Tuesday, September 30, 2014

Finding Quenching Field Magnitude Using Levitation Force: Lab Book 2014_09_29


Summary:  Working more on using the superconductor to detect its own quenching field.  The initial setup is shown below.  The quenching test is described in the following.  A YBCO superconductor is placed between the poles of a very uniform magnet and then cooled into its superconducting state.  The field frozen into the sample at the state transition opposes the fringing fields on the magnet.  However, had the magnetic field been strong enough to quench the superconductor, the results would have been the pendulum swinging freely beyond the pole pieces' diameter until it encountered a field less than its critical field at which point, it would have re-entered the superconducting state and frozen in those field lines, suspending itself.  There's another realization of this process that will be tested today.  The pendulum is again suspended in a uniform field and the field is slowly increased.  It is suspected the sample will be deflected until the quenching field is reached at which point, the pendulum will fall back to its equilibrium vertical position.  By measuring the angle of the pendulum, the levitation field could also be determined.




The superconductor is placed in a Styrofoam cup supported on a wood plank wedged between the two poles of the magnet.   The magnet gap was set at 2 and 9/16 inches.  This could be much smaller for the sample used here, I just need to find a smaller reservoir.

There are two movies.  The first contains the cooler alarm going off.  After the alarm went off, the magnet current supply was slowly ramped down, and water was added to the reservoir after the cooler was switched off.  The cooler alarm did not start again after it was turned back on, nor when the magnet supply was ramped up to 49 amps. 

The second movie detailed the superconductor not moving while the reservoir slipped out from underneath it. 
We’re measuring the magnetic field with a F. W Bell 5180 Hall Effect Gauss meter.
7.32 – 7.35 kG at a 2 and 9/16 inch gap.

12.8 kG at the gap setting, 1 and 1/8 inch gap setting.
A small Dewar was carved from blue Styrofoam to fit in the smaller gap space, see the first picture below.  The Dewar was suspended as a pendulum between the poles of the magnet as shown in the second picture below.  Dental floss was used to support the Dewar pendulum from the upper yoke of the electromagnet.

The quenching test was as follows:
A YBCO superconductor is placed between the poles of a very uniform magnet and then cooled into its superconducting state.  The field frozen into the sample at the state transition opposes the fringing fields on the magnet.  However, had the magnetic field been strong enough to quench the superconductor, the results would have been the pendulum swinging freely beyond the pole pieces' diameter until it encountered a field less than its critical field at which point, it would have re-entered the superconducting state and frozen in those field lines, suspending itself.  There's another realization of this process that will be tested today.  The pendulum is again suspended in a uniform field and the field is slowly increased.  It is suspected the sample will be deflected until the quenching field is reached at which point, the pendulum will fall back to its equilibrium vertical position.  By measuring the angle of the pendulum, the levitation field could also be determined.




The fringe field produced at the edge of the electromagnet pole, mentioned above, is shown in the diagram below from Lawrence's cyclotron patent application.  Note that the 'magnetic lines of force' become less uniform as the edge of the pole piece is approached.



The small YBCO sample did not quench at this gap and field setting.
Note in the video that at 35 amps during the ramp down, the sample seems to be drawn to the pole piece .  At 10 amps, the lower right corner of the Dewar relaxes.
Link to quenching test video









Friday, September 19, 2014

Writing Activity Metric Tracking

I'm playing around with tracking metrics on my writing activities today.  Clearly I need to enhance my charting presentation skills, but the information here is kind of interesting to me.  It's about me, so of course it is, but it's interesting to think about in terms of why a writing log is useful as well.  Here's what I learned  As the semester has ramped up, I've been doing more writing on EM homework and less on EM notes in preparation for class.  That's not a sustainable model.  Work on the hray presentation an proposal has been ramping up nicely.  I need more detail on what aspects of each project I'm working on and more tracking towards defined goals.


Tuesday, September 9, 2014

Proper Velocity!!! and Getting Index Notation Worked Out: EM II Notes 2014_09_09

Summary:  It looks like I'll finally get a good understanding of the gamma notation for moving proper velocities to lab velocities and back.  It'll be nice to know it inside and out, but a little irksome given all that can be done with the hyperbolic notation we're not using.  I want to maintain my fluency in both.

There may be a subtle second notation for inverted Lorentz transforms.  As it turns out, the subtle notation difference of moving around indices in the top and the bottom with spaces is meant to keep track of which index comes first when you go back to side by side notation.


First, we cover Lorentz transforms, (which are not in fact tensors), and contractions and arrive at the interesting result in equation 1.99:

$\Lambda^\mu_\rho \Lambda^\sigma_\mu T^\rho_\sigma = \delta^\sigma_\rho T^\rho_\sigma$

Which indicates the transpose of the Lorentz transform times itself follows a sort of orthogonality rule making use of contravariant indices.

Q:  Does this obviate the need for the $\eta$ metric?

A:  But wait!  There's so much more!  This is a way to write things without the $\eta$ cruising through everywhere, but it also explains the oddball spacing of the indices.  Maybe I just don't remember this from MacConnel?  First, the Lorentz transform is not a tensor.  Now, we hve that down.  The next bit is how to arrive at the above expression.

$\eta_{\mu\nu} \Lambda^\mu_{\;\rho} \Lambda^\nu_{\;\sigma} = \eta_{\rho\sigma}$

The next step is to plow the $\eta$ on the l.h.s. in and lower the $\nu$ on the second transform.

$\Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta_{\rho\sigma}$

We then raise the sigma

$\eta^{\sigma\lambda} \Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta^{\sigma\lambda}\eta_{\rho\sigma}$

$\eta^{\sigma\lambda} \Lambda^\mu_{\;\rho} \Lambda_{\mu\sigma} = \eta_\rho^\lambda = \delta_\rho^\lambda$

$\Lambda^\mu_{\;\rho} \Lambda^{\;\lambda}_\mu = \eta_\rho^\lambda = \delta_\rho^\lambda$

Now, since we can see that the first two indices are $\mu$s, we can call the above statement a transpose.  The index locations matter to get the transpose to be a transpose, making sure the rows and columns are handled properly.

NOTE:  This doe arise in MacConnel.  His notation is slightly different.  Instaed of $\Lambda^\mu_{\;\rho}$, he writes $\Lambda^\mu_{.\rho}$

We're going to need the delLambertian soon and it's important to note that it is

$\Box = -\partial_0\partial_0 + \partial_i\partial_i = \partial^\mu \partial_\mu$

A few notes follow on why the D'Alambertian is written with one index up and one down.  It has to do with the negative sign in the first entry of the Minkowski metric.  As it turns out, the index up version of the Kronecker delta is the same as the index down version, but not so for $\eta$ because of the $-1$ $00$ entry.  This is all much simpler if you never start writing your indices in the 'wrong' location in the first place.


OK, now for the interesting stuff.  First, with the choice of signature for the Minkowski metric here, we wind up having to write down $d\tau$ as \

$d\tau = -ds^2 = dt^2 - dx^2 - dy^2 - dz^2$

Given $d\tau$ we define the four velocity to be

$U^\mu = \dfrac{dx^\mu}{d\tau}$

Which is technically mixing frames, but then, that's what proper velocity does.  Intersting that we're starting with four velocity, proper velocity here.  It's really nice to get this out of the way early on and benefit from it.

Picture of the Day
Presenting the Takeno metric line element.  Hopefully one of the payoffs of all this will be understanding this more fully, and possibly using it to explain the circular Unruh effect.  However, a far more productive use is in explaining the Thomas Precession.



Saturday, September 6, 2014

Stepping Back Up With Classes: Lab Book 2014_09_06

Summary: Classes started this week.  They're a lot of fun, but they take time.  Consequently, the lab work is moving a little more slowly.  I'm looking into what we can accomplish with a YBCO superconductor sample.  The advantage is that we can test our experimental techniques using relatively cheap liquid nitroogen intead of iquid helium.  The downside is that with the size of YBCO sample we have, the expected maximum energy is only 3 deV which is kind of low without a specialized detector.

For more background on the experiment, please scroll to the bottom of the post.

The percolator peak does not appear when the detector is initially turned on.  The attenuator does however appear to create a rather copious amount of noise.

Suppose we used YBCO as a sample.  The energy we could expect doing a back of the napkin calculation is 3.8 keV.  The flux is approximately 230 photons for our sample size.  For a 25 square mm detector that may be available, this gives a total flux of 57 particles if we’re only 0.5 cm away.  This may work.

Another step would be to cycle the permanent magnet at a few Hertz to quench the superconductor repeatedly and increase the total flux.  The inside of the power supply control box is shown below.


The two potentiometers that control the supply are shown in the table below.  The coarse control is on the left and the fine control is on the right.

One technique might be to replace these potentiometers with a voltage controlled resistor.  The control circuit is shown below.



Background
Hirsch's theory of hole superconductivity proposes a new BCS-compatible model of Cooper pair formation when superconducting materials phase transition from their normal to their superconducting state[1].  One of the experimentally verifiable predictions of his theory is that when a superconductor rapidly transitions, (quenches), back to its normal state, it will emit x-rays, (colloquially referred to here as H-rays because it's Hirsch's theory).

A superconductor can be rapidly transitioned back to its normal state by placing it in a strong magnetic field.  My experiment will look for H-rays emitted by both a Pb and a YBCO superconductor when it is quenched by a strong magnetic field.
This series of articles chronicles both the experimental lab work and the theory work that’s going into completing the experiment.

The lab book entries in this series detail the preparation and execution of this experiment… mostly.  I also have a few theory projects involving special relativity and quantum field theory.  Occasionally, they appear in these pages.

Call for Input
If you have any ideas, questions, or comments, they're very welcome!

References
1.  Hirsch, J. E., “Pair production and ionizing radiation from superconductors”, http://arxiv.org/abs/cond-mat/0508529

Thursday, September 4, 2014

Showing that SpaceTime Intervals are invariant: EM II notes 2014_09_03

Summary:  Continuing notes on the tensor version of the Lorentz tranform.  It's time to start on the second set of examples.


The interval in four space is invariant under Lorentz transforms and is called the Lorentz scalar.


The Lorentz transform also applies to differential distances as,

$dx^{\prime\mu} = \Lambda^\mu_\nu x^\mu$


We were asked in class to work out $x^2+y^2+z^2-t^2 = x^{\prime 2}+y^{\prime 2}+z^{\prime 2}-t^{\prime 2}$

The transforms we'll use are:

$x = \gamma\left(x^\prime + vt^\prime\right)$

$t = \gamma\left(t^\prime + vx^\prime\right)$

Substituting these into the l.h.s. gives

$\gamma^2\left(x^\prime + vt^\prime\right)^2 - \gamma^2\left(t^\prime + vx^\prime\right)^2 = x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} +2vtx + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + 2vxt+v^2x^{\prime 2}\right)= x^{\prime 2} - t^{\prime 2}$

$ = \gamma^2\left(x^{\prime 2} + v^2t^{\prime 2}\right) - \gamma^2\left(t^{\prime 2} + v^2x^{\prime 2}\right) = x^{\prime 2} - t^{\prime 2}$

$\gamma^2 x^{\prime 2} - \gamma^2v^2x^{\prime 2} = \gamma^2 x^{\prime 2}\left(1 - v^2\right) = x^{\prime 2}$

Similarly

$\gamma^2 v^{\prime 2}t^{\prime 2} - \gamma^2 t^{\prime 2} = -\gamma^2 t^{\prime 2}\left(-v^2 + 1\right) = -t^{\prime 2}$

So

$x^{\prime 2} - t^{\prime 2} = x^{\prime 2} - t^{\prime 2}$

Done


Picture of the Day:
Here's a throwback Thursday pic from 1946.  These are the infamous Carter Boys.  My dad is the twin on the right.