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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 ...


17

Calculating the effect of acceleration in special relativity is straightforward, but I suspect the algebra is a bit much at high school level. See John Baez's article on the Relativistic Rocket for a summary, or see Chapter 6 of Gravitation by Misner, Thorne and Wheeler for a more detailed analysis. When you're first introduced to SR you tend to be told ...


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 ...


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. ...


7

If you define $x^0=ict$, then I assume one takes $x_\mu=x^\mu$ so that the metric is actually $\eta_{\mu\nu}=\text{diag}(1,1,1,1)=\delta_{\mu\nu}$, i.e. you're dealing with a Euclidean metric. Then $$ds^2=\delta_{\mu\nu}dx^\mu dx^\nu$$ gives the usual outcome : $$ds^2=-c^2dt^2+d\vec{x}^2$$ The usual conventions are as follows: Option one: One defines ...


7

as you wrote, the spacetime invariant can be expressed as: $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ and from that we normally get: $$ds^2=-c^2dt^2+dx^2+dy^2+dz^2$$ This is not because of some arbitrary imaginary time unit, this is because the metric ($g_{\mu\nu}$) is a diagonal matrix with the coefficients of each term of the $ds^2$ equation: ...


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 ...


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 ...


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 ...


5

There is some discussion about this in the question How does a photon experience space and time?. You'll commonly hear it said that photons don't experience time, but this is somewhat misleading. Observers moving at different velocities have different coordinate systems, and these systems are related by the Lorentz transformation. If you apply the ...


5

The problem with your idea is that each time the light reflects off the mirror it transfers some of its energy to the mirror (to increase the mirror's kinetic energy) and is red shifted as a result. So the thrust would fade as the light red shifts away to nothing. For obvious reasons the light can only transfer as much energy to the mirrors (in the form of ...


5

The addition of relative velocities is $$\frac{a+b}{1+ab}$$ so $\frac{.75+.75}{1+.{75}^2}$ = .96 c. Consider that you are travelling almost the speed of light (1-x) c and you see a space ship pass you at an equal speed. Combining (1-x) c with (1-x)c. This gives $$\frac{2-2 x}{2-2 x+x^2}$$ Since $2-2 x < 2-2 x+x^2$, we have $\frac{2-2 x}{2-2 x+x^2} ...


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 ...


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 ...


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 ...


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 ...


4

There is no way to be 100% sure, but we can put upper limits on the mass. Massless particles don't have a rest frame, so it doesn't make sense to talk about time dilation in the photon's frame. A massive photon would have a rest frame, so you could eventually catch up to it and move alongside it. List of experimental limits on photon mass more ...


4

No the speed of light in vaccuum is an absolute constant $c$ = 299 792 458 m/s The way to add up relativistic speeds is: $u' = \frac{u-v}{1-\frac{uv}{c^2}}$ to account for the constancy of the speed of light You cannot simply add them up. Edit: This also applies to normal everyday speeds. The reason we don't use this formula is because the speeds we are ...


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, ...


3

Imagine a stealthed1 alien vessel is approaching the earth with relative velocity $\beta = \frac{v}{c}$. At a distance of two light-years as measured in the Earth's frame2 the craft switches on its radio and send the Earth a ultimatum. We call this moment time zero in the Earth's frame. Then it turn the radio off until it reaches a distance of one ...


3

We don't normally answer homework question in full, but I'll answer because I'm very eager to make the point that the best, by far, method of answering questions like this is to choose relevant spacetime points in $S$, apply the Lorentz transformations and find where the points are in $S'$. This normally makes the answer obvious. In this case we have a pair ...


3

There is one formula relating the speeds of any two "platforms" (say $P$ and $Q$) between each other: $$V_{P}[ Q ] = V_{Q}[ P ].$$ And there's of course the well known symbol for "speed of light (in vacuum)", as determined of light signals exchanged by members of any one platform between each other: $c$. The speed of any one platform ($Q$) as determined ...


3

The short answer to (1). $F^\mu{}_\nu$ and $F^{\mu\nu}$ are related by $F^{\mu\nu} = g^{\nu\rho}F^\mu{}_\rho$ where $g^{\mu\nu}$ is the metric ($g^{\mu\nu} = \operatorname{diag}(1, -1,-1,-1)$ in Minkowski spacetime). Since the metric is invertible, either of $F^\mu{}_\nu$ and $F^{\mu\nu}$ uniquely determines the other. You can pick whatever version you ...


3

In relativity, in order for something to be at rest with respect to something else, both of them must be particles with nonzero mass in their respective rest frames, which are represented by time-like four-vectors. Saying that two photons are at rest one w.r.t another does not make sense, because there exists no inertial frame in which a photon can be at ...


3

If you are the observer and you are observing a car move in front of you, then the clock in the car will appear to move slower than the clock in your reference frame and the length of the car will appear to be contracted to you in the direction of motion. All these effects get magnified as the velocity of the car approaches that of light. In the same way, ...


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 ...


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 ...


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 ...


2

In the diagram you've drawn, any line with a gradient of greater than one is timelike and any line with a gradient of less than one is spacelike. As your diagram shows, the gradient of all radial lines (i.e. outwards from the time axis) on the hyperboloid have a gradient of less than one. So to the observer whose light cone it is, any object following one of ...


2

I claim that if the transformation between frames is homogeneous and differentiable, then it is affine (homogeneity is not strictly speaking sufficient for linearity since the full transformation between frames is actually a Poincare transformation which is affine, not linear) For a mathematically precise proof, we need a mathematical definition of ...



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