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Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
But if the physicist calls anything that transforms under rotations properly a scalar, then they would call this a scalar, and this would disagree with the math definition, is that not correct? Would it make sense to say that if a physicist is only interested in rotations of a certain coordinate system then there is scalar in the math sense which describes the coefficient of the volume form, namely the scalar 1, and this is what they mean?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
What about the coefficient of the volume form $dx\wedge dy\wedge dz\,.$ This transforms under rotations as a scalar does (but not under parity, but let's ignore that) but not under scalings, as far as I understand this. Obviously it's not a scalar in the mathematician's sense because the coefficient of a volume form isn't even well-defined, but if a physicist calls anything that transforms under rotations and parity a scalar (then ignoring that this happens to not transform properly under parity) why wouldn't the physicist call this a scalar?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
But doesn't this fact (about general coordinate transformations, or change of basis, or whatever) follow from the mathematician's definition? So how could they be equivalent?...
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
But wouldn't a physicist have to check all possible coordinate transformations in order to know if the quantity was a scalar? Why just rotations and reflections? I know they are isometries, but what if I do some weird transformation that gives me something I wouldn't expect?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
@DavidZ Ok that's my question, are the definitions equivalent?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
@David But is this not weird? Group actions have nothing to do with scalars in the mathematicians' sense.
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
So suppose an object satisfies 1. but not 2., then it can't be interpreted as a thing which doesn't vary with coordinate changes because otherwise it would have to satisfy 2...but what about every other possible coordinate transformation I can talk about that you haven't mentioned in 1. and 2.? What if object x is a scalar under 1. and 2., hence a scalar by the mathematicians definition, but then I introduce condition 3. which x doesn't satisfy? Then it wouldn't be a scalar under 1. 2. and 3., but then how could it be a scalar with the math definition?
Jun
17
comment Clarification on meaning of scalar in math and scalar in physics
@aaaaaa I think it's just a terminology issue.
Feb
20
comment Variational Principle to find Energy Eigenfunctions
$a=H\psi\,,b=\psi\,.$ Note that $\langle H\psi,H\psi\rangle=\|H\psi\|^2\,.$ So we have $|\langle H\psi,\psi\rangle|\le\|H\psi\|\|\psi\|=\|H\psi\|\,.$
Feb
20
comment Variational Principle to find Energy Eigenfunctions
The Cauchy-Schwarz inequality says that $|\langle a,b\rangle|\le\|a\|\|b\|$ with equality iff $a=\lambda b$ where $\lambda$ is a scalar.
Feb
2
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
Ok thanks. One more question: Is there a sense in which quantum fields satisfy the Klein-Gordon equation? I mean since they are operators. I've read that particles satisfy the Klein-Gordon equation, something about their mass being on-shell, but it's not really clear to me how to go from a field to a particle, some confusion resulting by the presence of real particles and virtual particles. Though I know that light (or electromagnetic waves) satisfies the Klein-Gordon equation with $m=0\,.$
Feb
1
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
Without forces? The analogous classical mechanics Lagrangian is $\frac{\dot{x}^2}{2}-kx^2\,,$ equivalently $F=kx\,,$ so doesn't the term involving $m$ result in a force of some kind? Is the definition of noninteracting "linear"?
Feb
1
comment Every Relativistic Field Satifies the Klein-Gordon Equation?
What's the definition of a free field? Is it defined to be a field with that Lagrangian density? This is how I've seen it defined. Which would make it a tautology...
Jan
3
comment Is conservation of energy a set of principles that is inevitable in any 'possible world'?
You can look up the variational complex, or look at Peter Olver's book "Applicatons of Lie Groups to Differential Equations". It's similar to how you determine if a function is the gradient of a scalar actually. I've written a bit about it, but it's not the best exposition.
Dec
29
comment Is conservation of energy a set of principles that is inevitable in any 'possible world'?
This is something that I see frequently, but I find it a bit disingenuous as presented. Conservation of energy doesn't merely follow from the fact that the laws of physics should have time symmetry, it must further be imposed that the laws of physics come from an action principle. This is highly nontrivial and one needs to know the laws in order to actually verify that this is true. However given Newton's laws it is easily checked that energy is conserved and one needs not invoke Noether's theorem.
Aug
30
comment How was the formula for kinetic energy found, and who found it?
@user17574 There are much much simpler proofs, all of these conserved quantities, momentum, energy, angular momentum could be proved rigorously just from Newton's laws. Also considering the Lagrangian involves the kinetic energy, I don't think this helps.
Jul
18
comment The Hole Argument
Well I'm glad you see what I mean. So then are the two spacetimes distinguishable? I mean, can we tell which one we "live" in? I think part of my confusion arises from the fact that the coordinates don't seem to be defined before finding for the metric.
Jul
18
comment The Hole Argument
But they only mention the coordinate transformation thing later, on page two they seem to be literally considering different metrics on the same manifold. They even write $G(r)$ and $G'(r)$, and since they are using the same letter I can only assume that $r$ represents the same point on the manifold, resulting in two different metrics. Which part of this isn't right?
Jul
18
comment The Hole Argument
I don't understand that, why do you call them the same thing if they are two different things? You seem to have two different metrics on the same manifold, hence should be able to use the same points.
Jul
18
comment The Hole Argument
So take two points $p,p'$ on $M$. Using $g, g'$ you measure two different distances between them, but since they are isometric there is no observable difference in the physics?