instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=m{c}^{2}$.

Landon Mckinney
2022-05-13
Answered

instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=m{c}^{2}$.

You can still ask an expert for help

Superina0xb4i

Answered 2022-05-14
Author has **17** answers

These two concepts are not related at all, so it's not possible to deduce anything either way.

Special relativity in its bare form talks only about space-time, i.e. geometry. You don't have any energy or mass there without first somehow postulating what it is.

For the reverse direction: you can't really deduce anything about SR from $E=m{c}^{2}$ because $E$ and $m$ could be anything (in particular you don't know whether $E$ is a component of a four-momentum $p$ or something completely different). And even if you knew what $E$ and $m$ was that would still give you no clue about what $c$ is (and that it should be constant in every inertial frame).

Special relativity in its bare form talks only about space-time, i.e. geometry. You don't have any energy or mass there without first somehow postulating what it is.

For the reverse direction: you can't really deduce anything about SR from $E=m{c}^{2}$ because $E$ and $m$ could be anything (in particular you don't know whether $E$ is a component of a four-momentum $p$ or something completely different). And even if you knew what $E$ and $m$ was that would still give you no clue about what $c$ is (and that it should be constant in every inertial frame).

asked 2022-05-08

This is the common problem of a charged particle moving in a static electric and magnetic field.

Say $\mathbf{\text{E}}=({E}_{x},0,0)$ and $\mathbf{\text{B}}=(0,0,{B}_{z})$

In the inertial frame of reference, the equation of motion is (1):

$\frac{d\mathbf{\text{v}}}{dt}=-\frac{q\mathbf{\text{B}}}{m}\times \mathbf{\text{v}}+\frac{q}{m}\mathbf{\text{E}}$

We can find equations for ${v}_{x}$ an ${v}_{y}$ and see that the resulting motion is a circular orbit with a constant drift velocity ${v}_{d}=\frac{{E}_{x}}{{B}_{z}}$.

Surely I should get the same answer if I solve the problem in a rotating frame of reference?

I know that (2):

$\frac{d\mathbf{\text{v}}}{dt}{|}_{Inertial}=\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}+\mathit{\omega}\times \mathbf{\text{v}};$

If I use Eq. (1) as the LHS of Eq. (2), and choose $\mathit{\omega}=-\frac{q\mathbf{\text{B}}}{m}$, then I get (3):

$\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}=\frac{q}{m}\mathbf{\text{E}};$

How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field E=$\mathbf{\text{E}}=({E}_{x},0,0)$ change form in the rotating frame?

Say $\mathbf{\text{E}}=({E}_{x},0,0)$ and $\mathbf{\text{B}}=(0,0,{B}_{z})$

In the inertial frame of reference, the equation of motion is (1):

$\frac{d\mathbf{\text{v}}}{dt}=-\frac{q\mathbf{\text{B}}}{m}\times \mathbf{\text{v}}+\frac{q}{m}\mathbf{\text{E}}$

We can find equations for ${v}_{x}$ an ${v}_{y}$ and see that the resulting motion is a circular orbit with a constant drift velocity ${v}_{d}=\frac{{E}_{x}}{{B}_{z}}$.

Surely I should get the same answer if I solve the problem in a rotating frame of reference?

I know that (2):

$\frac{d\mathbf{\text{v}}}{dt}{|}_{Inertial}=\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}+\mathit{\omega}\times \mathbf{\text{v}};$

If I use Eq. (1) as the LHS of Eq. (2), and choose $\mathit{\omega}=-\frac{q\mathbf{\text{B}}}{m}$, then I get (3):

$\frac{d\mathbf{\text{v}}}{dt}{|}_{Rotational}=\frac{q}{m}\mathbf{\text{E}};$

How do I obtain a constant drift velocity (as mentioned before) from this? Have I used any formula incorrectly? Does the electric field E=$\mathbf{\text{E}}=({E}_{x},0,0)$ change form in the rotating frame?

asked 2022-05-19

Looking for specific Relativity example

The example had to do with two people walking along a sidewalk in opposite directions, and an alien race on a planet millions of light-years away planning an invasion of the Solar System. The example showed that in one walker's reference frame the invasion fleet had departed, but in the other reference frame the fleet had not.

At the time, the explanation made perfect sense, but I have forgotten the details and have never run across this example again.

Does anybody know where this was, or have the text of the explanation?

The example had to do with two people walking along a sidewalk in opposite directions, and an alien race on a planet millions of light-years away planning an invasion of the Solar System. The example showed that in one walker's reference frame the invasion fleet had departed, but in the other reference frame the fleet had not.

At the time, the explanation made perfect sense, but I have forgotten the details and have never run across this example again.

Does anybody know where this was, or have the text of the explanation?

asked 2022-05-10

Take the following gedankenexperiment in which two astronauts meet each other again and again in a perfectly symmetrical setting - a hyperspherical (3-manifold) universe in which the 3 dimensions are curved into the 4. dimension so that they can travel without acceleration in straight opposite directions and yet meet each other time after time.

On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?

So the question is: Who will be older? And why?

And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...

How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT

On the one hand this situation is perfectly symmetrical - even in terms of homotopy and winding number. On the other hand the Lorentz invariance should break down according to GRT, so that one frame is preferred - but which one?

So the question is: Who will be older? And why?

And even if there is one prefered inertial frame - the frame of the other astronaut should be identical with respect to all relevant parameters so that both get older at the same rate. Which again seems to be a violation of SRT in which the other twin seems to be getting older faster/slower...

How should one find out what the preferred frame is when everything is symmetrical - even in terms of GRT

asked 2022-07-16

Why do we say that irreducible representation of Poincare group represents the one-particle state?

Only because

1. Rep is unitary, so saves positive-definite norm (for possibility density),

2. Casimir operators of the group have eigenvalues ${m}^{2}$ and ${m}^{2}s(s+1)$, so characterizes mass and spin, and

3. It is the representation of the global group of relativistic symmetry,

yes?

Only because

1. Rep is unitary, so saves positive-definite norm (for possibility density),

2. Casimir operators of the group have eigenvalues ${m}^{2}$ and ${m}^{2}s(s+1)$, so characterizes mass and spin, and

3. It is the representation of the global group of relativistic symmetry,

yes?

asked 2022-07-22

on the Dirac equation, expands the ${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}$ term as:

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}+\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

where $\overrightarrow{\gamma}=({\gamma}^{1},{\gamma}^{2},{\gamma}^{3})$, but to my knowledge,

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{\mu}{\eta}_{\mu \nu}{\mathrm{\partial}}^{\nu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}-\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

using the convention ${\eta}_{\mu \nu}=\mathrm{diag}(+,-,-,-)$.

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}+\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

where $\overrightarrow{\gamma}=({\gamma}^{1},{\gamma}^{2},{\gamma}^{3})$, but to my knowledge,

${\gamma}^{\mu}{\mathrm{\partial}}_{\mu}={\gamma}^{\mu}{\eta}_{\mu \nu}{\mathrm{\partial}}^{\nu}={\gamma}^{0}\frac{\mathrm{\partial}}{\mathrm{\partial}t}-\overrightarrow{\gamma}\cdot \overrightarrow{\mathrm{\nabla}}$

using the convention ${\eta}_{\mu \nu}=\mathrm{diag}(+,-,-,-)$.

asked 2022-04-07

Why the log? Is it there to make the growth of the function slower?

As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\mathrm{\infty})$ to $(-\mathrm{\infty},\mathrm{\infty})$ (For a particle emitted along the beam axis after collision $\theta =0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\mathrm{\infty})$ to $(-\mathrm{\infty},\mathrm{\infty})$ (For a particle emitted along the beam axis after collision $\theta =0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

asked 2022-05-09

Formula for the Bekenstein bound

$S\le \frac{2\pi kRE}{\hslash c}$

where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.

Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?

$S\le \frac{2\pi kRE}{\hslash c}$

where $E$ is the total mass-energy. That seems to imply that the presence of a black hole in the region is dependent on an observer's frame of reference. Yet, my understanding is that the Bekenstein bound is the maximum entropy that any area can withstand before collapsing into a black hole.

Does this mean that the existence of black holes is observer dependent? Or that even if an observer does not report a black hole in their frame, one is guaranteed to form there in the future?