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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 ...

6

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 ...

6

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. ...

4

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 ...

3

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 ...

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 $t_2$. ...

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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 ...

2

"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 ...

2

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 ...

2

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 ...

2

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 ...

2

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 ...

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} ... 1 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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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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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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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 ...

1

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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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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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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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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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 \$ ...

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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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