Homework Statement
A beam of spin-1/2 particles scatters off of a target consisting of spin-1/2 heavy nuclei. The interaction between the particle and nucleus is given by $$V(\vec{ r})= V_0~\delta (\vec{r})~ \vec{S}_1. \vec{S}_2$$
1) Averaging over initial spin states, find the differential...
Indeed, ##\Lambda^{-1}=g^{-1}\Lambda^T g ~##. But, for the Minkowski metric tensor, ##g^{-1}=g##, since ##g^2=1.##
Ok, so my basic confusion, distilled to its essence is thus:
Using the earlier derivation, we see that ##(\Lambda^{-1})^\rho{}_\beta=\Lambda_\beta{}^\rho##. Tung's definition of...
Oops, sorry! I meant,
$$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies\delta^\rho{}_\nu=g^{\rho\mu}g_{\mu\nu}=g^{\rho\mu}g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies \delta^\rho{}_\nu=\Lambda_\beta{}^\rho \Lambda^\beta{}_\nu$$
Now we write...
Yes, I'm aware of that and therein lies the contradiction. Starting from
$$g_{\mu\nu}=g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies\delta^\alpha{}_\nu=g^{\alpha\mu}g_{\mu\nu}=g^{\alpha\mu}g_{\alpha\beta}\Lambda^\alpha{}_\mu\Lambda^\beta{}_\nu\\
\implies...
I see what you are saying and that makes sense. But, if ##(\Lambda^T)_\mu {}^\nu = \Lambda^\nu{}_\mu## is true, then, ##(\Lambda^T)^\mu {}_\nu = \Lambda_\nu{}^\mu## is also true, which can be seen by lowering and raising indices by the metric (treating##\Lambda## as a tensor). But, we know that...
So, the way Tung defines his transpose is indeed inconsistent with my convention, right?
Also, I was wondering where this freedom to "define" transpose and inverse comes about. Why is there a need for a convention, at all? Given a matrix, its transpose and inverse are uniquely defined. Is it...
Let ##\Lambda## be a Lorentz transformation. The matrix representing the Lorentz transformation is written as ##\Lambda^\mu{}_\nu##, the first index referring to the rows and the second index referring to columns.
The defining relation (necessary and sufficient) for Lorentz transforms is...
Say, we have two Hilbert spaces ##U## and ##V## and their duals ##U^*, V^*##.
Then, we say, ##u\otimes v~ \epsilon~ U\otimes V##, where ##'\otimes'## is defined as the tensor product of the two spaces, ##U\times V \rightarrow U\otimes V##.
In Dirac's Bra-Ket notation, this is written as...
Yes, but where I got confused was that the entries of the column vectors themselves are vectors.
$$v_j=\sum_jS_{ij}w_i,~~~~\text{for all}~ j.$$
The ##v_j~'s ## and ##w_i~'s## in the equation above are not numbers but vectors.
But, I think, I have figured it out.
Let, U be a linear...
Say, we have two orthonormal basis sets ##\{v_i\}## and ##\{w_i\}## for a vector space A.
Now, the first (old) basis, in terms of the second(new) basis, is given by, say,
$$v_i=\Sigma_jS_{ij}w_j,~~~~\text{for all i.}$$
How do I explicitly (in some basis) write the relation, ##Uv_i=w_i##, for...
I have offers from UMass Amherst Physics and Case Western Astronomy for their respective Phd programs. I'm having a tough time deciding between the two. Leaving aside the differences due to astronomy vs physics, what other factors are important to consider while making a decision?
UMass Physics...
The given options only makes sense if by 'system' they mean ensemble and by 'subsystem' they mean members of the ensemble.
Then, ##P(E_1)= \cfrac{e^{-\beta E_1}}{e^{-\beta E_1}+e^{-\beta E_2}}##
Again, ##P(E_1)= \cfrac{N_1}{N_0}. ## Therefore, ##N_1 = N_0\left(\cfrac{e^{-\beta E_1}}{e^{-\beta...
Homework Statement
A system in thermal equilibrium at temperature T consists of a large number of subsystems, each of which can exist only in two states of energy and , where . In the expressions that follow, k is the Boltzmann constant.
For a system at temperature T, the average number of...
Yes, i think i get where i was going wrong. I was equating the thermodynamic notion of internal energy to the internal energy of one individual system, instead of the ensemble average. Thanks for the reply.
Just to be clear, there are multiple microstates with different U which correspond to...