Jared DeWitt

May 31, 2012
3:34 pm
by: Jared DeWitt

Bell’s Inequality

John Bell was a smart dude. He looked a little like the guy from Breaking Bad. Who, on a side note, decided to the use Heisenberg for his alias in the show. I think he should have used Bell personally since he looked like him).

The Problem
The 2nd probability I explained in the previous post was hard to swallow by a lot of people, including Einstein. So much in fact that Einstein got together with a few students of his (Podolsky and Rosen) and wrote a paper outlining that this kind of probability was nonsense and there must be something that’s just not being measure (A hidden variable). The paper is referred to as the Einstein, Podolsky, Rosen paper (EPR). The paper itself is long and I’ve never actually read it all the way though. I’m not smart enough. But here’s the gist of it as I understand it.

EPR
Heisenberg’s Uncertainty principle says that we cannot know the position and momentum of a particle with 100% certainty. Mathimatically it’s shown as the following if $x = position$ , $p = momentum$ and $\hbar = Plancks's Constant$

$\Delta{x}\Delta{p} \approx \hbar$

The EPR attacks this directly. The EPR proposed the following.

Suppose two particles (A and B) are shot at each other at very high speeds. These two particles would pass so close to each other, that they would pick up their traits. Now the two particles are zipping away from each other with the exact same momentum and position in relation to the point where they passed each other. Now take a measurement of particle A’s momentum and position. This would be an indirect and exact measurement of particle B’s momentum and position.

This seems pretty logical if you’re thinking about the problem from a Newtonian point of view. What a quantum theory point of view would suggest is that as soon as you make a measurement on particle A, particle B would then alter it’s own properties as if you had measured it directly.

So we’re left with two possibilities. Either Einstein is right and there is a hidden variable that we don’t yet understand, or the quantum physics is correct and there is a spooky “action at a distance” as it where to be later explained.

The Solution
So arises the question… “How do we test this?”. John Bell came up with a way do this. He started by outlining a theoretical experiment. This experiment involves sending two entangled quantum particles down two separate mediums, and filters (polarizer) at then end of these mediums. After the quantum particle passes through the filter… they could then be measured by a coincident counter what the resulting particles properties are. This could then tell them with if the particles are just acting by themselves (hidden variable) or if they influence each other (quantum entanglement).

Bell came up with a very simple inequality and graph to use.

$1 > x > -1$

He suggested that the experimenters do the experiments and plot it out. If the plot looks like the straight lines on the graph, then that would suggest there is a local hidden variable. If the plot took the shape of tiny curves, then that would suggest that there is an action at a distance and two particles are “entangled”.

This experiment has been done multiple times over history and the majority of the time, it appears that quantum entanglement is the winner.

Conclusions
It’s hard to make a conclusion in quantum physics. It’s a field that rarely gives “answers”. It seems like it just gives information in the form of more questions, which some could argue is true about most studies in science and specifically in theoretical physics. You also have to be careful not to get in the habit of trying to “figure out what’s going on”. You can’t picture it in your head like you can with classical Newtonian physics. All you can really do is show what the data is telling you. And in this case, the data is suggesting that making a change to one particle can and will instantly affect the other particle if the two are entangled. This is kind of crazy stuff and really challenges our ideas of what a particle really is. Leading to more questions, more math, and more experiments.

Posted in Jared's blog, Quantum Electrodynamics, Quantum Physics

May 29, 2012
10:28 pm
by: Jared DeWitt

Probability

So before moving in to such things as Heisenberg’s Uncertainty Principle and Bell’s Inequality I thought I should talk about probability first as I understand it.

There are two types of probability. The first kind is generated by our lack of knowledge. This is the kind of probability that we’re used to.

We can think of the roll of a single dice for a simple example. The probably of rolling a six is 1 in 6, or $\frac{1}{6}$ . Theoretically, if we knew the exact way the dice was thrown, the exact configuration of the table pits that it bounced on, and the air currents, and etc etc etc… we could predict how the dice would land (provided we had enough processing power to handle all those variables). So this $\frac{1}{6}$ probability is only generated because we don’t know all these variables.

The other type of probably is one that is a little harder to comprehend. Even Einstein didn’t really believe it at first. It’s the probability built in to quantum mechanics. A simple example pops up when trying to measure the exact position of an election. It seems to jiggle around a bit. You would think that if we knew everything there is to know about the system, we could predict with 100% certainty where it would be found. The fact that they never could do this suggested to Einstein that there was something we didn’t know about the system. He called this a Hidden Variable. He refused to believe it and was quoted saying something to the affect of “God doesn’t play dice”. Through many years of experiments, this was found to be false. He does play dice. And we can now transition in to Bell’s inequality.

Posted in Jared's blog, Quantum Physics

May 29, 2012
10:05 pm
by: Jared DeWitt

Einstein’s Twin Paradox

I know I said I was going to talk about Einstein’s Twin paradox, but that was before I realized there was no paradox at all. I’ll keep it short and explain why.

Here’s the scenanio

Two twins are both born on Earth. When they get older, one of the twins gets in a spaceship and flies away from the earth and near the speed of light. The other twin remains on Earth.

The Twin in the spaceship is now aging more slowly than the twin on earth. He will be “younger” as it would suggest.

Now the Twin in the spaceship turns it around and accelerates back to Earth at yet again near the speed of light.

When the two twins meet up again on Earth, the one who traveled in the spaceship would be “younger” as the one on Earth.

Here’s the Paradox.

It’s been argued that you could think of it a slightly different way. Once the twin in the spaceship has stopped and turn back around. You could think of the Earth as the spaceship. So therefore, the twin still on Earth would now be the one traveling near the speed of light and when the two met back up, they would once again be the same age. It’s as if the twin on Earth “caught up” to the twin in the spaceship.

This is wrong. You cannot think of the twin on Earth as the traveler because he didn’t experience acceleration. Only the twin in the spaceship felt it when he stopped. And felt it when he turned around, and then again felt the acceleration as he sped back to Earth.

Conclusion

Einstein was right and there is no paradox.

Posted in Jared's blog, Relativity

January 18, 2012
3:55 pm
by: Jared DeWitt

Time Dilation

The last post showed an example of space dilation. We showed that, with the Michelson Morley experiment, that things didn’t add up unless we adjusted by the Lorentz factor ( $L_{||}=\frac{L_0}{\sqrt{1-v^2/c^2}}$ ). With time dilation we’ll see that this still holds true ( $\Delta t' = \frac{\Delta t}{\sqrt{1-v^2/c^2}}$ )

We’ll go through this post and show how we arrive at this. Let’s take a look at the following figure. It represents a “light clock”. This clock consists of two mirrors and a light beam bouncing back and forth between the two. Each “tick” of the clock happens when the light beam hits a mirror. At rest, our clock looks like Fig 1.

Fig 1

We can find the change in time ( $\Delta t$ ) very easily.

$\Delta t = \frac{2L}{c}$

We now have our light clock and it functions really well. Now let’s give it to Bob. Bob now boards a train heading to work. Bob uses his new light clock on the train and observes exactly what is shown in Fig 1. If he didn’t, he would then be able to calculate how fast he is traveling. That would violate what we found in the last post. You can’t tell your absolute velocity!! So… no matter how fast the train travels, he still observes his light clock behaving like Fig 1.

Now let’s go hang out with Alice, who is sitting on a bench at the platform down the track. To her, Bob and his train are approaching. As Bob in his train passes Alice, she observes his light clock behaving like Fig 2.

Fig 2

The light would appear to Alice as moving in more of a zig-zag path. Let’s try and find the change in time as observed by Alice ( $\Delta t'$ )

Alice observes $\Delta t' = 2D/c$ . This is similar to what Bob observers, $\Delta t = 2L/c$ . We need to find what $D$ is in terms of L so we can compare the two.

So back to using the pythagorean theorem ( $a^2 + b^2 = c^2$ ). We can see that…

$(1/2v\Delta t')^2 + L^2 = D^2$

Solving for $D$ gives us…

$D = \sqrt{(1/2v\Delta t')^2 + L^2}$

Let’s now substitute our $D$ back in to our initial $\Delta t' = 2D/c$

$\Delta t' = \frac{2\sqrt{(1/2v\Delta t')^2 + L^2}}{c}$

Solving for $\Delta t'$ gives us…

$\Delta t' = \frac{\Delta t}{\sqrt{1 - v^2/c^2}}$

So what does this all mean? It means, that Bobs freaking cool light clock seems to be running more slowly for Bob than for Alice (as observed by Alice). You could even go so far as to say that time itself moves more slowly for Bob. This is quite the assumption. People might say… “Well, just because the light clock moves slower doesn’t actually prove that time is moving slower”. Well, actually it does. Time itself is measured by ticks. Whether that is generated by light ticks on a mirror, or by a quarts crystal in a modern day clock, it all behaves this way.

All of this is in Einstein’s relativity. As the speed of an object approaches the speed of light, time will slow down for the object itself as observed by an object at rest. Hence why a photon doesn’t decay. To decay would mean that you can experience time. If a photon had a watch slapped on his wrist… the watch would never move!! Crazy. Fig 3 shows how much time slows down as an object approaches the speed of light.

Fig 3

In the next post I’m going to talk about Einstein’s twin paradox, which is super fun. And NO MATH!!! woot.

Posted in Jared's blog, Quantum Physics, Relativity

January 11, 2012
9:50 am
by: Jared DeWitt

The Lorentz Transformation

I’m going to deviate a little bit here because I’m bouncing between a few books. I’m trying to finish up Six Not so Easy Pieces by Feynman. Then after that, I’m going to work on Entanglement. So I want to make sure I have a little bit of a grasp on relativity before moving on.

Let’s start with a the Michelson Morley experiment. Basically, these two guys wanted to see if light could be used to find the absolute velocity of a moving object. In their experiment, they tried to see if they could find the absolute velocity of the Earth through, at the time was thought of as, the “ether”. They theorized that light moved through a luminescent ether, and Earth traveled through this ether.

They ended up creating an apparatus to try and “catch” light moving through this ether. Fig 1 shows the apparatus.

Fig 1

That’s a lot of lines in a picture. Let’s go over what it’s doing.

You start with a light source (A). Light moves along and hits a partially silvered mirror (B). Since the mirror is partially reflective, some of light will get reflected up towards towards a full mirror (C) and some of light will pass through B to hit another full mirror (E). When the light gets reflected back from C and E, you can see that some of light will then pass through B again and make it’s way to down. What we’re left with are two different beams of light (D and F).

At rest, assuming $l_1 = l_2$ , the two light beams should be in phase and amplify each other.

Now let’s assume that the entire apparatus is moving, as it would be if on the surface of the Earth. Let’s say it moves by u distance. We get a new position for B, C, D, E, and F. We will just label these B’, C’, D’, E’, and F’. The thought was that if the apparatus was in motion in by u, then the two light beams would be out of phase and we would see some interference pattern emerge.

There is a slight mechanical problem to this that we should address. The degree of accuracy to make $l_1 = l_2$ would be quite challenging. In the end though, it doesn’t matter because you could rotate the apparatus and see an interference pattern. This suggests that the experiment is sensitive enough to discover what we are looking for.

If we can show that the time it takes for light to go from B→E and back is different than the time it takes to go from B→C and back… than the light beams should be out of phase and we’ll see an interference.

First we need to find the time it takes the light to get from B→E. We’ll label this as $t_1$ . Since we’re in motion, the distance from B→E is $ut_1$ .

To find $t_1$ we start with the following.

$ct_1 = l_1 + ut_1$

then we see…

$t_1 = \frac{l_1}{(c-u)}$

Now we need to find the time it takes to get from E -> B, which we’ll label as $t_2$

$ct_2 = l_1 - ut_2$

which we then see

$t_2 = \frac{l_1}{(c+u)}$

Now let’s add them up to find $t_1 + t_2$

$t_1 + t_2 = \frac{l_1}{(c-u)} + \frac{l_1}{(c+u)}$

Simplified down to

$t_1 + t_2 = \frac{2{l_1}c}{c^2-u^2}$

We will rewrite this as the following to make comparing easier later on.

$t_1 + t_2 = \frac{2l_1/c}{1-u^2/c^2}$

We now need to find $t_3$ . To this we’ll use the pythagorean theorem ( $a^2 + b^2 = c^2$ ).

Back on Fig 1, you can see the light moves in an angle. $c$ is the hypotonus of a right triangle $ul_{2}c$ . To find $t_3$

${ct_3}^2 = {l_2}^2 + {ut_3}^2$

Solving for $t_3$

$t_3 = \frac{l_2}{\sqrt{c^2-u^2}}$

coming back from C→B, we see the distance should be the same… so we double it to find $2t_3$

$2t_3 = \frac{2l_2}{\sqrt{c^2-u^2}} = \frac{2{l_2}/c}{\sqrt{1-u^2/c^2}}$

We now have the time it takes to go from B→E and back( $t_1 + t_2$ ), and from B→C ( $2t_3$ )and back.

$t_1 + t_2 = \frac{2l_1/c}{1-u^2/c^2}$
$2t_3 = \frac{2{l_2}/c}{\sqrt{1-u^2/c^2}}$

As you can see, $t_1 + t_2 \neq 2t_3$ . From that we should expect to see interference in our experiment.

The end result was that they saw no interference. Sad day No matter in which direction they rotated it, no interference pattern showed up.

The experiment was thought to be a bust. This was puzzling to say the least. Lorentz was the first to try and give a solution. He suggested that all material bodies in motion would shrink in the direction of their motion. A modern way of looking at this idea, is to think of your car being shorter in length when it is moving down the freeway than if it was sitting at a stop light. Crazy. He called the length, “length parallel” ( $L_{||}$ ).

He proposed the following

$L_{||} = L_0\sqrt{1-u^2/c^2}$

We now have what’s called the Lorentz Transformation, and this will be important in a later post if I do it.

Now, when using $L_{||}$ for $l_1$ we can find $t_1 + t_2$ again. Turns out to be

$t_1 + t_2 = \frac{2{l_2}/c}{\sqrt{1-u^2/c^2}}$

We now see that $t_1 + t_2 = 2t_3$ and thus verifies what we see in the experiment (no interference).

To sum up what this tells us. If an material object is in motion, it should shrink by the Lorentz transformation. From what we observe nature doesn’t allow us to measure our velocity through space (absolute velocity). It only allows us to measure our velocity compared to something else (relative velocity). It “tricks” us by shrinking stuff as they move.

Nature’s a tricky little bitch.

Posted in Jared's blog, Relativity

January 4, 2012
5:21 pm
by: Jared DeWitt

Beta Decay

Beta decay is a type of radioactive decay that yields a beta particle. A beta particle can be either be an election (e^-) or a positron (e⁺), a positron being the antiparticle of a an electron.

In β^- decay, we see a neutron turning in to a proton. It does this by converting one of it’s down (d) quarks to an up (u) quark. Neutrons consist of an up quark and two down quarks (udd) and protons have two up quarks and one down quarks (udu). The below Feynman diagram shows how this happens.

At the bottom left you see a neutron with an up quark and two down quarks. When one of it’s down quarks changes to an up, it emits a weak negative boson (W^-). The W^-decays rather quickly because it’s not stable. It decays in to an election and a election antineutrino. (sorry I can’t find a good way to show an election antineutrino in html)

n → p + e^- + ¯v_e

In β⁺ decay we see an proton turn in to a neutron by emitting a positron (e⁺) and an election neutrino. This is a little different because a neutron weighs more than a proton. So we have to add energy in to our system to account for the addition mass.

Energy + p → n + e⁺ + v_e

I can’t find a Feynman diagram to show this one. Maybe I’ll just edit this post later.

Posted in Jared's blog, Particle Physics, Quantum Electrodynamics

January 4, 2012
4:50 pm
by: Jared DeWitt

Feynman Diagrams

Before moving on, I thought I’d better go over Feynman diagrams (FD). What they are, why we use them, and how to read them.

Richard Feynman thought them up to help explain in a visual manner of how particles interact. He used them heavily in explain quantum electrodynamics in a simplified way. The following picture is an example of a Feynman diagram.

Fig 1: This diagram in is showing, if i’m not mistaken, an electron (e^-) releasing a photon(γ), changing direction. Then the photon gets absorbed in to a different election changing its momentum.

You can think of a FD as a 2d graph. In a normal 2d graph, you would label the vertical axis Y, and the horizontal axis X. In a FD, the vertical axis is usually time, and the horizontal axis is space.

In Fig 1, the first election starts out by moving to our right, then emits a photon and ends up moving to the left. The electron on the right, starts out by moving to the left. Then absorbs the photon and is now moving to our right.

This is actually an interesting thought. In my last post, I mentioned that photons carry the electromagnetic force. If you think about this diagram, all it’s showing is two electrons getting close together. Then they repel, like two magnets. The “repulsion” is carried by the photon! Someone please correct me if I’m wrong, this is just how I understand it.

We’ve now seen a simple FD. In the next post, we’ll see a more complex one while I try to explain beta decay of a neutron in to a proton.

Posted in Jared's blog, Particle Physics, Quantum Electrodynamics

January 3, 2012
4:54 pm
by: Jared DeWitt

Let’s start with the Standard Model

There are two main categories of elementary particles. Fermions and Bosons.

Fermions

Fermions are divided in to two categories.

Quarks
Leptons

Quarks

As you can see in the diagram, there are 3 generations of quarks.

up/down (u/d)
charm/strange (c/s)
top/bottom (t/b)

up/down are usually found is Hadrons (Neutrons/Protons/Pions). I’m still not sure what s/c and t/b usually do. I’ll probably have to go dig deeper and edit this post.

I’m trying to keep this short, so let’s move on to Leptops.

Leptons

Leptons are charged fermions and, like quarks, have 3 generations as well. Each getting progressively heavier.

election (e)
muon (μ)
tau (Τ)

Bosons

Bosons are particles that carry a force.

photon (γ)
charged Weak bosons (W^-/+)
zerocharge weak boson(Z)
gluon(G)
higgs (H) – Theoretical!

Photons

The photon carries the electromagnetic force.

Charged Weak Bosons

Charged weak bosons carry the weak nuclear force. Depending on the charge, it will either carry a – or a +. I’ll go more in to how these play a role in beta decay in the next post.

Zerocharge Weak Bosons

Zerocharge weak bosons are very similar to charged, but they just don’t carry a charge. They’re mainly responsible for carrying kinetic force. I’ll also go more in to detail on the next post about how these interact.

Gluons

Gluons carry the strong nuclear force. The strong nuclear force is what holds the quarks together.

Higgs

Higgs have never been observed. But they are thought to be what gives all matter their mass. I might go in to detail about these when one is observed. I will try to explain how they are possibly detected in a later post.

Posted in Jared's blog, Particle Physics

January 3, 2012
2:02 pm
by: Jared DeWitt

Why I ramble about this stuff

Since I was about 11 years old, I’ve been reading books about the cosmos. I read “A Brief History of Time” in 6th grade. I didn’t understand too much of it, but at least I tried. This book kick started my interest in the cosmos. My love for the cosmos, has now developed in to the love of physics. So much so, that after finishing my degree in computer science I decided to try graduate work in the physics. I failed horribly. I didn’t even finish a semester. I quickly found out that working full while going to school worked well for undergraduate degrees, but not so well for gradute work. And I don’t have the discipline needed for it as well. I made the decision to keep learning as much as I could on my own. While keeping my full time job in the technology industry.

I’ve found that a great way for me to understand something better, is to try and teach it to someone else. While my wife loves to listen to me ramble about this stuff, she isn’t always around. So I will be doing at minimum two posts about some things that I want to clear out of my head. If I like it, I will post more. If not, it will fade.

So that’s it. More to come soon hopefully.

Posted in Jared's blog, Particle Physics, Quantum Chomodynamics, Quantum Electrodynamics, Quantum Physics

January 27, 2011
9:38 pm
by: Jared DeWitt

The Tennis concert

We’re sitting here at the tennis concert. I’ve never hears of these guys but they’re kinda rad. It’s freaking freezing in here.

So I’m sitting here and to my delight, I see this glimmering machine. Could it be? A retro PAC man?! I slowly approach. What do I see? A miss PAC man…. Gay

Posted in Jared's blog

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