Symmetry of the $3times 3$ Cauchy Stress Tensor

When presenting the stress tensor (say in a non-relativistic context), it is shown to be a tensor in the sense that it is a linear vector transformation: it operates on a vector $n$ (the normal to a surface), and returns a vector $t_n$ which is the traction vector. It is then shown that conservation of angular momentum leads to symmetry of the matrix.

However, tensors are more more naturally presented a multilinear functions. I wonder:

• What type of tensor is the Cauchy stress tensor? Is $n$ a vector or a co-vector? What about $t_n$?


• Is there a way to understand the symmetry when thinking of the stress tensor as a function of two vectors (or two co-vectors), under which it will seems intuitive why $\sigma(A,B) = \sigma(B,A)$?

Edit: To clarify, let's look, for example, at the 1st coordinate of the traction vector $t_n$ of an arbitrary normal $n$: This is $\left$. From symmetry, this is equivalent to $\left$ - the inner product of $n$ with the traction vector of a surface orthogonal to $e_1$. Mathematically, I understand why this is correct. But is there any intuitive meaning as to why these two quantities are the same?

thermodynamics - Why don't Wien displacement law curves cross?

In the above image, the curves for different temperature dont intersect anywhere. Stefan-Boltzmann law and Wien displacement law dont preclude the intersection.

Is it because if, for example, they cross each other at some longer wavelength somewhere (after the peak), then that would imply that the body with the lower temperature emits more power than that with the higher temperature for the wavelengths after the crossing point? If yes, then why is that not possible?

Answer

Here are two reasons they can't cross, one from statistical mechanics and one from pure thermo.

• Microscopically, each frequency of light is produced by an independent mode. So asking why the curves don't cross is just asking why the amount of energy in a mode goes up as the temperature goes up, i.e. why the heat capacity is positive. That's true almost all the time, and systems for which it's not true aren't even thermodynamically stable.


• Imagine placing two blackbodies right next to each other. The hotter one must transfer energy to the cooler one, or else we've violated the Second Law of Thermodynamics, so hotter objects must overall emit more. But we can also place a filter in between the blackbodies that only lets a very narrow band of frequencies though. So to avoid breaking the Second Law, a hotter object must emit more at all frequencies.

A more direct proof is to go all the way to Planck's law, but that's not necessary: we can prove the curves can't cross with general principles, without knowing much about the details.

research level - Large gauge transformations for higher p-form gauge fields

Question: What is the large gauge transformations for higher p-form gauge field on a spatial d-dimensional torus $T^d$ or a generic (compact) manifold $M$? for p=1,2,3, etc or any other integers. Is there a homotopy group to label distinct classes of large gauge transformations for p-form gauge field on $d$-dimensional torus $T^d$ or any $M$ manifold ? (shall we assume the theory is a topological field theory, or not necessary?) References are welcome.

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Background: Large gauge transformation has been of certain interests. The Wiki introduces it as

Given a topological space M, a topological group G and a principal G-bundle over M, a global section of that principal bundle is a gauge fixing and the process of replacing one section by another is a gauge transformation. If a gauge transformation isn't homotopic to the identity, it is called a large gauge transformation. In theoretical physics, M often is a manifold and G is a Lie group.

1-form: The well-known example is a connection $A$ as Lie algebra value 1-form. We have the finite gauge transformation. $$ A \to g(A+d)g^{-1} $$ An example of a large gauge transformation of a Schwarz-type Chern-Simons theory, $\int A \wedge dA$, on 2-dimensional $T^2$ torus of the size $L_1 \times L_2$ with spatial coordinates $(x_1,x_2)$ can be $g=\exp[i 2\pi(\frac{n_1 x_1}{L_1}+\frac{n_2 x_2}{L_2})]$. This way, for the constant gauge profile $(a_1(t),a_2(t))$ (constant respect to the space, satisfying EOM $dA=0$), the large gauge transformation identifies: $$ (a_1,a_2)\to (a_1,a_2)+2\pi (\frac{n_1}{L_1},\frac{n_2 }{L_2}) $$

This seems the two $\mathbb{Z}^2$ integer indices $(n_1,n_2)$ remind me the homotopy group: $\pi_1(T^2)=\pi_1(S^1\times S^1)=\mathbb{Z}^2$.

2-form: If we consider a 2-form $B$ field for a Schwarz-type TQFT, do we have the identification by $\pi_2(M)$ on the $M$ as the based manifold? (Note that $\pi_2(T^d)=0$ - ps. from the math fact that $\pi_2(G)=0$ for any compact connected Lie group $G$.) Is this the correct homotopy group description? How does large gauge transformation work on $T^d$ or $M$?

3-form: is there a homotopy group description on large gauge transformation? How does its large gauge transform on $T^d$ or $M$?

quantum electrodynamics - Only one electron?

If it is true that an electron can be anywhere in the cosmos at any given time, then is it even theoretically possible that there is only one electron, instead of multiple electrons in the cosmos? If it isn't, I'd appreciate it if someone could point out where I may have misunderstood something as far as the fundamental properties of electrons are concerned.

Answer

If it is true that an electron can be anywhere in the cosmos at any given time,

No it is not a true statement. The true quantum mechanical statement is that

there exists a probability that an electron can be anywhere in the cosmos at any given time

This probability is infinitesimally small, and thus the probability that atoms could evolve is practically zero. So it is not even theoretically possible . Your misunderstanding comes from ignoring probabilities coming from the quantum mechanical basic framework.

electromagnetism - How can length be a vector?

Length and current both are not vectors. Then how can we assign the vector $l$ to the length of a wire carrying current while calculating for a current carrying conductor in a magnetic field. Also why in Biot—Savart law do we take small length element $dl$ as a vector?

Why is length sometimes a vector, sometimes not, whereas current always is a scalar?

terminology - The meaning of covariant but not manifestly covariant

What is the most general meaning of the expression covariant, but not manifestly covariant? Suppose I have a general (local) change of coordinates, $x^{\prime} = f(x)$, on an $(n+1)$-dimensional smooth manifold on which some classical fields are defined, say $A_{\alpha}(x_0,x_1,...,x_n)$, which transform into $A_{\alpha}^{\prime}(x^{\prime})$. Suppose the fields $A_{\alpha}(x)$ satisfy some equations of motion, where $x_0 = t$. How should these EOMs look like to be covariant with respect to the given change of coordinates, but not manifestly covariant? Could you explain in plain words the difference between the 2 forms of the EOMs?

If possible, can you write down a practical example of such a situation encountered in physics?

Thx.

Answer

In my experience, we usually call an expression manifestly covariant under some transformation if all the objects which appear in the transformation transform as tensors (or tensor fields) under the transformation.

For example, Maxwell's equations are not manifestly covariant under Lorentz transformations when written in terms of $\vec E$ and $\vec B$ fields, because these do not transform at tensors. On the other hand, Maxwell's equations written in terms of the electromagnetic field strength tensor $F_{\mu\nu}$, 4-vector current $j^\mu$, and '4-derivative' $\partial_\mu$ are manifestly covariant (under Lorentz transformations), because these objects transform as 2-tensors, vectors, and 1-forms, respectively.

Note that it's important to specify which transformations we're talking about. For example, Maxwell's equations in terms of $\vec E$ and $\vec B$ fields are manifestly covariant under rotations.

I don't think the story changes much for local transformations.

homework and exercises - Gravitational field of sphere containing a spherical cavity

I'm just struggling a little with this question:

A uniform sphere, of radius $R$, contains a spherical cavity of radius $\frac14R$, whose centre is $\frac38R$ from the surface. The diameter passing through the centres of the sphere and cavity meets the surface at points $X$ and $Y$. Find the ratio of the gravitational field at $X$ and $Y$.

My attempt at the solution goes something like this:

Using the superposition principle, the gravitation field due to the whole mass is equal to the sum of the gravitational fields due to the remaining mass and the removed mass.

The gravitational field due to a uniform solid sphere is zero at its centre. Therefore, the gravitational field due to the removed mass is zero at its centre.

The gravitational field due to the solid sphere is equal to the gravitational field due to the remaining mass. Now we know g acts towards the centre of the sphere. As such, both the gravitational field of the combination of the sphere and removed mass and the gravitational field of the sphere only act in the same direction, so we can use the scalar form of the equation.

Therefore the gravitational field is given by $g=\frac{GMr}{R^2}$.

Then insert $r=-R$ and $R$ for the gravitational field at $X$ and $Y$.

But this doesn't seem to be correct as it is just the same as if the removed mass wasn't there..... Have I gone wrong in my logic somewhere?