If a rocket ship is traveling at .99c for 1 year, and is streaming a video at 30 frames/sec to earth, how would the earth feed be affected? Would it show the video at a much slower rate, would it remain constant, or would it be sped up?

bruinhemd3ji
2022-11-08
Answered

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dilettato5t1

Answered 2022-11-09
Author has **25** answers

So let's say the rocket has a camera that's supposed to take a frame every $1/30$-th of a second. From the earth's point of view, this frequency will be reduced (by how much exactly is left for you as an exercise). Let's just for the sake of simplicity say that on earth it looks like the camera takes just one frame every (earth-)second.

Next comes the transmission of the stream back to earth. Let's say the rocket sends each frame individually in a single specially shaped pulse of light. Then in earth time, it sends out a frame every second. But during that second, it'll have flown another $.99c\cdot 1s$ further away from earth. Hence, the time between pulses is actually $1s+0.99=1.99s$.

So instead of receiving 30 frames per second, earth only receives $1/1.99\approx 0.5$ frames per seconds, or about 1 frame every 2 seconds.

Now of course depending on your age you remember the good old days where it actually took longer to download a video than to watch it. What you'd do then, of course, is first wait until enough data is loaded so you can resume playback (the good old buffering). If we wait long enough, we get a bunch of frames from the rocket, and we can then play them back at any speed we like, including the original supposed speed of 30 fps. The video will then look totally normal to us.

Next comes the transmission of the stream back to earth. Let's say the rocket sends each frame individually in a single specially shaped pulse of light. Then in earth time, it sends out a frame every second. But during that second, it'll have flown another $.99c\cdot 1s$ further away from earth. Hence, the time between pulses is actually $1s+0.99=1.99s$.

So instead of receiving 30 frames per second, earth only receives $1/1.99\approx 0.5$ frames per seconds, or about 1 frame every 2 seconds.

Now of course depending on your age you remember the good old days where it actually took longer to download a video than to watch it. What you'd do then, of course, is first wait until enough data is loaded so you can resume playback (the good old buffering). If we wait long enough, we get a bunch of frames from the rocket, and we can then play them back at any speed we like, including the original supposed speed of 30 fps. The video will then look totally normal to us.

asked 2022-09-22

Suppose we have a train moving. When the origin of train's frame coincides with the origin of observers frame; the the time is set to zero. At that very instant, a photon is emitted from train towards the direction train is moving. After time $t$ measured by observer at rest the photon will be at a distance $ct$ and the train at a distance $vt$; but the photon is at a distance $ct-vt$ from the train, now the time that the passenger on train measures for the photon to reach that point is the distance divided by the velocity of light ( which of course is $c$) so

time measured by observer on train is ${t}^{\prime}=\frac{ct-vt}{c}=(1-\frac{v}{c})t$ . where $t$ is time measured by observer at rest.

Where is the mistake in this reasoning?

time measured by observer on train is ${t}^{\prime}=\frac{ct-vt}{c}=(1-\frac{v}{c})t$ . where $t$ is time measured by observer at rest.

Where is the mistake in this reasoning?

asked 2022-08-16

Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field ${\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}$, which is equal to $(\frac{m}{2},\frac{n}{2})$ representation of the Lorentz group. There is the hard provable theorem:

${\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}$ realizes irreducible representation of the Poincare group, if

$({\mathrm{\partial}}^{2}-{m}^{2}){\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}=0,$

${\mathrm{\partial}}^{\alpha \dot{\beta}}{\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}=0.$

Can this theorem be interpreted as connection between fields and particles?

${\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}$ realizes irreducible representation of the Poincare group, if

$({\mathrm{\partial}}^{2}-{m}^{2}){\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}=0,$

${\mathrm{\partial}}^{\alpha \dot{\beta}}{\psi}_{\alpha {\alpha}_{1}...{\alpha}_{n-1}\dot{\beta}{\dot{\beta}}_{1}...{\dot{\beta}}_{m-1}}=0.$

Can this theorem be interpreted as connection between fields and particles?

asked 2022-08-28

struggling to derive Lorentz transformations for a sine wave, which is traveling at random direction. I started by prooving that phase $\varphi $ is invariant for relativity and that equation $\varphi ={\varphi}^{\prime}$ holds.

By using the above equation i am now trying to derive Lorentz transformations for angular frequency $\omega $, and all three components of the wave vector $k$, which are ${k}_{x}$, ${k}_{y}$ and ${k}_{z}$.

This is my attempt:

$\begin{array}{rl}{\varphi}^{\prime}& =\varphi \\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+{k}^{\prime}\mathrm{\Delta}{r}^{\prime}& =\omega \mathrm{\Delta}t+k\mathrm{\Delta}r\\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+[{{k}_{x}}^{\prime},{{k}_{y}}^{\prime},{{k}_{z}}^{\prime}][\mathrm{\Delta}{x}^{\prime},\mathrm{\Delta}{y}^{\prime},\mathrm{\Delta}{z}^{\prime}]& =\omega \mathrm{\Delta}t+[{k}_{x},{k}_{y},{k}_{z}][\mathrm{\Delta}x,\mathrm{\Delta}y,\mathrm{\Delta}z]\\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+{{k}_{x}}^{\prime}\mathrm{\Delta}{x}^{\prime}+{{k}_{y}}^{\prime}\mathrm{\Delta}{y}^{\prime}+{{k}_{z}}^{\prime}\mathrm{\Delta}{z}^{\prime}& =\omega \mathrm{\Delta}t+{k}_{x}\mathrm{\Delta}x+{k}_{y}\mathrm{\Delta}y+{k}_{z}\mathrm{\Delta}z\end{array}$

$\begin{array}{c}{\omega}^{\prime}\gamma (\mathrm{\Delta}t-\mathrm{\Delta}x\frac{u}{{c}^{2}})+{{k}_{x}}^{\prime}\gamma {\textstyle (}\mathrm{\Delta}x-u\mathrm{\Delta}t{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \gamma ({\omega}^{\prime}\mathrm{\Delta}t-{\omega}^{\prime}\mathrm{\Delta}x\frac{u}{{c}^{2}})+\gamma {\textstyle (}{{k}_{x}}^{\prime}\mathrm{\Delta}x-{{k}_{x}}^{\prime}u\mathrm{\Delta}t{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \gamma ({\omega}^{\prime}\mathrm{\Delta}t-{{k}_{x}}^{\prime}c\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c\mathrm{\Delta}t\phantom{\rule{thinmathspace}{0ex}}\frac{u}{{c}^{2}})+\gamma {\textstyle (}{{k}_{x}}^{\prime}\mathrm{\Delta}x-\frac{{\omega}^{\prime}}{c}u\frac{\mathrm{\Delta}x}{c}{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \mathrm{\Delta}t\phantom{\rule{thinmathspace}{0ex}}\gamma {\textstyle (}{\omega}^{\prime}-{{k}_{x}}^{\prime}u{\textstyle )}+\mathrm{\Delta}x\phantom{\rule{thinmathspace}{0ex}}\gamma {\textstyle (}{{k}_{x}}^{\prime}-{\omega}^{\prime}\frac{u}{{c}^{2}}{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\end{array}$

From this I can write down the Lorentz transformations.

$\begin{array}{rl}\gamma {\textstyle (}{\omega}^{\prime}-{{k}_{x}}^{\prime}u{\textstyle )}& =\omega \\ \gamma {\textstyle (}{{k}_{x}}^{\prime}-{\omega}^{\prime}\frac{u}{{c}^{2}}{\textstyle )}& ={k}_{x}\\ {{k}_{y}}^{\prime}& ={k}_{y}\\ {{k}_{z}}^{\prime}& ={k}_{z}\end{array}$

What am i doing wrong?

By using the above equation i am now trying to derive Lorentz transformations for angular frequency $\omega $, and all three components of the wave vector $k$, which are ${k}_{x}$, ${k}_{y}$ and ${k}_{z}$.

This is my attempt:

$\begin{array}{rl}{\varphi}^{\prime}& =\varphi \\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+{k}^{\prime}\mathrm{\Delta}{r}^{\prime}& =\omega \mathrm{\Delta}t+k\mathrm{\Delta}r\\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+[{{k}_{x}}^{\prime},{{k}_{y}}^{\prime},{{k}_{z}}^{\prime}][\mathrm{\Delta}{x}^{\prime},\mathrm{\Delta}{y}^{\prime},\mathrm{\Delta}{z}^{\prime}]& =\omega \mathrm{\Delta}t+[{k}_{x},{k}_{y},{k}_{z}][\mathrm{\Delta}x,\mathrm{\Delta}y,\mathrm{\Delta}z]\\ {\omega}^{\prime}\mathrm{\Delta}{t}^{\prime}+{{k}_{x}}^{\prime}\mathrm{\Delta}{x}^{\prime}+{{k}_{y}}^{\prime}\mathrm{\Delta}{y}^{\prime}+{{k}_{z}}^{\prime}\mathrm{\Delta}{z}^{\prime}& =\omega \mathrm{\Delta}t+{k}_{x}\mathrm{\Delta}x+{k}_{y}\mathrm{\Delta}y+{k}_{z}\mathrm{\Delta}z\end{array}$

$\begin{array}{c}{\omega}^{\prime}\gamma (\mathrm{\Delta}t-\mathrm{\Delta}x\frac{u}{{c}^{2}})+{{k}_{x}}^{\prime}\gamma {\textstyle (}\mathrm{\Delta}x-u\mathrm{\Delta}t{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \gamma ({\omega}^{\prime}\mathrm{\Delta}t-{\omega}^{\prime}\mathrm{\Delta}x\frac{u}{{c}^{2}})+\gamma {\textstyle (}{{k}_{x}}^{\prime}\mathrm{\Delta}x-{{k}_{x}}^{\prime}u\mathrm{\Delta}t{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \gamma ({\omega}^{\prime}\mathrm{\Delta}t-{{k}_{x}}^{\prime}c\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}c\mathrm{\Delta}t\phantom{\rule{thinmathspace}{0ex}}\frac{u}{{c}^{2}})+\gamma {\textstyle (}{{k}_{x}}^{\prime}\mathrm{\Delta}x-\frac{{\omega}^{\prime}}{c}u\frac{\mathrm{\Delta}x}{c}{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\\ \mathrm{\Delta}t\phantom{\rule{thinmathspace}{0ex}}\gamma {\textstyle (}{\omega}^{\prime}-{{k}_{x}}^{\prime}u{\textstyle )}+\mathrm{\Delta}x\phantom{\rule{thinmathspace}{0ex}}\gamma {\textstyle (}{{k}_{x}}^{\prime}-{\omega}^{\prime}\frac{u}{{c}^{2}}{\textstyle )}+{{k}_{y}}^{\prime}\mathrm{\Delta}y+{{k}_{z}}^{\prime}\mathrm{\Delta}z\end{array}$

From this I can write down the Lorentz transformations.

$\begin{array}{rl}\gamma {\textstyle (}{\omega}^{\prime}-{{k}_{x}}^{\prime}u{\textstyle )}& =\omega \\ \gamma {\textstyle (}{{k}_{x}}^{\prime}-{\omega}^{\prime}\frac{u}{{c}^{2}}{\textstyle )}& ={k}_{x}\\ {{k}_{y}}^{\prime}& ={k}_{y}\\ {{k}_{z}}^{\prime}& ={k}_{z}\end{array}$

What am i doing wrong?

asked 2022-07-15

If an object has very energetic particles in it like that of the sun then wouldn't its mass be higher hence making its gravity greater than that of the still state ones ?

asked 2022-05-19

Suppose we have a sphere of radius $r$ and mass m and a negatively charged test particle at distance d from its center, $d\gg r$. If the sphere is electrically neutral, the particle will fall toward the sphere because of gravity. As we deposit electrons on the surface of the sphere, the Coulomb force will overcome gravity and the test particle will start to accelerate away. Now suppose we keep adding even more electrons to the sphere. If we have n electrons, the distribution of their pairwise distances has a mean proportional to $r$, and there are $n(n-1)/2$ such pairs, so the binding energy is about ${n}^{2}/r$. If this term is included in the total mass-energy of the sphere, the gravitational force on the test particle would seem to be increasing quadratically with $n$, and therefore eventually overcomes the linearly-increasing Coulomb force. The particle slows down, turns around, and starts falling again. This seems absurd; what is wrong with this analysis?

asked 2022-09-03

Imagine you are one lightyear away from a photon sensitive (light sensitive) switch. So it is obvious that light would take one year to reach to the switch. Now I have a one lightyear long plank. I simply point the plank towards the switch and press it. Now I just did work which light would take 1 year to do in a matter of seconds.

Now the question is, did I break the laws of physics?

Now the question is, did I break the laws of physics?

asked 2022-09-27

Einstein's Special Theory of relativity postulates that the speed of light is same for all frames.

Suppose a neutrino is there moving at the speed of light. Then will that neutrino also be flowing with same speed for all frames or is this a special property of EM waves?

Suppose a neutrino is there moving at the speed of light. Then will that neutrino also be flowing with same speed for all frames or is this a special property of EM waves?