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## Hot answers tagged special-relativity

6

You can get an exact solution for $t(p)$, although it involves a rather nasty integral that I'm not sure can be written in closed form. Here's how: The equations of motion are $$\frac{dp}{dt} = -kx \qquad \frac{dx}{dt} = \frac{1}{m} \frac{p}{\sqrt{1 + p^2/m^2 c^2}}.$$ This second equation can be obtained by taking the equation $p = m v /\sqrt{1 - ... 4 There's no one-to-one relationship. With zero rest mass, a particle must always be observed to move at$c$. A particle with nonzero rest mass, on the other hand, can move at any speed in$[0,\,c)$(note the closed-open interval). At the risk of putting words in your mouth, I think I can recall the exact same question in my mind and it went something like ... 3 One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if$x^0 = y^0$the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ... 3 First: Maxwell's equations predict that the speed of light is absolute. The whole motivation for the special theory of relativity is to reconcile this with the notion that all motion is relative. In other words, you're worried about exactly the same thing that troubled Einstein. You just haven't understood how he solved it. The key to your confusion is ... 3 By my reckoning, if all speed is relative, then no mater how fast you go light should always race away from you at the same apparent speed. I.e. there should be no speed limit. If an invariant speed$c$exists, then if an entity has speed$c$relative to an inertial reference frame (IRF), the entity has speed$c$relative to all IRFs. That's what ... 3 It sounds like your confusion is coming from taking paraphrasing such as "everything is relative" too literal. Furthermore, this isn't really accurate. So let me try presenting this a different way. Nature doesn't care how we label points in space-time. Coordinates do not automatically have some real "physical" meaning. Let's instead focus on what ... 2 Gennaro Tedesco's fantastic answer shows how the speed$c$comes to mean the maximum speed that cause-effect links can propagate, relative to any observer. To, to sum Gennaro's answer up to answer your title question, the velocity concerned is the speed of cause-effect propagation relative to the observer's rest frame. It measures how long it takes to ... 2 Light has a frequency of approx. 1e15Hz. Can light be transmitted in a hollow copper tube? Yes. No need to go relativistic. Can objects move at near the speed of light in a coax cable with inner conductor? No. They can't move in there, at all, not even at walking speed. Does any of this has anything to do with photons? No. Your experiment does have a ... 2 It is an experimental fact that light moves at the same speed in every reference frame, no matter the underlying theory: see the experiment of Michelson and Morley. Every kinematic and dynamical quantity depends on the reference frame except the speed of light, which is the same for every observer. Besides the experimental result there is something deeper ... 2 Special relativity and general relativity have different views about inertial frames, but in some ways the general relativity take on them is (perhaps surprisingly) easier to explain. So I'll start with GR then extend the description to SR. In general relativity there are usually no global inertial frames i.e. it is impossible to construct a frame that ... 2 SECTION A : Non-relativistic conservation of energy The work done by the non-relativistic force$\:\mathbf{f}\:$per time unit, that is the power produced or consumed, on a particle moving with velocity$\:\mathbf{v}=d\mathbf{r}/dt\:is \begin{align} \dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}=&\dfrac{d\mathbf{p}}{dt}\circ \mathbf{v}=\\ ... 1 There is a subtle difference between saying(2,2)$and$2\otimes 2$. In the latter case we are thinking of both reps as transforming under the same element of the group$SU(2)$. In the former case we are thinking of$(2,2)$as transforming under the Lorentz group, which contains two distinct copies of$SU(2)$. Call one copy the$L$copy and the other the ... 1$E = pc$is only true for massless particles. For massive particles you have the mass-shell relation:$E^2 = m^2c^4+p^2c^2$After you use$E=T+mc^2$and you can find$p$1 In Newtonian physics, neither are inertial frames. In relativity, only the first one is an inertial frame. Edit: Let me clarify. An inertial reference frame is a frame that "follows" an uniform motion, i.e. a motion where there could be said not be affected by any net force. In Newtonian physics gravity is a force and objects in free fall are thus ... 1 The reason why the commutation relations between a field and its conjugate at equal times are of the form $$\left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y})$$ is only to mirror and copy the canonical hamiltonian commutation relations$[q_i,p_j]=i\hbar\,\delta_{ij}\$. No causality is involved, rather it is somehow ...

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