supersymmetry - T-Duality between Type HE String theory and Type HO string theory

My question is regarding T-Duality between the 2 Type H string theories.

I know that the Type II String theories are T-dual to each other because T-Duality changes the sign of the Gamma Matrix so $$\operatorname{T}:{{\cal P}}_{{\mathop{\rm GSO}\nolimits} }^ - \leftrightarrow {{\cal P}}_{{\mathop{\rm GSO}\nolimits} }^ + $$

Since the Type IIB String theory employs the same GSO Projections on the left and right movers while Since the Type IIA String theory employs the different GSO Projections on the left and right movers, the 2 theories are T-Dual to each other.

However, when one considers T-Dualities on the Type HE string theory and the Type HO String theories, why are they T-Dual to each other?

I presume that maybe T-Duality switches $\operatorname{Spin}(32)/\mathbb Z^2$ and $E(8)\times E(8)$, so therefore, the theories to T-Dual to each other. But if so, then why? Why does T-Duality switch $\operatorname{Spin}(32)/\mathbb Z^2$ and $E(8)\times E(8)$?

Thanks in advance.

Answer

One shouldn't imagine the T-duality between the two heterotic strings to be a $Z_2$ group, like in the case of type II string theories' T-duality. In type II string theory, there is only one relevant scalar field, the radius of the circle producing T-duality, and it gets reverted $R\to 1/R$ under T-duality.

In the heterotic case, it's more complicated because more scalar fields participate in the T-duality. Instead of a $Z_2$ map acting on one scalar field, one must correctly adjust the moduli, especially the Wilson lines generically breaking the 10D gauge group to $U(1)^{16}$, and find an identification between the points of the moduli space of the two heterotic string theories: there is one theory at the end.

A fundamental reason why the T-duality holds is that one may define the heterotic string theories in the bosonic representation, using 16-dimensional lattice $\Gamma^{16}$ which is the weight lattice of $Spin(32)/Z_2$, and $\Gamma^{8}\oplus \Gamma^8$ which is the root lattice of $E_8\times E_8$.

One may also describe the compactification on a circle in terms of lattices. It corresponds to adding ($\oplus$) the lattice $\Gamma^{1,1}$ of the indefinite signature to the original lattice. The extra 1+1 dimensions correspond to the compactified left-moving and right-moving boson (of the circle), respectively.

Now, the key mathematical fact is that the 17+1-dimensional even self-dual lattice exists and is unique which really means $$\Gamma^{16}\oplus \Gamma^{1,1} = \Gamma^{8}\oplus \Gamma^8\oplus \Gamma^{1,1}$$ Even self-dual lattices in $p+q$ dimensions (signature) exist whenever $p-q$ is a multiple of eight and if both $p$ and $q$ are nonzero, the lattice is unique.

It's unique up to an isometry – a Lorentz transformation of a sort – and that's how the identity above should be understood, too. So there is a way to linearly redefine the 17+1 bosons on the heterotic string world sheet so that a basis that is natural for the $E_8\times E_8$ heterotic string gets transformed to the $Spin(32)/Z_2$ string or vice versa. The compactified boson has to be nontrivially included in the transformation – the 17+1-dimensional Lorentz transformation that makes the T-duality manifest mixes the 16 chiral bosons with the 1+1 boson from the compactified circle.

A different derivation of the equivalence may be found e.g. in Polchinski's book. One may start with one of the heterotic strings and carefully adjust the Wilson lines to see that at a special point, the symmetry broken to $U(1)^{17+1}$ is enhanced once again to the other gauge group.

Vacuum energy of a real Klein-Gordon field

Hamiltonian for a Klein-Gordon field can be written as - $$H= \int \frac{d^3p}{(2\pi)^3 \omega_{\vec{p}}}[a^{\dagger}_\vec{p}a_\vec{p}+\frac{1}{2}(2\pi)^3\delta^{(3)}(0).\tag{1}]$$ In one of my lecture notes on QFT, It is written that - In absence of gravity, we can neglect the second term in above equation which will lead us to- $$H= \int \frac{d^3p}{(2\pi)^3 \omega_{\vec{p}}}a^{\dagger}_\vec{p}a_\vec{p}.\tag{2}$$ It is obvious that second term in the first equation will make vacuum energy infinite. But Neglecting this term is the best we can do? Something is infinite and just for our convenience, we are putting it to zero. How is this logically and mathematically justifiable? Secondly, What is the role of gravity in neglecting this term? Why can we not neglect this term if gravity is present?

newtonian mechanics - In perfect vacuum (no external force), if I apply a small amount of force to a body, will it continue to accelerate forever?

According to the equation $F = ma$, the greater the force applied to a body, the greater its acceleration. But in a place where there is no gravity or any other kind of external force, if I apply a force to a body, will it continue to move forever with a uniform velocity, or will it continue to accelerate forever?

Answer

The object accelerates while the force is applied to it. When the force stops acting on the object the object maintains it's current momentum (therefore it's velocity) until another force is applied to the object.

Because of e=mc^2 however, the amount of force required to for a given amount of acceleration will become greater and greater until you reacha point where you need infinite force to acclerate an object past a certain point (the speed of light).

electromagnetism - What does a Galilean transformation of Maxwell's equations look like?

In the 1860's Maxwell formulated what are now called Maxwell's equation, and he found that they lead to a remarkable conclusion: the existence of electromagnetic waves that propagate at a speed $c$, which turns out to be the speed of light, implying that light is an electromagnetic wave. Now the fact that Maxwell's equations predict speed of light is $c$ suggested to Maxwell and others that Maxwell's equations are not actually true in all frames of reference. Instead, they thought, Maxwell's equations only exactly true in one frame, the rest frame of the aether, and in all other frames they would have to be replaced by other equations, equations that were invariant under Galilean transformations in order to conform to the principle of relativity. These other equations implied that the speed of light in other frames was actually $c+v$ or $c-v$, where $v$ is the speed of the aether. But then the Michelson-Morley experiment, which was intended to find the speed $v$ of the aether, ended up showing that the speed of light was $c$ in all frames, apparently contradicting the principle of relativity. But Einstein showed that this doesn't contradict the principle of relativity at all, it's just that you need to rethink your notions of space and time.

But my question is, what are the equations that people thought were true in frames other than the aether frame? To put it another way, what are the equations you obtain if you apply a Galilean transformation to Maxwell's equations? (As opposed to a Lorentz transformation which leaves Maxwell's equations unchanged.)

I've actually seen the equations obtained before. They were formulated by some 19th century physicist, maybe Hertz or Heaviside, and they involve adding velocity-dependent terms to the Ampere-Maxwell law and Faraday's law. (Dependent on the velocity of aether, that is.) But I don't remember the details.

Answer

I'm no expert on the historical development of the subject, however I will offer a derivation.

Consider two frames of reference $S$ and $S'$, and suppose that $S'$ moves with speed $\textbf v$ with respect to $S$. Coordinates in $S$ and $S'$ are related by a Galileian transformation: $$\begin{cases} t' = t \\ \textbf x' = \textbf x-\textbf vt\end{cases}$$ To find how the fields transform, we note that a Lorentz transformation reduces to a Galileian transformation in the limit $c \to \infty$. In fact, under a Lorentz transformation the fields transform like: $$ \begin{cases} \textbf E' = \gamma (\textbf E + \textbf v \times \textbf B) - (\gamma-1) (\textbf E \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\ \textbf B' = \gamma \left(\textbf B - \frac{1}{c^2}\textbf v \times \textbf E \right) - (\gamma-1) (\textbf B \cdot \hat{\textbf{v}}) \hat{\textbf{v}}\\ \end{cases}$$ Taking the limit $c\to \infty$ so that $\gamma\to 1$, we obtain the Galileian transformations of the fields: $$ \begin{cases} \textbf E' = \textbf E + \textbf v \times \textbf B\\ \textbf B' = \textbf B\\ \end{cases}$$ We can then invert the transformation by sending $\textbf v \to -\textbf v$: $$ \begin{cases} \textbf E = \textbf E' - \textbf v \times \textbf B'\\ \textbf B = \textbf B'\\ \end{cases}$$ By the same reasoning, can obtain the Galileian transformation of the sources: $$ \begin{cases} \textbf J = \textbf J' + \rho' \textbf v\\ \rho = \rho'\\ \end{cases}$$ We know that the fields and sources satisfy Maxwell's equations in $S$: $$ \begin{cases} \nabla \cdot \textbf E = \rho/\epsilon_0\\ \nabla \cdot \textbf B = 0\\ \nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\ \nabla \times \textbf B = \mu_0 \left(\textbf J +\epsilon_0 \frac{\partial \textbf E}{\partial t} \right)\\ \end{cases}$$ Replacing the fields and sources in $S$ with those in $S'$ we obtain: $$ \begin{cases} \nabla \cdot \textbf (\textbf E' - \textbf v \times \textbf B') = \rho'/\epsilon_0\\ \nabla \cdot \textbf B' = 0\\ \nabla \times \textbf (\textbf E' - \textbf v \times \textbf B') = -\frac{\partial \textbf B'}{\partial t}\\ \nabla \times \textbf B' = \mu_0 \left(\textbf J' + \rho' \textbf v +\epsilon_0 \frac{\partial (\textbf E' - \textbf v \times \textbf B')}{\partial t} \right)\\ \end{cases}$$ As a last step, we need to replace derivatives in $S$ with derivatives in $S'$. We have: $$\begin{cases} \nabla = \nabla' \\ \frac{\partial }{\partial t} = \frac{\partial }{\partial t'} - \textbf v \cdot \nabla\end{cases}$$ Substituting and removing the primes and using vector calculus, we obtain: $$ \begin{cases} \nabla \cdot \textbf E + \textbf v \cdot (\nabla \times \textbf B) = \rho/\epsilon_0\\ \nabla \cdot \textbf B = 0\\ \nabla \times \textbf E = -\frac{\partial \textbf B}{\partial t}\\ \nabla \times \textbf B = \mu_0 \left(\textbf J + \rho \textbf v +\epsilon_0 \frac{\partial}{\partial t}( \textbf E - \textbf v \times \textbf B) - \epsilon_0 \textbf v \cdot \nabla (\textbf E - \textbf v \times \textbf B) \right)\\ \end{cases}$$

In a vacuum, we can take the curl of the fourth equation to obtain: $$c^2\nabla^2 \textbf B = \frac{\partial^2 \textbf B}{\partial t^2} + (\textbf v \cdot \nabla)^2 \textbf B - 2 \textbf v \cdot \nabla \left(\frac{\partial \textbf B}{\partial t}\right)$$ Substituting a wave solution of the form $\textbf B \sim \exp{i(\textbf k \cdot \textbf x -\omega t)}$ We obtain an equation for $\omega$, which we can solve to obtain: $$\omega = -\textbf v \cdot \textbf k \pm c |\textbf k|$$ Therefore the speed of propagation is the group velocity: $$\frac{\partial \omega}{\partial \textbf k} = -\textbf v \pm c \hat{\textbf{ k}}$$ which gives you the expected $c\pm v$ with an appropriate choice of $\textbf v$ and $\textbf k$.

history - Save our WWII planes

This is an actual piece of real history, and an illustration of smart thinking.

During WWII, Allied airforces were having a difficult time getting their bombers to withstand German artillery. Their planes desperately needed some armour to increase the chances of them surviving bomb runs. Unfortunately armour is heavy, so it could only be applied in specific areas where it was absolutely needed.

A mathematician named Abraham Wald examined the surviving planes, making a note of where the artillery damage was most severe. The military felt these damaged areas were obviously the most vulnerable parts of the plane and therefore should should have the armour applied to them. Wald, however, recommended that armour be added to the areas that hadn't been damaged by the artillery fire.

Why?

Answer

Because ...

... the examined planes made it home more or less safely. That means that the holes weren't critical.

On the other hand, the planes that were shot down and therefore couldn't be examined were hit in other places — the places that could probably do with some extra armour.

mathematics - Use 2, 0, 1 and 8 to make 109

Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order that equals 109. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(\sqrt{8*2})!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $109$ will get plus one from me.

Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get 109, but will not mark them as correct.

Here are some examples to this problem:

many thanks to the authors of these questions for inspiring this question.

Answer

I think...

$\sqrt{\frac{12!}{8!} + 0!} = \sqrt{11881} = 109$

Capacitors and external electric fields

I am grading questions from the lab manual provided by the professor to review with the students. I am trying to wrap my head around one of the questions about what an external electric field would do.

Assume two parallel plates, with a positive charge on the right, and negative/ground on the left.

If there is an external field aligned with the capacitor plates (positive on right), that field will result in a greater negative charge on the left plate, thus raising the voltage, and thus the charge, correct?

Then in the opposite case, the external field has its positive source on the left side of the capacitor. It will repel positive charge on the right plate, thus decreasing the voltage potential between the plate, thus lowering the "capacitance".

Is "lowering the capacitance" the correct term, since "capacitance" is a fixed property of the component geometry? Or does capacitance also mean the amount of charge that is held by the plates? When voltage goes down, the amount of charge held goes down, thus "capacitance" goes down, too?

Thank you for your feedback.

Answer

Assuming you mean you have a full electrical circuit and you're wondering what an external field does to it:

The external electrical field would change the voltage, and thus the charge on the capacitor. This is not changing the capacitance.

Generally, though, capacitors are very small, and so the potential difference across the capacitor would be very small for reasonable electrical field magnitudes anyway.