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5

The answer is - velocities just do not add like that when we are getting close or even equal to the speed of light $c$. The whole body of Special theory of relativity concerns itself with what are the consequences of postulating the speed of light to be constant in all frames of reference. That is, we find transformations between frames of reference by ...


5

Your teacher is correct. The link you give, gives the correct definitions of the various "mass" concepts in special relativity. m_0 here characterises an entity that is being described by a four momentum vector in the special relativity framework. In vector spaces, the vector has an invariant length, (otherwise mathematically it would not be a vector ...


5

It's a matter of terminology. You can define an object's mass to be proportional to its energy in the reference frame you observe it from, which includes kinetic energy in the mass, or you can define it to be proportional to the object's energy in its own rest frame, which excludes kinetic energy. Or you can do both, which is what physicists originally did ...


4

The relationship between nuclear masses and mass differences and binding energies has been confirmed by many decades of careful nuclear spectroscopy. It's possible to measure an atom's mass by purely mechanical means: you ionize the atoms, accelerate them to a known energy, and use a magnetic field to measure their momentum. This lets you come up with an ...


4

The key sentence is "from the frame of reference of the accelerator, the circumference must stay the same." So from the accelerator frame, no length is changing, just accelerator's time passes by more quickly by a factor of gamma than in the particle's frame. This is what allows the particles to "travel extra distance" than would normally be possible in ...


3

A personal point of view is that you may consider that Lorentz transformations apply primarily on momenta, and not primarily on (infinitesimal or not) space-time coordinates. This is, of course, a "strong" postulate. If you assume (some additional postulates are needed there) that transformations are linear, and that there is a rotation invariance, you ...


2

Consider $$M_{31} = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} \text{ and } P_0 = -i \begin{pmatrix} 0 & 0 & 0 & 0 & 1 ...


2

In what sense does the answer make sense? Does it describe some physical phenomenon? You've done the algebra right, and gotten a result, but I don't think your inputs (imaginary velocity) mean anything physically. So the answer can't be expected to mean anything physically.


2

Rob's explanation of how we know is bang on, but I wanted to address a part of your question that might point to a basic misunderstanding. What is special relativity inside the nucleus? Everything is always relativity. Everything. Always. All those Newtonian equations like $T = \frac12 m v^2$ for the kinetic energy can be properly understood as ...


1

An analogy (possibly from Griffiths' textbook): imagine you're out in desert. Many miles away you see a truck moving on the highway. In which case is it easier to judge its speed: if it's moving radially towards or away from you, so that you have only its change in apparent size to go on, or if it's moving laterally across your line of vision?


1

If all the lengths in this diagram are as measured by the observer on set 1 then only lamp 1 will light up, but not lamp 2. When the left "touchers" on each set align in frame 1 then the right touchers will too, because they are both 10 meters separated from the left ones. However, the touchers on set 2 are further than 10 meters apart according to observer ...


1

"The speed of light is constant, no matter how an observer moves relative to the light." This is true. The light is moving independently, thus an observers actions of movement, make no changes to it. However, the interesting part is that the speed of light is always "Measured" as the speed of light no matter how an observer is moving relative to that ...


1

The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to ...


1

First, from the point of view of $O$. The lightning strikes at points $A$ and $B$ happen simultaneously. Light propagates away from those points, and since $O$ is halfway between $A$ and $B$, the light fronts reach him at the same moment (equal distances and equal velocities gives equal times). Now, for things as $O'$ sees them. The important thing to ...


1

Actually, "unitary representation" is meant with respect to the spinors, which do not form a finite-dimensional space and therefore allow a unitary representation of the proper Lorentz group. The action is defined by $D(\Lambda)\psi(x)=U(\Lambda)\psi(\Lambda^{-1}x)$, and you can simply calculate that this is unitary on your spinor space. However, this does ...



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