New answers tagged

1

The notation $(\vec S_1\otimes 1) \cdot (1\otimes \vec S_2)$ is interpreted as a contraction over spatial indices. Hence, $\vec S_1\cdot \vec S_2=\sum_{ij} \delta_{ij}S_{1i} \otimes S_{2j}$. By contrast, $(S_{1x}+S_{1y}+S_{1z})\otimes(S_{2x}+S_{2y}+S_{2z})=(\vec S_1\cdot [1,1,1])\otimes (\vec S_2\cdot [1,1,1])$.


1

How do we formally define vectors in physics? An excerpt from chapter one, page 12 of "Mathematics of Classical and Quantum Physics" Originally, we introduced a vector as an ordered triple of numbers. The rule for expressing the components of a vector in one coordinate system in terms of its components in another system tells us that if we ...


1

A physical quantity is a vector if it transforms in the same way as a position vector when the coordinate system undergoes a transformation.


1

Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3). For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), ...


1

Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...


1

There are no easy to verify criteria which will work in general. One way to see it is by noting that for every classical StatMech model, we can define a PEPS with the same correlation functions (for which the tensors can be easily constructed from the StatMech model), see http://arxiv.org/abs/quant-ph/0601075. On the other hand, for StatMech models it is ...


1

As a mathematician, I would like to point out the following technicality: What we (can) represent as a matrix are tensors with one index up and one index down, that is elements of $V\otimes V^*$ (where $V$ is a vector space, or a vector bundle over a manifold). A tensor with two indexes up is a bilinear form on $V$, and one with two indices down a bilinear ...


7

Slogan: Matrices are a tool to compute sums; tensors tell you which sums make sense. When you convert between rank-2 tensors and matrices, the decision as to which index of the tensor labels the rows and which one labels the columns is purely conventional. Matrix multiplication is no more than a convenient way to write products of the form $$K(i,k) = ...


4

As @AccidentalFourierTransform said, it's a mistake to try and think of tensors as matrices, as you rapidly end up going mad when you need to compute $A^{\alpha\beta\gamma\delta}B_{\beta\delta}$ or something. That said $A^{\alpha\beta}B_{\alpha\gamma}$ 'looks like' $A^TB$. The right way to think of tensors is as multilinear functions, because that is what ...


2

Yes, their outer product is defined as you said. Further, the product of operators is given by $$ (A \otimes 1_B)(F \otimes 1_B) = (AF) \otimes (1_B1_B) = (AF) \otimes 1_B $$ Therefore, $$ [A \otimes B, F \otimes 1_B] = [A,F] \otimes [B,I_B] = [A,F]\otimes {\bf 0} = 0 $$


-1

The coordinate invariant volume element/measure on a manifold with metric $g$ is $$ d^d x \sqrt{|g|} $$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$ d^d x' ...


1

Take for example $q$ a vector field: $$ q^a=A^\mu $$ where $a=\mu$ is a vector index. The conjugate momentum is $$ \frac{\partial\mathcal L}{\partial A^\mu} $$ and, as it is an upper index in the denominator, it makes sense to write it as $\pi_\mu$. Also, you can use the definition of vectors and covectors to prove that $\pi_\mu$ transforms as a covector, ...


0

It seems to me that there's a confusion about how the flavour is acting. In your example $R_1 \sim \bar{\bf{2}}$ and $R_2 \sim {\bf{2}}$. Therefore, by multiplying these representations we expect a singlet state ($\ell =0$) and a vector ($\ell =1$), i.e., $R_1 \otimes R_2 = \bf{1} \oplus \bf{3}$. Now, if we identify the fundamental doublet as: $R_2 = ...


5

Your mistake is that you left out the summation over double indices (which would give another delta). Since you have to sum over all possiblities of e.g. $c$, one of these possibilities is when c=d and only this term remains. I'll give a simplified example: $$\frac{\partial}{\partial{x^d}}({Q_{bc}}x^c) = \frac{\partial}{\partial{x^d}}({Q_{b0}}x^0) + ...


2

It seems that you confused the Jaumann derivative $\overset{o}{{S}}$ (in your notation $\overset{\bigtriangleup}{{S}}$) with the time derivative ${\dot{S}}$ $$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} -{S} \cdot {w} +{w} \cdot {S}$$ See how it is derived in "http://www.continuummechanics.org/cm/corotationalderivative.html". Using the argument that $w^T ...


1

Yes. That's how the tensor calculus is formulated. For two observers on different frame of reference, all they have to see is the same physical laws. On this basis we formulate our physical laws so that it will be valid for any observers irrespective of their reference frames. Let's take a tensor of rank one (a vector) as example. Suppose two observers on ...


0

Yes; he is saying that a tensor encodes geometric information. A vector is a special case, and it's direction and legth are unchanged under a change of basis, though the components do change. A more dramatic example is the determinant, which gives the volume of the paralellopiped whose edges are defined by three coterminous vectors. A change of basis ...


0

Depends on conventions of the author, usually yes, though, $\langle A,B|=\langle A|\langle B|=\langle A|\otimes\langle B|$. If $|A\rangle$ is from $\mathcal{H}_1$ and $|B\rangle$ is from $\mathcal{H}_2$, then an operator that acts on $|A\rangle\otimes|B\rangle$ is a linear operator on $\mathcal{H}_1\otimes\mathcal{H}_2$, however if you have an operator, say ...


0

Multiplying matrices is not correct here. You need to work these things out using the definition, $(\bar{\sigma}^\mu)^{\dot{\alpha}\alpha}=\epsilon^{\alpha\beta}\epsilon^{\dot{\alpha}\dot{\beta}}(\sigma^\mu)_{\beta\dot{\beta}}$. We have, for example, $$(\bar{\sigma}^0)^{11}=\epsilon^{12}\epsilon^{12}(\sigma^0)_{22}=(\sigma^0)_{22}=1,$$ while ...


12

The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field $\mathbb K$ (usually $\mathbb R$ or $\mathbb C$). A formal definition is given below (there are many equivalent approaches). First, if $V$ is a vector space, $V^*$ denotes its algebraic dual space, namely the vector space of the linear ...


15

$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ...


2

You are, perhaps, a little confused by the fact that Minkowski space (call it $M$) looks a lot like the tangent space to $M$ at a given point $E$. (Call that tangent space $T_E$.) Although they look a lot alike, $M$ and $T_E$ are not the same thing. For example, $T_E$ has a distinguished point called "zero", and $M$ does not (you can, of course, always ...


0

Spacetime is a manifold Let me provide a more pure-math perspective on your problem - maybe it'll be lifted afterwards. Let's first have a look at what a manifold is: A manifold is a collection of points that looks, locally, like a vector space. For example, a sphere (i.e. the surface of a ball) is such an object. Our spacetime in Newtonian or relativistic ...


2

thus consider it as a tensor of rank {0 1} ? In this context, $f(t,x,y,z)$ can be thought of as function on spacetime, $f(\mathcal{P})$, assigning a number to each event $\mathcal{P}$ where the coordinates of $\mathcal{P}$ are, in some coordinate system, $t,x,y,z$, While we can write $f(\mathbf{r})$, where $\mathbf{r}$ is understood to be the 'vector' ...


1

The definition of one-form is simply not "function of a vector". A one-form must be linear transform in a special way under a change of coordinates. That is, a one form is something that looks like $\alpha=\alpha_i \textrm{d}x^i$. It has components, $\alpha_i$. If you go from coordinates $x^i$ to coordinates $y^i$, the components change, such that the ...


-3

The constant is only $8\pi$ when you're working in units where $G=1$. The number $G$ was originally defined as the constant of proportionality in Newton's $$F=G\frac{m_1m_2}{r}.$$ So in some sense the only thing we can do is to pass to the Newtonian limit and calculate the force between two point masses.


6

To add more details to jimjo's answer, I would like to explain the "at most" in my comment Vector, at most, can be covariant. Three-vectors are only covariant under rotations, but if you include boosts then three-vectors transform in a non-covariant way. Therefore, Newton's second Law is non-covariant under the full Lorentz Group. To get a covariant ...


7

Newton's second law is covariant, as it does not change its form if we switch to another frame of reference. As already explained by @AccidentailFourierTransform in his comment, Newton's second law is a vector law. This means the quantities in the law are vectors, which have different values in different frames of reference. Only scalars, by definition, do ...



Top 50 recent answers are included