# Tag Info

## Hot answers tagged special-relativity

10

But we can not (even theoretically) consider length shorter than Planck length. This is a popular misconception. Treating Planck units as special is really more numerology than anything else. For example, the Planck mass is about the mass of a single biological cell. Does that mean physics doesn't apply to anything smaller (or is it larger?) than a ...

10

First I just want to point out that saying that the four velocity $u_\mu$ satisfies $u_\mu u^\mu=-1$ is a convention, it is not a requirement. It amounts to a choice of the parameterization $\tau$. However, it is a very useful parameterization, it's not common to use other choices. In this parameterization, the four velocity takes the form ...

7

Why is the scalar product of four-velocity with itself -1 The scalar product is invariant In the coordinate system in which the object is (momentarily) at rest, the only non-zero component is the temporal component. See that, in the rest frame, $\gamma = 1$ thus $d\tau = dt$. Then, (setting $c = 1$) we have \frac{dx^0}{dt} = 1,\,\frac{dx^i}{dt}=0 ... 4 The derivatives with respect to \tau very much are numbers, but they are not all 1. Consider your worldline as a curve \gamma parameterized by \lambda. We have \begin{align} \gamma : \mathbb{R} & \to \mathbb{R}^4 \\ \lambda & \mapsto (x^0, x^1, x^2, x^3). \end{align} At any point in your worldline you have a position (x^0, x^1, x^2, x^3), ... 4 No, for several reasons. First, the idea of time "slowing down" is a little bit of a misnomer. If you were traveling at relativistic speeds, you would not perceive the passage of time any differently than you do right now. It's only when you compare your clocks to an observer in another reference frame (let's me, sitting in my living room, at rest with ... 3 Suppose we play a racing game. I scatter a little bit of dust around space, then you come by me in your spaceship at some speed v. Let's start with v = c/2, just so we're not contentious. Right as you pass, I fire a really bright laser pulse in the direction you're going. You're racing the laser light. The dust means that you see reflections of it, so ... 2 Short answer : the same light goes at the same speed (c) relative to any observer. There is no grid. This is counter-intuitive : if you're standing in a bus traveling at 30 mph and you walk at 3 mph towards the driver, you walk at 30 + 3 = 33 mph relative to the road. But that doesn't work for light : if the bus travels at c/2 and you shine a flashlight ... 2 To my knowledge, other than red/blue shifting light, the frequency cannot be changed after it's emission. However you may be able to somehow set up some other material which would emit x-rays when infrared light is absorbed. Due to the nature of light and how it acts like a particle when interacting with particles so the energy gained doesn't accumulate when ... 2 It's relative to all inertial reference frames--in special relativity the coordinates of one inertial frame are related to the coordinates of another by the Lorentz transformation, and this transformation has the property that anything with a coordinate speed (change in coordinate position divided by change in coordinate time) of c in one inertial frame will ... 2 You can draw this yourself. Make the vertical axis time and the horizontal axis space. The lines of simultaneity for various observers are all the lines with slopes of less than 45 degrees (in absolute value). So pick any two points that are connected by a line of slope greater than 45 degrees (or equivalently, draw any line with slope greater than 45 ... 2 In special relativity, an object at any non-zero velocity (within the universal speed limit) experiences a length contraction. This isn't actually correct. The object does not experience length contraction since the object is at rest with respect to itself. It is correct to say that, in an inertial reference frame (IRF) in which the object is ... 2 The coordinate velocity does indeed change discontinuously, but only if the acceleration changes discontinuously i.e. the jerk is infinite. Since for any physical system none of the time derivatives of position can be infinite, in a physical system the coordinate velocity can't change discontinuously. But let's ignore this for now and examine why we get a ... 2 Things will be bigger/smaller/slower with speed as \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. Thus, as v \rightarrow c, Mass (or energy) goes to infinite, Time goes to zero. But it is only a limit, which is unreachable (at least in the special relativity), because of (1). 2 Its not possible to stop time but using relativity it can be thought of to be slowed down . Nothing can be faster than the speed of light so its not possible . Even when we near it , energy tends to become infinity . 1 All clocks obey the time dilation formula, therefore all durations measured by a clock obey the time dilation formula. All time differences definitely do not obey the time dilation formula. To figure out the time that it takes for something to travel some distance you might use two clocks. In that case you are not measuring the time with a clock, you are ... 1 Well, by massive, I assume you mean objects that have non-zero rest mass. In that case, it would take infinite energy for that object to reach the speed of light. However, their speed would get closer and closer to the speed of light as more energy is put in, until their speed was practically (but not exactly) the speed of light. Additionally, the smaller ... 1 Given that you imposed t= \gamma t', the obtained relation between L and L' is \tag 1 L = \gamma L'.$$Your reasoning is now, if I understand, that (1) is inconsistent with the phenomenon of length contraction because L is the distance measured in the reference frame \mathcal O and L' is the one measured in the reference frame \mathcal O' ... 1 Several misconceptions in your question. Let's focus on one statement: "For the twins paradox to be plausible, one of the twins must reach "99.995% the speed of light" (Lorentz factor of γ = 100 ) without being atomized." No. That's not true, for several reasons. Firstly, it is a RELATIVE speed, and relative to the frame of reference of a cosmic ray ... 1 Did physicists immediately realize Newtonian mechanics was incorrect after special relativity was published? Of course not. Physicists did not immediately realize that Einstein's description of electromagnetism (the second part of his 1905 paper on special relativity) was correct after special relativity was published. I can't think of a single ... 1 I've come across an answer in Peskin's Introduction to Quantum Field theory where he looks at how to make the 3-momentum delta function invariant that might satisfy you. Look at a boost in the p_3 direction so that p'_3= \gamma(p_3+\beta E),E'=\gamma(E+\beta p_3).$$ \delta^3(p-q)= \delta^3(p'-q')\frac{dp'_3}{dp_3} = ...

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