# Tag Info

18

Photons don't have a rest frame, since in all inertial frames they must go at the speed of light. So the following statement: By that logic, photons don't age in a vacuum state as, to us, the time stops for them. is meaningless because one really can't talk about proper time for a photon. However, in a medium, their speed decreases, Nope. The ...

7

I will turn my comment into an answer, because the question in the header: Do photons age? is very anthropomorphic , and physics is a discipline that discourages interpreting data by use of the anthropic principle. The photon is an elementary particle. Aging is not a verb to be used with elementary particles in general because a) they have no ...

6

Did he knew about the Michelson-Morley experiment? He just knew the name of the experiment not any details. The experiment didn't play any role in the formulation of STR by Albert Einstein. The context is taken from the book: Special Theory of Relativity by V. A.; Atanov, Yuri (Trans.) Ugarov (Author) Art: Was Michelson's experiment "decisive" for ...

5

You seem to have overcomplicated this question quite a bit, you don't need to go further than the theory of relativity to find the answer. Heat is simply chaotic movement on the molecular and submolecular level, and as relativity dictate that any object moving relative to an observer will to that observer appear to have greater mass with greater relative ...

5

With the first question you are correct. Any "thing" with nonzero mass cannot achieve light speed. From this equation you can see why $$m=\frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $m_{0}$ is the rest mass of the body (i.e. the mass it has when its speed is zero). As you can see from the equation, when $v=c$, the right hand side will blow up to ...

5

It is nothing but a problem with real quadratic forms. You have a pair of vectors $v,v' \in R^4$ with, respectively, components $(\Delta t, \Delta x, \Delta y, \Delta z)$ and $(\Delta t', \Delta x', \Delta y', \Delta z')$. Actually these components describe the same vector in spacetime (describing the difference of events) but referring to two different ...

5

Another way to say this: Speed of photon, graviton, gluon all equal to c? or Whether all massless particles necessarily have the same speed? You must not have been introduced to the concept of a virtual particle: In physics, a virtual particle is a transient fluctuation that exhibits many of the characteristics of an ordinary particle, but that ...

4

If the twins never meet, but just continue travelling in a straight line at constant velocity then each twin will see the other as being younger. The *paradox*$^1$ only occurs if one or both of the twins is accelerated, which of course is necessary for the twins to meet again. $^1$ it's not a paradox of course, just an unintuitive result!

4

Yes. More specifically, if $d$ is the distance between the planets in their rest frame, then in the astronaut's frame the distance between the planets will be $\frac{d}{\gamma}$ so the travel time as measured from his frame will be \begin{align} t_\mathrm{astro} = \frac{d/\gamma}{v} = \frac{1}{\gamma}\frac{d}{v} \end{align} Notice that the quantity $d/v$ ...

4

If you say that earth's velocity around the sun is 67,000 mi/h, your reference point is the sun itself, which makes the aeroplane's velocity 68,000 mi/h, not 1000. Using special relativity only, and (A) observing from the sun, a clock on the plane would seem to run slower than a clock on earth. A person (B) on earth would measure also measure an ...

4

So I assume you actually need to prove Poincare invariance of $\int d\tau A_\mu\dot{x}^\mu$ for a particle trajectory, rather than invariance of $A_\mu\dot{x}^\mu$, but the former expression is equal to $\int_a^b dx^\mu A_\mu$, where $a$ and $b$ are the initial and the final points of the trajectory, and Poincare invariance is indeed almost obvious for this ...

4

In general, uniform motion in one reference frame implies uniform motion in a different reference frame. Suppose that frame $K'$ is moving at a constant velocity $\mathbf{v}$ relative to frame $K$. The transformation from frame $K$ to $K'$ must be linear, so it must be true that $$ds'^2=a\,ds^2\tag{1}$$ where $a$ depends on the relative motion of $K$ and ...

4

The affine Galilean structure is assigned by the first principle of Newtonian dynamics, i.e. by giving the class of inertial reference frames in the spacetime $G^4$. On the one hand it assigns the structure of an affine space to the spacetime, on the other hand it selects a subclass of permitted transformations between reference frames. A reference frame ...

3

Here is the proof taken from Landau & Lifshitz' "Classical Theory of Fields": Take the complex (3)-vector: $$\mathbf{F} = \mathbf{E}+i\, \mathbf{B}.$$ Now consider the behavior of this vector under Lorentz transformations. It is easy to show that Lorentz boosts correspond to rotations through the imaginary angles, for example boost in $(x,t)$ plane: ...

3

I think it is correct. However, just as a pedantic remark, you should prove $U(\Lambda)a^{s\dagger}_\mathbf{p}U^{-1}(\Lambda)$ and $\sqrt{\frac{E_{\Lambda\mathbf{p}}}{E_{\mathbf{p}}}} a^{s\dagger}_{\Lambda \mathbf{p}}$ act on all the vectors in the same way, not just on the vacuum $|0\rangle$, you need a slight modification of your proof: $$\sqrt{2 ... 3 It is incorrect to say that$$ U(\Lambda) \left|p,\sigma\right> \propto |\Lambda p, \sigma\rangle~~~~~~ \text{WRONG!!} $$Here is the correct logic. Consider the state U(\Lambda) |p,\sigma\rangle. We have just shown that (in eq. 2.5.2) that this state has a momentum eigenvalue \Lambda p. Now, there are a whole bunch of states with momentum \Lambda ... 3 Note that if you lower an index of the Kronecker delta, it becomes the metric: \eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho} And in your last step you got a wrong index. It should be \omega_{\rho\sigma}, not \omega^{\rho}_{\sigma}. Then, the metric terms cancel and you neglect cuadratic terms. That should be enough to solve it. 3 The answer is that a decrease in temperature does decrease the mass, though in most cases that change is exceedingly small. Temperature is a macroscopic phenomenon, so you can't really talk about the temperature of a single string or an atom. However consider the following analogy: If you have an isolated string (or atom) in some excited state, then to ... 3 Muons are single-particle excitations (states) of the e-\mu-\tau quantum field, except that these states don't have definite values of energy (they are in a superposition of states that have definite energy). Because states with different energies change at different rates, this superposition changes with time. After some time has elapsed, the ... 3 If basic symmetry and homogeneity assumptions about the Universe hold, then yes, all massless real particles (see Anna V's answer for virtual particles must travel at a universal constant c, the speed of a massless particle, in all frames of reference. Given these basic symmetry and homogeneity assumptions, one can derive the possible co-ordinate ... 2 Let$$ \eta_{\mu\nu}={\rm diag}(+1,-1,-1,-1) \qquad \bar\eta_{\mu\nu}={\rm diag}(-1,+1,+1,+1) $$with corresponding Lorentz force laws (in units where mass equals charge)$$ \ddot x^\mu=\eta_{\nu\lambda}F^{\mu\nu}\dot x^\lambda \qquad \ddot{\bar x}^\mu=\bar\eta_{\nu\lambda}\bar F^{\mu\nu}\dot{\bar x}^\lambda  As the trajectories $x^\mu, \bar x^\mu$ should ...

2

It always helps to draw the right picture. This picture assumes that Boxguy is standing next to the lamp, and that the flash leaves the lamp just as it passes PlatGirl. (If, for example, BoxGuy were standing next to the mirror, the picture would look a little different.) The black vertical line is Platgirl's worldline, and any black horizontal line is ...

2

In his original work, Fermi considered only vectors $f^{\mu}$ which are orthogonal to the curve $f^{\mu} v_{\mu} = 0$. His analysis is relevant to the spin or photon polarization vectors which are orthogonal to the four-velocity by definition. Walker generalized Fermi's work to vectors which are not necessarily orthogonal to the velocity. (Thus the ...

2

Time dilation (and also length contraction) always occurs with respect to an observer in a different frame of reference. You, in your own inertial frame, will not notice any difference. However, when you compare your measurement to that of an external observer, you will see a discrepancy in the results. If you enter a spaceship and go on a journey through ...

2

Wiki says : "Einstein and Fokker observed that the Lagrangian for Nordström's equation of motion for test particles, $L = \phi^2 \, \eta_{ab} \, {u}^a \, {u}^b$ , is the geodesic Lagrangian for a curved Lorentzian manifold with metric tensor $g_{ab} = \phi^2 \, \eta_{ab}$ ." [Remark : here $u^a = \frac{dx^a}{ds}$, so no "dot" here on the $u^a$, I think ...

2

I struggled with this one as well and once I found I have written it in LaTeX which I will copy here below. Do note that I am using slightly different conventions than P&S, however it should still work out the same. \begin{aligned} S^{\mu \nu} & = - \frac{i}{4}[\gamma^\mu,\gamma^\nu] \\& = - \frac{i}{4}(\gamma^\mu \gamma^\nu - ...

2

What if without meeting they send a light pulse to each other, such that they can know each other's age The result will still be the same - each twin judges the other twin to be ageing more slowly than themselves. However, sending a light pulse to each other involves other factors that must be taken into account such as time of flight and ...

2

Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get ...

2

The answer to your question: Would it be correct to assume that the particle has a stronger gravitational field [...]? is no, it would not be correct. Here is why. Comparing gravitational field in special relativity to its Newtonian limit means trying to take an ill-defined limit. If one wants to include relativistic corrections such as relativistic ...

2

Is there any significance in saying an observer as an imaginary entity? Yes. From Wikipedia: Physicists use the term "observer" as shorthand for a specific reference frame from which a set of objects or events is being measured. Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is ...

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