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Is there a situation in which two photons/particles violate Lorentz transformation?

What you are not taking into account is the relativisic velocity addition. When calculating the gamma factor we are calculating it for a single reference frame moving relative to our reference frame. ...
KDP's user avatar
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Is there a situation in which two photons/particles violate Lorentz transformation?

The frequency of the photon changes with your relative velocity. that's because time slows down at relativistic speed. But you know that. It is the frequency that experiences the Lorentz shift, not ...
Miss Understands's user avatar
1 vote

Is there a situation in which two photons/particles violate Lorentz transformation?

What you did was introduce a third inertial frame in which the relative velocity between the two particles is indeed higher than $c$. This does not violate Special Relativity. If you go in the ...
Damian's user avatar
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An electromagnetic twist on Ehrenfest's paradox

First, lets examine the linear analogue without the complication of circular motion. Imagine we two very long rods with the charges evenly distributed such that the combined charge of the two rods is ...
KDP's user avatar
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3 votes

Meaning of general Lorentz transformations

Let's look at the bigger picture first. A Lorentz transformation maps a set of coordinates $(t,\mathbf r)$ to another set of coordinates $(t',\mathbf r')$. More specifically, if an observer $\mathcal ...
AccidentalTaylorExpansion's user avatar
2 votes

Meaning of general Lorentz transformations

I will call these observers $S$ and $S'$. The condition $\mathbf{r}_{\parallel}=\mathbf{v}t$ implies that $S$ is measuring a point in space that travels with velocity $\mathbf{v}$, with the same ...
agaminon's user avatar
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1 vote

When is the Lagrangian a Lorentz scalar?

An obvious, kind of dumb, answer is that the Lagrangian corresponding to a given Hamiltonian will be a Lorentz scalar if the Hamiltonian has the form, $$ \mathcal{H} = \pi^a \frac{\partial}{\partial t}...
Josh Newey's user avatar
1 vote

When is the Lagrangian a Lorentz scalar?

As far as I know, there are no good ways of stating what conditions on the Hamiltonian will cause the Lagrangian to be a Lorentz scalar other than to just say the Hamiltonian must be derived from a ...
Travis's user avatar
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4 votes
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Some confusion about understanding the relativistic quantum mechanics

The group is still the Poincare group (Lorentz+ translations). The tricky thing is that we need to find a way for that group to act on the Rays in such a way that it preserves the probability. There ...
Josh Newey's user avatar
0 votes

Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?

This considers a more general case in which both the source and the receiver are in motion in the lab frame. This calculation (using a spacetime diagram and its geometry) will obtain the Doppler ...
robphy's user avatar
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Parallel axes between inertial frames in Special Relativity

See: wikipedia:.... $\mathbf{r}=\mathbf {r} _{\perp }+ \mathbf {r} _{\parallel }\;\;, \;\mathbf{r'}=\mathbf {r'} _{\perp }+ \mathbf {r'} _{\parallel }$ then the transformations are:$$\begin{cases} t'=\...
The Tiler's user avatar
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-2 votes

Parallel axes between inertial frames in Special Relativity

A Lorentz boost is just linear acceleration in a spatial direction. An amazing fact that a lot of folks don't know: all linear acceleration in N dimensions is rotation in N + 1 dimensions. Think of a ...
Miss Understands's user avatar
16 votes
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Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?

The relativistic Doppler factor has to be the same whether it is the receiver or the source that is considered to be stationary, because in relativity there is no way to determine if it is the ...
KDP's user avatar
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2 votes

Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?

For sound waves propagating in a medium, there is an obvious absolute frame of in which the sound wave exists with a frequency that does not depend on the motion of the source (Tx) or receiver (Rx). ...
JEB's user avatar
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24 votes

Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?

You have to include a factor of $\gamma$ in both effects. The terms that are the same are $\gamma (1-\frac v c)$ and $\frac 1 {\gamma \left(1 + \frac v c \right)}$. This is because $\displaystyle \...
gandalf61's user avatar
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6 votes

Wikipedia states that the relativistic Doppler effect is the same whether it is the source or the receiver that is stationary. Can this be true?

But your final two expressions are approximately the same in the limit $v\ll c$. Use the binomial expansion, $$ \begin{align} (1+\epsilon)^n&= 1+n\epsilon+\frac{n(n-1)}{2!}\epsilon^2+\cdots \\&...
rob's user avatar
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Resolution of Ehrenfest paradox using only special relativity

Let's say we have a turntable where any point on the perimeter has a instantaneous tangential velocity such that the gamma factor is 2. A ruler placed on the perimeter will length contract to half its ...
KDP's user avatar
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2 votes
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Resolution of Ehrenfest paradox using only special relativity

Instead of a solid disk, we might as well think of a circular train traveling on a circular track. Alice sits in the station. Bob is riding on the train. The Question: The track is shorter in Bob's (...
WillO's user avatar
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Null vector space in Minkowski space

The question is: What would the linearly independent null vectors of this space be? Since the null vectors do not form a vector space, the question only makes sense if "this space" is ...
mma's user avatar
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3 votes

Question on special relativity

No, there is no way to get a satisfactory theory using $$\tag1(c\Delta t)-\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}=(c\Delta t^\prime)-\sqrt{(\Delta x^\prime)^2+(\Delta y^\prime)^2+(\Delta z^\...
naturallyInconsistent's user avatar
1 vote

Question on special relativity

Using $\Delta$anything is going to increase confusion. Relativity is about measuring the same events in different coordinates. So for light propagating a distance $L$, you have 2 events: Emission (Tx):...
JEB's user avatar
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7 votes

Question on special relativity

With some algebra, we can see eq. 1 and 2 are equivalent, \begin{align} &c\Delta t - \sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2} = 0 \;\;\;\; (1) \\ \Rightarrow\;\;\; &c\Delta t = \sqrt{(\...
Aiden's user avatar
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Interpretation of degenerate metrics

My favorite example of the occurrence of degenerate metrics in general relativity is on null surfaces. Take for example Minkowski spacetime with Cartesian coordinates $(t,x,y,z)$. Then the plane $t-x =...
Níckolas Alves's user avatar
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Interpretation of degenerate metrics

When the determinant of the components of the metric is zero, then one cannot write its inverse-metric. This makes the metric tensor "degenerate". With a non-degenerate metric tensor, one ...
robphy's user avatar
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Proper Time Along a Trajectory with Changing Velocity

There is another way of looking at the problem: we replace the coordinates by the speeds in the Lorentz transformations (LT), i.e. $v'=\gamma(v-at)\;\;,\;\;t'=\gamma(t-\frac{ac}{a_{l}^{2}})$ we assume ...
The Tiler's user avatar
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Derive Minkowski metric from Lorentz transformation

The problem is over here in the RHS: $\mathbf{x'^T x} = \mathbf{(Lx')^T (Lx)}$ This is not what is meant when we say that dot product of a vector remains invariant between reference frames. You don't ...
morpheus's user avatar
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4 votes

Time dilation in Einstein's train example (lightning strike)

The time dilation formula applies to measurements by a single clock that is at rest in the "moving" frame. In this case there are two clocks involved, at the front and back of the train. An ...
Eric Smith's user avatar
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3 votes
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Time dilation in Einstein's train example (lightning strike)

Consider the situation : Say the lightning stike occurred at $t=0$ and when both the observers ( on platform, $O_p$ and on train , $O_t$ ) were at $x=0$ , the train being traveling along positive $x$-...
CP of Physics 's user avatar
0 votes

Why the $E$ of the time component 4-momentum is the total energy and not another?

Posting my own explanation as I was very confused and disturbed by this. I ran into this as I was watching these lectures. Right around 55:00 the speaker says given the 4-momentum vector of a particle:...
morpheus's user avatar
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2 votes

Would Doubly Special Relativity or Mach's Principle allow faster-than-light travel to occur without violating causality?

I'm not sure why you think that these theories imply a preferred reference frame. They do not. Doubly special relativity, in particular, is explicitly intended to preserve the principle of relativity (...
Eric Smith's user avatar
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0 votes

Does aether not exist or is it simply superfluous?

Why was the aether not observed in the MM experiment? Oliver Heaviside in "The Electrician" series showed that the Electric field of a moving charge will appear to be pancaked. This led ...
Pradyuman's user avatar
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Length contraction in Michelson-Morley experiment

The pancaking of electric field first described by Oliver Heaviside lead George Francis FitzGerald to predict the length contraction, around same time Michelson-Morley experiment was being conducted ...
Pradyuman's user avatar
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1 vote

Proper Time Along a Trajectory with Changing Velocity

Below my original answer, find UPDATED calculations of some features of the problem and an attempt to address your calculations. Original answer: This spacetime diagram with the ticks drawn might ...
robphy's user avatar
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1 vote

Can a big mass defect make the mass negative?

This question leads to deep insights into what constitutes a particle relative to the vacuum. If you were to have a system of particles with binding energy less than the vacuum, the universe would ...
Xerxes's user avatar
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2 votes

Testing the twins problem with an accelerator

It has been performed even by/on humans. Check Hafele-Keating experiment from 1971 (https://en.wikipedia.org/wiki/Hafele%E2%80%93Keating_experiment), where they took a pair of atomic clocks: one on ...
John's user avatar
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3 votes

Proper Time Along a Trajectory with Changing Velocity

Your assumption about instantaneous acceleration is entirely harmless. In Alice's frame: Betty leaves at time $0$, arrives at Alpha Centauri at time $5$, having aged $3$ years, and returns to earth ...
WillO's user avatar
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6 votes

Can a big mass defect make the mass negative?

TL;DR: No, quantum field theory does not allow this. The general situation you describe is not at all exotic, and in fact this is why we have (meta)stable bound states. What you have termed 'mass-...
SethK's user avatar
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Proper Time Along a Trajectory with Changing Velocity

Your picture says v=0.8c so time for Betty is 0,6 times time of Alice. what you calculated for v I don't understand. the velocity of Betty is always ±0.8c from E to F its 4 Ly in 5y same as from D ...
trula's user avatar
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10 votes
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Can a big mass defect make the mass negative?

If you apply no rules to the possible interactions then what you describe is possible. But the negative bound state mass can be avoided if we assume two things: A force field that has always positive ...
Jos Bergervoet's user avatar
9 votes

Testing the twins problem with an accelerator

Yes. This was studied by Bailey et al. in "Measurements of relativistic time dilation for positive and negative muons in a circular orbit," Nature 268 (July 28, 1977) pg 301. They used muons ...
Dale's user avatar
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Is a photon truly massless?

Mass is not an inherent property of energy. It is a property energy can have, but does not always have. Specially, confined energy has associated mass, while free energy does not. A freely traveling ...
RC_23's user avatar
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5 votes

Is a photon truly massless?

I like pictures, and the relevant picture is: You can ignore the formulae, they are just high school trig, tho, so nothing prohibitive. (The do look bad, though, I think it's because they are crammed ...
JEB's user avatar
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11 votes

Is a photon truly massless?

Although everyone has heard of the famous equation: $$ E = mc^2 \tag{1} $$ few realise that this is a special case and applies only in limited circumstances. Specifically it applies only to a massive ...
John Rennie's user avatar
9 votes

Is a photon truly massless?

The correct version of your syllogism is: $E=mc^2$ for a particle at rest. For a photon, $E>0$. For a photon, $m=0$. The correct conclusion is that a photon can never be at rest.
WillO's user avatar
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6 votes

Is a photon truly massless?

Light has inertia: it takes a force to change the direction in which they travel, and if you have a box with light bouncing around inside of it, it takes more force to change the speed of the box than ...
g s's user avatar
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What effect would relativity have for an objective with absolutely no momentum?

Momentum is actually truly relative. Neither mass nor velocity is absolute. Well, time slows and mass increases due to the lorentz transformation, (A part of General Relativity). As v tends to c, the ...
Ritzthephysibeast's user avatar
4 votes

What effect would relativity have for an objective with absolutely no momentum?

Momentum is relative, not absolute. Relative to a passenger on the train, the train's momentum is zero. Relative to someone on the ground it is very large. So your question doesn't make sense, it ...
Eric Smith's user avatar
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2 votes

Physical intuition for the Minkowski space?

I don’t think that people can give you intuition. Intuition is something that you develop through experience. Since we basically live non-relativistic lives, such experience is gained through working ...
Dale's user avatar
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3 votes
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Given a representation $(n, m)$ of the Lorentz group, is the little group representation just the tensor product $n \otimes m$?

Consider the restricted Lorentz group $SO^+(3,1;\mathbb{R})$ and its complexification $$SO(3,1;\mathbb{C}).\tag{1}$$ Picking the COM frame the massive little group becomes the 3D rotation group $SO(3,\...
Qmechanic's user avatar
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2 votes
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Clocks are not synchronized in moving system (book of David J. Griffiths)

It might help if you create a simple thought experiment and forget about clocks altogether, because it is actually time that is out of synch, and clocks are just a tool for keeping track of time. Try ...
Marco Ocram's user avatar
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