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12

Inconsistency between two theories just means that there are statements that one theory says are true, and the other says are false. An easier example than the one you're asking about is the inconsistency between Newtonian mechanics and special relativity. Newtonian mechanics says that if you keep applying a force to a material object, it will eventually go ...


11

First, the non-relativistic equation $$ E= mc^2 + \frac{mv^2}{2} $$ is equivalent to its second power, $$ E^2 = (mc^2)^2 + m^2 c^2 v^2+ \frac{m^2v^4}{4} $$ If $v/c\ll 1$, then the last term is much smaller than the previous two, and the first two terms on the right hand side are equivalent to the correct relativistic $$ E^2 = (mc^2)^2+ (pc)^2 $$ which ...


7

Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. Yes, it would be nice if we could find an equation that was "universal". In the case of mass m, energy E, and momentum p, there is such a universal equation: $$E^2 = (mc^2)^2 + (pc)^2$$ Einstein, always wanted to ...


6

I am hoping someone could explain, rather in-rigorously, the use of the word inconsistent. Essentially, in this context, inconsistent means the two theories give different, incompatible answers to the same question. In this specific case, the question is: If light (an electromagnetic wave) is measured to propagate at speed $c$ in an inertial frame ...


5

If you put $p = \gamma mv $ where $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ and $\beta = \frac{v}{c}$ in Einstein's equation, you get $E^2 = (pc)^2 + (mc^2)^2 = (\gamma mvc)^2+ (mc^2)^2 $ $= (\frac{c^2}{c^2 - v^2})v^2(mc)^2 + (mc^2)^2= (\frac{v^2}{c^2 - v^2})(mc^2)^2 + (mc^2)^2$ $ = (\frac{c^2}{c^2 - v^2})(mc^2)^2 = (\gamma mc^2)^2$ $ \therefore E = ...


4

If the body's speed $v$ is much less than $c$, then the equation reduces to That is an imprecise wording, and the cause of your confusion, I believe. The correct description is that the second equation is an approximation of the first, and it will be closer to accurate the lower $v$ is. The approximation is important for understand the connection ...


3

First, your findings are correct and are found by Einstein. However the fact that a new term appears is only showing that our previous knowledge was incomplete. The new term expresses the Energy related to the rest mass of the body which was not considered before. And it makes sense because before, in equating the Energy of a body only were included those ...


3

I) OP wrote (v2): I know that $\mathcal{L}$ is a functional not a function. Actually for local theories the Lagrangian density $$ \mathcal{L}(\phi(x), \partial\phi(x), \partial^2\phi(x), \ldots, ;x) $$ is a function of $\phi(x)$, $\partial\phi(x)$, $\partial^2\phi(x)$, $\ldots$, and $x$. In contrast, the action $S[\phi]=\int \!d^dx~\mathcal{L}$ is a ...


2

The ground state of Dirac sea is full of particles with negative energy, according to Dirac equation. When a particle/electron absorbs energy and excites to a positive energy state with energy level $E$, it will leave a hole with energy $E-E\prime$ (if the total energy absorbed is $E$) , which is positive according to energy conservation. So the hole seems ...


2

If the curve is a geodesic then in the coordinate system of an observer moving along the geodesic coordinate time and proper time are the same. That's because in the freely falling observer's coordinates $dx = dy = dz = 0$ and therefore $ds^2 = -c^2dt^2 = -c^2d\tau^2$. This makes proper time a natural way of parameterising the curve because it's just the ...


2

Special equations of the kind you mention are useful as they elucidate limiting behaviors. Your example is no different from a statement like: the hypotenuse $c$ of a right triangle with legs $a$ and $b$ with $b \ll a$ is given by $c = a + \frac12 b^2/a$. The Pythagorean relation $c^2 = a^2 + b^2$ is valid for any right triangle, but the special case $b ...


2

A direct answer: Next, why does this not work in every aspect? surely a equation should be "universal" and should still work with any values given. The equation $E^2 = (mc^2)^2 + (pc)^2$ does work for all values. As other answers have explained, at low speeds (or equivalently, low momenta), $E = \frac{1}{2}mv^2 + mc^2$ gives approximately the same ...


2

[...] my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame? What information does observer A have about observer B? If all A knows about B is B's state of motion, and if he assumes that B is going to choose coordinates in which he is at rest at the origin, ...


2

It is usual to make a distinction between proper rotation as opposed to all rotations (which include those that change handedness). The matrices of proper rotations have determinate +1 while those of improper rotations have determinate -1. In terms of groups the proper rotations are $\mathrm{SO}(3)$ while all rotations taken together are the orthagonal ...


1

The traveling astronaut is younger. The situation is not reversible between both astronauts because the traveling astronaut is submitted to an effect which is similar to acceleration because he is following the curvature of space in the fourth dimension. The solution must consider the geometric/ topological constellation. And topologically, the traveling ...


1

All matter is made up from atoms. While it's very hard to accelerate a spaceship to nearly the speed of light we can collide two atoms (well, two nuclei) at speeds approaching the speed of light. This experiment is done at the RHIC and also at the LHC in its heavy ion mode. So what happens when we collide nuclei at nearly the speed of light? Well, the ...


1

The above answers are all good, but I want to add something else. You can derive the energy-momentum relation from at least principle, the action is $A=-m\int \sqrt{1-v^2}{\rm d}t$ . Lagrangian is $\cal L$$=-m\sqrt{1-v^2}$, then you can get momentum $p$ from derivative with respect to $v$ (this is the real momentum, not $mv$). Energy $H=E=pv-\cal L$ , just ...


1

Poincare invariance simply means that the Lagrangian does not change under a Poincare transformation: $$\mathcal{L}(x)\rightarrow\mathcal{L'}(x)=\mathcal{L}(\Lambda^{-1}x).$$ Note that I have only written down Lorentz transformations, for simplicity. We now say that $\mathcal{L}$ is invariant if $$\mathcal{L}(x)=\mathcal{L}'(x).$$ This does not mean ...


1

If I remember well, this section of Peskin & Schroeder uses the Wick rotation to solve integrals of type $$\int \frac{1}{k^n + ...} \cdot ... $$ Where $k^n = k \cdot k \cdot k ... (\rm n \, times)$. By performing the Wick rotation we suddenly get a spherically symmetric problem which enables us to use the well known tricks for such a case. But notice ...


1

You could say that $\partial_{\mu}\mathcal{L}=0$ only if the fields it applies to are treated as independent variables and not as functions of $x^\mu$. I.e. the Lagrangian does not explicitly refer to coordinates, but if you were to consider the fields as $\phi = \phi(x^\mu)$, then obviously $\partial_{\mu}\mathcal{L}\neq 0$. The more rigorous way to say ...



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