Applying Final Year Physics Project during Lincolnshire SCITT Internship

Recently two of our Physics with Philosophy BSc students Jodie and Olivia have completed a teacher trainings internship in a Lincolnshire school, where she delivered a lesson based around an experiment from her third year undergraduate dissertation project. Read more about their experience (written by Jodie) below.


When the Lincolnshire SCITT (School centred initial teacher training) came to talk about their internship, I didn’t expect to be able to use my final year physics project during the internship to take a lesson in a school.

The internship allows undergraduate students to see what teaching would be like in their chosen subject, in my case Physics. Consisting of three weeks in a school, the internship provided opportunities to observe lessons, act as an informal teaching assistant and the chance to take a lesson. I was placed with fellow student Olivia at William Farr C of E school in Welton. The staff in the science department were eager to provide as many experiences and insights into the teaching life as possible, which without their open support and encouragement, we would never have had such an immersive and rewarding experience.

Returning to the subject of university however, my final year project was called “Visualising Acoustic resonance with lasers”. The brief was to design an outreach activity for A-Level students to teach them about the physics concepts involved in a particular experiment, linking elements of the A-Level curriculum to the activity in order for students to apply the knowledge attained in the classroom to a fun alternative experiment. The experiment in question consisted of a small mirror adhered to an elastic membrane to which reflected a laser onto a nearby wall. A speaker was placed under the elastic membrane so that when frequencies were played the membrane would vibrate and cause the laser to create visual patterns on the wall.

With the opportunity to take a lesson during the internship, it was the perfect moment to put the project to the test.

Ultimately the project wasn’t used with A-Level students and instead with a lively Year 7 class. They provided Olivia with the opportunity to introduce the basic concepts and terminology of waves before handing over to me to take them through the outreach activity I had amended in line with a KS3 Science textbook borrowed from one of the Physics teachers.

The students were very enthusiastic and had lots of questions to ask. At the end of the lesson, the students filled out feedback sheets which asked them if they enjoyed the activity, what they had learnt and what they would improve. I’m pleased to say that all students selected yes for the first question (thankfully) and they were all able to state something they had learnt, so we must have done something right! As for their suggestions for improvements they were all very kind and creative in their ideas and provided lots of food for thought going forward.

Classical logic and Pinocchio

Last week we hosted a work experience student, Dexter Harland-Hackenschmidt, who is interested in mathematics and logic. We tasked him with a logic problem that we had seen on YouTube. The following is his report in the style of a blog post. Enjoy!


In the video “Can you pass this logic test from Brazil?” the following question is proposed:

Assume the following sentences are true:

(1) Pinocchio always lies;

(2) Pinocchio says, “All my hats are green”

We can conclude from these two sentences that:

A) Pinocchio has at least one hat.

B) Pinocchio has only one green hat.

C) Pinocchio has no hats.

D) Pinocchio has at least one green hat.

E) Pinocchio has no green hats.

The video goes on to provide counter examples to options B, D and E, proving that these options cannot be concluded, as they are not true in all cases. After narrowing it down to options A and C, the truth table for implication (→) is invoked to argue that C would make (2) vacuously true (instead of false), and therefore that Pinocchio is not lying and thus the only viable answer, if any, is A. However, it is not made clear why the truth table for implication was used, and the video fails to explain other parts of the question. What does it even mean to conclude something in a logical sense?

An introduction to logic
In order to fully understand the proof for this question we must first understand the basics of classical logic, which the video used in the proposed solution, and first order logic which allows the use of predicates which will be explained later. Classical Logic is built up of 3 core elements: propositions, logical connectives and ‘laws of inference’. Propositions are statements in classical logic which can only be true or false – not both.

Logical connectives connect propositions together and allow us to construct logical statements out of them. The first of the logical connectives used in this proof is conjunction, which is denoted by (ꓥ) and means “and”. For example, the two propositions “it is raining” (R) and “it is cloudy” (C) can be combined using the conjunction connective to form the statement “it is raining and it is cloudy” (R ꓥ C), and for this to be true both the individual propositions must be true as well. The second connective we shall introduce is disjunction, which is denoted by (V) and means “or”. For example, if the propositions “Steve has an apple” (A) and “Steve has an orange” (O) were connected by disjunction to form “Steve has an apple or Steve has an orange” (A V O), then if either of the individual propositions is true then their disjunction is true. The third connective we must define is negation, which is denoted by (¬) and means that if a proposition S is true then the negation of S (¬S) is false. The logical connective referred to as implication is denoted by (→) and means “if … then …”. For example “if Steve has oranges, then he has apples” (O→A). Finally, “if and only if” is denoted by (↔) and means that the propositions on either side have logical equivalence and can be used interchangeably. Basic sentences built out of atomic (i.e. irreducible) propositions A and B with the above connectives return different truth values depending on the truth values of the atomic propositions they act on. These can be summarised into a logical tool called a “truth table” (see Table 1).

Rules of inference allow for simple logical conclusions to be gathered from truth tables (shown below) for example Modus Tollens (if A→B is true and B is false then A is also false). In order to properly apply these connectives to the question being analysed in this blog post, first order logic must also be applied, since this helps in constructing more precise statements to rewrite the question in a logical form. This requires the use of two tools within first order logic, quantifiers such as ꓯ (for all) and ꓱ (there exists), and the use of predicates such as “green” to describe the properties of objects (i.e. hats). This can be written as g = green and h = hat such that g(h) means a hat that is green. These can be constructed into logical sentences such as ꓯh g(h), meaning all hats are green.

Table 1: Truth table of basic sentences built from atomic propositions Q and R and the basic connectives of propositional logic (negation, conjunction, disjunction and implication). The value 0 signifies False and the value 1 signifies True. The truth values of any sentence of propositional logic can be obtained from these basic ones.

Applying first order logic to the problem
In order to apply first order logic to the problem of Pinocchio, we must first rewrite the question using formal notation. In order to do this, we begin by defining the universe or domain of discourse which in this case would be all hats, whether they are owned by Pinocchio or not. Defined as such, let the set of all hats hi be H= {hi} (the subscript ‘i’ refers to the indices with which we label the hats; hi could be any hat within the set, i.e. i = 1, 2, 3, …, N where N is the number of hats). We must also take care to define what the question means by “conclude”. We interpret this to mean that if an option can be concluded, then when Pinocchio makes the statement X the negation of that statement should imply that option for all possible cases. When this is the case, it is called a tautology. An example of a tautology within propositional logic is the logical inference Modus Ponens, which states that if Q → R is true and Q is true, then R is also true. The proof for this statement is demonstrated in truth table below (Table 2).

Table 2: Truth table evaluating the truth values of sentence corresponding to Modus Ponens in all possible cases (Q True, R True or Q True, R False or Q False, R True or Q False, R False). We see that the Modus Ponens sentence is true in all cases thereby making it a tautology i.e. a sentence true in all cases.

The final column in Table 2 is Modus Ponens and since it is true (blue) in all cases, then as described above it is a tautology. We will apply this to Pinocchio’s hats later on but first we must define some of the notation that we will use. Pinocchio will be referred to as ‘P’ and hats will be referred to as ‘h’. The predicate ‘g’ means “green” and the predicate p means “owned by P”.

We can then rewrite the question in formal logic…

Assume the following sentences are true:

  1. If P says X then ¬X; ‘Pinocchio always lies
  2. P says “ꓯhi p(hi) → g(hi)”     ‘Pinocchio says, “All my hats are green” ‘       

(Initially we attempted to use conjunction “p(hi) ꓥ g(hi)” however writing the statement in that way meant that Pinocchio owned all hats in the universe!)

We can conclude from these two sentences that:

A) ꓱhi p(hi)      ‘There exists a hat that is owned by Pinocchio

B) ꓱhi ꓯhj : g(hj) ↔ i=j  ‘There exists precisely one green hat that is owned by  Pinocchio’

C) ¬ꓱhi p(hi)    ‘There does not exist a hat that is owned by Pinocchio’

D) ꓱhi p(hi) ꓥ g(hi)    ‘There exists a hat that is both owned by Pinocchio and is green’

E) ¬ꓱhi p(hi) ꓥ g(hi)    ‘A hat that is both owned by Pinocchio and is green does not exist

(We use X to refer to “ꓯhi p(hi)→g(hi)” for the sake of brevity)

The approach we took in our work was to note that if Pinocchio says X then X is a lie and thus ¬X is true, and if the negation of X implies any of the options is a tautology then the corresponding statement is a valid conclusion from (1) and (2) above. We created a test universe comprising of only two hats and examined the truth table (Table 3) to see if any of the statements were false in any of the models.

We therefore need to negate Pinocchio’s statement X (ꓯhi p(hi)→g(hi)). The first thing to mention is that when negating a sentence that contains the quantifier ꓯ, the ꓯ is replaced with ꓱ. Then, the negation of an implication Q→R can be written as follows:

¬(Q→R) ↔ Q ꓥ ¬R

So, the negation of Pinocchio’s statement reads:

¬X ↔ ¬(ꓯhi p(hi) → g(hi)) ↔ ꓱhi p(hi) ꓥ ¬g(hi)

Table 3: Truth table evaluating the truth value of statements of the form ¬X→Y, where Y represents the proposed conclusions A, B, C, D and E. Only ¬X→A is true in all possible cases and is therefore a tautology. Thus A is the only valid conclusion from the premises (1) and (2).

From Table 3 we can gather that (¬X→B), (¬X→ C), (¬X→ D), and (¬X→E) are not tautologies, since they contain some falsehoods (they are false in some rows or ‘models’). However, (¬X→A) is a tautology in this universe consisting of two hats as it contains only truths in every row (in every ‘model’).

Is this hold if we increase the number of hats arbitrarily? Yes! The way in which we can prove that ¬X→A is a tautology for any number of hats is with the use of the truth table for implication (see the first 3 columns of Table 2). Notice that whenever statement Q is false the statement Q→R is still true. This was mentioned in the video by Presh Talwalker but perhaps not explained fully. What this means is that when the ‘antecedent’ Q is false the ‘consequent’ R is vacuously true; because the statement holds no information it is impossible to prove false and is thus considered true within classical logic. We can apply this to the case of ¬X→A as it is only false in the case of ¬X being true and A being false. This allows us to prove A is a tautology in the following way. A is only false when P owns no hats, however, in that case ¬X is also false because P has to own a hat for ¬X to be true. Therefore whenever A is false ¬X is also false and the implication is true for all possible cases and A can be deemed true.


Although we came to the same logical conclusion as Presh did in his video, we have accomplished it in what we think is a more logical way. We have delved further into formal logic and have applied first order logic which was left unmentioned in the video. If, like us, you were initially confused by the proposed solution, hopefully this blog post has made both the question and the solution more clear.

Undergraduate physics research is on the rise

Today our undergraduate Physics students presented their summer research results at the University’s UROS poster conference (UROS is Undergraduate Research Opportunities Scheme). Physics team was one of the largest in the university while we are the smallest school so far: 3 posters were on Experimental Physics and 3 – on Computational Physics.

Our Physics superheroes of today are:

  • Aaron Adams
  • Christopher Dickens
  • Hannah Thurlbeck
  • Niall Garry
  • Robert Sharp
  • Tom Vale

Their supervisors were: Drs Matt Watkins, Marco Pinna, Fabien Paillusson, Matt Booth and Professor Waqar Ahmed.

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Christine Stokes

Centre for Computational Physics

Christine in MontanaChrissystk - FlickriverChrissystk - Flickriver

I am Christine Stokes, a former Undergraduate student of the University of Central Lancashire where I obtained a first class honours MPhys in Physics. I joined the Computational Physics group on completion of my second year of my degree (2008) when I was awarded an internship through the University. The research involved simulating thin films of Diblock copolymer with an external force of shear applied. The results were presented at a poster conference at the University which it was one of the eight commended. Following this I further presented the research orally at the 24th National Conference for Undergraduate Research in Montana US. In the summer of 2009 I got to experience international research at the Leiden University, funded by UCLan’s International travel bursary Scheme, doing a project for one month. Also over the summer of 2009, Dr Marco Pinna and I was one of the 350 people awarded, across…

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Xun Wu

Centre for Computational Physics

I’m Xun Wu, a Chinese student at the University of Central Lancashire. I majored applied Physics and it’s my third year study now. When in my physics learning process, I have a strong desire to apply my physics knowledge to solve some practical problem and material science is one of my interesting field. It is fortunate that I have such a good opportunity to do my project in Computational Physics. In my final year, I will research the BCP in confinement. Even though computational physics is a brand new field for me, I’m full of confident in doing it well. I’m looking forward to participate the different kinds of activity organized by computational physics group.

I graduated in the summer of 2011 and am currently studying for  MSc in Composites at the Imperial College London.

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Changing the laws of physics research

Centre for Computational Physics

UCLan reports on an undergraduate student success:

The ground-breaking research of an undergraduate Applied Physics student has been published in a top scientific journal.

The work of Ludwig Schreier, an Erasmus exchange student, has appeared on the back cover of the March 2009 issue of Soft Matter, a world leading science journal linking the disciplines of physics, chemistry and biology.

In his third year BSc project, Ludwig predicted how nano-scale material structures could be manipulated with the help of an ordinary electric field. The work was undertaken with the help of Marco Pinna, a research student within UCLan’s Computational Physics Group.

Ludwig comments: “I’d heard good things about UCLan from my fellow students at the University of Applied Sciences in Wiesbaden (Germany) and so I decided to experience it for myself.”

“I was delighted to take up the opportunity offered through the Erasmus exchange program and have really enjoyed my study…

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