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No, the age of the universe doesn't depend on the observer. What depends on the observer is the "perceived" time that has passed since the Big Bang. What you are asking is if the conformal time and the age of the universe are the same and the answer is negative as you can see in that link.

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Graphene is considered to be a two dimensional material. Technically, it's one atom thick, but that is as thin as anything can be. There's also carbon nanotubes (CNTs) which are considered to be one dimensional fibers, and fullerenes, which are structurally spherical but considered to be zero dimensional because all three dimensions are less than 1 nm.

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It seems relevant to mention the importance of distinguishing between explicit, implicit, and total time-dependence. The Lagrangian $L=L(q,v,t)$ depends implicitly on time via the position $q$ and the velocity $v$. The total time derivative of the Lagrangian $L=L(q,v,t)$ is $$\underbrace{\frac{dL}{dt}}_{\text{total t-derivative}}~=~\underbrace{\frac{\... 1 The Lagrangian only depends on the potential energy and the kinetic energy. What the statement you quoted means is that if both the potential and kinetic energies are constant w.r.t. time, then so is the Lagrangian. This makes a lot of sense. Usually, we have:$$\mathscr{L}=K(x,t)-P(x,t)$$Where K and P are the kinetic and potential energies. But if ... 1 No, there is not because, contrary to your last sentence: The amount of water and the container itself I am not concerned with, strictly some sort of equation I could use to get a measurement of time. the boiling time is not a well defined quantity unless you take these things into account carefully. To calculate the rate of temperature rise you need ... -1 So, I'm not completely sure if the question I answered was the one you were asking (please make your question clearer!) Here's my answer, though: Newton thought that time could only go one way and had a starting point. Newton also thought time was absolute, and was the same for every observer. Einstein, on the other hand, thought time wasn't absolute (that ... 2 The quick answer is: no. The Hamiltonian operator is a unitary operator that maps state vectors to other state vectors in a given Hilbert Space, regardless of time. Lubos's answer in this thread discusses this distinction very clearly: Why \displaystyle i\hbar\frac{\partial}{\partial t} can not be considered as the Hamiltonian operator? Another point ... -3 I would like to make the argument that time is a force. But I would first like to address some of the items mentioned above. A force does not, by definition, need to have a counterforce as a requisite for existance (F=MA). It exists independently from any counterforce. The "counterforce" is independent and incidental. If you drop a ball, the force of ... 1 The noboundary condition means there is no boundary that marks the end of space or time. With respect to time one might think of the lines of longitude on a globe as representing the time direction at different point in a spatial manifold modeled as the lines of latitude. As one looks further to the north, which is the big bang that eventually you look north ... 0 Short answer: position momentum uncertainty exists because their operators do not commute. Likewise the time and energy operators do not commute. Longer answer: First: Physics is an empirical science so "proof" must be an experiment. Sometimes thought experiments will be allowed because they are useful for building and testing models. Mathematical proofs ... 24 The main problem is, as you say, that time is no operator in quantum mechanics. Hence there is no expectation value and no variance, which implies that you need to state what \Delta t is supposed to mean, before you can write something like \Delta E \Delta t\geq \hbar or similar. Once you define what you mean by \Delta t, relations that look similar ... 4 This is the case. The uncertainty relationship with energy and time is a matter of Fourier analysis. In fact the relationship \Delta\omega\Delta t \simeq 1 was know in classical EM and electrical engineering before quantum physics. The use of Fourier analysis in electrical engineering had much the same uncertainty relationship as the reciprocal ... 2 The universe is a four dimensional object i.e. to locate any point within it you need four numbers. Most commonly we use a coordinate system (t, x, y, z) and the four numbers give location of the spacetime point in this coordinate system. You ask: But how can this be, if time is relative and dependent on speed of reference frame? and the answer is ... 3 Time is that which the clock shows. Any one clock. Clocks do not all show the same time but their readings are related to each other and that relation is what the theory describes. In non-relativistic theory any two (perfect) clocks can only differ by a constant time difference but they all progress at the same rate. In relativity any two clocks that are in ... 1 You're confusing concepts, as far as I can tell. Being a "dimension" doesn't imply all values of that dimension (or any of them) are arbitrarily reachable or even exist physically at a given time. It doesn't mean that the intuitive sense of all the baggage a more usual dimension comes with, are applicable to time. It doesn't imply time travel or multiple ... 1 Time is not a combination of a past, present, and future dimensions, but rather a one-dimensional axis, where the past and future are dependent on the present for definition, and the present is a particle's position in time. In the theory of relativity, a particle has position (x,y,z,t), where (x,y,z) is the particle's position in space, and t is the ... 4 Itzhak Bars proposes a two-time physics. This means the spacetime(s) metric is of the form$$ ds^2 = dt^2 + du^2 - dx^2 - dy^2 - dz^2. $$The anti-de Sitter spacetime is a subspace of this type of metric as well, where a constant surface is a hyperboloid. A hyperboloid in 5 dimensional space times is given by$$ t^2 + u^2 - x^2 - y^2 - z^2 = 1.  The ...

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Suppose you are standing upright aligned with the $z$ axis and facing along the $y$ axis, then the $x$ axis is to the left of you, at your position and to the right of you. But we don't say there are three $x$ axes. The $x$ axis simply comes in from $-\infty$ from your left, through your position and then off away to $+\infty$ on your right. In exactly the ...

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