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1

The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow ...


3

One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike. Alternatively you can work from the definition where two events are spacelike separated if (and ...


0

So, here's the deal. "Time is relative" means a lot of different things to a lot of different people. In order to make a solid step forward, Einstein and company basically needed to clarify what they were trying to say. What they were trying to say looks something like this: "if you see a train passing by you, you're going to see things happen in slightly ...


-3

Time doesn't really flow according to how we usually think of time flowing. Time is just our mind's understanding of the changing environment around us. It is the changing environment that determines how fast time is moving. Assuming that the expansion of the universe is tied to the energy therein, so that if expansion were to stop the energy also would be ...


-4

If you think about time as we know it does not actually exist/flow - it is our mental manifestations of the world around us that we think of as time. For example, what we see is not actually there how we view it. Whatever the object is sends us light-waves (only a small portion possibily of what the object really is), our eyes then have to decode the light ...


0

Assuming constant total energy, $E^2 = (m_0^2c^2)^2 + (pc)^2 = \gamma^2 m_0^2c^4$ leads to the equation $E^2 = \frac{m^2c^6}{c^2-v^2}$ and then $m = \frac{E}{c^3} \sqrt{c^2 - v^2}$, which also yields the quarter of an ellipse: $v^2 +\frac{c^6}{E^2}m^2 = c^2$, producing the following graph (ignore the number): where the vertical axis is rest mass which ...


0

I was trying to see what results I would get if I were to incorporate relativistic corrections into the case of a harmonic oscillator in one dimension. [...] $ \gamma^3~m_0~\ddot{x} = -k x $ Since your question is specificly about harmonic motion, we might (instead) insist on this solution $$x[~t~] := x_{\text{max}}~\text{Sin}[~\omega~(t - t_0)~],$$ ...


4

There's no one-to-one relationship. With zero rest mass, a particle must always be observed to move at $c$. A particle with nonzero rest mass, on the other hand, can move at any speed in $[0,\,c)$ (note the closed-open interval). At the risk of putting words in your mouth, I think I can recall the exact same question in my mind and it went something like ...


-1

Only equation I know is $$m = \frac{m_0}{\sqrt{1 - (\frac{v}{c})^2}}$$


0

That is a very interesting question, but you have to pay close attention. The current value of the speed of light in a vacuum (c) is 299 792 458 m/sec. Assuming that 1 g is exactly 9.8 m/$sec^2$, then the time to accelerate from rest to c is just 299792458 / 5 * 9.8, or 6 118 213.4 seconds, or just under 71 days. But. Let's put it this way - as an ...


6

You can get an exact solution for $t(p)$, although it involves a rather nasty integral that I'm not sure can be written in closed form. Here's how: The equations of motion are $$ \frac{dp}{dt} = -kx \qquad \frac{dx}{dt} = \frac{1}{m} \frac{p}{\sqrt{1 + p^2/m^2 c^2}}. $$ This second equation can be obtained by taking the equation $p = m v /\sqrt{1 - ...


2

Light has a frequency of approx. 1e15Hz. Can light be transmitted in a hollow copper tube? Yes. No need to go relativistic. Can objects move at near the speed of light in a coax cable with inner conductor? No. They can't move in there, at all, not even at walking speed. Does any of this has anything to do with photons? No. Your experiment does have a ...


1

There is a subtle difference between saying $(2,2)$ and $2\otimes 2$. In the latter case we are thinking of both reps as transforming under the same element of the group $SU(2)$. In the former case we are thinking of $(2,2)$ as transforming under the Lorentz group, which contains two distinct copies of $SU(2)$. Call one copy the $L$ copy and the other the ...


2

SECTION A : Non-relativistic conservation of energy The work done by the non-relativistic force $\:\mathbf{f}\:$ per time unit, that is the power produced or consumed, on a particle moving with velocity $\:\mathbf{v}=d\mathbf{r}/dt\:$ is \begin{align} \dfrac{dW}{dt}=\mathbf{f}\circ \mathbf{v}=&\dfrac{d\mathbf{p}}{dt}\circ \mathbf{v}=\\ ...


2

Special relativity and general relativity have different views about inertial frames, but in some ways the general relativity take on them is (perhaps surprisingly) easier to explain. So I'll start with GR then extend the description to SR. In general relativity there are usually no global inertial frames i.e. it is impossible to construct a frame that ...


1

In Newtonian physics, neither are inertial frames. In relativity, only the first one is an inertial frame. Edit: Let me clarify. An inertial reference frame is a frame that "follows" an uniform motion, i.e. a motion where there could be said not be affected by any net force. In Newtonian physics gravity is a force and objects in free fall are thus ...


1

$E = pc$ is only true for massless particles. For massive particles you have the mass-shell relation: $E^2 = m^2c^4+p^2c^2$ After you use $E=T+mc^2$ and you can find $p$


1

I will reply to this because the checked answer is not answering the question.The question is about photons, the answer is about light. It is as if the question were about atoms and the answer is about density of material. The question is asked about photons, the quantum mechanical framework is relevant to it. The checked answer is about light which is in ...


1

Although it has been said in other comments and answers, it bears repeating succinctly: photons (as far as any experiment can tell) are massless and therefore always move at the universal, invariant speed of light. There is NO non-relativistic description of the photon. Even the "classical" description of light - Maxwell's equations - can be interpreted as ...


2

People are addressing the speed question, but just to be clear: a photon can be very low energy. For instance, radio waves are much lower energy than gamma rays, even though both are made of photons (and, in vacuum, both travel at the speed $c$). What determines the energy of a photon is the frequency of the excitation (frequency of the corresponding light ...


1

Refractive Index is when light travels more slowly in a medium. Here is an example of light being slowed down to 38 miles per hour. The speed of a photon does not affect its energy. It has zero mass, therefore zero kinetic energy. The energy it has is due to its frequency (color), and nothing else. (However, it does have momentum!)


0

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 ...


0

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 ...


0

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: ...


0

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 $k^2={k^0}^2 - k_ik^i$ ...


1

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}) = ...


2

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 ...


1

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 ...


3

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 ...


0

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 ...


7

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 ...


0

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 ...


0

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 ...


1

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 ...


0

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, ...


0

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 = ...


1

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 ...


2

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 $\sqrt{v_{(\parallel)}^2+\frac{v_{0,(\perp)}^2}{\gamma^2}}$, ...


-1

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 ...


2

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 $$ \frac{E_\gamma E_\gamma'}{c^2} (1 - \cos \theta) = (E_\gamma + E_e - E_\gamma') m_e - m_e^2c^2 = E_\gamma - E_\gamma'$$


0

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 ...


1

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 ...


6

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 ...


0

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.


11

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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