Hot answers tagged parity
15
Good question! First you need to know that parity refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two "kinds" of parity:
If $f(x) = f(-x)$, we say the function $f$ has even parity
If $f(x) = -f(-x)$, we say the function $f$ has odd parity
Of course, for most ...
9
No, it really is arbitrary. The reason we use the right hand rule today (although it may have been chosen for different reasons of convenience in the past) is simply that our coordinate system of choice is right-handed. Mathematically, this means that we define the directions of the axes so that you have to use the right-hand rule to evaluate this cross ...
8
Our spacetime cannot be unorientable.
That's because the laws of physics describing our spacetime are not left-right symmetric. We say that they break the P-symmetry (parity) or that they are "chiral" (derived from a Greek word for the hand which is either left or right.) For example, a left-handed neutrino would turn into a right-handed neutrino if you ...
7
Presumably you are asking about the communication ambiguity in physics: can we unambiguously specify what we mean by "a right handed coordinate system" to a correspondent far away without a pre-arrnage communications channel (i.e. using SETI)?
For a long time the answer seemed to be "no", but the discovery of parity violation in 1957 changed the answer to ...
5
There is no mathematical difficulty of a non-time-orientable spacetime in GTR, and they can be generated by taking a quotient space rather easily, as you suggest. On the other hand, the same kind of action can be in some sense be undone as well. For any spacetime $M$, let $M' = (p,o)$, where $p\in M$ and $o$ is a time orientation. Then this double ...
5
First, an assignment of the parities.
The parity of fermions is a bit ambiguous because one may always redefine parity by
$$ P \to P (-1)^{2J}, P(-1)^L, P(-1)^{3B}, P(-1)^{3Q} $$
or one may add the product of several factors of this kind because the second factor is a multiplicatively conserved sign. By this definition, one gets another parity that is still ...
3
No, it's not true.
Suppose I'm floating in outer space (presumably in a space suit or something else to keep me alive). I'm still me, and I still know that, for example, my left hand is the one on the left, and my right hand is the one I can write with.
Even on Earth, we don't need environmental clues to distinguish left from right; it's more a matter of ...
3
I) Consider an arbitrary coordinate transformation
$$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$
Let
$$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$
denote the corresponding Jacobian.
Traditionally in physics,
a scalar $\sigma$ transforms as
$$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$
a pseudo-scalar ...
3
If you have two coordinate systems with the same origin, you can represent a (linear) transformation of coordinates from one to another as a matrix. This matrix has either positive or negative determinant. This sign of the determinant is what gives the transformation its parity. (All this applies to any number of dimensions, not just 3.)
If you compose ...
3
No, actually it implies that the original particle has positive parity. That's because parity is a multiplicative quantum number, which means that when you want to find the overall parity of a system that consists of multiple parts, you multiply the parities of the individual parts. If you have two decay products each with parity -1, then you multiply them ...
3
You prove the equality of operators by applying them to a function, we have
$$
H = - \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)
$$
Ergo:
$$
HP f(x) = H f(-x) = (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(-x) = - \frac{\hbar^2}{2 m} f''(-x) + V(x) f(-x)
$$
and
$$
PH f(x) = P (- \frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + V(x)) f(x) = P (- ...
3
Dear Pie86, the emission of particles in a weak decay is a complicated reaction, and Gell-Mann's totalitarian principle applies: every process or effect that is not prohibited by a symmetry will occur at a nonzero probability. The asymmetry or the spin-momentum correlation for the electrons is such an effect.
In this case, it is infinitely unlikely that the ...
2
I haven't read that book, but I did read Feynman's discussion of (sounds like) exactly the same thing. Easy: Tell the aliens how to build a telescope, then describe the configuration of some galaxies near them. OK OK, but suppose we rule that out: We can't see any objects in common. Easy: Send them circularly-polarized radio waves (thanks @Anonymous Coward). ...
2
I assume OP means an even potential $V(x)~=~V(-x)$, e.g., a finite square well potential $V(x) ~\propto~ \theta(|x|-a) $.
Then the answer to the question(v1) is No.
Sketched proof: Under the assumption that $V$ is even, the Hamiltonian
$$H= \frac{p^2}{2m}+V(x)$$
then commutes with the parity operator $P$. So the operators $H$ and $P$ can be ...
1
I think David is being a bit harsh, because I had to read the Wikipedia article a couple of times to see what they were getting at.
As the article states at the beginning, a parity transformation is the flip of a single spatial co-ordinate. In effect it's like looking in a mirror: when you look in a mirror your height and width co-ordinates are unchanged bu ...
1
1) I thought parity is an intrinsic property of a particle, and does not depend on the angular momentum. However, I seem to be wrong. There seems to be an additional factor of (-1)^L.
Since the omega is a vectormeson, it has spin 1. Because J=1 for the omega, L must be 0.
The pion has J=0 and S=0, so L=0.
The rho has J=1 and S=0, so L=1.
Now, if that is ...
1
Left and Right can be used in different contexts. In most cases they work equally well in space but not quite always. Often when people talk about left or right they mean something relative to a person, e.g my left hand or my right eye or the book to my right. This works equally well in space because it is defined relative to the position of a person, not ...
1
Dear lurscher, the quote is the kind of C-physics described by the C-word which is a favorite word of mine but is discouraged on this server, so I won't use it - but you have used it. You don't misunderstand anything - quite on the contrary, you're right on the money.
These comments about a non-existent test of parity in the equivalence principle are due to ...
1
I read in a book recently a theoretical conversation between an alien and and an earthling where the earthling was trying to explain which hand is the right hand (without any visual contact) and which the left. The alien would have to perform experiments to determine a reference left or right. It turns out after a whole chapter showing how there isn't ...
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