If time in systems moving with different speed goes differently, does speed of entropy change differ in these systems?

Hanna Webster
2022-11-13
Answered

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Lillianna Salazar

Answered 2022-11-14
Author has **22** answers

Entropy is an invariant $S\to {S}^{\prime}=S$, whereas time is not, in general. Therefore the rate of change of entropy (this is the correct term) is a frame dependant quantity.

$\frac{\mathrm{d}S}{\mathrm{d}t}=\gamma \frac{\mathrm{d}{S}^{\prime}}{\mathrm{d}{t}^{\prime}}=\gamma \frac{\mathrm{d}S}{\mathrm{d}{t}^{\prime}}$

with $\gamma $ the time-dilation factor.

$\frac{\mathrm{d}S}{\mathrm{d}t}=\gamma \frac{\mathrm{d}{S}^{\prime}}{\mathrm{d}{t}^{\prime}}=\gamma \frac{\mathrm{d}S}{\mathrm{d}{t}^{\prime}}$

with $\gamma $ the time-dilation factor.

asked 2022-11-01

On a relativistic train a kitchen timer is set to give a signal after 4.95 minutes. You are standing next to the rail road tracks. In your reference frame it takes 8.95 minutes for the kitchen timer to signal. What is the speed of the train?

asked 2022-04-07

While the speed of light in vacuum is a universal constant (c), the speed at which light propagates in other materials/mediums may be less than c. This is obviously suggested by the fact that different materials (especially in the case of transparent ones) have a particular refractive index.

But surely, matter or even photons can be accelerated beyond this speed in a medium?

But surely, matter or even photons can be accelerated beyond this speed in a medium?

asked 2022-09-30

How much does a proton weigh when it is going around the LHC at CERN?

Considering that speed increases weight and the proton is going at almost the speed of light, I would like to know how much a speeding proton would weigh in the LHC.

Considering that speed increases weight and the proton is going at almost the speed of light, I would like to know how much a speeding proton would weigh in the LHC.

asked 2022-07-16

Three events $A,B,C$ are seen by observer $O$ to occur in the order $ABC$. Another observer $O\prime $ sees the events to occur in the order $CBA$. Is it possible that a third observer sees the events in the order $ACB$?

I could draw spacetime diagram for three arbitrary-ordered events for $O$ and $O\prime $. But I couldn't draw spacetime diagram for third observer.

I could draw spacetime diagram for three arbitrary-ordered events for $O$ and $O\prime $. But I couldn't draw spacetime diagram for third observer.

asked 2022-11-05

Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,Suppose we want to construct a wave function for 2 free (relativistic) fermions. As we are dealing with fermions the total wave function has to be antisymmetric under interchange of the coordinates,

$\mathrm{\Psi}({x}_{1},{x}_{2})=-\mathrm{\Psi}({x}_{2},{x}_{1})$

If we assume that we can factorize the wave function in terms of single particle wave functions we can write

$\mathrm{\Psi}({x}_{1},{x}_{2})={\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})-{\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})$

which fulfills the anti-symmetry requirement. The plane wave single particle states are given by,

${\psi}_{\mathbf{k},{m}_{s}}(x)={u}_{\mathbf{k},{m}_{s}}(s)\varphi (\mathbf{k}\cdot \mathbf{r})$

So expect the total wavefunction to be

$\begin{array}{rl}\mathrm{\Psi}({x}_{1},{x}_{2})& ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\\ & ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\end{array}$

However

$u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})$

If I'm not mistaking one cannot freely change the order of the Dirac spinors $u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\ne u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})$ so these expressions seem to be uncompatible. What would the correct expression look like?

$\mathrm{\Psi}({x}_{1},{x}_{2})=-\mathrm{\Psi}({x}_{2},{x}_{1})$

If we assume that we can factorize the wave function in terms of single particle wave functions we can write

$\mathrm{\Psi}({x}_{1},{x}_{2})={\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})-{\psi}_{1}({x}_{1}){\psi}_{2}({x}_{2})$

which fulfills the anti-symmetry requirement. The plane wave single particle states are given by,

${\psi}_{\mathbf{k},{m}_{s}}(x)={u}_{\mathbf{k},{m}_{s}}(s)\varphi (\mathbf{k}\cdot \mathbf{r})$

So expect the total wavefunction to be

$\begin{array}{rl}\mathrm{\Psi}({x}_{1},{x}_{2})& ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\\ & ={u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{1}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{2})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-{u}_{{\mathbf{k}}_{1},{m}_{{s}_{1}}}({s}_{2}){u}_{{\mathbf{k}}_{2},{m}_{{s}_{2}}}({s}_{1})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})\end{array}$

However

$u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{1})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{2})-u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})\varphi ({\mathbf{k}}_{1}\cdot {\mathbf{r}}_{2})\varphi ({\mathbf{k}}_{2}\cdot {\mathbf{r}}_{1})$

If I'm not mistaking one cannot freely change the order of the Dirac spinors $u({\mathbf{k}}_{1},{m}_{{s}_{1}})u({\mathbf{k}}_{2},{m}_{{s}_{2}})\ne u({\mathbf{k}}_{2},{m}_{{s}_{2}})u({\mathbf{k}}_{1},{m}_{{s}_{1}})$ so these expressions seem to be uncompatible. What would the correct expression look like?

asked 2022-09-29

How does one tell whether an expression is covariant or not? I get it for a single tensor, but how is it defined when there is no overall up/down index to base it on?

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?