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3

A Lorentz transformation lets you compute an object's properties in one inertial frame, given its properties in another inertial frame. Inertial frames, by definition, do not accelerate. An accelerating object is always instantaneously at rest in some inertial frame. Whether such-and-such is a law of nature is an experimental question. We have no ...


2

If you were going to search for a violation of the equivalence of frames, you'd want to test a proposed special frame against some other one (rather than two randomly selected frames). The local rest frame of the CMB is probably the best bet going (just like the heliocentric frame was the best bet for Michelson and Morley at the dawn of the 20th century). ...


-2

GPS satellite navigation system doesn't use, doesn't need and doesn't prove Einstein's General Relativity. The GPS satellites use classical (Newtonian) relativistic principles to work. These are the same relativistic principles that make sense in the everyday world, that most people equate with 'common sense'. GPS calculates positions based on geometric ...


0

This is a fun, high-quality qualifying exam question. The algebra is not hard; the physical insight takes some real thought; there are many ways to be partly right. Here's my take on it. Acceleration From Einstein's equation $E^2 = p^2 + m^2$ we have for each photon $E = p = hf_0$ (in the reference frame of the laser). We can use the power of the laser ...


0

Your question covers numerous topics, but I will try to answer several of them here. For a proton (I am most familiar with this case), the proton asymptotically becomes a black disk Ref. The disk part is because it is Lorentz contracted at the collision point. The same process happens for nuclei as well (see the bottom left of here for example). The "black" ...


5

Covariant notation is a simple way to say how something transforms under Lorentz. An object with an index, e.g. $z^\mu=(z^0,\vec z$), transforms under Lorentz as, $$ {z^{\prime}}^{\nu} = {\Lambda^\nu}_\mu z^\mu $$ where $\Lambda^\mu_\nu$ is the matrix you have written down in your question. $z^\mu$ is called a four-vector. This transformation property is ...


1

One shoul think as c as a kind of space-time conversion constat, massless energy travel at this speed. Light and gravity are kinds of massless energy. The idea of E=mc^2 is that mass converts to energy like this.


1

The causal structure of spacetime is determined by the lightcones at every point of it, basically. From this you can construct such things as the chronological past (all events that influenced this point), the chronological future (all events that will be influenced by this point), the Cauchy horizon (the limits of what you can predict in the future from a ...


3

by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ...


1

Partial answer: In the context of general relativity, it is conventional to make use of conformal diagrams (a.k.a. Penrose diagrams to enable us to visualize a spacetime. These bring the boundaries of the spacetime (at infinitey) to a finite coordinate value, and keep lightlike worldlines at 45 degrees. As such, they're the generalization of Minkowski ...


5

You can understand the expression by attempting the limit for $v\rightarrow c$ and $m\rightarrow0$. Notice that $\gamma\rightarrow\infty$ when $v\rightarrow c$. Therefore $m v \gamma$ is an undetermination of the form $0\cdot\infty$. From your expressions, you cannot say that $E=0$. The limit does not exist, and this implies that this expression is not valid ...


6

That's because the relation $p=\gamma mv$ doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ...


3

Just to add to Danu's Answer, which I believe to be right. The relative scalings of the "electro" and "magnetism" parts of the unified electromagnetism whole are somewhat arbitrary; we're only required to ensure that $c=\frac{1}{\sqrt{\mu_0\,\epsilon_0}}$ to achieve a valid set of Maxwell equations. As we change these relative scalings, we change the ...


8

Although you might not like to hear it, the answer really DOES lie in the definition of $\mu_0$ (and $c$). $\mu_0$ is defined to be exactly $4\pi *10^{-7}\ \text{H m}^{-1}$. Similarly, $c$ is defined as exactly $299792458\ \text{ms}^{-1}$. It immediately follows from the relation $$\epsilon_0=\frac{1}{\mu_0 c^2}$$ that $\epsilon_0$ also has no uncertainty. ...


1

The scalar potential and magnetic vector potential are combined into a four-vector, $A_{\mu}=(\phi,\vec{A})$ which is a gauge field, and in the language of differential geometry, a 1-form. The Lagrangian of the field theory (i.e. Maxwell theory) is, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ where $F_{\mu\nu} = \partial_{[\mu}A_{\nu]}$ is the ...


1

1 - It is false! If $E = mc^{2}$ is true only for an object that isn’t moving, the mass never changes (is a "Lorentz invariant"). 2 - Can you rephrase it, please? 3 - Energy and mass are not at all the same thing; an object’s energy can change when its motion changes, but its mass remains the same. 4 - In Special Relativity, time can be variable, its ...


2

The speed of light in a vacuum is invariant: it is the same no matter what point you pick as "stationary". So if I'm on a train, and you're on the ground, and we both measure $c$, we'll get exactly the same number. The speed of light does not depend on the wavelength. Gamma rays travel at the same speed $c$ as radio waves. The frequency $f$ and wavelength ...


6

The speed of light is there for much more than to look cool, and in fact there are a number of derivations of mass-energy equivalence that shows why $c$ is present; I will say that one basic reason is that the units of mass and energy are different, so we require at least some sort of constant factor to make the units work. I'll also say that we often use ...


0

"Now imagine the observer that is looking at the light which is traveling away from him in a direction he is looking." "What's wrong with this reasoning?" Well, this observer will see nothing - not even a point. It's just physically impossible to see light moving away from you.


2

Critical mass is actually more about 'the right number of nuclei in a specified space'. As we are talking about solid matter this equivalently translates to a given number of atoms (or molecules depending on your matter). And this furthermore translates to our everyday mass. But it's 'not about mass', it is just a practical way to specify the quantity. So ...


0

Sure, flash lights sent within the train cannot be observed by the observer on the platform. Nobody (or no device) can see light that is not sent directly to its eyes (receiver). Observer on the platform can see light sent within the train only if it is resent to him after it travelled locally (i.e. unseen directly from the platform). Therefore, there will ...


1

The bomb doesn't "care" what its mass might look like to an observer in another frame. If you calculate critical mass you don't worry how big it might seem to observers located in billions of other possible frames of reference. Local frame of reference is the only one valid for making calculations concerning the occurence of local phenomena.


2

As you have discovered proper time, $\Delta\tau$, can be either real or imaginary. However, this means that it does not necessarily reflect something measurable with a clock. When it is imaginary, as in the case of a space-like relation of two events, then there is no single clock that can be present at both events. To do so would require having a velocity ...


0

You probably know already that there is no derivation of the velocity transform in Wikipedia, or anywhere else, along with other transforms' derivations for special relativity (at least I did not find any). The sub-section on the principle of relativity states, however, that $v' = -v$. So, according to Wikipedia, you should assume that the apparent velocity ...


0

I have been thinking more about this problem. Of course, my understanding of the application of Lorentz transformations was substantially enhanced by the two answers kindly provided. I was able to solve it using proper time. Also, due to the nature of the problem, I can solve it using just the time equation of Lorentz transformations: $t' = \gamma (t - ...


5

Here's my two cents worth. Why Lie Algebras? First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine. The Lorentz Lie Algebra It turns out that the ...


0

It is immediate from the definition of the cross product. Write $$\mathbf{B} = B_x \mathbf{e_x} + B_y \mathbf{e_y} + B_z \mathbf{e_z}$$ and use that $$\mathbf{e_x} \times \mathbf{e_y} = \mathbf{e_z}$$ and $$\mathbf{e_x} \times \mathbf{e_z} = -\mathbf{e_y}$$ If you can't get an identity by applying the physics it's sometimes useful to make sure you ...


2

Firstly, what book is this? It will help greatly if I can reference it myself. It is highly likely that when he says $\mbox{SO}(1,3)$ [or $\mbox{SO}(3,1)$!] that he means $\mbox{SO}(1,3)-\uparrow$, which is absolutely not the same! But most people are very lazy about this. Here you're picking out the simply-connected region of $\mbox{O}(1,3)$ ...


1

$F_{\mu\nu}$ is a Lorentz tensor, easy to see by $\partial_\mu A_\nu - \partial_\nu A_\mu$, which is a 2-form. Contractions of Lorentz tensors are Lorentz tensors. $\tilde{F} = \star F$ is the Hodge dual of $F$, which is also a 2-form, hence a Lorentz tensor, therefore the same applies about its contractions. By these definitions, they are also tensors in ...


6

The first equation is only valid for massive particles. If you see the formula of the Lorentz factor: $$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ If $v=c$ (the case of massless particles), it is undefined. You can also see as $v \rightarrow c$, $\gamma \rightarrow \infty$, which "compensates" for $m_0=0$. The second doesn't imply zero energy, because ...


0

An important note to the other answers so far: the actual mass of a single particle (the rest mass as it used to be commonly referred to) does not change if the temperature rises (by temperature I mean that of the environment since temperature is a macroscopic quantity). The (rest) mass of a particle is the $m$ in $$E = \sqrt{|\mathbf{p}|^2c^2+m^2c^4}.$$ ...


-1

photon has very light mass almost negligible amount, so temperature does not effect mass of the object, but you take large object with high temperature it indicates measurable amount of mass change. example sun.


1

Why do we set x0=ct instead of x0=t? Nobody holds us to set x0=t and measure the time in centimeters of the travel of the clock's pointer. Or measure it by the change in the circumference of the earth during the day night cycle, or... It is a matter of calibrating time to centimeters in terms that are universal and unique. When the Lorenz ...


8

But that's exactly the deeper meaning! Setting things up so that all coordinates are in the same units (besides being a reasonable requirement for $x^\mu$ to be considered a four-vector) is a constant reminder that time is really not that different from space. In fact, if it weren't for that sign in the metric, spacetime would be completely symmetric in its ...


4

The equation of motion for a scalar massless relativistic point particle is $$ \tag{A} \dot{x}_{\mu}\dot{x}^{\mu}~\approx ~0, $$ where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is $$ \tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, ...


21

If (and that's a big if) tomorrow we had a $70\sigma$ detection in a repeatable experiment of a particle that travelled faster than $c$, then one of several things would be true. 1) We would be forced to conclude that $c$ is not, in fact, the limiting speed of information transfer; everything based on this assumption would have to be scrapped (pretty much ...


-3

Particle with zero mass has to have also zero electric charge, otherwise the Lorentz formula for EM force acting on it cannot be used to find its acceleration according to $$ m\mathbf a = q\mathbf{E}_{ext} + q\frac{\mathbf v}{c} \times \mathbf B_{ext}. $$ However, particle with zero mass and zero charge has trivial equation of motion $$ 0=0 $$ and has zero ...


5

But what if tomorrow we happen to observe a particle X that travels with a speed V>c? We would have made the first observation of a tachyon. In special relativity, a faster-than-light particle would have space-like four-momentum, in contrast to ordinary particles that have time-like four-momentum. It would also have imaginary mass. Being ...


0

(Still can't comment!) How are you doing a course on CED without knowing (Classical) field theory?! Do you mean to say, you are doing a course on Electromagnetism? Assuming that part of the aim of writing the essay is to learn things along the way, and given that you don't yet know Classical field theory, this would be exactly a good thing to write about! ...


0

For a massless particle we can not have a centre-of-mass frame. Unfortunately I cannot yet add comments. Are you studying Classical Field Theory (CFT) or Quantum Field Theory (QFT)? My guess is CFT since this looks like a line out of a few lectures into a CFT course when you start to make find equations of motion. In that case, for the (massless) photon, ...


0

It is incorrect to state that there is a photon with energy $E$ in its rest frame, but it would be correct to talk about a photon with energy $E$ in some frame. In this case, the equation that you wrote would be valid to express the energy of that photon observed from a different frame (that you choose to characterize with a prime) that moves away with ...


0

Unfortunately, this result was later found to be caused by faulty electronics, according to the CERN press release.


1

So, you are correct, that it is not quite as simple as that. Have you studied Special Relativity (SR) yet? One of the most fundamental ideas is the limiting speed of light. Nothing moves faster than c, and this means that velocities must add differently. Think of the reverse, if you were on the rocket-ship traveling at $0.5c$ and shot a laser-beam at $c$, ...


0

As noted within my other post at Time dilation in special relativity, here I have basically the same situation yet the box becomes a 300,000 km long spaceship. The basics are explained. However, the only absolute measure that exists within it, is that all objects are constantly traveling at the speed of light within the 4 dimensional environment known as ...


0

First, to make things simpler to understand, imagine that the spaceship is 300,000 km's long. Now let's assume the spaceship is at rest in space and a clock at one end of the spaceship is synchronized with another clock located at the opposite end. Here, if we send a burst of light from one end to the other at say 12:00 noon, if done in either direction it ...


1

I want to add my personal understanding of the concept of reference frame. In the articles: Marmo, G., Preziosi, B. (2006). The Structure Of Space-Time: Relativity Groups. International Journal of Geometric Methods in Modern Physics, 03(03), 591-603. Marmo, G., Preziosi, B. (2005) Objective existence and relativity groups. Symmetries in Science XI, ...


0

Einstein says: "If, relative to K, K' is a uniformly moving co-ordinate system devoid of rotation, then natural phenomena run their course with respect to K' according to exactly the same general laws as with respect to K." Apparently, "uniform movement", i.e. one without accelerations, is the assumption behind the whole SR Theory that allows us to compare ...


1

What you are calculating is the invariant mass, for a photon the invariant mass is 0. This transformation is only 1 component of the Energy-Momentum 4-vector. Use the full form and you will get the result you want. The full form is given right at the end of this page


0

Notice that the movement of light must always be treated as a local phenomenon, which in your thought experiment means the light is travelling only within the moving box. It is not travelling from the moving box to a stationary observer. Therefore, throughout the whole experiment the box is stationary for the light and no wall is moving toward or from it. ...


1

"... my questions is now one of why and how (mathematically) do we obtain the Minkowski Metric Signature. More specifically the one element with a different sign." Well, If you check Einstein's "Relativity: The Special and General Theory", you will find in the Appendix I (just before equation (10)) that Einstein simply started off with the ...



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