If some kind of source was able to supply an infinite amount of energy, does that imply that it also must have an infinite mass? Is the contrary also true?

Aleseelomnl
2022-08-11
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asked 2022-08-14

Angular momentum is defined from linear momentum via $\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}$, and is conserved in a closed system. Since energy is the time part of the linear four-momentum, is there a quantity defined from energy that's also conserved?

asked 2022-09-27

What is the minimal voltage between anode and cathode in a linear accelerators to achieve speeds where relativity starts to show?

asked 2022-08-29

The momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as

$\lambda =\frac{h}{p}$

and

$p=\frac{mv}{\sqrt{1-{v}^{2}/{c}^{2}}}$

when $v\to c$, $p\to \mathrm{\infty}$, so is it say that the wavelength is ZERO?

$\lambda =\frac{h}{p}$

and

$p=\frac{mv}{\sqrt{1-{v}^{2}/{c}^{2}}}$

when $v\to c$, $p\to \mathrm{\infty}$, so is it say that the wavelength is ZERO?

asked 2022-07-14

If there is a non-zero expectation value for the Higgs boson even in "vacuum", since the Higgs boson has a mass unlike photons, then I would expect it to have a rest frame.

So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry?

So why doesn't a non-zero expectation value for the Higgs boson not only break electroweak symmetry, but also break Lorentz symmetry?

asked 2022-05-08

Imagine a space shuttle traveling through space at a constant velocity close to c. As the shuttle passes earth, a previously set-up camera starts broadcasting from earth to the shuttle. Since radio waves travel at the speed of light, the shuttle is receiving a constant transmission feed, assuming the camera is broadcasting 24/7.

Now, from what I have understood of special relativity so far, time will flow slower for the astronaut than for the earthlings. Hence, assuming $v=0.8c$, the astronaut will after 30 years have received a video transmission 50 years long!

Is my reasoning correct, that even though the transmission is live, the astronaut would actually be watching things that happened many years ago, while still receiving the "live" feed, which would be stored/buffered in the shuttles memory, thus making it possible for the astronaut to fast-forward the clip to see what happened more than 30 years after passing the earth?

Now, from what I have understood of special relativity so far, time will flow slower for the astronaut than for the earthlings. Hence, assuming $v=0.8c$, the astronaut will after 30 years have received a video transmission 50 years long!

Is my reasoning correct, that even though the transmission is live, the astronaut would actually be watching things that happened many years ago, while still receiving the "live" feed, which would be stored/buffered in the shuttles memory, thus making it possible for the astronaut to fast-forward the clip to see what happened more than 30 years after passing the earth?

asked 2022-07-16

Special relativity says that anything moving (almost) at the speed of light will look like its internal clock has (almost) stopped from the perspective of a stationary observer. How do we see light as alternating electric and magnetic fields?

asked 2022-09-25

consider a particle, $A$ receiving energy from a second one,particle $B$ in a one dimensional collision.

${E}^{2}={p}^{2}+{m}_{0}^{2}$

$EdE=pdp$

For particle $A$:

${E}_{A}d{E}_{A}={p}_{A}d{p}_{A}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(1)$

For Particle $B$:

${E}_{B}d{E}_{B}={p}_{B}d{p}_{B}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(2)$

Now

$\mid d{E}_{A}\mid =\mid d{E}_{B}\mid \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(3)$

From Conservation of linear momentum we have:

$\overrightarrow{{p}_{A}}+\overrightarrow{{p}_{B}}=\overrightarrow{K}$

where $\overrightarrow{k}$ is a constant vector. Now,

$d\overrightarrow{{p}_{A}}+d\overrightarrow{{p}_{B}}=0$

Or,

$d\overrightarrow{{p}_{B}}=-d\overrightarrow{{p}_{A}}$

Or,

$\mid d\overrightarrow{{p}_{B}}\mid =\mid d\overrightarrow{{p}_{A}}\mid \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(4)$

Applying relations $(3)$ and $(4)$ to $(1)$ and $(2)$ we have:

$\frac{{E}_{A}}{{E}_{B}}=\frac{\mid \overrightarrow{{p}_{A}}\mid}{\mid \overrightarrow{{p}_{B}}\mid}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(5)$

A pair of particles cannot interact unless relation $(5)$ is satisfied. Can we conclude that that relation $(5)$ to be a restriction for $1D$ collisions?

Now let's move to the general type of $3$d collisions between a pair of particles $A$ and $B$.

A frame is chosen where the particle $B$ is initially at rest in it.

${E}_{B}d{E}_{B}=\overrightarrow{{p}_{B}}\cdot d\overrightarrow{{p}_{B}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(6)$

If the particle $B$ is initially at rest the RHS of $(6)$ is zero. But the LHS cannot be zero unless $d{E}_{B}=0$.

How does one get round this problem?

${E}^{2}={p}^{2}+{m}_{0}^{2}$

$EdE=pdp$

For particle $A$:

${E}_{A}d{E}_{A}={p}_{A}d{p}_{A}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(1)$

For Particle $B$:

${E}_{B}d{E}_{B}={p}_{B}d{p}_{B}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(2)$

Now

$\mid d{E}_{A}\mid =\mid d{E}_{B}\mid \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(3)$

From Conservation of linear momentum we have:

$\overrightarrow{{p}_{A}}+\overrightarrow{{p}_{B}}=\overrightarrow{K}$

where $\overrightarrow{k}$ is a constant vector. Now,

$d\overrightarrow{{p}_{A}}+d\overrightarrow{{p}_{B}}=0$

Or,

$d\overrightarrow{{p}_{B}}=-d\overrightarrow{{p}_{A}}$

Or,

$\mid d\overrightarrow{{p}_{B}}\mid =\mid d\overrightarrow{{p}_{A}}\mid \phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(4)$

Applying relations $(3)$ and $(4)$ to $(1)$ and $(2)$ we have:

$\frac{{E}_{A}}{{E}_{B}}=\frac{\mid \overrightarrow{{p}_{A}}\mid}{\mid \overrightarrow{{p}_{B}}\mid}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(5)$

A pair of particles cannot interact unless relation $(5)$ is satisfied. Can we conclude that that relation $(5)$ to be a restriction for $1D$ collisions?

Now let's move to the general type of $3$d collisions between a pair of particles $A$ and $B$.

A frame is chosen where the particle $B$ is initially at rest in it.

${E}_{B}d{E}_{B}=\overrightarrow{{p}_{B}}\cdot d\overrightarrow{{p}_{B}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}(6)$

If the particle $B$ is initially at rest the RHS of $(6)$ is zero. But the LHS cannot be zero unless $d{E}_{B}=0$.

How does one get round this problem?