New answers tagged

1

Are situation (1) and (2) equivalent when viewed from twin A's frame? No. The equivalence principle says that the outcome of any local experiment is the same if you are at rest in a gravitational field or accelerating uniformly. Situation 1 and 2 are therefore not equivalent. First, the experiment is not a local one per the meaning used in the equivalence ...


0

You have asked a very central question, and yet one that physics really can't answer. Physics describes how the universe works. The reason why one law is true can only be that it is a consequence of some simpler law. The simplest laws (and many more complex laws) are known to be true because they have been experimentally verified. Nobody knows why the ...


3

It's just a direct consequence of the geometry of spacetime. This question is similar to asking, "What aspect of traveling on the surface of a globe makes you return to your starting point after traveling long enough in a particular direction?" It's simply the fact that moving along the curved surface eventually changes your direction of travel in such ...


0

There are a few things to remember here. First, that your bodily activities act as your biological clock. Second that time runs slower for the moving twin, and the twin who switches direction comes back younger. Now If you have experienced to have aged by 5 years, depending on your speed your twin must have aged by more than 5 years. Your internal clocks ...


4

Time dilation affects, but is not a consequence of, any processes, microscopic or macroscopic. It is a consequence of the Minkowskian geometry of the spacetime in which all processes occur and has absolutely nothing to do with the details of their dynamics.


0

The perpendicular line $x=1$ ($x=0$) is the world-line of the end (start) of the meter-stick. You have to follow that line from $Q$ up to $P$ because that is the point that is simultaneous with $O$ in the primed frame, that is, both $O$ and $P$ are at $t'=0$. So $\bar{OP}$ defines the object in the primed frame; $\bar{OP} < \bar{OQ}$, but that is not ...


1

It seems to me that the $T_{\alpha,\gamma}$-expression is in arbitrary coordinates. However, the $v_{ij}$-expression is in Cartesian coordinates... and has likely implicitly used the metric to raise and lower indices.


1

Motivated by breaking up "relativistic" energy ($\tilde{H}$) into rest mass energy (M$c^2$) and non-relativistic kinetic energy (H) we add the following redefinition $\tilde{H} \rightarrow M c^2 + \frac{1}{2}H.$ And further, motivated by the equivalence of mass and energy in relativity, we define a mirrored redefinition for $\tilde{M}$ given by $\tilde{M} \...


1

First, we need to describe what length of an object is? The length of a ruler is the distance between the two coordinates of endpoints of the ruler which are measured simultaneously. Now, 2 simultaneous events happening in one frame might not be simultaneous in other and that's where length contraction comes in. Lorentz transformation is given by: \begin{...


5

As explained in the beginning of section 2.2.2 in Ref. 3 let us distinguish between 3 algebras: The (centrally extended) Poincare algebra in $n$ spacetime dimensions, whose generators we will adorn with a hat. The redefined relativistic algebra with $c$-dependent structure constants. Its generators we will adorn with a subscript $c$. Its $c=\infty$ limit,...


7

Edit: I just noticed that Rob's much more succinct answer basically says the same thing. This answer is mainly for people who'd like to derive how the electric and magnetic fields transform under a boost. The short answer is that when you have a charge moving in space, there is a charge density as well as a current density, though it's not quite as simple ...


1

I believe your (fairly complex) question is answered in a general way on Kevin Brown's Mathpages, in this article: https://www.mathpages.com/rr/s6-05/6-05.htm It compares the elapsed times for two different inertial motions in the Schwartzschild spacetime that intersect at beginning and end. Applying to your specific question is an exercise for the reader!


8

The Lorentz force is not a Lorentz invariant, so if you get the same total force in S' as S, then you are doing something wrong. In the stationary frame of the electrons, the force on one electron due the other is given by a pure Coulomb force $$ {\bf F'} = -e{\bf E'} = \frac{e^2}{4\pi \epsilon_0z'^2}\ {\bf \hat{z}} ,$$ where $z'$ is their separation along ...


0

A point charge $q$ can be thought of as the charge density $\rho(x^\mu) = q\ \delta^4(x^\mu)$ if you want. But in the case of point charges you could apply the transformation directly to the Lorentz force $$f = \frac q{m_0}\iota_p F,$$ where $p$ is the 4-momentum and $F$ is the electromagnetic tensor. Indeed, this last expression alone, which in ...


2

Your friend was more wrong than right, as others here have said. I am just adding some thoughts that emerged after long experience with relativity. This becomes a question of practicalities. Basically there are pros and cons to natural units (units where $c=1$ among other things). Pros reduce clutter in formulae and in long derivations train one's mind to ...


0

The four momentum: $$ p^{\mu} = (E/c, \vec p) $$ transforms between frames as a Lorentz 4-vector, so energy (scaled by $1/c$) is the time component and 3-momentum is the space component. What that means kinematically can be understood by looking at the relationship with 4-velocity $$ u^{\mu} \equiv (\gamma c, \gamma \vec v) $$ which is: $$ p^{\mu} = mu^...


0

The electric and magnetic fields are always perpendicular to each other and the direction of wave travel. In all frames of reference. However, the strength of the fields and the frequency/wavelength can change.


2

What am I missing? You are missing the fact that a gauge transformation transforms $\psi$ as well. It changes the complex phase of the spinor at each point in spacetime. See if you can figure out how much of a phase change will leave the equation unchanged. It’s this local phase change of the matter field that explains why the gauge group for ...


4

You are missing the phase transformation of $\psi$. The Dirac equation is indeed not invariant to the gauge transformation $$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\Lambda$$ alone. But it is invariant to the combined gauge/phase transformation $$\begin{align} A_{\mu}&\rightarrow A_{\mu}+\partial_{\mu}\Lambda \\ \psi &\rightarrow \exp\left(-\frac{...


0

I would say that not only are the definitions not equivalent, they are not even correct. A frame of reference means the reference matter which is used to define a coordinate system. For example we talk of the Earth frame, or if we are in a car, we will use the car as a reference frame. Norton's definition does not even refer to reference matter, but only to ...


9

The rule (1) contains an implicitly written angular $k$-integration: $$ \int_{\mathbb{S}^{d-1}}\! d^{d-1}\Omega ~ k^{\mu}k^{\nu} ~=~\int_{\mathbb{S}^{d-1}}\! d^{d-1}\Omega ~ k^2 \delta^{\mu\nu}/d.\tag{1'}$$ (Schwartz is working with $d=4$ dimensions and Euclidean signature.) The rule (1) is of course not true without the angular integration/averaging in $...


6

Pedagogically, I agree with @tparker's answer in that I do think it is not a wise thing to rush to $c=1$ before a student is relativistically mature enough to not misunderstand it. However, ultimately, I think your friend is $100\%$ right, you can omit that $c$, and not just that, it's a bit silly to write that $c$ as an adult. ;) A Wheelerean Delight In ...


0

You should be aware of the derivation of these equations and what the different terms mean. The first fraction in your last equation is the classical Doppler effect. The square root expression stems from the relativistic time dilation. This time dilation is between the reference frames of the source and the receiver. If their relative velocity is small ...


7

This is basically a philosophical question, but I'm going to take what will probably be an unpopular position that your friend's reason is basically wrong in the context of an introduction to special relativity. Sure, you can work in units where c = 1, and then the equation $E = m c^2$ reduces to $E = m$. But that fact alone is kind of vacuous: you can also ...


1

Since the OP asks to find the, and I quote here, formula for relativistic momentum under the requirement of conservation of momentum for inertial frames (the last word being my guess which makes most sense), we do the following. We first define the particle orbits as functions $x^\mu(\tau)$ in spacetime, where $\tau$ is an arbitrary Lorentz-invariant ...


-1

Momentum is only conserved if there is no external force on your system. Since force is the time derivative of momentum, momentum is conserved if the external force on a particle is zero. Let me make this more mathematical: $0 = \vec f = d\vec p / dt$. This holds for the non-relativistic and the relativistic case. Of course one can write down a lagrangian ...


2

Conservation of energy-momentum is a fundamental principle in relativity; it is "built in" to Einstein's equation $$G^{ab}=8\pi G T^{ab}.$$ It can be proven for interactions in quantum field theory, but that is a heavy duty proof. Otherwise, it is best to take it as a fundamental principle (it can also be proven from Noether's theorem, but that depends on ...


0

In a collision, the conservation of 4-momentum can be described by a polygon (just like a free-body diagram of forces on an object in static equilibrium): $$\sum_i \tilde P_{i,\rm before} - \sum_j \tilde P_{j,\rm after}=\tilde 0.$$ Then, as @stackoverblown says, Lorentz Transformations are linear transformations (just like Euclidean rotations and Galilean ...


0

Conservation of momentum is mingled with the conservation of energy when you go to relativistic speed. $$E^{'} = \gamma (E - v p)$$ $$p^{'} = \gamma (p - \frac{v E}{c^2})$$ Now if you have conservation $E_1+E_2 = E_3+E_4$ and $p_1 + p_2 = p_3 + p_4$ then because Lorentz transformation is linear, they will just transform into $E^{'}_1 + E^{'}_2 = E^{'}_3 + E^...


0

Your friend is right. If you adopt the length unit l = 299 792 458 metres then c=1 l/s. This can be convenient because in these units $E^2=m^2+p^2$ instead of $E^2=m^2c^4+p^2c^2$.


2

In Special Relativity, we use something known as a spacetime diagram. It looks like this: Now, here we are only looking at 1 dimension of space and dimension of time. The green dot represents and event, that takes place at some point in space, and at some point in time. In this diagram, every object can be viewed as a worldline (see the orange line), and ...


0

In the Natural Units, the speed of light in vacuum i.e. $c$ is taken to be the fundamental speed of the universe. Under this system, all the fundamental physical constants are defined in such a way that their value is just 1 (e.g. $\hbar=k_B=1$). However, in the end one has to include the numerical values when switching from one system of units to another, ...


0

It is very common to "omit" constants, especially in theoretical physics. Actually, there is a system based only on physical constants ($c,\hbar,\epsilon_0,G ...$). Physical units in this system are called natural units and we can "normalize" those constants by choosing a system in which their value is 1, they are called purely natural units (or just ...


1

A Lorentz invariant is what we call a Lorentz scalar. Scalars don't have any upper or lower indices. For example, the 4-momentum $p^\mu$ is a Lorentz vector because it transforms the way a vector $V^\mu$ transforms, i.e $p'^\mu=\Lambda^\mu_\nu p^\nu$. A Lorentz scalar on the other hand stays the same when performing a Lorentz transformation $m'=m$. Here $m$...


0

Just combine tensors in a way that all Lorentz indices are contracted e.g $a_\mu a^\mu$, $a_\mu a_\nu T^{\mu\nu}$, $T^{\mu\nu}T_{\mu\nu}$ etc.


1

Actually, $E = mc^2$ describes the relation between energy of an object at rest. $E$ is the rest energy and $m$ is the rest mass. The actual relation between energy and momentum in Special Relativity is $$E^2 = (mc^2)^2 + (\vec{p}c)^2$$ Here the $\vec{p}$ is the momentum, and $E$ is the energy of the object. This is the more universal law and applies to all ...


1

The change in mass of a composite system due to the kinetic or potential energy of its components is very real. The amount of kinetic and potential energy of the components of the system affects how the composite system accelerates when you apply a force to the system as a whole, and it also affects the amount of gravitational force it exerts on other masses....


-2

Why is the mass of the system larger when the mass points are far apart? The mass of the system is not larger when the mass points are far apart or are close together. Since there is no interchange between mass and energy during this process. During nuclear binding process, as the process you described, some energy is stored as gravitational potential ...


0

Noether's theorem states that conservation of momentum follows from homogeneity of space (i.e., the invariance of tha physical laws in respect to translations in space), whereas the conservation of energy follows from the homogeneity of time. In QM this interpretation becomes obvious, since the momentum and the energy operators are just the generators of ...


0

The reason we construct a mathematic representation where we describe the same physics in two different coordinate systems is because we want a mathematical description that obeys the Equivalence Principle. Essentially, the fundamental idea of relativity is that physics works the same way for all intertial frames of reference. It would be very strange if you ...


2

The construction of the S'-axes for Minkowski spacetime is analogous to the S'-axes for Euclidean geometry. So, it is not "unique to special relativity". In other words, first ask your questions about Euclidean geometry [and how you might prove it] then see if the answers can be applied to Minkowskian geometry by analogy (possibly with some generalizations ...


1

"Why ...?" By construction. The way we plot the axis of the second coordinate frame (wiki), i.e. the way we map $S'$ onto the piece of paper, is chosen in such a way as to allow for every point allready plotted on paper, as being in $S$, to also be the corresponding event in $S'$. You may ask wether such construction is non-contradictory, i.e. whether you ...


1

The answer to this question, is similar to the equivalence principle The Equivalence Principle The Equivalence Principle For a small enough system, there is no difference between a gravitational field and uniform acceleration. Note the equivalence principle as given by Einstein doesn't talk of the sizes of the system, the one stated above is the ...


0

I found that part of Schwartz's derivations on the next page explains this retroactively. As preliminaries, note that $(\gamma^0)^2 = \mathbf{1}$ and $\gamma^{0\dagger} = \gamma^0$, as well as $$\gamma^0 (S^{\mu\nu})^\dagger \gamma^0 = S^{\mu \nu}.$$ These can be derived from the Dirac algebra or checked explicitly with the Weyl representation. Then we can ...


0

You know that $m = 1 MeV/c^2$. Your calculation of the initial total energy of 1.25 MeV is correct. The initial momentum should be 0.75 MeV/c. You can now find 2 equations for the momenta of the 2 photons, one equation from conservation of energy, the other equation from conservation of momentum. Note that energy is a positive scalar, but momentum is a ...


0

Why $\Lambda_s^\dagger \gamma^0 \Lambda_s = \gamma^0$ for general $\Lambda_s$? Expanding infinitesimal form of $\Lambda_s = \exp{\left(i \theta_{\mu\nu}S^{\mu\nu}\right)}$ and keeping only first order terms gives this: \begin{align} \Lambda_s^\dagger \gamma^0 \Lambda_s &= \left(1-i \theta_{\mu\nu}S^{\mu\nu}\right) \gamma^0 \left(1+i \theta_{\mu\nu}S^{\...


1

You need to apply conservation of energy & momentum. Based on your comment it seems the difficulty is that "photons are massless". They are indeed massless, but they can still carry momentum. Once you know the momenta of the photons, it's easy to convert to energy via $E = pc$.


-2

In general, $\overline{AB}=\bar B \bar A$. Hence $\overline{\Lambda_{s}\psi}=\bar \psi \Lambda ^{-1}_{s}$ since $\Lambda_{s}$ is real and $\Lambda^{t}_{s}=\Lambda^{-1}_{s}.$


1

This paper seems to fit what you are looking for precisely: it starts with a derivation of the Euclidean distance formula with respect to two dimensional creatures who live in a three-dimensional world and can measure the travel they go through in the third dimension using altimeters. It then naturally extends this argument to our case, that of three ...


0

You can make any theory formally covariant but the point of the principle of relativity is that you have to make your theory invariant among all inertial frames, i.e., the laws in one inertial frame should not refer to its velocity with respect to another inertial frame. This can be achieved if and only if the metric is invariant. The only metric that is ...


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