Does a relativistic version of quantum thermodynamics exist? I.e. in a non-inertial frame of reference, can I, an external observer, calculate quantities like magnetisation within the non-inertial frame?

measgachyx5q9
2022-05-19
Answered

Does a relativistic version of quantum thermodynamics exist? I.e. in a non-inertial frame of reference, can I, an external observer, calculate quantities like magnetisation within the non-inertial frame?

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Kharroubip9ej0

Answered 2022-05-20
Author has **10** answers

As a simple example the transformation law for Temperature is stated as: $T=\sqrt{(}1-{v}^{2}/{c}^{2}){T}_{0}$ when changing to a Lorentz moving frame.

Another example is that "entropy density" $\varphi $ is introduced, which is also subject to a Lorentz transformation. Finally this becomes a scalar with an associated "entropy 4-vector" in GR. The Second Law is expressed using these constructs.

Another example is that "entropy density" $\varphi $ is introduced, which is also subject to a Lorentz transformation. Finally this becomes a scalar with an associated "entropy 4-vector" in GR. The Second Law is expressed using these constructs.

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 2022-05-02

A Newtonian homogeneous density sphere has gravitational binding energy in Joules $U=-(3/5)(G{M}^{2})/r$, G=Newton's constant, M=gravitational mass, r=radius, mks. The fraction of binding energy to gravitational mass equivalent, $U/M{c}^{2}$, is then (-885.975 meters)(Ms/r), Ms = solar masses of body, c=lightspeed.

This gives ratios that are less than half that quoted for pulsars (neutron stars), presumably for density gradient surface to core and General Relativity effects (e.g., billion surface gees). Please post a more accurate formula acounting for the real world effects.

Examples: 1.74 solar-mass 465.1 Hz pulsar PSR J1903+0327, nominal radius 11,340 meters (AP4 model), calculates as 13.6% and is reported as 27%. A 2 sol neutron star calculates as 16.1% and is reported as 50%. There is an obvious nonlinearity.

This gives ratios that are less than half that quoted for pulsars (neutron stars), presumably for density gradient surface to core and General Relativity effects (e.g., billion surface gees). Please post a more accurate formula acounting for the real world effects.

Examples: 1.74 solar-mass 465.1 Hz pulsar PSR J1903+0327, nominal radius 11,340 meters (AP4 model), calculates as 13.6% and is reported as 27%. A 2 sol neutron star calculates as 16.1% and is reported as 50%. There is an obvious nonlinearity.

asked 2022-07-21

What reference frame is used to measure the velocity, $v=\frac{distance}{time}$?

$\mathbf{p}=\frac{mv}{\sqrt[2]{1-\frac{{v}^{2}}{{c}^{2}}}}$

Suppose the momentum of an electron is being observed in a lab. It could be observed in a lab that as the speed of electron increases, it becomes harder to accelerate it. To measure the velocity of electron, we need to measure distance or displacement and time.

$\mathbf{p}=\frac{mv}{\sqrt[2]{1-\frac{{v}^{2}}{{c}^{2}}}}$

Suppose the momentum of an electron is being observed in a lab. It could be observed in a lab that as the speed of electron increases, it becomes harder to accelerate it. To measure the velocity of electron, we need to measure distance or displacement and time.

asked 2022-05-08

Suppose we are given a mechanical frame consisting of two points. How can we prove that assuming any initial conditions there is an inertial frame of reference in which these points will be in a static plane?

asked 2022-05-18

"You are traveling in a car going at a constant speed of 100 km/hr down a long, straight highway. You pass another car going in the same direction which is traveling at a constant speed of 80 km/hr. As measured from your car’s reference frame this other car is traveling at -20 km/hr. What is the acceleration of your car as measured from the other car’s reference frame? What is the acceleration of the other car as measured from your car’s reference frame?"

Shouldn't they both appear to have an acceleration of zero, because both velocities are constant? I can imagine sitting in the faster car and watching the slower car, its speed would not appear to change, only its position?

Shouldn't they both appear to have an acceleration of zero, because both velocities are constant? I can imagine sitting in the faster car and watching the slower car, its speed would not appear to change, only its position?

asked 2022-07-17

Suppose $\varphi (x)$ a scalar field, ${v}^{\mu}$ a $4$-vector. According to my notes a quantity of form ${v}^{\mu}{\mathrm{\partial}}_{\mu}\varphi (x)$ will not be Lorentz invariant.

But explicitly doing the active transformation the quantity becomes

${\mathrm{\Lambda}}_{\nu}^{\mu}{v}^{\nu}({\mathrm{\Lambda}}^{-1}{)}_{\mu}^{\rho}{\mathrm{\partial}}_{\rho}\varphi (y)={v}^{\nu}{\mathrm{\partial}}_{\nu}\varphi (y)$

where $y={\mathrm{\Lambda}}^{-1}x$ and the partial differentiation is w.r.t. $y$. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.

I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that $v$ transforms nontrivially under the active transformation?

But explicitly doing the active transformation the quantity becomes

${\mathrm{\Lambda}}_{\nu}^{\mu}{v}^{\nu}({\mathrm{\Lambda}}^{-1}{)}_{\mu}^{\rho}{\mathrm{\partial}}_{\rho}\varphi (y)={v}^{\nu}{\mathrm{\partial}}_{\nu}\varphi (y)$

where $y={\mathrm{\Lambda}}^{-1}x$ and the partial differentiation is w.r.t. $y$. This seems to suggest that the quantity is a Lorentz scalar, so could be used to construct a Lorentz invariant first order equation of motion.

I'm clearly making a mistake here. But I don't see what I've done wrong. Am I wrong to think that $v$ transforms nontrivially under the active transformation?

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 ?