# Tag Info

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Given the context of the question, the fact that it seems to be about $E=mc^2$ specifically, and that the OP says he's having a hard time understanding it, i'm going to try and give a simple answer in plain english without a load more complicated formulae. I am no physicist, and although the concept may not be that easy, the formula is pretty simple, maybe ...

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I will try to answer this question with my basic understanding of special relativity: Is matter condensed energy? It kind of is, but a better way to phrase it would be that everything that has energy, (behaves like it) has mass. Imagine you have a hollow box with the insides covered with perfect mirrors and you put it on a scale. If you shone a light ...

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The famous equation $E = mc^2$ is actually just a special case of the relativistic equation for the total energy: $$E^2 = p^2c^2 + m^2c^4 \tag{1}$$ where $p$ is the relativistic momentum and $m$ is the (constant) rest mass: $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$ For an object that isn't moving $p=0$ and equation (1) becomes: $$E = mc^2$$ which is ...

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does this equation mean masses are just condensed energy? No, it means that mass is just another form of energy, just like heat, motion, electric attraction, etc. For example, the energy of a charged sphere is $$E=\frac{3}{5}\frac{Q^2}{R}$$ This equation doesn't mean that charge is just condensed energy; it means that charged objects have energy. ...

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The two are indeed related. The relativistic expression for the total energy is: $$E^2 = p^2c^2 + m^2c^4 \tag{1}$$ where $p$ is the relativistic momentum: $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$ and $m$ is the rest mass. If the object isn't moving then $p=0$ and equation (1) becomes: $$E = mc^2$$ which is of course the well known expression for the ...

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The link given by LLlAMnYP for the Ehrenfest paradox gives the classical physics rational : Any rigid object made from real materials that is rotating with a transverse velocity close to the speed of sound in the material must exceed the point of rupture due to centrifugal force, because centrifugal pressure can not exceed the shear modulus of ...

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Some cellular automata, like the basic rule 110 are universal, i.e., Turing complete. What this means is that you can simulate/emulate any mathematics on them, including any physical theory, including non-local ones. Many people make the mistake of thinking that because cellular automata have local and discrete rules they are limited to simulate only those ...

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To be clear, Maxwell's equations are known as "Lorentz-invariant" equations, which means that they take the same form in every Lorentz-transformed frame of reference. Special relativity actually came about from studying Maxwell's (classical) equations without charges or currents. Then we get: $$\nabla \cdot \mathbf{E}=0$$ $$\nabla \cdot \mathbf{B}=0$$ ...

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Yet another way. Consider a wave of a form $f(kx-\omega t)$ in the emitting frame, being observed as some other wave $f^{\prime} (k^{\prime} x^{\prime}-\omega^{\prime} t^{\prime})$ in a frame moving at $v$ along the x-axis. We actually don't care what $f^{\prime}$ is, only the argument of the function matters. So use the Lorentz transforms to change $x$ and ...

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Classical electromagnetism is perfectly compatible with special relativity. In classical E&M, light is an electromagnetic wave and there is generally no useful formulation in terms of particles. The most widely used technique to combine quantum mechanics with special relativity is relativistic quantum field theory. The relativistic QFT that ...

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Neither origins nor emitters nor receivers are required: consider a photon with momentum $\vec{p}$: it has 4-veclocity $p_{\mu}=(||p||/c, \vec{p})$. To compute what another observer moving at $\vec{v}$ sees, do a Lorentz transform: $p'_{\mu} = \Lambda_{\mu}^{\ \nu}p_{\nu}$. From this you'll get the relativistic (time-dilated) Doppler effect and stellar ...

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Is important to realize time dilation and the Doppler effect are 2 frame-dependent parts of a single relativistic phenomenon: the constancy of the magnitude of four-velocity (which is always ||c||). Requiring that all inertial observers see a four-velocity with magnitude ||c|| leads to frame dependent observations of time dilation and Doppler effect. For a ...

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It depends what exactly you mean by "coordinate". If your Lagrangian/Hamiltonian is time-independent, then you may consider time to be purely a parameter parametrizing e.g. the integral curves of the vector field associated to the Hamiltonian on phase space. If your Lagrangian/Hamiltonian is time-dependent, you should indeed properly consider your theory on ...

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I think this issue is best clarified by closely looking at the way time is mixed into coordinate frame transformations in Classical Mechanics as opposed to Relativistic Mechanics. Let's take the case of an observer, Alice, moving at velocity $v$ in the positive $x$ direction away from her friend Bob. Both Alice and Bob are looking at an object situated at ...

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In special relativity there are two major assumptions: -the laws of physics are the same in all inertial frames -the speed of light that you observe is always the same, (thus independent of the relative motion between the light source and the observer). From this two assumptions follows the famous Lorentz transformations. In these Lorentz transformations ...

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The set of transformations that leaves the speed of light unchanged is the Lorentz group. Representation theory enables us to investigate the irreducible representations of the Lorentz group. The lowest-dimensional representations act on scalars four-vectors However, take note that usually we consider representations of the corresponding Lie algebra ...

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Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $\psi_{a}$ and $\psi_{b}$, both of which correspond to energy $E_{0}$. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...

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To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $\pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ... 0 In this nice reference the autor assumes the relativity principle + homogeneity + isotropy and deduce the general coordinate transformations which contains both Lorentz and Galileo transformations. Further he imposes the postulate of the constancy of the speed of light, restricting the transformations to be the Lorentz type. 0 The principle of relativity says that there is no experiment that can determine absolute motion. So all observers, regardless of relative motion, need to agree on the outcome of any experiment. Because to the relativity of observers' measuring devices, they may not numerically agree on the measurements. By applying the laws of relativity they will be able ... 0 A few comments before doing the calculation: In the CM frame, there is only an attractive force, while in the given frame, there is both an attractive and a repulsive force. This is no more mysterious than the fact that a vertical object in my frame can look tilted to somebody with rotated axes. Going from the CM frame to your frame mixes the electric ... 2 Lorentz Transformations Suppose we call the lab frame the K-frame and a frame moving at velocity, \mathbf{v}, relative to the K-frame called the K'-frame. Then we can express the electromagnetic fields in the K'-frame in terms of the K-frame fields as:$$ \begin{align} \mathbf{E}' & = \gamma \left( \mathbf{E} + \boldsymbol{\beta} \times \mathbf{B} ...

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The observer moving with the CM will measure that the force of repulsion the electrons is given by $F=\frac{e^2}{4 \pi \epsilon_0 d^2}$ ($d$ is their separation), he can only make measurements in his reference frame (that is moving with speed $v$), and will not be able to be determine this speed.

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In a sense, the two definitions you mention are the same. One of the postulates of special relativity is that the speed of light is the same in all reference frames, so the definition of "the coordinate transformation according to the postulates of special relativity" is the same definition as "the coordinate transformation under which the speed of light is ...

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There are lots of ways of approaching special relativity. My own preferred approach is the invariance of the line element. Suppose you move a small distance in spacetime $(dt, dx, dy, dz)$ then the length of the line element $ds$ is defined by: $$ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 \tag{1}$$ This equation is known as the metric equation and is derived ...

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The point is exactly that agreeing on a particular value for one measured quantity causes other quantities to have different measured values, for the values in question of distance, time and velocity (any one of which can be calculated from the other two). The limitation we're stuck with is, at its root, that we have no way to measure time passively. While ...

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"What Einstein was really looking for was a new way to transform between reference frames that would keep the laws of electrodynamics invariant inasmuch as the Galilean transformations keep the laws of Newton invariant?" It wasn't so much about finding the transformations, because the Lorentz transformations had been known for a while at the time, since ...

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How do we formally define vectors in physics? An excerpt from chapter one, page 12 of "Mathematics of Classical and Quantum Physics" Originally, we introduced a vector as an ordered triple of numbers. The rule for expressing the components of a vector in one coordinate system in terms of its components in another system tells us that if we ...

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A physical quantity is a vector if it transforms in the same way as a position vector when the coordinate system undergoes a transformation.

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I think the purpose of that example is to show how observers from different frames of reference can disagree on what they actually observe, or measure. The example you provided can be solved with a simple galilean transformation. This example is crucial because it shows you how different observers can disagree on the same event, which seems pretty intuitive ...

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Since the train has no relativistic velocity, you will not see any effect such as time dilation, length contraction, lack simultaneity and so on. The importance of this example as a prelude of Special Relativity is that it shows that even Galilean Relativity has some physical quantities which are not absolute. Namely (in this example) the length of the path ...

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Start by taking the curl of Maxwell's third equation (for vacuum), and substituting $\vec{B}=\mu_0\vec{H}$ one can obtain, $$\nabla^2.\vec{E} = \mu_0 \epsilon_0 \frac{\partial^2}{\partial t^2} \vec{E}$$ Similarly, by taking curl of Maxwell's fourth equation, substituting $$\nabla\times\vec{E} = -\frac{\partial}{\partial t}\mu_0\vec{H}$$ one can ...

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This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...

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The concept of 'straight' is a bit ill defined in GR and has no real definition. In fact in a sense the geodesics themselves be seen as 'straight' lines; they are the shortest paths connecting 2 points (this is what in normal Euclidean space would be a 'straight line') In the LC connection they are the integral curves of some vector field $V$ with $... 3$I$has a clear physical meaning if$I\lt 0$– which is a significant percentage of the spacetime, so to say: $$I = -c^2\Delta t_{\rm proper}^2$$ where$\Delta t_{\rm proper}$is the time measured by clock that moves by a constant velocity (without acceleration); and that visits the point$(x_1,y_1,z_1)$at time$t_1$and$(x_2,y_2,z_2)$at time$t_2$. ... 0 The finding of such a particle would overturn Einstein's theories of relativity, which state that light-speed is a cosmic constant and cannot be passed by anything in the universe. Such superluminal particles would also have the ability to time travel (hypothetically). This would be caused since as a particle approaches the speed of light, time slows down. ... 0 The obvious difference is that Newton's equations retain their form for all inertial reference frames when Galileo's Principle of Relativity is used, but Maxwell's equations are not invariant under this transformation. Instead one must use the Lorentz transform, which recognizes that there is a fixed speed for light,$c$. This limit was recognized by ... 0 Two of Maxwell's equations combine to yield a wave equation with a fixed wave velocity, the speed of light$c$, for both of two observers in relative motion to one another, contrary to the behavior of waves in Newtonian mechanics. 2$|q\phi|\ll mc^2$means that the rest mass of the particle is way larger than the electrostatic potential energy. This is the usual assumption of non-relativistic mechanics, which is equivalent to$v\ll c$. You can see this with the virial theorem: $$\frac{1}{2}mv^2\sim q\phi$$ which means that$q\phi\ll mc^2$iff$v\ll c$. ... 4 In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ... 0 Yes it will do. You can take the potential energy to be zero when the spring is neither compressed or stretched. In special relativity the total invariant mass of the system would then include a contribution from the potential energy /$c^2$. The concept of mass in general relativity is quite subtle, but for weak gravitational fields, the Newtonian limit for ... 1 And regarding why it's called a "free" theory, it's not specific to a momentum-space formulation. It's "free" because the Lagrangian is quadratic in the fields, and therefore the equations of motion (what you get from plugging the Lagrangian into the Euler-Lagrange equation) are linear in the fields. Therefore you can superpose different classical ... 2 Yes. You are correct. A non-relativistic theory would be invariant under the Galilean group. Lorentz invariance (specifically, invariance under Lorentz boosts) is what defines a relativistic theory. 0 If the webcam is directed onto a watch, you would see it ticking more slowly. If the webcam is directed on the outside, you will see the normal speed of happenings, e.g. it would show you the same number of rounds as you have seen by looking at the vehicle. Only it would have to be a very fast webcam, otherwise you would see less frames per second. And ... 0 "Presumably, the usual method for speeds v≪c is to consider a Newtonian n-body approach and simply integrate the equations of motion with the gravitational inverse square law, using a numerical scheme such as Runge-Kutta or, something more sophisticated, like a symplectic integrator?" It might work for non-relativistic but for more than two bodies it is ... 1 There is no problem using length contraction, other than that it may take a couple of extra lines to arrive at the same result. Just to have a reference, starting from$\Delta x = \gamma (\Delta x′ + v \Delta t′)$and denoting${\bar u}$the velocity of the missile relative to the rocket,$u$the velocity of the missile relative to Earth, and$L$the ... 0 Well if we neglect the hidden momentum the conservation law of momentum in electromagnetism is simple: The momentum can be stored in static fields ($D\times B$); the mechanical momentum ($mv$) + electromagnetic momentum ($D\times B$)$= constant\$. The similar formula is valid for angular momentum (where it is not hidden momentum) See Feynman's Lectures ...

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This might or might not be responsive to the question you intended to ask: Suppose you've got a meter stick. Over a period of time, I apply identical forces to the front and back ends of that meter stick, causing them to accelerate in the same direction. Therefore the entire stick, being a rigid body, accelerates in that direction. After a while, the ...

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Despite their superficial similarity Lorentz contraction and time dilation are different things and this is why there isn't a distance version of the twin paradox. To see the difference you need to understand that a clock is a form of odometer. Suppose you start at the origin and travel 100 metres, then the odometer you carry will show the total distance ...

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