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1

Events can only be said to happen in a particular order if they are space-like events, i.e. $(\Delta r)^2 > c^2(\Delta t)^2$. If this is true, the there is no universal agreement about the order in which the events happen. If, however, $(\Delta r)^2 ≤ c^2(\Delta t)^2$ holds true, then every observer will agree on the order.


0

In Newtonian physics, neither are inertial frames. In relativity, only the first one is an inertial frame.


0

Let me describe a geometrical way to approach your question. We start by drawing a spacetime diagram with distance along the horizontal axis and time on the vertical axis. We'll also take the speed of light to be $1$, so on our diagram a light ray travels along the line $x = t$ i.e. at 45º. If we have some other observer moving at a speed $v$ relative to ...


1

The answer is, "as long as the light from the front of the train has not hit the back of the train first, then yes, there exists some reference frame which thinks of both as simultaneous." In special relativity, there is a number which everyone agrees on: take any two events that are separated by a vector (in your coordinates) $\vec r$ and time interval ...


0

The situation is quite possible.As you have not mentioned any Exact value of acceleration or time between the two lights opening it is difficult to tell exact values which will satisfy your event.But I have made a similar thing which may help you understand. Imagine AE is the train and B,C,D are points on it such that AB=BC=CD=DE.Now AE is moving with ...


1

Let me recall what Newton's laws are, to start with: 1) In the universe exists at least one reference frame (that we call inertial) where $\textbf{v}= \textrm{const.}$ whenever no external interactions act on the particle. All other reference frames (if any) moving at constant speed wrt this very one will be inertial as well. 2) In the above reference ...


-1

Yes the first and second law are indeed similar. However you fail to take into account of the period of which he proposed the laws. During that time, it was obvious that if one were to roll a ball on a smooth surface, the ball would eventually stop although no forces were seemingly acting on it. The first law states that an object will continue moving in a ...


3

You are asking us for the distance of the trip in the rest frame of the photon. The problem with asking that is that there is no rest frame of a photon. A photon can never be at rest, so it has no rest frame. This is like asking what a bowl of petunias thinks about its existence as it falls to the surface. A bowl of petunias doesn't think, therefore we can't ...


1

And while it is philosophically acceptable to just "know" that the speed of light is constant but it not to just "know" that it is invariant. Fixed constant values such as the mass of an electron or the spin set of an electron are things one can accept as given. Kind of: be careful. We can fix some of the fundamental constants by choosing covarying unit ...


0

I'm not sure I understand your question, but I'll try anyways. It's true that constancy (not changing in time) and invariance (same in all reference frames) are different. SR and GR assume a constant speed of light, but lets imagine a theory where the speed of light changes in time. The passage of time is different for different observers. If my clock ...


10

Back before Copernicus (Or rather, before his view was accepted), we used to think the earth was the center of our solar system. Therefore, if you search for those models, you can find examples such as: This is, of course, based on observations rather than calculations, but it represents the complication of the solution nonetheless. (Image taken from ...


11

You certainly could define your origin of coordinates to be the center of the Earth. It would be a little tricky, because this would no longer be an inertial frame of reference, so there would be fictitious forces (or Coriolis forces). That is, your equations of motion would no longer look the same. One reason the standard barycenter frame of reference is ...


2

This is an interesting philosophical question(IMHO). In physics, we don't prove theories to be right, but we do prove theorems about the math used to hold together our theories. Which theories are right is determined by mere phenomena. We have to develop them together out of observation and mathematical logic. The key logic behind Special Relativity was ...


4

That the speed of light is a fixed constant in all inertial reference frames is a consequence of Maxwell's equations of electromagnetism (assuming that two other standard constants, $\mu_0$ and $\epsilon_0$ are, in fact, non-zero constants). The math goes like this: Consider $\nabla\times B = \mu_0 J + \mu_0\epsilon_0\frac{\partial E}{\partial t}$. In a ...


0

The subtlety here is that the lengths are defined at a given time. And in special relativity, "simultaneity is relative". We'll see what this means by working it out. We assume that the object is at rest in the frame $S'$, and has length $L' = x'(b)-x'(a)$ in this frame. We take $L = x(b) - x(a)$, with say $t(b) = t(a) = 0$. We then use the Lorentz ...


1

Your problem is that $E=mc^2$ is false here. That's only rest energy: the complete formula would be $E=mc^2 + \frac12 mv^2 + mgh$ (in the non relativistic limit).


3

Let's do a thought experiment. We'll prove by contradiction, that the speed of light must be the same in all reference frames. scenario: you're sitting in a train travelling at the speed of light. There's a mirror in front of you, inside the cabin of the train. question: Will you, or will you not see your reflection in the mirror? Answer: It's a logical ...


3

It is that the speed of light is the same for all inertial frames that causes Special Relativity, not the other way around. As to why it is the same, nobody knows. Or why it is a finite value. There is some speculation about quantum gravity and a magical æther called spacetime, but nothing has been proven yet.


1

Also Qmechanic gave the correct answer, I believe it is overloaded, because there is actually no need to use motion equations (Euler-Lagrange equations) to answer to the second part of the OP question at least. Actually you can simply just generalize original Landau approach to this issue for an answer, so I will mention it here in details: Lets suppose ...


0

Actually, Mach's Principle is not much applied to linear motion, but rather to rotational motion. Consider that, for linear motion, there is no such thing as a privileged frame of reference. In general, there is no way to tell (in a sealed box) whether the box is stationary or moving with a constant velocity. This is not true of a rotating box. Detecting ...


2

You cannot just combine the Lorentz boosts in that way - the boosts do not form a subgroup of the Lorentz group, the successive application of two boosts is, in general, not a boost, but a boost followed by a rotation, as you may see by explicitly writing down the $4\times 4$-matrices corresponding to the boosts and computing their product, which simply ...



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