proud of him (I don't say that in jest either). There's an abstract mathematical concept that a physicist was able to apply to models of our physical reality. It doesn't invalidate Bohr's postulates. It's a perfect example of indirect collaboration between the fields. Lucky for them, they didn't have the internet to argue complacently about their fields on. They just did good work in their fields and contributed to a significant explosion in science and technology along with a lot of other mathematicians and physicists. I'd personally love to work with a mathematician on a project... probably not an arrogant one who saw physics as inferior though. Doesn't sound too constructive.
 
  • #111
A resent discussion about complex conjugates in some other threads reminded me of one strange thing with the Dirac's field. I'll first show why the Lagrangian density

<br /> \mathcal{L}(\psi) = \overline{\psi}(i\gamma^{\mu}\partial_{\mu} - m)\psi<br />

with the action principle implies the Dirac equation. Let \xi(x) be some arbitrary variation, that vanishes at the end points of some time interval [t_1,t_2]. We then demand

<br /> 0 = D_{\alpha} \int d^4x\; \mathcal{L}(\psi + \alpha\xi)\Big|_{\alpha=0}<br /> \;=\; D_{\alpha} \int d^4x\;\Big((\overline{\psi} \;+\; \alpha\overline{\xi})\big(i\gamma^{\mu}\partial_{\mu} \;-\; m\big)(\psi \;+\; \alpha\xi)\Big)\Big|_{\alpha=0}<br />

<br /> = \int d^4x\;\Big( \overline{\xi}(i\gamma^{\mu}\partial_{\mu} \;-\; m)\psi \;+\; \overline{\psi}(i\gamma^{\mu}\partial_{\mu} \;-\; m)\xi\Big)<br /> = \int d^4x\;\Big(i\gamma^{\mu}(\overline{\xi}\partial_{\mu}\psi \;+\; \overline{\psi}\partial_{\mu}\xi) \;-\; 2m\textrm{Re}(\overline{\xi}\psi)\Big)<br />

When we perform integration by parts to move the second derivative like this \overline{\psi}\partial_{\mu}\xi\to -(\partial_{\mu}\overline{\psi})\xi, we obtain

<br /> = 2\int d^4x\;\textrm{Re}\Big(\overline{\xi}\big(i\gamma^{\mu}\partial_{\mu}\psi - m\psi\big)\Big)<br />

and we get the Dirac's equation

<br /> (i\gamma^{\mu}\partial_{\mu} - m)\psi = 0.<br />

However, this is not the way physicists derive the Dirac's equation. Physicists way goes like this. We first derive the Euler-Lagrange equation

<br /> \partial_{\mu}\frac{\partial\mathcal{L}(\phi)}{\partial(\partial_{\mu}\phi)} - \frac{\partial\mathcal{L}(\phi)}{\partial\phi} = 0<br />

for real valued fields, and then assume, that we can use it for complex fields, by assuming that the complex conjugate \phi^* is constant with respect to \phi. This assumption enables one to use derivation formulas like D|z|^2 = D (z^* z) = z^* which seem to be so ridiculously wrong, that one might think it is impossible to ever get correct results with them. However, this turns out to be a false assumption. See how IT WORKS!

We have

<br /> \frac{\partial\mathcal{L}}{\partial\psi} = -\overline{\psi}m,<br /> \quad\quad\quad<br /> \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)} = \overline{\psi}i\gamma^{\mu}<br />

so the Euler-Lagrange equation is

<br /> \partial_{\mu}\overline{\psi}i\gamma^{\mu} + m\overline{\psi} = 0<br />

By taking the complex conjugate of this, we get

<br /> -i(\gamma^{\mu})^{\dagger}\gamma^0 \partial_{\mu}\psi + m\gamma^0\psi = 0<br />

Here the knowledge (\gamma^0)^{\dagger} = \gamma^0 was used. By multiplying with -\gamma^0 from left, and using identities \gamma^0 \gamma^0 = 1 and \gamma^0(\gamma^{\mu})^{\dagger}\gamma^0 = \gamma^{\mu} we get

<br /> (i\gamma^{\mu}\partial_{\mu} - m)\psi = 0<br />

which is the Dirac's equation again. Okey, so it works...

Why do physicists want to do it this way? Often the answer is that they want to do things the easier way. I'm not fully convinced the physicists way is easier in this case, it's a matter of opinion. For example you need to know some properties of gamma matrices, which you don't need to know when you do the variations manually.

It is somewhat puzzling how the assumption that complex conjugates can be assumed to be constants works like this. According to my intuition it shouldn't work for anything, because it is plain wrong. But surprisingly, I find it comforting that actually this assumption doesn't usually work! Suppose we used the Lagrangian

<br /> \mathcal{L} \;=\; \frac{i}{2}\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi \;-\; \frac{i}{2}(\partial_{\mu}\overline{\psi})\gamma^{\mu}\psi \;-\; m\overline{\psi}\psi\quad \Big(= \textrm{Re}\big(i\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi \;-\; m\overline{\psi}\psi\big)\Big)<br />

The action principle gives precisely the same equation of motion from this Lagrangian, but the assumption of \overline{\psi} being constant and the Euler-Lagrange equations do not, because you get the factor 1/2 wrong.

(ARGH! EDIT EDIT: I just noted I made a mistake here. It is still working! Noooooo... :cry:)

So in oder to get the correct equation of motion the physicists way, you first need to "choose" the "correct form" of the Lagrangian!

Peskin & Schroeder describe the derivation of Dirac's equation like this

The correct, Lorenzt-invariant Dirac Lagrangian is therefore
<br /> \mathcal{L}_{\textrm{Dirac}} = \overline{\psi}(i\gamma^{\mu}\partial_{\mu} - m)\psi.<br />
The Euler-Lagrange equation for \overline{\psi} (or \psi^{\dagger}) immediately yields the Dirac equation in the form (3.31); the Euler-Lagrange equation for \psi gives the same equation, in Hermitian-conjugate form:
<br /> -i\partial_{\mu}\overline{\psi}\gamma^{\mu} - m\overline{\psi} = 0.<br />

I disagree with the statement a little bit. The Euler-Lagrange equation for \overline{\psi} doesn't immediately imply the correct result because you have

<br /> \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\overline{\psi})} = 0.<br />

You need to first, with integration by parts, to move the derivative from \psi onto the \overline{\psi}, and then apply the Euler-Lagrange equations. If you give a physicists an exercise where he is supposed to derive the Dirac's equation like P&S instruct, he will first do the integration by parts, and then use the Euler-Lagrange equations, and then get the correct results. If I ask "why do you do the integration by parts first?", the answer is "because otherwise you don't get the correct result".

You cannot derive the equation of motion the physicists way unless you already know the result before!

BUT WHY?

Wouldn't it be so much easier for everyone if we just wrote 0=D_{\alpha} \int dt\; L(\psi+\alpha\xi)\Big|_{\alpha=0}, and understood what we are doing? Why do physicists want to start making this simple thing into so much more complicated?

It seems that the authors of the physics books think, that the readers are so intelligent, that it doesn't matter what they write into the books, because the readers are not going to develop major misunderstanding in anyway. This has been a fatal mistake IMO. I once tried to ask about this from one physicist, and in the end he was explaining to me, that similarly as \textrm{Re}(\phi) and \textrm{Im}(\phi) are independent variables, also \phi and \phi^* are independent variables. In other words, he was explaining, that if x and y can be chosen independently, then components of (x,y) do not fix the components of (x,-y). Despite this, he also understood, that components of \phi^* actually do fix the components of \phi, but still on the other hand, they did not fix each others components, because they were describing "physical dynamics". They are always these "physical arguments", where I get lost.
 
  • #112
jostpuur said:
for real valued fields, and then assume, that we can use it for complex fields, by assuming that the complex conjugate \phi^* is constant with respect to \phi. This assumption enables one to use derivation formulas like D|z|^2 = D (z^* z) = z^* which seem to be so ridiculously wrong,
Two things:
(1) The variables z and z* are, in fact, analytically independent: the only analytic function satisfying f(z,z*)=0 is the zero function
(2) I believe it is fairly common (although I couldn't tell you where I first encountered it) when dealing with nonanalytic functions, to define the operator

\frac{\partial}{\partial z} = \frac{1}{2} \left( \frac{\partial}{\partial x} - i \frac{\partial}{\partial y} \right)

and similarly for the derivative w.r.t. z*. This does generalize the complex derivative, and gives us the identity that a function f is analytic if and only if \partial f(z) / \partial z^* = 0
 
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  • #113
does anyone have more information on the topics explain mentioned, specifically the "black magic" in QFT?
 
  • #114
jostpuur said:
A resent discussion about complex conjugates in some other threads reminded me of one strange thing with the Dirac's field. I'll first show why the Lagrangian density

<br /> \mathcal{L}(\psi) = \overline{\psi}(i\gamma^{\mu}\partial_{\mu} - m)\psi<br />

with the action principle implies the Dirac equation. Let \xi(x) be some arbitrary variation, that vanishes at the end points of some time interval [t_1,t_2]. We then demand

<br /> 0 = D_{\alpha} \int d^4x\; \mathcal{L}(\psi + \alpha\xi)\Big|_{\alpha=0}<br /> \;=\; D_{\alpha} \int d^4x\;\Big((\overline{\psi} \;+\; \alpha\overline{\xi})\big(i\gamma^{\mu}\partial_{\mu} \;-\; m\big)(\psi \;+\; \alpha\xi)\Big)\Big|_{\alpha=0}<br />

<br /> = \int d^4x\;\Big( \overline{\xi}(i\gamma^{\mu}\partial_{\mu} \;-\; m)\psi \;+\; \overline{\psi}(i\gamma^{\mu}\partial_{\mu} \;-\; m)\xi\Big)<br /> = \int d^4x\;\Big(i\gamma^{\mu}(\overline{\xi}\partial_{\mu}\psi \;+\; \overline{\psi}\partial_{\mu}\xi) \;-\; 2m\textrm{Re}(\overline{\xi}\psi)\Big)<br />

When we perform integration by parts to move the second derivative like this \overline{\psi}\partial_{\mu}\xi\to -(\partial_{\mu}\overline{\psi})\xi, we obtain

<br /> = 2\int d^4x\;\textrm{Re}\Big(\overline{\xi}\big(i\gamma^{\mu}\partial_{\mu}\psi - m\psi\big)\Big)<br />

and we get the Dirac's equation

<br /> (i\gamma^{\mu}\partial_{\mu} - m)\psi = 0.<br />

However, this is not the way physicists derive the Dirac's equation. Physicists way goes like this. We first derive the Euler-Lagrange equation

<br /> \partial_{\mu}\frac{\partial\mathcal{L}(\phi)}{\partial(\partial_{\mu}\phi)} - \frac{\partial\mathcal{L}(\phi)}{\partial\phi} = 0<br />

for real valued fields, and then assume, that we can use it for complex fields, by assuming that the complex conjugate \phi^* is constant with respect to \phi. This assumption enables one to use derivation formulas like D|z|^2 = D (z^* z) = z^* which seem to be so ridiculously wrong, that one might think it is impossible to ever get correct results with them. However, this turns out to be a false assumption. See how IT WORKS!

We have

<br /> \frac{\partial\mathcal{L}}{\partial\psi} = -\overline{\psi}m,<br /> \quad\quad\quad<br /> \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\psi)} = \overline{\psi}i\gamma^{\mu}<br />

so the Euler-Lagrange equation is

<br /> \partial_{\mu}\overline{\psi}i\gamma^{\mu} + m\overline{\psi} = 0<br />

By taking the complex conjugate of this, we get

<br /> -i(\gamma^{\mu})^{\dagger}\gamma^0 \partial_{\mu}\psi + m\gamma^0\psi = 0<br />

Here the knowledge (\gamma^0)^{\dagger} = \gamma^0 was used. By multiplying with -\gamma^0 from left, and using identities \gamma^0 \gamma^0 = 1 and \gamma^0(\gamma^{\mu})^{\dagger}\gamma^0 = \gamma^{\mu} we get

<br /> (i\gamma^{\mu}\partial_{\mu} - m)\psi = 0<br />

which is the Dirac's equation again. Okey, so it works...

Why do physicists want to do it this way? Often the answer is that they want to do things the easier way. I'm not fully convinced the physicists way is easier in this case, it's a matter of opinion. For example you need to know some properties of gamma matrices, which you don't need to know when you do the variations manually.

It is somewhat puzzling how the assumption that complex conjugates can be assumed to be constants works like this. According to my intuition it shouldn't work for anything, because it is plain wrong. But surprisingly, I find it comforting that actually this assumption doesn't usually work! Suppose we used the Lagrangian

<br /> \mathcal{L} \;=\; \frac{i}{2}\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi \;-\; \frac{i}{2}(\partial_{\mu}\overline{\psi})\gamma^{\mu}\psi \;-\; m\overline{\psi}\psi\quad \Big(= \textrm{Re}\big(i\overline{\psi}\gamma^{\mu}\partial_{\mu}\psi \;-\; m\overline{\psi}\psi\big)\Big)<br />

The action principle gives precisely the same equation of motion from this Lagrangian, but the assumption of \overline{\psi} being constant and the Euler-Lagrange equations do not, because you get the factor 1/2 wrong.

(ARGH! EDIT EDIT: I just noted I made a mistake here. It is still working! Noooooo... :cry:)

So in oder to get the correct equation of motion the physicists way, you first need to "choose" the "correct form" of the Lagrangian!

Peskin & Schroeder describe the derivation of Dirac's equation like this



I disagree with the statement a little bit. The Euler-Lagrange equation for \overline{\psi} doesn't immediately imply the correct result because you have

<br /> \frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\overline{\psi})} = 0.<br />

You need to first, with integration by parts, to move the derivative from \psi onto the \overline{\psi}, and then apply the Euler-Lagrange equations. If you give a physicists an exercise where he is supposed to derive the Dirac's equation like P&S instruct, he will first do the integration by parts, and then use the Euler-Lagrange equations, and then get the correct results. If I ask "why do you do the integration by parts first?", the answer is "because otherwise you don't get the correct result".

You cannot derive the equation of motion the physicists way unless you already know the result before!

BUT WHY?

Wouldn't it be so much easier for everyone if we just wrote 0=D_{\alpha} \int dt\; L(\psi+\alpha\xi)\Big|_{\alpha=0}, and understood what we are doing? Why do physicists want to start making this simple thing into so much more complicated?

It seems that the authors of the physics books think, that the readers are so intelligent, that it doesn't matter what they write into the books, because the readers are not going to develop major misunderstanding in anyway. This has been a fatal mistake IMO. I once tried to ask about this from one physicist, and in the end he was explaining to me, that similarly as \textrm{Re}(\phi) and \textrm{Im}(\phi) are independent variables, also \phi and \phi^* are independent variables. In other words, he was explaining, that if x and y can be chosen independently, then components of (x,y) do not fix the components of (x,-y). Despite this, he also understood, that components of \phi^* actually do fix the components of \phi, but still on the other hand, they did not fix each others components, because they were describing "physical dynamics". They are always these "physical arguments", where I get lost.

Physicist do use least action principle :b...

but if we want to write down dirac equation in the way he did discovered we have to use both math and physical principles...

i don't really understand where this topic is going? :(
 
  • #115
ice109 said:
does anyone have more information on the topics explain mentioned, specifically the "black magic" in QFT?

It's called black magic by people that read Peskin & Schroeder since they present it poorly (sweeping infinities under the rug). The black magic tricks are physically motivated. Only seeing half of the logic, mathematicians typically jump to the wrong conclusion that physicists are just doing bad math.

I will illustrate with one example-- infrared divergence. There is no simple throwing away an infinity here. One must realize that the theory is not valid for all energies, there are assumptions that are being violated. Issues such as those of quantum gravity would creep in. Based upon common sense we know that any field theory we right down is only valid within a finite energy range. Introducing a cutoff is nothing more than making explicit what was before an implicit assumption of the theory. The solutions now depend on a cutoff, but it's not a trick to get rid of an infinity, it's assigning a variable to represent the failure of the theory beyond that point. Every theory has limits of validity.

When mathematicians rant about the terrible math that physics does, they commonly make implicit the assumption that the mathematical model is an exact representation of nature (which of course it's not). They do offer valid criticism as well, which I'll get to in my next post.
 
  • #116
I've seen something in this thread that's very common-- math methods textbooks and classes are terrible! There really is no amount of apologizing that can be done for cramming in partial differential equations, complex analysis, group theory, approximate and numerical methods etc etc in a one semester course. Methods are taught with either no reason, or silly hand waving arguments.

If you are a theorist, get that math degree as a second major instead. If you are an experimentalist you probably need to be more familiar with the broad concepts and not the specific methods. Just say no to math methods courses!
 
  • #117
For what it is worth, my proffessor in Math Methods of Physics class often said if a mathematician were looking at this he or she would slap me but in physics, it works so we use it.

Also, I know there are a lot of strange exceptions and rules we learned in math for certain theorems (Stokes comes to mind) but when we did it in physics, we said not to worry about them becuase they were either unimportant or not physical. Or, sometimes rather than go into the rules, we said something to the affect that use the Right hand rule as a rule of thumb (pun not intended) for the sign.
 
  • #118
DavidWhitbeck said:
There really is no amount of apologizing that can be done for cramming in partial differential equations, complex analysis, group theory, approximate and numerical methods etc etc in a one semester course.

This is the very reason why the University of Buffalo does not teach a math methods course at the undergraduate level, they make their students take a two semester sequence in applied math which covers advanced multi-variate calculus and PDE's only... If you want to learn complex variables, group theory etc... they are separate courses. By the way, the grad course in Math Methods used to be applied math, Fourier series, Differential Equations, etc in the first semester, then a full semester of Linear algebra and Group Theory.
 

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