9 votes

Will a distant observer really see an object that has fallen close to a black hole freeze in time?

One reason the Schwarzschild metric is so well known, is because it was the first exact solution found for a gravitational body in General Relativity and even Einstein did not think an exact solution ...
KDP's user avatar
  • 1,858
7 votes

Do the Lorentz transformations lead to negative amounts of time?

The word 'event' here is a misnomer, because it carries with it the connotation of time, and the two events are simultaneous in the unprimed frame The word event is correct, not a misnomer. The fact ...
Dale's user avatar
  • 98k
6 votes

Will a distant observer really see an object that has fallen close to a black hole freeze in time?

Tristan Diotte wrote: "if we switched to the "Eddington-Finkelstein" coordinate, this singularity would disapear" That doesn't help the far away observer, since the time ...
Yukterez's user avatar
  • 11.4k
4 votes

Do the Lorentz transformations contradict reality by implying time had no beginning?

Take t=0 to correspond to the first moment in time. Recall that the Lorentz transform applies locally, meaning it applies in a small enough region of spacetime such that spacetime curvature is ...
Dale's user avatar
  • 98k
4 votes
Accepted

Antisymmetry and coordinate invariance

How can we prove that $\phi(y_1,y_2) = -\phi(y_2,y_1)$? You cannot. Consider $$f_1(y_1,y_2) = \frac{1}{2}(y_1 + y_2),\qquad f_2(y_1,y_2) = \frac{1}{2}(y_1 - y_2).$$ Then, let us take $x_2=0$. In ...
mdi's user avatar
  • 353
4 votes

Equation of Motion Invariance in Galilean Mechanics

Well here minimizing the action is the same demanding zero variations both on $L'$ and $L$. That's good. But now when you use write the Euler-Lagrange equations you'll have one equation in $\...
ssm's user avatar
  • 184
4 votes
Accepted

Will a distant observer really see an object that has fallen close to a black hole freeze in time?

Choice of coordinates does not change what is measured in physics. A distant observer will see a falling object slow down as it approaches the event of horizon. They will not see it "freeze" ...
ProfRob's user avatar
  • 129k
3 votes

Do the Lorentz transformations lead to negative amounts of time?

The order of some events (those that are space like separated i.e. for which $c^2 \delta t^2 - \delta x^2 - \delta y^2 - \delta z^2 < 0$) is undetermined. In some frames event A comes before event ...
Eric Smith's user avatar
  • 8,571
3 votes

Is it possible to label particles of a continuum body?

Is it true to say $p \in R$ (real numbers) ? Yes, or more generally $p \in R^n$ i.e. $p$ is a tuple of real numbers. For example, one way to label infinitesimal fluid parcels in a continuum model is ...
gandalf61's user avatar
  • 50.5k
3 votes

Equation of Motion Invariance in Galilean Mechanics

Briefly speaking, OP's mistake in eq. (4) [as compared to eq. (3)] is to prematurely use the equations of motion (EOM) in the Lagrangian, i.e. $L^{\prime}$ is not necessarily $L$ (modulo a total time ...
Qmechanic's user avatar
  • 200k
3 votes

Equation of Motion Invariance in Galilean Mechanics

You have very nicely proven Noether's Theorem here. If a given infinitesimal transformation, $q_i \rightarrow q_i^\prime = q_i + \varepsilon \, Q_i\left(\vec{q},t\right)$, is a symmetry of the ...
Ben H's user avatar
  • 905
2 votes

Do the Lorentz transformations lead to negative amounts of time?

In SR the time order of events can be reversed by a change of coordinates, provided the events are sufficiently far apart that light from one can't reach another. In that case, none of the events can ...
Marco Ocram's user avatar
2 votes

Do the Lorentz transformations contradict reality by implying time had no beginning?

You've assumed that $a \ne 0$ when $t = 0$, which is also a contradiction to reality (if by "reality" you mean the best physical model we have right now of the beginning of the universe, in ...
Eric Smith's user avatar
  • 8,571
2 votes

How was the reference system taken here?

You are right in that the lecturer chose a reference such that $h=0$. Which is good sense, since this choice simplifies the problem; no need to have an $h$ floating around everywhere. When stuck, it ...
Albertus Magnus's user avatar
2 votes

Variation of action under coordinate transformations

Before I get to your questions I need to do a quick derivation. Recall the Taylor expansion of a function f(x) around zero is given by $$ f(x) = f(0) + \frac{df}{dx}\Bigg{|}_{0}x + \frac{1}{2}\frac{d^...
Wintermute's user avatar
2 votes

Deaccelerating to the Speed of Light

This regime does not make physical sense as a worldline. This would be a worldline of a particle moving faster than $c$. In general, a simple rotation by 90 deg on a spacetime diagram will not ...
Dale's user avatar
  • 98k
2 votes

Physical interpretation of the two possible roots for the isotropic Schwarzschild coordinate $r'$

The isotropic coordinates satisfy a simple isometry. If your map $r'\mapsto \frac{R_s^2}{16r'}$ you obtain the exact same expression for the metric. Consequently, both the $r'>R_s/4$ and $R_s/4 >...
TimRias's user avatar
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1 vote
Accepted

Derivation of lagrange equation in classical mechanics

Applying the derivative (don't forger the inner derivative): $$\frac{\partial}{\partial\dot{q}_{j}}\left(\frac{1}{2}m_{i}v_{i}^{2}\right)=\frac{1}{2}m_{i}\frac{\partial}{\partial\dot{q}_{j}}v_{i}^{2}=...
ssm's user avatar
  • 184
1 vote
Accepted

How is the trajectory of a star found relative to the Sun?

What you're calling "space/true velocity" is velocity relative to the Sun. You're using observations in the solar reference frame without adjustment to another frame. Velocity is always ...
John Doty's user avatar
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1 vote
Accepted

Value of $p^{2}$ for little groups

Weinberg writes four-vector in the order $1,2,3,0$ with $p^2:= \vec{p}^2-(p^0)^2$ (c.f. p. xxv "Notations"). (c) $p=(0,0,\kappa,\kappa)$ with $\kappa \gt 0$ $\; \Rightarrow \;$ $p^0=\kappa \...
Hyperon's user avatar
  • 4,592
1 vote

Non-orthogonal frames

However, the choice of basis vectors is not necessilarly orthogonal. Therefore, although the physics that the vector signifies should not change, i.e its length and direction remain invariant, the ...
AccidentalTaylorExpansion's user avatar
1 vote

Non-orthogonal frames

The basis does not necessarily have to be orthogonal, correct. For that matter you don't have to choose 3 space like and 1 time like vector. You just need 4 independent vectors. But note that the form ...
Eric Smith's user avatar
  • 8,571
1 vote
Accepted

Coriolis Acceleration in Local Cartesian Coordinates

I think your question needs further clarification, if you are interested in the mathematical description, then the reason for $u$ to appear in the second and third component of the resulting vector is ...
Extraherby's user avatar
1 vote
Accepted

On the distinction between frame of reference and observer

Not necessarily. You can pick any reference frames you like. However, if you are considering things in relative motion, it usually simplifies the problem if you pick a reference frame in which one of ...
Marco Ocram's user avatar

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