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I'm afraid you're overcomplicating things, aepryus. Yes, most modern clocks use electromagnetic phenomena, but your pendulum clock employs gravity in much the same fashion as your water-drop clock. The clock rate doesn't depend on gravitational potential, it depends on the first derivative of potential, the "slope" as it were. The force of gravity. And this ...


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I know light's speed in vacuum is constant Correct. Specificly, we speak of "vacuum" (and evaluate "refractive index" value $n = 1$) in the context of signal exchange if phase speed and group speed of the "signal carrier" are equal to the signal front speed $c_0$; where $c_0$ is just a particular (non-zero) symbol which appears in the chrono-geometric ...


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First of all, polarization vectors ${\epsilon^\pm_i}^\mu$ can be shifted by gauge transformations such that a quantity proportional to the corresponding Minkowski momentum is added to it: $${\epsilon^\pm_i}^\mu \rightarrow ({\epsilon^\pm_i}^\mu+A_i p_i^\mu)$$ It is easy to show that the above equation is invariant under any such gauge transformations: ...


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Start from $d^4p = dp_{\mu}\,dp^{\mu}$ which is manifestly Lorentz invariant. In order to obtain the actual measure to integrate against you have to pair this up with the mass-shell condition for the particle to have positive energy and lie on the mass-shell hyperboloid $$ dp_{\mu}\,dp^{\mu}\,\theta(k^0)\,\delta(k^2-m^2). $$ Use now $k^2={k^0}^2 - k_ik^i$ ...


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The result given to you by your professor is OK. Special relativity allows you to solve the problem in any reference system, and then going back to the original reference system. So the easiest way to solve your problem is going to the reference frame where the two electrons are at rest. There you have only a electric field $$\vec{E}'(\vec{r}) = ...


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A Lagrangian can easily be written down for a relativistic particle in a curved spacetime (i.e., under the influence of gravity.) Specifically, the "action" is the proper time between two events along a particle's world-line, and the particle's trajectory will extremize the proper time between these events: $$ S = \tau = \int \sqrt{ - g_{\mu \nu} dx^\mu ...


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Constantine, take a look at what Minkowski said in Space and Time: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy ...


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In a very real sense, the velocity of a light ray in a curved spacetime is constant, or at least as constant as it can be; this is because it follows a special path in spacetime called a geodesic. The problem with defining a "constant" vector on a curved surface (the surface of the Earth, say) is that you can't easily compare tangent vectors at two ...


2

Yes light does have different directions in different frames. Two observers with different velocities will see the same photon traveling in different directions. One observer standing still at noon sees light traveling vertically downward. Light that strikes the top of his head would also strike his toes. An observer running forward see light slanted ...


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I know light's speed in vacuum is constant, but what about its velocity? The speed of light in vacuum is not constant, and because of this light curves, hence its vector-quantity velocity varies. Have a look at the Einstein digital papers and you can find Einstein talking about it: This is what John Rennie was referring to. Think about the room you're ...


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Light can obviously travel in any direction, but the magnitude of its velocity (in vacuum) is always $c$. The magnitude of the velocity is a scalar i.e. just a number, but the velocity is a vector. To specify the velocity we need to choose some axes. For example I might choose the Cartesian axes $x$, $y$ and $z$. In that case light approaching me from the ...


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The very useful solutions of the Shrodinger equation that are usually taught in beginning quantum mechanics are not Lorenz invariant and therefore paradoxes with respect to special relativity may be constructed. The relativistic equations of Dirac: the Dirac equation is a relativistic wave equation derived by British physicist Paul Dirac in 1928. In its ...


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The answer is an emphatic yes! We can find the actual rest mass of things on earth. How the earth is moving with respect to the sun, galaxy,etc., is irrelevant. By saying, "on earth," the frame of reference is specified (the earth), so whatever mass a thing has on earth, (as long as it is not moving with respect to the earth), is its actual rest mass! In ...


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Comment on the answer of @Michael: The short answer is that you need antiparticles is false. In Quantum Field Theory you have perfectly working solutions also without antiparticles, i. e. for real fields. Even if you do want to consider antiparticles, always have in mind that despite the misleading name they are in fact different particles from the ...


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The best way to describe a rotating reference system is via a simple change of coordinates. If you have a description of a phenomenon in some set of inertial coordinates $\{t, x, y, z \}$, then you can obtain a description of its motion in a rotating reference frame with coordinates $\{t', x', y', z'\} $ by making an appropriate substitution. For example, ...


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I will try to give you here the proof of work-energy theorem. We have that $\bar p = {m \bar u \over \sqrt{1- {u^2 \over c^2} } } $ the relativistic momentum. We define the force as $\bar F ={dp \over dt} $. So the work is: $W= \int \bar F \cdot d \bar l =\int {d \bar p \over dt} \cdot d \bar l = \int {d \bar p \over dt} {\cdot d \bar l \over dt } dt = ...


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In the entangled system we do not have two separate particles. Instead we have a single wavefunction describing a single system. When you interact with the wavefunction you are not interacting with particle $A$ or with particle $B$, you are interacting with a single wavefunction and causing it to change as a result. So the statement measuring $A$ affects ...


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You are correct that the protons' linear charge density doesn't change because any Lorentz contraction happens perpendicular to the wire. You are also right that the relativistic velocity addition formula predicts that the total speed of the electrons isn't simply $\sqrt{v^2+v_0^2}$. It is actually $\sqrt{v_{(\parallel)}^2+\frac{v_{0,(\perp)}^2}{\gamma^2}}$, ...


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Am I missing something here? Well, you (among others) seem to be missing that to measure "momentum" is defined through the application of the gradient of the translation operator $\nabla \hat T_{\mathbf r}[~] := \frac{d}{d \mathbf r_{\mathcal S} }[~]$ to what's given through observational data (e.g. concerning a particular object $A$ under ...


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You say energy conservation gives $$ E_{\gamma}' + E_e' = E_{\gamma} $$ I think it should be $$ E_{\gamma}' + E_e' = E_{\gamma} + E_e $$ where $E_e = m_e c^2$. This would make your second to last equation $$ \frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma + E_e - E_\gamma') m_e - m_e^2c^2 = E_\gamma - E_\gamma'$$


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Let's say you have previously agreed with your friend doing the other measurement that if the combination of spins is up, down you go to the cinema and if it is down, up you go eat a pizza. Now you do the measurement and he does, both just before going out of the lab. You now have excluded two possible future states, i.e., not meeting because one went to the ...


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Am I missing something here? Yes. What you're missing is "the mass of a body is a measure of its energy-content". Read Einstein's original paper, and take note of this: "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c²". Next, imagine your body is a massless photon in a gedanken mirror-box. It isn't actually at rest ...


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They're both saying the same thing: the relativistic momentum is given by $$ \mathbf{p}=\gamma(v)\,m\mathbf{v} $$ The confusion, it seems, is that you are using Feynman's $m=\gamma m_0$ as equivalent to the $m$ in Resnick & Halliday's text; the actual correlation is Feynman's $m_0$ to Resnick's $m$--both of these terms are the (invariant) rest mass. ...


1

The way to define momentum in a special relativistic context is the following: Start with the trajectory of the particle parametrized by its proper time $x^{\alpha} (\tau)$; define the four-momentum by $p^\alpha = m \frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}$, where $m$ is the mass of the particle (note that I'm only using one mass, not distinguishing ...


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Basically, yes. You can always speed up in your own reference frame, you'll just experience more and more time dilation to keep $c$ constant. You could use the Lorentz transformations to calculated the perceived change in speed. $u$ - spaceship velocity $v$ - reference frame velocity After Acceleration: observer reference frame $S$: $u = 0.9999c$. ...


2

Yes, basically. The spaceship can keep on increasing its velocity, though only in increasingly smaller increments. When dealing with relativistic speeds, there are two variables that are arguably better for picturing motion. One is the gamma factor $$ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} $$ which approaches infinity as $v$ approaches the speed of light. The ...


1

John, have a look at the simple inference of time dilation due to relative velocity. If you and I are identical twins, and you take a fast out and back trip, when you come back we agree that you've experienced less time than me. As you pointed out, we can relate this to the Lorentz factor and write: $$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ ...


0

When the object is an elementary particle or a charged ion we can use electromagnetic interactions to measure its rest mass, given the charge in an e/m experiment. One can get the charge with Milikan's oil drop experiment. Here is a setup for the lab.


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Earth moves around the Sun and the Sun moves around the galaxy and the galaxy moves with unknown speed and direction. We have speed so the mass of us all altered. The relativistic mass is altered, but this is a somewhat archaic term these days, and is said to be a measure of energy. Nowadays when we say mass without qualification, we tend to mean rest mass. ...


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The rest mass of an object, by definition is the total energy of an object as measured by an inertial observer who is at rest relative to the object. If the object is not moving uniformly, then you can measure its rest mass from a momentarily co-moving freefall frame. This rest mass is also the constant in Newton's law, i.e. the inertial mass as well. So ...


2

We do have a rough idea of the relative speed and direction of our galaxy, with respect to the other galaxies around us, the so called local group. In general relativity, which is our best theory of the universe to date, there is no such thing as absolute speed, as it depends on which frame of reference you use to measure things in. Our Earth and the ...


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There is a nice thought experiment that explains this phenomenom, usually called the wire argument. I think you can find it in Feynman's lectures though I don't know who thought of it first. The set up is the following: let's consider a wire with zero total charge but with a current going through. For simplicity's sake, let's assume that the charge carriers ...


0

Secondly, if we consider Maxwell's equations, which are not invariant under Galilean transformations, but we require them to hold in all inertial frames, doesn't it immediately follow that the speed of light has the same constant value in all inertial frames from this assumption (given that Maxwell's equations imply a constant speed of light). Why is it ...


1

Why is it given as an axiom of special relativity? In a nutshell: because of the Michelson-Morley experiment. This ought to have detected a variation in the speed of light caused by our motion through space. But it didn't. So Einstein reasoned that speed = distance / time, and that if the speed didn't change, your the time did. And then if your time ...


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You can think of the fuel being stored in barrels, and observers in all frames will agree on how many unused barrels are left when the ship reaches its destination, since discrete quantities aren't affected by the Lorentz boost. If you're wondering how this makes sense with the time dilation, imagine that on board the ship, it appears that the barrels are ...


3

To being able to reach a high fraction of the speed of light, you need also a significant fraction of your equivalent payload mass as mass-energy. According to the formula for energy (rest and kinetic) $$ E = \gamma m c^2$$ so, to reach $0.1c$, that means your gamma is $\gamma = (1-\beta^2)^{-1/2} = 0.99^{-1/2} = 1.01$, so that means that for each kilogram ...


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The spaceship and the Earth start out in the same frame, so the amount of fuel agrees in both frames.


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You are asking us for the distance of the trip in the rest frame of the photon. The problem with asking that is that there is no rest frame of a photon. A photon can never be at rest, so it has no rest frame. This is like asking what a bowl of petunias thinks about its existence as it falls to the surface. A bowl of petunias doesn't think, therefore we can't ...


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first the speed of light is related to the permittivity and permeability of the medium. changing either one of those values changes the speed of light. copper has different values then free space. the speed of light through copper is 2/3rds that of free space. about 1 foot per nanosecond. slightly faster in aluminum, slower in iron. this can be measured ...


1

The photon does not have an rest frame because that would imply it's velocity is 0, when by postulate it must be moving at C in all frames. Rather pointless to wax on about not experiencing any time or what not.


1

And while it is philosophically acceptable to just "know" that the speed of light is constant but it not to just "know" that it is invariant. Fixed constant values such as the mass of an electron or the spin set of an electron are things one can accept as given. Kind of: be careful. We can fix some of the fundamental constants by choosing covarying unit ...


0

I'm not sure I understand your question, but I'll try anyways. It's true that constancy (not changing in time) and invariance (same in all reference frames) are different. SR and GR assume a constant speed of light, but lets imagine a theory where the speed of light changes in time. The passage of time is different for different observers. If my clock ...


1

As John Duffield pointed out in his answer, there is no reference frame in which the photon is at rest. This is the simple reason why most parts of your statements do not make sense. Still, there is an interesting perspective to the issue you raise: Normally one considers the events or states "photon emitted", "photon is traveling", "photon absorbed" as ...


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Let's start with your assumptions: 1) Photons travel at the speed of light. Right. No problem with that. 2) From the photon's reference spacetime is contracted to 0 length in the direction of photon travel. Wrong. The photon doesn't have any kind of reference frame. To appreciate why, imagine you're travelling at the speed of light. We know you can't ...


0

Your question has nothing to do with the (hypothetical) photon's reference frame. Even if the spacetime interval of lightlike movements is reduced to zero, the order of events is persisting - In the same way as when you cover one sheet of paper by another, you cannot reach the hidden sheet even if their respective distance approaches zero. Edit: By the ...


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The Minkowski vacuum is defined on the whole Minkowski space. The Rindler wedge covers only half of a given spacelike surface, the other half being covered by a different Rindler wegde. So that the $\tilde{\phi}$ field defined gives only half od the degrees of freedom corresponding to $\phi%$. Ehen you trace over the the DOF from the other Rindler wedge the ...


1

Mathematicians don't talk about rotations or isometries or whatnot because they already know if they're talking about a scalar or something else; a physicist has to determine whether a physical quantity has the properties of a scalar or a vector or something else. The easiest way to do that is to look at transformation laws--to devise some experiment (real ...


1

When talking about scalars, mathematicians usually use your definition, that is, something which doesn't vary with coordinate changes. (Basically, that there's some mapping to the actual points in space, in which the scalar is well defined) When physicists talk about scalars, we usually refer to Lorentz scalars, which requires two things: Invariance under ...


0

When traveling with the speed of light time does not "exist". Well sort of. If a body is traveling with the speed of light its origin and destination are one and the same. The question has no point because traveling with the speed of light means you have always been and always will be traveling with the speed of light. Of course from the viewpoint of an ...


3

Light is described well by the classical electromagnetic theory and Maxwell's equations. In this framework, the classical one, the speed of light is constant in vacuum. When light impinges on transparent materials, its speed, classically changes, and this is measured with the index of refraction of the material: where c is the velocity of light in vacuum ...



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