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1

By transforming into a frame moving in the $y$-direction with the proper speed (i.e., the speed of the particle in the $y$-direction), we can make the particle's $y$-momentum be zero. Working (like you) in units with $c=1$, the Lorentz transformation matrix for a boost to velocity $v\hat y$ looks like $$\begin{pmatrix} \gamma & 0 & -\gamma v & ...


1

a nonlinear transformation necessarily means that there is some point in space which is "special". If I translate $\bf{x}$ to $\bf{x}+\bf{x}_0$ and want to transform it, a nonlinear transformation will involve a product $\bf{x} \bf{x}_0$, which means that this translation depends on the specific point from which we measured everything. Think of ...


1

In this answer, Adam Zalcman describes how Roemer determined the speed of light using the orbit of one of Jupiter's moon called Io. Go through his answer (with the help of this section from Wikipedia, if needed). More specifically, I want you to understand how he obtained equation $(1)$ because it's a prerequisite to understanding my answer. $$\Delta t=t_1-...


0

Roll a wheel down the road. If it’s really rolling, the place where the wheel touches the road doesn’t move relative to the road: it’s just sets down as the wheel gets to a spot, then lifts up after. On the other hand, the top of the wheel is moving twice as fast as the axle. The skyhook is like that. It’s engineered to be rotating and orbiting at just the ...


2

you can start with the position vector of the ball bearing: $$\vec{R}_B=\begin{bmatrix} x+f_x(s) \\ f_y(s) \\ \end{bmatrix}\tag 1$$ where x is the car position and $f_x(s)\,,f_y(s)$ are arbitrary function that described the position of the ball bearing relative to the car. and the position vector of the car $$\vec{R}_C=\begin{bmatrix} x \\ 0 \\ \...


2

I do not know whether I should analyse the ball bearing in its equilibrium (i.e. A=constant,x′=constant) to obtain some expressions for $y$ I strongly believe that's what the question means. To be more precise, I would frame the question as follows: "The equilibrium position of the ball bearing relative to the ramp is used as a measure of the ...


1

Are Lorentz ether theory and special relativity fully compatible/interchangeable? Interchangeable in the sense that, at least in principle,there is either no different prediction the other doesn't make or if there is it could be settled experimentally (or has been) in favor of one of the two? Special relativity (SR) uses the Lorentz transform to make all of ...


1

A frame usually refers to a coordinate system that is moving in space. It can translate and rotate, and it is attached to ta rigid body. This way the inertial properties of the body remain fixed on this coordinate frame. In the example below the reference frame $A$ has origin $\vec{r}_A$, and unit direction vectors $\hat{x}$, $\hat{y}$ and $\hat{z}$. In ...


0

The wave-particle duality of photons (or other particles) is I think not relevant for this question. I won't try to answer the 'why' part of your question, but I will give a description that should help you intuitively grasp how it is possible that light emitted from a moving source appears to travel at the speed of light in any frame of reference. ...


3

Please note that I have limited scientific knowledge. This is only a popularized explanation that helped me understand that concept. Take this for what it is, it may be inaccurate or very simplified. The problem is that our real life definition of "speed" is flawed when talking about photons. Space-time has 4 dimensions, time being one of them. Every ...


18

Why is the speed of light independent of the speed of the source? There is nothing peculiar at all here. This is not where special relativity or quantum mechanics comes in. This is, in fact, the standard behavior of all wave phenomena, i.e., the speed of a wave depends only on the medium and not on the source. For example, the speed of the sound wave from a ...


9

You can see the mathematics about the velocity addition formula and say, wow this is how it works. But you are asking why? We use mathematics to describe the real world, and not the other way around. And you are correctly asking for a down-to-earth explanation why reality is like that. Your intuition tells you that if the light source moves in space at ...


1

It is one of Einstein postulates. Scientific answer on your question based on velocities additional formula: $$ V^\prime = \frac{V+u}{1+Vu/c^2} $$ If $V=c$ then $V^\prime = c$. To understand more deeply you need use Maxwell theory. As convenience of this theory, speed of light is constant in any system frame. This laws was discovered experimentally. Some ...


3

She receives it when her proper time was 4. She wants to figure out what was her proper time when he sent it. Unfortunately, this is undefined. The proper time for an observer is only defined for events on the worldline of that observer. So Alice's proper time is only defined for events on Alice's worldline. The event when Bob sent the signal is an event ...


3

It is not possible to reverse the structure of the atom. Electrons have a very low mass and (relatively) protons are quite massive. Because of the uncertainty relationship between position and momentum the more tightly confined a particle is the more certain its position and the more uncertain its momentum. As the uncertainty in momentum increases so does ...


1

Lorentz ether theory describes a world in which light moves through a medium called ether, and observers that are not at rest with respect to this ether see everything Lorentz transformed. In some philosophical sense, there is a preferred reference frame: that in which the ether is at rest. But the funny thing about Lorentz transformations is precisely ...


1

Physics is about things that really happen in a real universe. Coordinate systems and reference frames are tools we use to describe and understand those things. But they’re not inherent to that physical universe. We impose them. Now consider two things that happened at the same time in the same place. That’s a real thing that happened. Nothing we do with ...


1

I don’t know what counts as “following from the postulates”, but my approach would be: Take the space-time coordinates shared by the events in frame $S$, and transform them to frame $S’$, and note that the still share the same (transformed) coordinates. That argument is also the one I would use in Galilean relativity, BTW.


0

"I want to emphasize that light comes in this form - particles. It is very important to know that light behaves like particles, especially for those of you who have gone to school, where you probably learned something about light behaving like waves. I'm telling you the way it does behave - like particles. You might say that it's just the photomultiplier ...


0

There is no medium. The equations stand alone. The possibility that empty space may itself be a stationary medium for somebody has been disproved by experiment. Hence, any electromagnetic propagation in empty space always seems to any inertial observer to proceed at the universal constant speed irrespective of where it has come from or where else it might ...


1

Suppose you have three inertially-moving clocks A, B and C. Albert always stays half way (in his own frame of reference) between A and B, and confirms A and B always show him the same time. Barclay always stays half way (in his own frame of reference) between B and C, and confirms B and C always show him the same time. Albert and Barclay argue because ...


-4

Ether – less Special Relativity cannot be sound alternative to Lorentz theory. Special Relativity assumes isotropy of one – way speed of light in all relatively moving frames.This isotropy follows from Einstein synchrony convention, that makes the one-way speed equal to the two-way speed. Michelson - Morley experiment confirms, that in moving laboratory ...


1

When you consider different frames, something weird happens. Frames result in you measuring things in weird ways that cause problems for a Newtonian explanation of the world, with a single cartesian space with galilean transformation between frames, and a single universal time. The question is whether SR is the only way to resolve that. Any alternative ...


1

Why is the speed of light held constant here? You are missing one important point of that thought experiment: it is meaningful after considering the Michelson-Morley experiment. If our interpretation of the Michelson-Morley experiment is that the speed of light is constant with respect to the observer, then we can perform the thought experiment you ...


-2

You put considerable effort on merry go rounds (Even when seated) in order to subconsciously show there is no such thing/s. Or you are just wired up to face the rotating system you are in and your brain is taking charge. and $a=0$. You can't fool the Coriolis part is a lot more sneaky. Just look at that term with the angular velcoity vector crossed with ...


1

Would the mirror appear black to A or will the reflection appear after a delay? The reflection would appear after a delay. If the mirror moves at a speed infinitely close to $c$, this delay would approach infinity as seen by $A$. You can say that the mirror appears black only for when the speed of the mirror is exactly equal to light speed, which is ...


-2

The confusion is not realizing that real forces are real, and fictitious forces are not real. Therefore, they can not "cancel each other out" in real life. Using a rotating frame of reference and introducing fictitious forces is just adding and equal quantities (a force, and a mass $\times$ acceleration) to both sides of the math equations. It has nothing ...


-4

No. Such equivalence is inconsistent with Galilean principle: inertial observers are equivalent, and there's no physical processes lest you difference between beying at rest or moving with constant speed.


5

For a mass moving in circles, like a child in a merry-go-round, there has to be a centripetal force that we (I suppose) have to write. It is a real force, after all, making it move in circles. But, in that case, wouldn't the centripetal term and the centrifugal term cancel each other out every time we have a situation like this? If you are in a frame ...


1

When the bird hits the clock, the bird and clock will have the same frame of reference. So, sooner or later, everyone everywhere will see the clock reading the same when the bird hits it, regardless of their own clocks.


2

Why is it that if an event happens , and a clock at the same location that the event happened at says that its a certain time when the event happens, observers in EVERY reference frame agree that the clock said that time when the event happened? That clock records the event by making a record: punches a hole in paper with the hands takes a photograph of ...


4

According to Lorentz Ether Theory simultaneity is absolute and one – way speed of light is frame dependent. One-way speed of light is isotropic only in the preferred frame, or Ether; hence in all moving in the Ether laboratories one-way speed of light is anisotropic; but introduction of length contraction and time dilation for all phenomena in a “preferred” ...


16

There is a physics where time is constant and the speed of light is not: Newtonian with an ether. Galilean transformations in Newtonian physics have time as a fixed parameter that ticks away uniformly for all points and states of motion. For the speed of light to be frame dependent, one needs the ether in which it propagates at $c$. The ether then defines ...


26

Keep in mind that several other Einsteinian effects are hard to explain in an absolute-time scenario and are tested. For these purposes I would concentrate on: The existence of a speed limit for massive particles Looping accelerator systems based RF cavities only work because once the particles have enough energy their speed is effectively constant (and ...


10

No. You're neglecting that time dilation is only one effect explained by relativity. You'd also need to take into account length contraction or, more generally, Lorentz invariance. Your "absolute time" is not going to capture the fact that there are properties of spacetime (not just time) at play here. In light of the comments (pun intended!), a further ...


1

I confess I am not completely comfortable with this problem. (Witness my first attempt at an answer.) Pressing ahead regardless, I think that $\vec{r_1}$ and $\vec{r_2}$ are constrained by the requirement that the Lorentz transformation $\tilde{L}$ be a pure boost; that is, that spatial vectors perpendicular to the boost direction $\vec{\beta}/\beta$ are ...


0

Three good questions. The first is ambiguous so answering both senses: 1a) Because that's the way it is, only God knows why. 1b) We know because we know the speed of light is a physical constant unaffected by the movement of an observer. Time and space measurements must be distorted in three ways. Time is dilated, lengths are shrunk, and time is ...


-2

You write: Now, since earth is an inertial frame of reference, and since this frame of reference is not accelerating with respect to it then it must be the case that this frame of reference in which the cup is viewed is also an inertial frame of reference. This is very confusing! To say the least. The coffee in the cup, stuck to the floor simply starts ...


0

The general Lorentz transformation matrix is: $$L=\begin{bmatrix} \gamma & -\gamma\,\vec{\beta}^T \\ -\gamma\,\vec{\beta} & I_3+\frac{\gamma-1}{\vec{v}\,\cdot \vec{v}}\,\vec{\beta}\,\vec{\beta}^T \\ \end{bmatrix}$$ if you rotate the Lorentz matrix then: $\vec{\beta}\mapsto R\,\vec{\beta}$ where R is $3\times 3$ orthogonal Rotation Matrix $...


0

My friend, however, made a simple answer to this anomaly as to which I am not convinced. He claimed that EEP is applicable only for the observers inside the shuttle and on the planet. That is, a Schwarzschild observer is necessarily not able to use EEP for other noninertial observers from his own viewpoint. Do you think his answer is valid? Yes, your ...


6

The equivalence principle gets stated in all sorts of different ways, but it is basically the statement that the four-acceleration is given by the equation: $$ A^\alpha = \frac{\mathrm d^2x^\alpha}{\mathrm d\tau^2} + \Gamma^\alpha{}_{\mu\nu}U^\mu U^\nu $$ The first term on the right hand side is basically the coordinate acceleration, i.e. rate of change of ...


4

The answer by @Johan Liebert gives the Newtonian analysis, so this answer will give the general relativistic analysis (note, in comments the OP later clarified that he/she is interested in the Newtonian analysis, so this is not directly relevant to his/her question but I left it for others who may have a similar question about the relativistic analysis): In ...


8

$$\pmb {\underline {\text { Newtonian Picture}}}$$ The problem is in this statement the tension force, the gravitational force, the reaction force, all add up to zero Actually the only force acting on the cup is gravitational force and hence in your "inertial reference frame" the cup is seen to be accelerating downward due to a real force. So there is ...


1

EEP tells you that effect of gravitation is locally indistinguishable from that of an acceleration, so locally you can forget gravitational field and you can just use special theory of relativity (STR). Locality is important though. For example, you cannot simulate radial field by acceleration. Moreover you can only consider area so small, that even tidal ...


0

If they are massive objects travelling at $c$ then the Lorenz factor will diverge and they won't have a well defined energy. To find the energy at smaller but still relativistic speeds use $E=\gamma mc^2$ twice.


2

Rapidity: $$ \omega = \cosh^{-1}{\gamma} = \cosh^{-1}{\frac E m}$$ is additive, so: $$ \omega = \omega_1 + \omega_2 $$ solves the problem. Note that thinking about what newtonian physics says is not helpful, nor is being imprecise with statements like "moving at about the speed of light". Generally people say "ultra-relativistic" for that concept. ...


0

With no external forces acting the linear momentum of a system is constant. With your two masses one of which is moving let’s take out the linear momentum of the system of the two masses which cannot change. This is the linear momentum of the centre of mass of the system. You are then left with two particles in the centre of mass reference frame with a ...


1

Well, let's look at the math: Clearly when $\vec{v} = 0$, then $\gamma =1$ and so we expect neither a shift in time or space. That's pretty self-explanatory, after all, it's just a stationary frame isn't it? What about when we start increasing $\vec{v}$? Well, $\frac{d\gamma}{d\vec{v}} = \frac{d}{d\vec{v}}\frac{1}{\sqrt{1-\frac{\vec{v}^2}{c^2}}} = - \frac{...


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