Why are Physicists so informal with mathematics?

[TurtleKrampus]

I realize of course that this will probably not apply to all physicists, but at least every physicist in my university's math department is very unrigorous when it comes to mathematics. This is frustrating because some of the physics material seems genuinely interesting, but the lack of an axiomatic approach, proofs, and rigor makes it incredibly unappealing (I've skimmed the course notes).

I've had my first Physics class today, which is only taught to 3rd year pure math undergraduates, and it was incredibly disappointing. The professor said a lot of words to say so little, the first 30 minutes could've been resumed to ''we assume the position of our point particles at any given time can be described as a vector in a 3 dimensional Euclidean vector space". The professor was literally explaining what an inner product is, to a class that's had 2 semesters of linear algebra and 2 semesters of geometry.

Several of the times that the professor tried to formalize things mathematically were ought right incorrect, for example forgetting that sets are unordered and thus you can't both describe an N point system as a point in IR^(3N) and a set (simultaneously that is x = {r1,...,rN} in IR^3, unless you have a fixed ordering beforehand, which would be the same thing as considering IR^(3N)/S_N with a fixed set of representatives).
I also hate that the space we're considering at any given time is mostly not written. this drives me nuts.

Also what does " t = t' " actually mean, that is probably the worst way to write that time is absolute. I hate that the burden of formalizing everything falls on the students and not the professor...

 

[PeroK]

You ought to chill out and go with the flow. It's a bit like complaining about the grammar in a rock 'n' roll song.

Seriously, mathematics can be practically more powerful if you don't have to prove everything. Physics isn't a branch of pure mathematics, as you're discovering.

 

[kuruman]

There is a joke that dates back to the days when college dances were chaperoned.

A group of ladies in a room are lined up against one wall. A chaperone brings in a group of math and physics majors, lines them up against the wall facing the ladies and informs them that they can move towards the ladies by a distance that each time is half the current distance to the ladies' wall.

Upon hearing this, the math majors leave the room disgusted knowing that it will take them infinite time to reach the ladies. The physics majors start inching forward knowing that they will get close enough for all practical purposes.

 

[Vanadium 50]

It's a bit like complaining about the grammar in a rock 'n' roll song.

Or even pre-rock and roll. Is You Is, Or Is You Ain't My Baby

 

[kuruman]

Or even pre-rock and roll. Is You Is, Or Is You Ain't My Baby

It ain't necessarily so $\dots$

 

[symbolipoint]

Unrigorous?
One knows how to say what one needs to say, and chooses the parts of Mathematics which work. But let somebody else try to pick-apart the rigor (if any) in that statement.

 

[hutchphd]

One of the best quotes from my freshman physics text (Sears and Zemansky) was from Lewis Carol

(I believe he did some maths)

 

[andresB]

but the lack of an axiomatic approach, proofs, and rigor makes it incredibly unappealing (I've skimmed the course notes).

Do you already know what the axioms of nature are?

 

[Bandersnatch]

No way. A mathematician thinks physics butchers maths. This has to be the first.

 

[Haborix]

OP, while you feel this way, there are those of us that upon hearing "Lebesgue integral" all of a sudden start seeing a pencil for the weapon that it could be.

 

[strangerep]

OP, while you feel this way, there are those of us that upon hearing "Lebesgue integral" all of a sudden start seeing a pencil for the weapon that it could be.

Many years ago, a new D&D character type was introduced: the "Mathematician". S/he had amazing powers: a 10th level Mathematician could translate an opponent into Hilbert space with the mere wave of a pencil.

 

[strangerep]

I realize of course that this will probably not apply to all physicists, but at least every physicist in my university's math department is very unrigorous when it comes to mathematics. This is frustrating [...]

I totally agree -- that was my reaction when I started university physics. It seemed like we were expected to know all sorts of little tricks and shortcuts that were never really explained.

Later, in 2nd-year electromagnetism class, the lecturer had to teach us basic vector calculus (div, curl and all that) because the math courses were not adequately synchronized with physics. Of course, the lecturer "proved" stuff (like div curl = zero) by drawing lots of little circles on a boundary, (sigh). Later that year, a math lecturer went off his nut trying to hammer into us that this is not a proof and anyone who submitted such in an exam would receive a fail.

Fortunately, my applied math courses also covered some of the material, and they did it more carefully.

I've had my first Physics class today, which is only taught to 3rd year pure math undergraduates, and it was incredibly disappointing. The professor said a lot of words to say so little, the first 30 minutes could've been resumed to ''we assume the position of our point particles at any given time can be described as a vector in a 3 dimensional Euclidean vector space". The professor was literally explaining what an inner product is, to a class that's had 2 semesters of linear algebra and 2 semesters of geometry.

That's bad. One would expect the lecturer to have a reasonable idea of the capabilities of the students.

Several of the times that the professor tried to formalize things mathematically were ought right incorrect, for example forgetting that sets are unordered and thus you can't both describe an N point system as a point in IR^(3N) and a set (simultaneously that is x = {r1,...,rN} in IR^3, unless you have a fixed ordering beforehand,

Heh, heh, we get around that using notions of transformations and covariance.

Also what does " t = t' " actually mean, that is probably the worst way to write that time is absolute.

Do you understand that the $t$ and $t'$ denote time coordinates set up by measurements made by different physical observers? How would you explain that the passing of time, as measured by different observers (in Galilean relativity), is an invariant notion?

 

[martinbn]

I find your examples not very good. It seems that your complaint is about the level of exposition or the level of formality rather than rigour.

In any case I don't think that lack of rigour is a problem for mathematicians. The problem that they have with physics (this is just my opinion) is due to the difference of language.

ps You might find this interesting.

 

[TurtleKrampus]

Do you understand that the $t$ and $t'$ denote time coordinates set up by measurements made by different physical observers? How would you explain that the passing of time, as measured by different observers (in Galilean relativity), is an invariant notion?

Given two inertial reference frames (both being isomorphic to $\mathbb R^3$) $R_1,R_2$ there exists some $A \in SO(3)$ and $v \in C^2(\mathbb R, \mathbb R^3)$ with $v'' = 0$ such that the projection of the position of some point mass $r$ to $R_2$, which I'll denote $\pi_2(r)$, factors through $R_1$ from it's projection of its position in $R_1$, which I'll denote $\pi_1(r)$, in the following way:
$$\pi_2(r) = A\pi_1(r) + v$$
I believe that this enlarges into time being invariant across the reference frames, since our transformation is (in terms of $\pi_1(r)$) time invariant.
There are some minor things that I wrote, for example we can consider the domain of the translation to be an interval, but I wrote it for the sake of simplicity. I don't know if it's possible for reference frames to rotate through time, but if so we just replace A with a continuous function onto SO(3). This is a first thought into how I'd go about writing things, there are probably many errors here..

 

[PeroK]

This is a first thought into how I'd go about writing things, there are probably many errors here..

You might want to sink your teeth into the Poincare Group:

https://en.wikipedia.org/wiki/Poincaré_group

 

[gmax137]

Given two inertial reference frames (both being isomorphic to R3) R1,R2 there exists some A∈SO(3) and v∈C2(R,R3) with v″=0 such that the projection of the position of some point mass r to R2, which I'll denote π2(r), factors through R1 from it's projection of its position in R1, which I'll denote π1(r), in the following way:
π2(r)=Aπ1(r)+v
I believe that this enlarges into time being invariant across the reference frames, since our transformation is (in terms of π1(r)) time invariant.

Do you actually prefer this to t=t' ?

 

[martinbn]

Given two inertial reference frames (both being isomorphic to $\mathbb R^3$) $R_1,R_2$ there exists some $A \in SO(3)$ and $v \in C^2(\mathbb R, \mathbb R^3)$ with $v'' = 0$ such that the projection of the position of some point mass $r$ to $R_2$, which I'll denote $\pi_2(r)$, factors through $R_1$ from it's projection of its position in $R_1$, which I'll denote $\pi_1(r)$, in the following way:
$$\pi_2(r) = A\pi_1(r) + v$$
I believe that this enlarges into time being invariant across the reference frames, since our transformation is (in terms of $\pi_1(r)$) time invariant.
There are some minor things that I wrote, for example we can consider the domain of the translation to be an interval, but I wrote it for the sake of simplicity. I don't know if it's possible for reference frames to rotate through time, but if so we just replace A with a continuous function onto SO(3). This is a first thought into how I'd go about writing things, there are probably many errors here..

What is your definition of a reference frame, and how can it be isomorphic to $\mathbb R^3$?

 

[TurtleKrampus]

Do you actually prefer this to t=t' ?

Yes, t = t' makes no sense to me

 

[martinbn]

It seems that you don't want to answer my questions, so i will ask one more. What exactly are the projections? You say that they are to $R_2$ and $R_1$ but from where?

 

[PeroK]

Yes, t = t' makes no sense to me

Coordinates are, in a way, subtly different from the mathematical objects studied in pure mathematics. There's only one number line, or one version of $\mathbb R^3$. Then you study, say, mappings from $\mathbb R^3$ to itself. In physics, the fundamental object is spacetime; or space and time. But, Newtonian space is different from $\mathbb R^3$ in subtle ways. First, there is no definite origin. The same physical event can be described using different coordinate systems. This is where the concept of relativity comes from. Galilean relativity in Newtonian physics, as opposed to Special Relativity. These are the only two possibilities, given some basic properties of isotropy and homogeneity of spacetime.

Trying to retreat into pure mathematics is a mistake, if you really want to learn physics. There's also an operational aspect. Operationally, what $t = t'$ means is that two different reference frames assign coordinates (through physical measurements) to each event, but the $t$ coordinate always has the same numerical value as the $t'$ coordinate.

Note that there are some quantities where this is assumed to the point where it feels odd to question them: for measurements of the mass of a particle $m = m'$; electric charge: $q = q'$. These are physical, operational statements (or assumptions). The assumption $t = t'$ might be another obvious one. In Special Relativity, however, this assumption is abandoned in favour of $c = c'$ (where $c$ is the measured speed of light.)

In any case, the statement $t = t'$ makes physical sense in that it says that time has the same measured value in all reference frames. That's an important physical assumption.

 

[TurtleKrampus]

It seems that you don't want to answer my questions, so i will ask one more. What exactly are the projections? You say that they are to $R_2$ and $R_1$ but from where?

Sorry I just got home, I'm still thinking of how I want to verbally express myself.

What is your definition of a reference frame, and how can it be isomorphic to $\mathbb R^3$?

I'll start by what I mean by a reference frame. Some of these definitions will be a bit loose, but I hope they will suffice...

By reference frame I mean **both** an 3 dimensional Euclidean vector space (with an orthonormal direct basis) and a mapping from a set of point masses onto it, which can be seen as a way of adding coordinates onto it. Perhaps the name projection might be misleading, but in my view it seemed appropriate.
Regardless, we may wish that to add additional restraints on what a reference frame is, for example, a sensible restriction may be that it's *somewhat* like an isometry, i.e. given some physical way of measuring distances between point masses, this distance should be close to a "universal" positive constant * the distance of the projection of these points.
Though what I feel like is important here is to note is really the space + mapping pair.
It may also be more reasonable that the mapping be a continuous function onto that space, as opposed to a single point, because we like to model movement, but what I feel like is important is the idea itself.
Since 3 dimension Euclidean vector spaces are all isomorphic to $\mathbb R^3$ we can simply assume that it is exactly that, however with different mappings.

In this case, a change of referential can be seen as simply an **isometry** of $\mathbb R^3$, or rather a time dependent continuous function, whose codomain are isometries if we consider change in time.
That is, if we consider only the constant time changes of referential, then (by the classification of endomorphic isometries in Euclidean space) we have that a change in referential is of the form $\Phi:z \to Az + v$ for some $A \in SO(3)$ and $v \in \mathbb R^3$. This becomes more iffy if we don't consider constant time, where a change in referential will have the form of something like:
$$\Phi:C(\mathbb R; \mathbb R^3) \to C(\mathbb R; \mathbb R^3)$$

Where time independence may be expressed as $\Phi$ being expressible as something whose domain is $\mathbb R^3$, whose domain is expanded.

If we suppose time invariance, reference frames not rotating relative to each other, and objects are rest in one reference frame not accelerating in another, we'd get at what I initially wrote I believe.

Hopefully this makes a bit more sense, I'm trying to write what I mean clearer but still, there are probably many flaws about what I wrote, I am very grateful to everyone who may help.

 

[TurtleKrampus]

Coordinates are, in a way, subtly different from the mathematical objects studied in pure mathematics. There's only one number line, or one version of $\mathbb R^3$. Then you study, say, mappings from $\mathbb R^3$ to itself. In physics, the fundamental object is spacetime; or space and time. But, Newtonian space is different from $\mathbb R^3$ in subtle ways. First, there is no definite origin. The same physical event can be described using different coordinate systems. This is where the concept of relativity comes from. Galilean relativity in Newtonian physics, as opposed to Special Relativity. These are the only two possibilities, given some basic properties of isotropy and homogeneity of spacetime.

Trying to retreat into pure mathematics is a mistake, if you really want to learn physics. There's also an operational aspect. Operationally, what $t = t'$ means is that two different reference frames assign coordinates (through physical measurements) to each event, but the $t$ coordinate always has the same numerical value as the $t'$ coordinate.

Note that there are some quantities where this is assumed to the point where it feels odd to question them: for measurements of the mass of a particle $m = m'$; electric charge: $q = q'$. These are physical, operational statements (or assumptions). The assumption $t = t'$ might be another obvious one. In Special Relativity, however, this assumption is abandoned in favour of $c = c'$ (where $c$ is the measured speed of light.)

In any case, the statement $t = t'$ makes physical sense in that it says that time has the same measured value in all reference frames. That's an important physical assumption.

I'll read what you wrote in a while, sorry to everyone else as well to whom I haven't responded, I don't have a lot of free time at the moment that I can dedicate to this thread.

 

[haushofer]

What you want reminds me of a painter who first wants to understand the exact chemical structure of his paint before painting. That doesn't make a better painting.

Also, don't do quantum field theory. You'll get a stroke.

 

[symbolipoint]

What you want reminds me of a painter who first wants to understand the exact chemical structure of his paint before painting. That doesn't make a better painting.

Also, don't do quantum field theory. You'll get a stroke.

Some of what you say is what I was trying to say. You have a different way of saying it.
I say, Physics people need Mathematics as a workable language and plenty of Mathematics allows for this.

 

[weirdoguy]

This is frustrating because some of the physics material seems genuinely interesting, but the lack of an axiomatic approach, proofs, and rigor makes it incredibly unappealing (I've skimmed the course notes).

Well, if physics were taught the way you want it then probably more than 95% of physicists and people interested in physics would find it incredibly unappealing. Even I would be in that group, and I am a mathematical physicist (long live jet bundles!).

 

[gmax137]

Operationally, what t=t′ means is that two different reference frames assign coordinates (through physical measurements) to each event, but the t coordinate always has the same numerical value as the t′ coordinate.

Note that there are some quantities where this is assumed to the point where it feels odd to question them: for measurements of the mass of a particle m=m′; electric charge: q=q′. These are physical, operational statements (or assumptions). The assumption t=t′ might be another obvious one. In special relativity, however, this assumption is abandoned in favour of c=c′ (where c is the measured speed of light.)

In any case, the statement t=t′ makes physical sense in that it says that time has the same measured value in all reference frames. That's an important physical assumption.

Thank you, @PeroK . I thought I was the one missing the point of t=t'.

@TurtleKrampus , what happens to your preferred approach when, as @PeroK points out, we abandon t=t' in special relativity? I don't think one can sit in one's mind and deduce the properties of spacetime. Instead, we need experience in the world, experiments, to lead our thoughts. This is the opposite of an axiomatic approach.

 

[andresB]

Yes, t = t' makes no sense to me

Mathematical symbols are a shorthand for words and Ideas. t=t' seems to be a crystal clear sentence to me.

 

[TurtleKrampus]

Coordinates are, in a way, subtly different from the mathematical objects studied in pure mathematics. There's only one number line, or one version of $\mathbb R^3$. Then you study, say, mappings from $\mathbb R^3$ to itself. In physics, the fundamental object is spacetime; or space and time. But, Newtonian space is different from $\mathbb R^3$ in subtle ways. First, there is no definite origin. The same physical event can be described using different coordinate systems. This is where the concept of relativity comes from. Galilean relativity in Newtonian physics, as opposed to Special Relativity. These are the only two possibilities, given some basic properties of isotropy and homogeneity of spacetime.

Trying to retreat into pure mathematics is a mistake, if you really want to learn physics. There's also an operational aspect. Operationally, what $t = t'$ means is that two different reference frames assign coordinates (through physical measurements) to each event, but the $t$ coordinate always has the same numerical value as the $t'$ coordinate.

Note that there are some quantities where this is assumed to the point where it feels odd to question them: for measurements of the mass of a particle $m = m'$; electric charge: $q = q'$. These are physical, operational statements (or assumptions). The assumption $t = t'$ might be another obvious one. In Special Relativity, however, this assumption is abandoned in favour of $c = c'$ (where $c$ is the measured speed of light.)

In any case, the statement $t = t'$ makes physical sense in that it says that time has the same measured value in all reference frames. That's an important physical assumption.

While there might not be a definite origin, but don't we take a relative origin?
I'm going to have to be very brief with my response, I have to wake up early tomorrow so I'll give my best shot at a response.

Firstly I don't see the issue with talking about a vector space coordinate system, while sure categorically the morphisms in question (i.e. the change of reference frames I was mentioning) aren't typical vector space homomorphisms, but rather a special class of affine transformations, there's still no problem with endowing the space with a vector space structure, it's just that the operations won't be compatible with each other. But that's fine, they don't need to be compatible, it just aids with coordinates operations there doesn't have to be any physical significance to these operations.
I believe that, at least in finite dimensions, an affine space may always be endowed a vector space structure so there shouldn't be any loss in generality here.

I think that the idea $t = t'$, or more generally $f(t,X)=t'$, where $X$ is a finite list of variables relating to the points masses, may be written more clearly using the language of commutative diagrams. Though I, as of the time of writing this comment, wouldn't know which diagram to write, though if I ever think about it I may post it here.

Also, just to clear this up, I am not against something like time invariance being an assumption, I really just want to be presented with a clear model, which probably exists, for the topics I'm being exposed to.
Thank you for your comment, I appreciate it, hopefully this cleared some things up.

 

[strangerep]

Also, don't do quantum field theory. You'll get a stroke.

Heh, I was about to suggest the opposite. Maybe TurtleKrampus can finally prove whether quantum electrodynamics does/doesn't exist.

 

[strangerep]

[...]
Though what I feel like is important here is to note is really the space + mapping pair.

...which is exactly the meaning that physicists express using coordinates.

This becomes more iffy if we don't consider constant time, where a change in referential will have the form of something like: $$\Phi:C(\mathbb R; \mathbb R^3) \to C(\mathbb R; \mathbb R^3)$$

...and thus you approach the starting point of the (so-called) "1-postulate" derivations of Special Relativity. That "1-postulate" is the Relativity Principle: the mapping must be such that inertial frames map to inertial frames (i.e., zero acceleration is preserved).

 

[George Jones]

every physicist in my university's math department

At your university, why are physicists in the math department?

 

[Rive]

I really just want to be presented with a clear model, which probably exists, for the topics I'm being exposed to.

Physics has a lot of tradition and parallel layers of history still alive and often the absolutely necessary is used/displayed only (no sledgehammer for flies does apply).

You might need to opt for some advanced topics to get what you seek here. When advanced math becomes necessary it's customary to explain the less advanced parts again in the new context.

... though likely you'll still get math-by-phisics at higher level too, so better be prepared

 

[martinbn]

At your university, why are physicists in the math department?

In some countries theoretical physics is part of applied mathematics and in the math departments.

 

[TurtleKrampus]

At your university, why are physicists in the math department?

brain fart, meant professors in the physics department.

 

[dextercioby]

@TurtleKrampus There's a branch of physics called mathematical physics whose purpose is to formulate theories of physics by fully respecting the rigors of various branches of mathematics. It's not at all compulsory, or should I say it's not very common that universities have staff doing research and teaching in mathematical physics. Therefore, for lectures delivered to students, most of the staff use the "standard semi-rigor" of physics textbooks.

For example, none of the exposé here https://en.wikipedia.org/wiki/Ehrenfest_theorem uses the methods of mathematical physics. The theorem of Ehrenfest was proven within the realm of mathematical physics only in 2009 by Friesecke and Schmidt. arXiv:1003.3372v1.

Returning to your dismay, there are, if you want to learn theories of physics properly using mathematics, you can very well do so by opting to books instead of lecturers. For example, classical mechanics is neatly covered either by Abraham & Marsden or by V. Arnol'd, I don't think there's anything sketchy or less rigorous in their books.