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Jun
20
awarded  Enthusiast
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
awarded  Yearling
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.
Jun
17
asked Clarification on meaning of scalar in math and scalar in physics
Jun
15
awarded  Popular Question
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
19
revised Variational Principle to find Energy Eigenfunctions
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Feb
19
revised Variational Principle to find Energy Eigenfunctions
added 178 characters in body
Feb
19
revised Variational Principle to find Energy Eigenfunctions
added 178 characters in body
Feb
19
answered Variational Principle to find Energy Eigenfunctions
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"?