Hi, I have attached the question to this post. I understand on the process on getting to the answer in that you use $$\arrowvert 2, 2\rangle=\arrowvert 1,1\rangle \otimes \arrowvert 1,1\rangle$$ and apply the isospin-lowering operator to obtain $$\arrowvert 2,1 \rangle$$. Then I understand you...
Yes I think so, I have made an attempt on part b using a similar method we just used by using dimensionless variables into the Navier Stokes equation:
$$ \rho(\frac{\partial u}{\partial t} + u \cdot \nabla u)= \rho g- \nabla p+ \mu \nabla^2u$$
$$\hat{u}= \frac{u}{U}, \hat{x}=\frac{x}{L}...
Ok I think I understand it now, I rearranged to get this:
$$v_{z0}=\frac{H(0)}{R(0)}v_{r0}, v_{r0}=\frac{R(0)}{H(0)}v_{z0}$$
And since H(0)<<R(0), it means that the magnitude of the vz component is negligible compared with the vr term.
Yes I think that makes sense, so in a way your just relating the r and z terms ( and making them dimensionless) with the parameters within this problem?
Ok I did the substitution and I got this:
$$\frac{1}{\bar{r}R(t)}(\frac{\partial}{\partial r}(\bar{r}R(t)\bar{v_r}v_{r0}))+\frac{\partial}{\partial z}(\bar{v_z}v_{z0})=0$$
I am I right in thinking the $$\bar{r}R(t)$$ term cancels out with the other term? And also I wasn't sure if you could...
I am right in thinking that we can right the equation like this: $$v_r \delta r+v_z \delta z=0 $$. From this and the volume, only vr depends on r meaning the derivative with respect to z doesn't affect r meaning you can consider it as negligible.
I probably have, it's just I can't remember at the moment. I am I right in thinking you can reduce the equation to this: $$v_r \delta r+v_z \delta z=0 $$. And I believe this is dimensionless?
Ah right yes that's true it does factor out: $$\rho(\frac{1}{r}(\frac{\partial}{\partial r}(r v_r))+\frac{\partial}{\partial z}( v_z))=0 $$
Yes, I have learned about it I just haven't been told that I can use it in this situation. Am I right in thinking I need to make a dimensional matrix for...
Oh ok, I assume it's a steady flow therefore the first term goes to 0. I think this is the equation: $$\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{\partial}{\partial z}(\rho v_z)=0 $$
I'm using latex overleaf, I assume that works with this site?
Oh my bad this is the equation I believe: \frac{\partial\rho}{\partial t}+\frac{1}{r}(\frac{\partial}{\partial r}(r\rho v_r))+\frac{1}{r}(\frac{\partial}{\partial \theta}(\rho v_\theta))+\frac{\partial}{\partial z}(\rho v_z)=0 . For an incompressible fluid the density is constant and that the...
Ok I'll make sure to do that next time I post something on here.
I think it's this differential equation equation: δρ/δr+1/r(δ/δr(rρvr)+1/r(δ/δθ(ρvθ)+δ/δz(ρvz).
Hi sorry about the way I've posted I'm new to this site. Anyway basically I've been set this question which should be attached to this post, I have attempted to do this question but I'm having trouble in forming an equation in the first place. I'm unsure where to start, I understand I need to...