Consider a (classical) system of several interacting particles. Can it be shown that, if the Lagrangian of such a system is Lorenz invariant, there cannot be any space-like influences between the particles?

Baladdaa9
2022-07-22
Answered

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Mireya Hoffman

Answered 2022-07-23
Author has **14** answers

A counterexample:

1) Take two particles in a laboratory $S$

2) Set the clocks attached to the particles so that at laboratory time $t=0$ both clocks show ${\tau}_{1}=0$, ${\tau}_{2}=0$

3) Leave the particles to evolve under the following lorentz-invariant lagrangian:

$\mathcal{L}={\mathcal{L}}_{1}+{\mathcal{L}}_{2}+{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}\phantom{\rule{0ex}{0ex}}{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}={u}_{1}^{\text{}\mu}({\tau}_{1}){u}_{2\mu}({\tau}_{2}),$

where ${\mathcal{L}}_{1},{\mathcal{L}}_{2}$ are free-particle lagrangians. The total lagrangian is lorenz-invariant, but the physical system is not due to its initial setup, which allows the lagrangian to transmit the interactions in a space-like manner.

1) Take two particles in a laboratory $S$

2) Set the clocks attached to the particles so that at laboratory time $t=0$ both clocks show ${\tau}_{1}=0$, ${\tau}_{2}=0$

3) Leave the particles to evolve under the following lorentz-invariant lagrangian:

$\mathcal{L}={\mathcal{L}}_{1}+{\mathcal{L}}_{2}+{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}\phantom{\rule{0ex}{0ex}}{\mathcal{L}}_{\mathrm{i}\mathrm{n}\mathrm{t}}={u}_{1}^{\text{}\mu}({\tau}_{1}){u}_{2\mu}({\tau}_{2}),$

where ${\mathcal{L}}_{1},{\mathcal{L}}_{2}$ are free-particle lagrangians. The total lagrangian is lorenz-invariant, but the physical system is not due to its initial setup, which allows the lagrangian to transmit the interactions in a space-like manner.

asked 2022-05-17

if $u$ and ${u}^{\prime}$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by

$(l,m,n)=\frac{1}{|u|}({u}_{x},{u}_{y},{u}_{z})$

and

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{|{u}^{\prime}|}({u}_{x}^{\prime},{u}_{y}^{\prime},{u}_{z}^{\prime})$

are related by

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{D}(l-\frac{v}{u},m{\gamma}^{-1},n{\gamma}^{-1})$

and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.

$D=\frac{{u}^{\prime}}{u}(1-\frac{{u}_{x}v}{{c}^{2}})={[1-2l\frac{v}{u}+\frac{{v}^{2}}{{u}^{2}}-(1-{l}^{2})\frac{{v}^{2}}{{c}^{2}}]}^{\frac{1}{2}}$

Why is that better than the second expression?

Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ and $|u|=|({u}_{x},{u}_{y},{u}_{z})|=\sqrt{{u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}}$

$(l,m,n)=\frac{1}{|u|}({u}_{x},{u}_{y},{u}_{z})$

and

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{|{u}^{\prime}|}({u}_{x}^{\prime},{u}_{y}^{\prime},{u}_{z}^{\prime})$

are related by

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{D}(l-\frac{v}{u},m{\gamma}^{-1},n{\gamma}^{-1})$

and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.

$D=\frac{{u}^{\prime}}{u}(1-\frac{{u}_{x}v}{{c}^{2}})={[1-2l\frac{v}{u}+\frac{{v}^{2}}{{u}^{2}}-(1-{l}^{2})\frac{{v}^{2}}{{c}^{2}}]}^{\frac{1}{2}}$

Why is that better than the second expression?

Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ and $|u|=|({u}_{x},{u}_{y},{u}_{z})|=\sqrt{{u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}}$

asked 2022-07-16

Can one explain the relativistic energy transformation formula:

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

$E=\gamma \text{}{E}^{\prime},$

where the primed frame has a velocity $v$ relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation $E=h\nu $?

Imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period $T$.

Now A stays at rest whereas B is boosted to velocity 𝑣.

Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

From A's perspective B has extra kinetic energy by virtue of his velocity $v$. Relativistically A should use the energy transformation formula above.

But we should also be able to argue that B has more energy from B's perspective as well.

From B's perspective he is stationary and A has velocity $-v$. Therefore, due to relativistic time dilation, B sees A's oscillation period $T$ increased to $\gamma \text{}T$.

Thus B finds that his quantum oscillator will perform a factor of $\gamma \text{}T/T=\gamma $ more oscillations in the same period as A's quantum system.

Thus B sees that the frequency of his quantum system has increased by a factor of $\gamma $ over the frequency of A's system.

As we have the quantum relation, $E=h\nu $, this implies that B observes that the energy of his quantum system is a factor of $\gamma $ larger than the energy of A's stationary system.

Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

Is this reasoning correct?

asked 2022-05-18

Consider a person standing still on the ground and a car, which is moving with a constant speed $v$ in a straight line. From the frame of reference of the car the person is moving backwards with same speed $v$.

What causes to the person to move backwards in the frame of reference of the car?

What causes to the person to move backwards in the frame of reference of the car?

asked 2022-05-10

Observer A and B are at the same "depth" in a gravity well. Observer B then descends into the well. A will observe B's time as going slower than their own. B will observe A's time as going faster than their own.

What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?

What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree?

What happens if B were to ascend the well back to A's depth, would B's local time speed back up to the same rate as A's, but B would be younger (relative to A)?

What about the paradox caused by relative motion (ignoring gravity)? If A is moving relative to B, A and B will both observe the other's time as going slower. If A and B were together initially, then B moves away and returns, do their clocks agree?

asked 2022-07-22

Orbital velocity of a circular planet is $\sqrt{aR}$, where a is the centripetal acceleration, and $R$ is radius of the planet. With ${v}_{1}$ as the tangential velocity of the rotating planet at the equator.

On the non rotating body, suppose that the orbital velocity is ${v}_{0}$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of ${v}_{1}+{v}_{2}$ (the body and the object both going counterclockwise). Now, I half-hypothesized ${v}_{0}={v}_{1}+{v}_{2}=\sqrt{aR}$ and ${v}_{2}=\sqrt{{a}^{\prime}R}$ where ${a}^{\prime}$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as

${a}^{\prime}=a-\frac{{v}_{1}^{2}}{R}$

The logic was that from rotating body's reference frame, the object would be traveling at ${v}_{2}$, less than ${v}_{0}$ because of the centrifugal force, so ${v}_{2}$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.

Tried to solve for ${a}^{\prime}$ and comparing it to the value, got from rotating frame of reference, ending up with

${a}^{\prime}=a-(\frac{{v}_{1}({v}_{0}+{v}_{2})}{R})$

Something's not right, and if I had to choose, I would guess the ${v}_{0}=v1+{v}_{2}$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.

On the non rotating body, suppose that the orbital velocity is ${v}_{0}$, and, for an object launched on the rotating body's "equator", that the orbital velocity will be in the form of ${v}_{1}+{v}_{2}$ (the body and the object both going counterclockwise). Now, I half-hypothesized ${v}_{0}={v}_{1}+{v}_{2}=\sqrt{aR}$ and ${v}_{2}=\sqrt{{a}^{\prime}R}$ where ${a}^{\prime}$ is the "true" rotating body's acceleration, able to be calculated from the rotating frame of reference as

${a}^{\prime}=a-\frac{{v}_{1}^{2}}{R}$

The logic was that from rotating body's reference frame, the object would be traveling at ${v}_{2}$, less than ${v}_{0}$ because of the centrifugal force, so ${v}_{2}$ has to be the orbital velocity if the gravity was "weakened" by centrifugal force.

Tried to solve for ${a}^{\prime}$ and comparing it to the value, got from rotating frame of reference, ending up with

${a}^{\prime}=a-(\frac{{v}_{1}({v}_{0}+{v}_{2})}{R})$

Something's not right, and if I had to choose, I would guess the ${v}_{0}=v1+{v}_{2}$, that acceleration is not the same on those two planets, but I don't know how it would change, or why.

asked 2022-04-06

If an observer were to rotate around a point at near light speeds, what sort of length contraction would he observe the universe undergo?

asked 2022-05-19

Einstein said that the synchronization of two clocks is dependant on the velocity of the observer. But I feel a conceptual contradiction can be made:

There are two observers A and B. Observer 'A' faces direction X, and will be labeled "stationary." Another observer B faces direction X and is travelling rapidly in that direction as well on a collision course with 'A'. Both observers are holding two clocks; One in each hand, with hands held perpendicular to direction X. Both observers hit a "synchronization" button on the clocks before colliding.

My expectation in this case is that when the moving observer B halts to greet observer A - both observers will agree the B-pair clocks are synchronized and the A-pair clocks are synchronized (though all four clocks are not necessarily synchronized with eachother, and is not at issue in this question).

Bottom line: what was considered "synchronicity" by the speedy individual is accepted by the stationary individual. This seems to contradict special relativity.

Now if my scenario were modified slightly, then I believe special relativity would apply. If observer A were to hold the A-pair clocks with one held out before herself and the other held out behind herself (instead of the original left and right perpendicular angles), then finally I would expect there to be disagreement between the two observers when they stop to meet eachother.

All this suggests that position is an unaccounted for primary component of relativity, but from what little I know of SR, it doesn't hardly factor in at all. Can someone explain what I'm missing?

There are two observers A and B. Observer 'A' faces direction X, and will be labeled "stationary." Another observer B faces direction X and is travelling rapidly in that direction as well on a collision course with 'A'. Both observers are holding two clocks; One in each hand, with hands held perpendicular to direction X. Both observers hit a "synchronization" button on the clocks before colliding.

My expectation in this case is that when the moving observer B halts to greet observer A - both observers will agree the B-pair clocks are synchronized and the A-pair clocks are synchronized (though all four clocks are not necessarily synchronized with eachother, and is not at issue in this question).

Bottom line: what was considered "synchronicity" by the speedy individual is accepted by the stationary individual. This seems to contradict special relativity.

Now if my scenario were modified slightly, then I believe special relativity would apply. If observer A were to hold the A-pair clocks with one held out before herself and the other held out behind herself (instead of the original left and right perpendicular angles), then finally I would expect there to be disagreement between the two observers when they stop to meet eachother.

All this suggests that position is an unaccounted for primary component of relativity, but from what little I know of SR, it doesn't hardly factor in at all. Can someone explain what I'm missing?