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0

Here's an example I like of why entanglement doesn't let you violate relativity. Say you have two spaceships moving in opposite directions along a line, with constant velocity. At $t = 0$, they synchronize clocks and entangle two particles. They also decide, at some predetermined time $T$, to measure the spins of the particles (actually, ship 1 will measure ...

1

the emitter and the receiver move with constant velocities relative to an inertial frame and $v$ is the constant velocity of the receiver relative to the emitter and away from it. No, both the emitter and the receiver are accelerating, and the receiver has gained an extra velocity $v$ between the time the photon was emitted and the time it was received. ...

0

I myself overlooked it too but wikipedia actually happens to have a great such list at https://en.wikipedia.org/wiki/Quantum_gravity#Points_of_tension There are other points of tension between quantum mechanics and general relativity. First, classical general relativity breaks down at singularities, and quantum mechanics becomes inconsistent ...

0

I am 100% confused by this. I do not understand why the plane clock runs slower than the earths clock? Why the preference on the plane? Why not on Earth? Why does the plane lose time, and not the other way around? (I cant think why one body or the other gets the preference as from the perspective of one or the other they are both moving the same). From the ...

1

I don't know where you measured 39°. The blue angle is 36.87° (the sides of the red triangle are 3-4-5), as it should be.

0

Use the relativistic velocity composition formula: $$\frac{u+v}{1+uv/c^2}$$

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 ...

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 ...

1

I thought the same thing for a long time. I wondered why gluons don't fly out of the nucleus at the speed of $c$. The difference is that photons don't interact with other photons and gravitons don't interact with other gravitons. They can move around and pass through each other. On the other hand, gluons do interact with each other. In fact, gluons form ...

1

If you know about Lagrangian and Hamiltonian formalisms yo might try to find first the equations of motion. This is done in the paper Relativistic harmonic oscillator. In a nutshell, what is done is the following, a "relativistic" hamiltonian (for slow particles) is set up (we set c=1): $$H = \sqrt{p^2+m_0^2} + \frac{1}{2}k x^2$$ Then the evolution of a ...

0

The relativity theory goes further and implements rules how to add up speeds. In the case you described, one spceship will percieve the other one as being nearly the speed of light, but not above. The formula that has to be used to add up speeds correctly in special relativity in the described case is: $$s = \frac{v+u}{1+(vu/c^2)}$$ where $u$ and $v$ denote ...

1

Take the scalar product of $(1)$ and $(4)$ to get $$\mathbf{F\cdot U}= \gamma^2(u)\left(\frac{dE}{dt} - \vec f\cdot \vec u\right)$$ In the proper frame, this becomes the rate of change of the rest energy, so that for a rest-mass preserving force, $\mathbf{F\cdot U}= 0$. Hence the four-force in this case must be of the form $$\mathbf{F} = \gamma(u)(\vec f, ... 1 It's quite easy to get confused by differences in notation: in this post of mine I defined v_\parallel and v_\perp as the velocity components parallel and perpendicular to the coordinate acceleration \vec{a}. In your question however, you're interested in the angle between the velocity and the force vector \vec{f}, and that's a different angle: ... 0 Take units c=1. You have U_0^2-\vec U^2=1, that is \gamma^2(1-\beta^2)=1. With some basic transformations, you will get : \frac{\gamma - 1}{\beta^2}= \frac{\gamma^2}{\gamma + 1} Now, from your Wikipedia matrix, you have obvious term,  U_0 =\gamma , U_i =\gamma \beta_i You have (\gamma -1) \frac{\beta_i\beta_j}{\beta^2} = \frac{\gamma^2}{\gamma ... 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 ... 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 ... -1 Please correct me if I am wrong. But I think that the speed of light, measured distance and time from a frame of reference are concepts defined relatively to each other. In the sense that we fix the speed of light and define distance and time relatively to it. In particular I am not really convinced that the measured speed of light is the same in every ... 0 What happens after the blast does not depend on the speed of the object - it is the principle of relativity. Assuming that the object moves at 9/10 of the speed of light and that the blast accelerates the particles at 1/10 of the speed of light in all directions, the 'at rest' observer would see the particles in front of the object moving at a speed of:$$ ...

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 ...

1

Spinor algebra is very helpful in sorting out spinorial equations. In chiral representation the (four-component) wave function of the fermion field $\psi$ is considered as a formal sum of first rank spinor and first rank co-spinor fields: $\psi{}(x)=\left\{\xi{},\ \dot{\eta{}}\right\}=\ \left\{\xi{},\ 0\right\}+\left\{0,\ \dot{\eta{}}\right\}=\ ... 5 It is nothing but a problem with real quadratic forms. You have a pair of vectors$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')$. Actually these components describe the same vector in spacetime (describing the difference of events) but referring to two different ... 4 In general, uniform motion in one reference frame implies uniform motion in a different reference frame. Suppose that frame$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 ... 1 If you're familiar with differential forms, then akhmeteli's answer is great, especially if you want to generalize to curved geometries. Let's try to be notationally and mathematically precise without using forms and be as explicit as possible. Let a vector potential$A = (A^\mu) = (A^0, \mathbf A)$be given. Consider a parametrized path$x(\lambda) = ...

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 ...

-1

Yes, Einstein definitely knew about the (lack of) results of the Michelson-Morley experiment. It is like asking whether Stephen Hawking knows that gravitational waves have not been directly detected yet. That was simply the most exciting experiment of the time(I believe even regular newspapers published the results). No one would try to revolutionize physics ...

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 ...

0

"how does energy perceive what atoms it has to form" May i correct you here energy will not create new matter(or new atoms as you mentioned) in the case you mentioned. It will just increase the mass of the existing matter. For if you accelerates an electron from rest to a speed comparable to speed of light what you will get is the same electron with ...

1

Let's be clear. At the present time, you cannot go back in time and you cannot send a grocery list faster than light and there is little chance it will ever be true. Unless it's been done, the answer remains a very clear NO. Also, Einstein never meant his relativity formula to indicate that you could go back in time. The formula was only meant for velocities ...

1

"Superluminal Interactions" is a fancy way (and, IMO if you look at the Latin roots, a not too accurate way) of saying "faster than lightspeed interactions". It is the notion that objects or events at a spacelike separation in spacetime from one another can influence each other's physics or transfer information. In flat, Minkowskian spacetime, the "proper ...

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 ...

0

A particle's interaction(with anything it can interact with) can be thought of as ,it making a measurement (of the physical quantity associated with the interaction-eg electric field in case of charged particles interaction)and acting accordingly This isn't the correct way to think about interactions, and as you later point out this (kind of) violates ...

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 ...

0

The aeroplane is moving in the atmosphere of the earth, and so how can you say that the earth is moving faster, when viewed from outerspace. I think the aeroplane will still be faster.

1

Basically, the universe has a speed limit. No object can ever exceed the speed of light. Now imagine you decide to prove Einstein wrong by building a train capable of nearly reaching the speed of light, and then shooting a bullet forward in that train, so that the bullet will break the speed of light. In order to preserve this speed limit, time for you and ...

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 ...

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 ...

1

I wonder if you're getting mixed up with propagation of waves in a physical medium like a string. If you have a wave travelling on a string then it has a velocity along the string, but the string is also oscillating normal to its length. So if you stretched the string along the $z$ axis, as the wave travelled along the string (i.e. the $z$ axis) the string ...

0

A photon is the quantized unit of the electromagnetic field. If you have en electromagnetic wave propagating in the x-direction, this must consist of a magnetic field and an electric field oscillating perpendicularly to the direction of travel, and to each other, i.e. in the y and z directions. If you have a wave with a frequency of, as an example, 50Hz, it ...

0

You're confusing the process of quantization with the wave-nature of propagating electromagnetic fields. When you look at a Electromagnetic waves as photons, this means you don't look at their wave-characteristics, and you consider them as particles travelling with the speed of light, and those particle could "hit" electrons and knock them our of the atom ...

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 $d/v$ ...

1

Here is another proof. Let us assume that there is another invariant $I_3$ functionally independent of $I_1= E^2-B^2$ and $I_2=\mathbf{B}\cdot\mathbf{E}$. This would mean that There are pairs of vectors $(\mathbf{E},\mathbf{B})$ and $(\mathbf{E}',\mathbf{B}')$, which have the same $I_1$ and $I_2$ but which cannot be turned into one another by some ...

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: ...

1

I think the point is that the only invariant tensors (under proper Lorentz transformations) are $\epsilon^{\mu\nu\alpha\beta}$ and $\eta^{\mu\nu}$, so any invariant will contain some number of powers of $F_{\mu\nu}$ where the indices are contracted (raised) with these two invariant tensors. Because of antisymmetry and symmetry, $\eta$ can only act once on ...

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 ...

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 ...

1

How do we prove that any directions are orthogonal? [...] we can use the pythagorean theorem. This involves of course a definition of (how to measure or compare) "angle(s)" in the first place; such that one may comprehend statements about (distinct) angles being "equal" (or else: "not equal") for instance in Euclid's 4th axiom (on "right angles") or in ...

2

It is proven in Tong's QFT script http://www.damtp.cam.ac.uk/user/dt281/qft.html section 4.1. in a quite nice fashion.

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

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 ...

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