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 Apr 20 comment How can you tell if the work done by a force is negative? Work is positive when the object is moved in the same direction as the force, and negative when it's moved against the force. Apr 13 comment Acceleration of particle “held in place” at $x = 1$ @user12262 Those questions didn't help. Did you invent this notation yourself or did you get it from some other source? If the latter please forward it :). "Surely a particular value of a dimensionful quantity is independent of particular choices of units; and is not just any coordinate value either." What I'm saying is, the question as posed assumes that lengths are dimensionless. The function cosh(x) doesn't make sense otherwise. So you could say the question was expressed badly. If the question were posed better it would use cosh(Ax). Apr 12 comment Acceleration of particle “held in place” at $x = 1$ @user12262 I'm not familiar with the notation you're using in the first comment. "Well, that's not quite the same as applying $d/d \tau$ outright in any case." You're right, it's not. That definition of acceleration is chosen because the operator $d/d \tau$ does not map tensors onto tensors while the operator $\mathbf v \cdot \nabla$ does. "But, notably, re-introducing "c"... So what's missing?" The function cosh(x) implies dimensionless x. If you're working in more common units the function would need to be cosh(Ax) for constant A with dim. $[L]^{-1}$. Apr 8 comment Acceleration of particle “held in place” at $x = 1$ @user12262 In general: $$\mathbf a = (\mathbf v \cdot \nabla) \mathbf v$$ where $\nabla$ is the covariant derivative. When the metric is Minkowskian the operator $\mathbf v \cdot \nabla$ reduces to $d/d \tau$, however this is not true in general. The quantities appear dimensionless because in GR we tend to work in geometric units where c=G=1. Apr 7 comment Acceleration of particle “held in place” at $x = 1$ @user12262 The components $u^\mu = \frac{d x^\mu}{d \tau}$ are certainly coordinate dependent. The quantities $dx_\mu dx^\mu$ and $u_\mu u^\mu$ are coordinate invariant. On curvilinear/non-inertial coordinates the definition of the acceleration four-vector makes use of the connection via a covariant derivative: $$a^{\mu} := u^{\nu} \nabla_\nu u^{\mu}.$$ This is done to ensure that $a_\mu a^\mu$ is also invariant, and corresponds to the square of the magnitude of the acceleration the particle would actually experience. Thanks for the +1 :) Apr 7 comment Acceleration of particle “held in place” at $x = 1$ @user12262 I derived both the x-coordinate acceleration AND the physical (proper) acceleration (magnitude of 4-accel). They just happened to be equal due to the form of the metric ($g_{11}=1$). Apr 7 comment Acceleration of particle “held in place” at $x = 1$ @garyp Apologies. I've removed the final answer as well as the expression for four-acceleration. Apr 7 comment Derivatives involving four vectors I'm not clear on exactly what you're trying to do. Are you trying to "break up" the derivative into its temporal and spatial components? If so why? Could you give an example of where this would be useful? Apr 6 comment The Theoretical Minimum: Confusion Over Susskind's Reasoning for mutually orthogonal states @MonaLisaOverdrive Have a look at the answer I posted last night. Since the graphical/Cartesian representations were confusing you I did the problem explicitly using only bra-ket notation. Let me know if this helps or if you're still confused. Jan 9 comment How is complex permittivity measured? Rolled back edit after incorrect and unnecessary addition. Dec 24 comment What exactly does the Kretschmann scalar implies and how does it work? When you test whether or not a singularity is removable, you're basically just testing whether or not scalars make any sense there. The easiest way to do this is probably with the volume element. Or, more simply, the metric determinant. If det(g) is zero or infinite then volumes don't make sense, so the singularity is not removable. Dec 23 comment What exactly does the Kretschmann scalar implies and how does it work? Wouldn't it be easier to determine if a metric has a removable singularity by whether or not the volume element $\sqrt{det(g_{\mu \nu})} d^4 x$ makes sense? If it blows up or becomes zero at the singularity in question then it is not removable. Dec 23 comment Black hole and relativity They would appear to asymptotically slow down as they approached the event horizon, and the light coming from them would be asymptotically red-shifted until they were no longer visible. Note that this isn't what 'actually' happens to the person who's falling, as they would experience nothing unusual when they approached and passed the event horizon (besides some potentially very strong tidal forces if the BH is small enough). This is just what would appear to happen due to the BH's effect on light. Sep 27 comment $\hbar \rightarrow 0$ in quantum mechanics @UnlimitedDreamer We do the same sort of thing when we take $c \to \infty$ to get from relativistic equations to Newtonian equations. $c$ has a definite value, but by pretending it is infinite our equations become Newtonian. In the same way, when we pretend Planck's constant is zero our equations go from quantum to classical. Edit: Just realized this is an old question. Not sure why it popped up in my recent questions page. Mar 22 comment Why does the electric field escape a black hole? A field isn't a 'thing' that can be sucked into a BH. A field is a mathematical object that has value(s) at every point in space. There is no reason to ad hoc assume that the EM field outside of a BH is zero. Jan 22 comment Does potential energy always equal kinetic energy? @StanShunpike $T=V$ under special circumstances. What's important is that $T+V=constant$ over time. There may be some points on a particle's path where $T=V$, but it certainly won't be true everywhere. Jan 22 comment Does potential energy always equal kinetic energy? But $T \neq V$ in general. However $dT/dt = -dV/dt$ in general. Jan 22 comment Confusion about what the Euler-Lagrange equation says I'm not sure I understand what you're asking. Given a Lagrangian, the E-L equations tell you what the equations of motion are. Dec 30 comment Can one of Newton's Laws of motion be derived from other Newton's Laws of motion? @Timaeus x(t)=t^3 does not have a(t)=0. It has a(t)=x''(t)=6t. The only moment where a=0 is at t=0. It is implicit in my answer (and I have now explicitly said it) that I'm setting a(t)=0 for all t, not just at a single instant. a(t)=0 does in fact imply v(t)=const. Dec 30 comment Can one of Newton's Laws of motion be derived from other Newton's Laws of motion? @Timaeus Why the strawman? I clearly never said "a=0 at a single instant is enough to tell you v=const." When I say "set F=0" I'm implicitly doing this for all time, and that is indeed enough to tell you that v=const. You're arguing against a strange strawman that I clearly never stated nor implied in any way.