Tag Info

New answers tagged

1

The difference is that $dW$ is an infinitesimal ''quantity'', whilst $W$ is not. I assume the context here is thermodynamics, which make use of calculus. In calculus there is the concept of the infinitesimal. I suggest, for you, to concern yourself with the structure of calculus if you are to tackle thermodynamics.


2

I'm going to go a bit overboard here and give you the sketch of how vectors are geometrically constructed, since I think it's helpful to know. While writing this I found I was phrasing things very carefully, which means: you may need to reread parts of this in a quiet corner if it doesn't all make sense at first. Suppose that you have a set of scalar fields ...


2

Rule of thumb is that things with indices are tensors, the order being given by the number of free indices you have. By free index I mean any index for an object which is not repeated, hence not involved in a sum (or contraction). For example $$g_{\mu\nu}$$ indicates a covariant tensor of order (or rank) 2. If this represents a metric, its inverse is the ...


3

The derivative $\mathrm{d}\phi$ is a proper (co)vector: $$ \left(\begin{matrix}\frac{\partial\phi}{\partial x^1} \\ \frac{\partial\phi}{\partial x^2} \\ \frac{\partial\phi}{\partial x^3} \\ \frac{\partial\phi}{\partial x^4}\end{matrix}\right)$$ with components $\partial_\mu \phi$ and $\partial^\mu \phi = g^{\mu\nu}\partial_\nu \phi$. This works also in ...


1

Imagine two different particles, e.g. a proton and an electron, the former described by functions in the Hilbert space $\mathcal H_p$, and the latter by functions in the Hilbert space $\mathcal H_e$. Now assume a base states $\phi_1, \phi_2,...$ in $\mathcal H_p$ and a base $\chi_1, \chi_2,...$ in $\mathcal H_e$. Your pair of particles is (proton, electron), ...


1

$|\phi(1)\rangle \otimes |\chi(2)\rangle $ is a cumbersome notation to write ket corresponding to $\psi$ function $\phi(\mathbf r_1)\chi(\mathbf r_2)$, where $\mathbf r_i$ refers to coordinates of the $i$-th subsystem. That's why the order of factors in $\otimes$ product does not matter; the resulting ket corresponds to the same $\psi$ function and is thus ...


1

You are correct in that the tensor product does not commute in general. The ordering of vectors in some tensor product, say \begin{equation*} |\psi\rangle\otimes|\phi\rangle\otimes|\xi\rangle \equiv |u\rangle \end{equation*} implicitly refers to how the resulting Hilbert space is defined via the tensor product. To the above vectors, let us associate ...


1

It depends on the specifics what the notation means, but in general one might say: we pretend, mathematically, that we can label each particle $1, 2, \dots n$, and look at its individual distribution in space, which is (for pure states) some wavefunction $a_k(\vec r, t)$. The product state simply means "particle #1 is in state $a_1$, particle #2 is in state ...


2

$$g_{\mu\nu}=\begin{pmatrix}g_{00}&g_{01}&g_{02}\\g_{10}&g_{11}&g_{12}\\g_{20}&g_{21}&g_{22}\end{pmatrix}$$ $\mu,\nu=0,\ldots,N$ are the matrix indices of the metric (and of tensors in general) in $N+1$ dimensions.


7

Those Greek letters are indices indexing the components of $g$. Generally if one expresses a rank-2 tensor like $g$ as a matrix, the first index indexes the rows, the second the columns. In your example, we have $g_{rr} \equiv g_{11} = 1$, $g_{\theta\theta} \equiv g_{22} = r^2$, $g_{r\theta} \equiv g_{12} = 0$, etc. As you can see, we sometimes use numbers ...


3

The double inner product expands to be (for second rank tensors that you encounter in hydrodynamics): $$ \mathbf{a}\mathbf{:}\mathbf{b} = a_{ij}b_{ij} = a_{11}b_{11} + a_{12}b_{12} + ... $$ So it behaves just like you would expect a vector dot product to behave. You add up the product of all of the values with the same indexing. You can do the same ...


1

I suppose the right function to consider for $R$ is $R=\Vert\mathbf x_i-\mathbf x_j\Vert$ and the usual differentiation operators for the $\nabla$s. So, if $\mathbf x_i = (x_i,y_i,z_i)$, then $$R = \sqrt{(x_i-x_j)^2+(y_i-y_j)^2+(z_i-z_j)^2}$$ and $$\nabla_i = \left(\frac\partial{\partial x_i},\frac\partial{\partial y_i},\frac\partial{\partial z_i}\right)$$ ...


1

Two undistinguishable fermions, electrons in your case, are in an anti-symmetrical state. So, the state you wrote has to be corrected, see below how. Also if you use the notation of the tensor product $\otimes$, it is desirable to use it consistently. $$\frac {\langle r_1|a\rangle \otimes \langle r_2|b\rangle - \langle r_1|b\rangle \otimes \langle ...


2

When contracting with the metric $\eta_{\mu\nu}$ one has explicitly, $$\eta_{\mu\nu} \frac{\partial \mathcal L}{\partial (\partial_\mu \phi)} \partial^\nu \phi = \frac{\partial \mathcal L}{\partial (\partial_0 \phi)} \partial^0 \phi - \sum_{i=1}^3\frac{\partial \mathcal L}{\partial (\partial_i \phi)} \partial^i \phi$$ For the other term, one has, ...


6

Given the way that you've presented your table, I would personally put a "-" rather than a 1 in the units column. This to me would signify that units such as "g, km, s, A" etc. do not apply here. In terms of your symbols, in many branches of physics it is common to use a "hat", "tilde" or "star" notation above a symbol to indicate that it is a ...


5

Conventionally we use $1$ for dimensionless quantities, although it may cause some confusions. In additon, The International Committee for Weights and Measures contemplated defining the unit of 1 as the 'uno', but the idea was dropped. --https://en.wikipedia.org/wiki/Dimensionless_quantity


2

The only convention I am familiar with is to put a $1$ for dimensionless quantities. If you feel that this could give rise to confusion, you could explain the convention somewhere above or below the table.


3

Yes. In the unit column, put $1$.


0

I think your notation makes sense. You define a vector $ C \in \mathbb{R}^n$ as $ C= (c_1,c_2,...,c_n) $. The absorbance is a function of $n+1$ variables: $ \nu $ and $ c_{mol}$ for $mol=1,2,...,n $, than can be expressed in a more conpact way as $A(\nu, C)$. Your last equation it is perfectly valid since you have defined $c_{mol}$ as the $mol$-th component ...


2

$T_\mu^\nu = T_{\mu\sigma}g^{\sigma\nu}$


3

For completeness I would like to pose an alternative answer to those already given. When doing calculations in physics, in particular in particle physics where particles decay into other particles (but only when certain conditions are fulfilled, see below), one often comes across the (Heaviside) step function (which is also widely used in engineering) ...


8

If you want to make statements over (discrete) time, linear temporal logic may be worth looking at. For instance, $\qquad\displaystyle \Box (A \implies B)$ means that whenever $A$ holds, $B$ has to hold at the same time. $\qquad\displaystyle \Box (A \implies \Diamond B)$ means that whenever $A$ holds, $B$ will hold at some point in the future. Or yet ...


17

The statement "$A$ happens given that $B$" is equivalent to "If $B$, then $A$", which is symbolically represented as an implication $$ B \Rightarrow A $$ or, if you want to preserve the order of $A$ and $B$ in the original statement $$ A \Leftarrow B $$



Top 50 recent answers are included