# Tag Info

## New answers tagged time

0

Proper time is defined for an arbitrary time-like trajectory $x^\mu(t)$ and an arbitrary spacetime metric $g_{\mu\nu}(x)$ by the formula $$\tau=\int\sqrt{g_{\mu\nu}(x)\dot{x}^\mu\dot{x}^\nu} dt.$$ I would recommend to compare this with the general formula for the length of a curved path to get a feeling that this is something invariant, and does not depend ...

0

Entropy is monotonically increasing with time, but its rate of change can be anything from zero to an arbitrarily large value. Lets say you have two boxes, one filled with oxygen and another filled with hydrogen. There is a door between the two boxes but it is closed. In this state, the entropy has a relatively low value and the gases and entropy can stay ...

1

Reichenbach's original volume, "Axiomatization of the Theory of Relativity", appeared in 1924. It is one of a long string of works that periodically rediscover and/or explore the issue of non-Einstein synchronization in Special Relativity. See for instance this review on "Synchronization Gauges and the Principles of Special Relativity" and refs. therein ...

0

Derivative is defined at a point not for an infinitesimal interval. $$\text {If y=f(x)} \Longrightarrow \; y’(x)|_{x=a}=\frac {dy}{dx}|_{x=a}=f’(a)=\lim_{\Delta x\to 0}\large{\frac {f(a+\Delta x)-f(a)}{\Delta x}}$$ $\large{\frac {d\vec \Omega}{dt}}$ doesn’t represent a fraction. That represents derivative of the $\vec \Omega$ relative to $t$ at a time ...

5

The spacetime interval is a relativistic invariant, and is proportional to the travelers proper time. So in a since you are traveling one second per second, per your own wrist-watch. Every other measurement would be the speed of some other inertial reference system, measured with your clock. Let $s^2 = x^2 + y^2 +z^2- (ct)^2$, where $x$, $y$, $z$ are ...

2

Also, the fact that $\Delta t' \rightarrow 0$ if we formally let $v \rightarrow c$ can be interpreted as saying that no time at all passes for a particle moving at the speed of light. Photons cannot "age" or in any other way change over time.

0

The question is whether time is an operator in the sense of ${\hat T}|t\rangle~=~t|t\rangle$. This at first glance would seem to make sense because we do have a position operator ${\hat X}|x\rangle~=~x|x\rangle$. However, this does not work. This is a subtle question in many ways. Quantum mechanics is unitary. Consider a state vector $|\psi(t)\rangle$ ...

4

That formula holds for a simple pendulum of length $L$ in a gravitational field $g$ and released from rest with an amplitude $\theta_0$. Since this system is conservative its mechanical energy is constant and equals the gravitational potential energy when it is released. Setting the zero of potential energy at the fixed point of the pendulum, the mechanical ...

5

Is time made into one observable? No. It is known that an operator $T$ that satisfies $[H,T]=i\hbar$ is either self-adjoing and $H$ unbounded below or anti-self-adjoint. Therefore, the theory is either intrinsically flawed (arbitrary negative energy) or $T$ is not observable (anti-self-adjoint $\Rightarrow$ imaginary eigenvalues).

0

The theorem that time is not an observable is quite general, Unruh W., Wald R. prove this in "Time and the interpretation of canonical quantum gravity, Physical Review D Volume 40 issue 8 1989" in the following form: "... in the context of ordinary Schrodinger quantum mechanics, no dynamical variable in a system with Hamiltonian bounded from below can act ...

0

In that sense, in Relativistic Quantum Mechanics based on Dirac's Equation, is time made into one observable? No, but it is treated on equal footing to the space coordinates. Dirac's equation is $$(i \hbar \gamma^\mu \partial_\mu - mc )\psi$$ where $\psi$ is not the classical wavefunction, but a four-component spinor. The components of the spinor are ...

1

Only the mathematical formulas are time reversible. That does not mean that the law itself is reversible. You can understand it this way - Every law/formula has an implicit condition that says - "time flows only forward". Other way to understand can be that when the coffee cup falls, universe (gravity in this case) makes the cup fall. There is no force/law ...

3

It is not really correct to call them " reversible laws of physics" Laws of physics lead to mathematical models that describe observations. These models are usually differential equations of space and time. The solutions of these equations exist both for time going towards infinity as for time going towards minus infinity in most cases, for mechanics, ...

2

The standard reply is: get your video camera out, take a movie of dropping a (empty) coffee cup on the kitchen floor, then play the movie backwards. There is no physical way of telling which way time is running from your movie or the equations behind the event (Newtons Laws). Rather than thinking of an absolute arrow of time, so the coffee cup will break ...

0

Could the time for light to go that distance (4734 feet) be the same as the time difference caused by [relativity gravitational time dilation]? No, for two reasons. The units are not the same, one is nanoseconds per day. The other is nanoseconds. light travels 4734 feet in about 4813 nanoseconds not 20.

0

In special relativity (things are a bit different in general relativity) time dilation is always relative. There is no such thing as an absolute velocity so there is no such thing as an absolute time dilation. If you are travelling relative to me at some speed $v$ then the relative rate at which I observe time passing for you is: $$\frac{d\tau}{dt} = ... 0 I just had the same pendulum time dilation problem. Notice this oscillation does not depend on the mass (m) of the pendulum (Einstein in his booklet). Only mass Zero is not possible. Time Dilation (based on Schwarzschild Metric)$$T=2\pi \sqrt{\frac{lo}{g(r)}/ \left(1-\frac{2g(r)r}{c^2}\right)} the pendulum (lo) ticking at $T_0=1 sec$ "say high rate ...

0

1) Your question states that, until the compressor is switched off, a constant pressure of $5\; atm$ is maintained in the box. This answers the 1st part of your question. (However, perhaps you mean that the compressor delivers a certain amount of gas per second while the 2nd hole is open, and you wish to know the pressure of the gas in the box when ...

1

It depends what exactly you mean by "coordinate". If your Lagrangian/Hamiltonian is time-independent, then you may consider time to be purely a parameter parametrizing e.g. the integral curves of the vector field associated to the Hamiltonian on phase space. If your Lagrangian/Hamiltonian is time-dependent, you should indeed properly consider your theory on ...

1

I think this issue is best clarified by closely looking at the way time is mixed into coordinate frame transformations in Classical Mechanics as opposed to Relativistic Mechanics. Let's take the case of an observer, Alice, moving at velocity $v$ in the positive $x$ direction away from her friend Bob. Both Alice and Bob are looking at an object situated at ...

Top 50 recent answers are included