general relativity - How has the age of the Universe been derived from the observations made by the Planck mission?

The parameters of $\rm\Lambda CDM$ model have been determined to an amazing high precision from the measurements made by the Planck mission. In particular, the Hubble "constant" (the value of Hubble parameter at the present day) has been determined to be $H_0 = (67.3 \pm 1.2 )\, \text{km} \, \text{s}^{-1} \text{Mpc}^{-1}$. They have also given values for the density parameters at the present day (the density divided by the critical density) coming from cosmological constant, $\Omega_{\Lambda,0} = 0.6817$, and from matter, $\Omega_{m,0} = 0.3183$. The age of the Universe $t_0$ can be calculated from other cosmological parameters in this way

$$t_0 = \frac{1}{H_0}\int_0^1 \frac{a\,\text{d}a} { \sqrt{\Omega_{\Lambda,0}a^{4} + \Omega_{k,0}a^{2} + \Omega_{m,0}a + \Omega_{r,0}} },$$

where $\Omega_k = 1 - \Omega$ is the space curvature parameter, $\Omega = \Omega_{m,0} + \Omega_{\Lambda,0} + \Omega_{r,0}$ is the total energy density parameter (the energy density divided by the critical density) and $\Omega_r$ is the energy density parameter coming from radiation.

I have read that the age of the Universe has been established from the Planck mission measurements as $t_0 = 13.82 \times 10^9$ years. My question is: how is this value been calculated? I mean, it has been calculated assuming a flat space geometry, that is, assuming that $\Omega_k = 1 - \Omega = 0$? If not, what values for $\Omega_{r,0}$ and $\Omega_{k,0}$ have been used to perform the calculation?

Answer

As far as I have understood from this paper, they have given some observational limits to the value of $\Omega_{k,0}$, but this article concludes asserting that "there is no evidence from Planck for any departure from a spatially flat geometry". Taking $\Omega_{k,0}=0$ and the value for $\Omega_{r,0}$ given at this post, one can compute the above integral obtaining $t_0 = 0.947797 \, H_0^{-1}$, which, taking $H_0 = 67.3 \, \text{km} \, \text{s}^{-1} \text{Mpc}^{-1}$, gives $t_0 = 13.78 \times 10^9$ years.

electromagnetism - Alpha Particle moving through a magnetic field

How would I find the acceleration of an alpha particle moving through a magnetic field given the force of the magnetic field, the charge, the initial velocity and the strength of the magnetic field.

general relativity - why do x Schwarzschild radii equal time dilation effects of speed of light going y times faster than an object^2?

let me walk you through the math.

First you start with the gravitational time dilation formula where:

$$ T_1=T\sqrt{1-\frac{2GM}{rc^2}} $$

and rather than entering $r$ for the radius we replace $r$ with the Schwarzschild radius formula $(2GM/c^2)x$ with an $x$ at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:

$$ T_1=T\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}xc^2)}} $$

Which when simplified breaks down to:

$$ T_1=T\sqrt{1-\frac{1}{x}} $$

and if you make $T=1$ then you just get

$$ T_1=\sqrt{1-\frac{1}{x}} $$

This is very similar to the one in many physics books $=\sqrt{1-r_0/r}$, where $r_0$ is equal to the Schwarzschild radius and then $r$ equals the radius from the center. The formula above it just makes it slightly simpler due to making $r_0$ equal to 1 and $x$ equal to how many radii a point you are observing is from the center of the mass.

That is the gravitational time dilation side portion of this relationship. Now for the velocity time dilation side we use a similar methodology and start with:

$$ T_0=T\sqrt{1-\frac{v^2}{c^2}} $$

Now we make $T$ equal to 1, $v$ equal to one, and $c$ to $y$ because now we are going to make $c$ a variable.

$$ T_0=\sqrt{1-\frac{1}{y^2}} $$

What you see now "$1/y^2$" is showing the velocity as a constant 1 and $y$ represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as $1/5^2$ then this would be the same as saying an object is going at a velocity 1/5th the velocity of light. So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:

$$ \sqrt{1-\frac{1}{x}}=\sqrt{1-\frac{1}{y^2}} $$

We can simplify this to

$$ x=y^2 $$

What does this mean?

mathematical physics - Are there cases in which we should consider tensors as equivalence classes?

Usually in texts about Physics that uses tensors defines them as multilinear maps. So if $V$ is a vector space over the field $F$, a tensor is a multilinear mapping:

$$T:V\times\cdots\times V\times V^\ast\times\cdots\times V^\ast\to F.$$

In texts about multilinear algebra, however, a tensor is defined differently. They consider a collection $V_1,\dots,V_k$ of vector spaces over the same field, consider the free vector space $\mathcal{M}=F(V_1\times\cdots\times V_k)$, consider the subspace $\mathcal{M}_0$ genereated by vectors of the form

$$(v_1,\dots,v_i+v_i',\dots,v_k)-(v_1,\dots,v_i,\dots,v_k)-(v_1,\dots,v_i',\dots,v_k)$$

$$(v_1,\dots,kv_i,\dots,v_k)-k(v_1,\dots,v_i,\dots,v_k)$$

And then define the tensor product $V_1\otimes\cdots\otimes V_k = \mathcal{M}/\mathcal{M}_0$ and define tensors as elements of such space, which are equivalence classes of functions with finite support in $V_1\times\cdots\times V_k$.

Now, is there some cases in Physics where it's better to think as tensors as such equivalence classes rather than multilinear mappings? If so, how then we get some physical intuition behind those objects?

quantum mechanics - What is the difference between the Balmer series of hydrogen and deuterium?

In my quantum mechanics textbook, it claims that the Balmer series between hydrogen and deuterium is different. However, I was under the impression that the Balmer series

$$H_\alpha, H_\beta, H_\gamma$$ is related by the equation $$\lambda=C\frac{n^2}{n^2-4}$$ where $$ C=3646 \mathring{\text{A}} $$ $$n=3,4,5$$

Is there a equation relating the mass and the Balmer series?

Any hint would be appreciated

homework and exercises - 4-Momentum conservation for particle annihilation

Disclaimer: this is a homework question, so I am happy with just a hint or the expressions needed to proceed with my understanding.

I am working on the momentum conservation of a particle/anti-particle annihilation process, and I have been asked to show that the annihilation of a particle with a finite mass and its anti-particle cannot lead to the emission of only one photon.

I understand why this happens: the conservation of momentum. However, I would like show this in a more sophisticated 4-momentum proof...how would I go about showing that momentum is conserved for two photons but it is not conserved when the annihilation process creates just one photon...?

This may be a duplicate of: Proving the conservation of 4-momentum for a particle collision $A+B\to C+D$

Answer

There are many possible proofs. Here is one that involves some practice with four-vectors. I write with mostly-minus metric st $p^2=m^2$.

You can write four-momentum conservation as $$ p_1 + p_2 = a $$ Now Minkowski-square, finding $$ 2m^2 + 2 p_1 \cdot p_2 = 0 \Rightarrow p_1 \cdot p_2 < 0$$ Try to show that the latter inequality is impossible. Hint: evaluate the Minkowski product in the rest frame of $p_1$ or $p_2$.

Alternatively, as @Danu suggests, think about the centre of momentum frame, in which $\vec p_1 + \vec p_2 =0$. Can a photon have zero momentum but non-zero energy?

group theory - Rotation in Higher Dimensions

In a world of three spatial dimensions plus time, every atom rotates around a line, the axis of rotation.

In a world of $N$ spatial dimensions where $N$ is greater than 3, must every atom rotate, and if so does it rotate around a line, a plane, or a subspace of smaller number of dimensions?