Recent content by Mark M

If you wish to write the energy levels with ## \hbar ##, then the equation will look like this $$ E = \frac {n^{2} \hbar ^{2} \pi^{2}} {2m a^{2}} $$ The two expressions are equivalent. The way the equation is derived is by using the relation ## p =\hbar k ##, substituting in ## k = \frac {n...

That's the solution to the infinite square well problem in three dimensions. Here is a derivation for the solution in one dimension, you can generalize it to three. The first equation is the time-independent portion of the wavefunction, and the second line contains the boundary conditions...

No. The difference is that dark energy is heavily supported by evidence, but the steady state model was not.

As phinds said, this is correct. When we say that the universe is expanding, we mean that the distance in between these galaxies gets larger over time, without the galaxies themselves actually moving. In the very early universe, rather than a homogeneous distribution of galaxies, the...

Right, but considering it's generally accepted that dark energy is a cosmological constant (See here), we arrive at the conclusion that it doesn't make a contribution to the energy density.

No, literally, a cosmological constant is a constant energy density. If it wasn't that, it would be something else. For example, phantom energy or quintessence. The data from WMAP points towards dark energy being a cosmological constant. If it did not, then we would not be able to make...

Th definition of the cosmological constant. It takes a constant value ##\Lambda## and does not change. It is equivalent to a vacuum energy with a constant energy density of $$ \rho = \frac{\Lambda} {8 \pi G}$$ No, it must be the case in order to satisfy Lorentz invariance. Otherwise, you...

You cannot rigorously make such a statement in cosmology. See the FAQ.

The reason that ##w = -1## for a cosmological constant doesn't have anything to do with the above reasons. Since it maintains the same value over time, it must not have any dissipative properties such as heat conduction, viscosity, etc. So, it must be a perfect fluid. Therefore $$T_{\mu...

The fact that ## p = - \rho ## is based off of the fact that a cosmological constant has an equation of state of ##w = -1##. This then gives you the result that the energy density is independent of time. In an infinite universe, there is, as you say, no meaningful way to integrate the...

The cosmological constant doesn't make a difference. For a perfect fluid energy tensor, and a perfect gas equation of state, cosmic pressure and energy density are described by this energy conservation law $$ \dot \rho = -3H(\rho + p)$$ For a cosmological constant, ##p = - \rho##. So, the...

This is incorrect. There is no center to the universe, and the big bang wasn't an explosion. Is the collective name of the expansion of the early universe and the synthesis of the elements in the early universe. I think the problem is that when you hear 'expansion' you think of an expanding...

Because the surface of the Earth is a two-dimensional surface sitting in three spatial dimensions. The surface of the Earth is an analogy used to demonstrate one aspect of the universe, similar to to the the balloon analogy. So, don't look too deep into it (i.e. what is the universe embedded in?).

No, the problem is that the question isn't valid. There's no such thing as 'outside' the universe. The universe may be finite in size, but it certainly does not have an edge. Think of the surface of the Earth - finite, yet you won't find an edge. Take a look at this article for some...

A tensor's order (or rank or degree) is the number of free indices the tensor has. So, a tensor ##T^{\mu \nu}## has an order of two. So, no. A matrix is an order two tensor, since it has two indices (row and column). The rank of a matrix is the number of linearly independent row or column...