Transmission Line in Accelerating Frame: Energy Conservation Puzzle

[pervect]

While I believe I have an answer to this problem, I think it's an interesting one and counter-intuitive. I think it might spark some interesting discussions.

The bow of a Born rigid accelerating spaceship has a proper acceleration of g. There is a transmission line running from the stern to the bow of proper length d. There is a battery at the stern of the ship supplying power to the bow via the transmission line. The received power at the bow of the space-ship is 1 watt, steady-state direct current, with no time variations, as measured locally by an observer at the bow. What is the transmitted power drawn from the battery at the stern of the spaceship as measured by a local observer?

I don't believe it is necessary to make other assumptions to solve the problem, I believe that energy conservation arguments should be sufficient. But if it makes the problem simpler, feel free to make or suggest additional simplifying, clarifying, or convenient assumptions.

 

[Ibix]

I think this is a complete answer, so spoilered.

Spoiler

Use the current at the bow to power a laser pointing back down the ship and use a photocell in the stern to recharge the battery. Assuming 100% efficiency everywhere this is a closed loop, so the answer to the question is equal to the power received from the laser.

In the lapse $\Delta t$ of bow time the laser emits energy $P\Delta t$, where $P=1\mathrm{W}$ is the laser power. This is gravitationally blue shifted by some factor $f$ so arrives with energy $fP\Delta t$, and does so in the lapse $\Delta t'=\Delta t/f$ of stern time. Thus, if I'm not mistaken, the power received is $f^2$ times the power emitted.

I don't know $f$ off the top of my head. If the rocket is short (length $l\ll c^2/g$) it will be $\approx 1+gl/c^2$. I'd need to do some maths for an exact answer - it's the ratio of proper times along the bow and stern Rindler hyperbolae.

 

[Sagittarius A-Star]

Thus, if I'm not mistaken, the power received is $f^2$ times the power emitted.

I agree. Reason: $P=I^2R$, with $I=dq/d\tau$.

If the rocket is short (length $l\ll c^2/g$) it will be $\approx 1+gl/c^2$. I'd need to do some maths for an exact answer

I see no reason, why this should not be also the exact answer (at least on front of the Rindler horizon), reason:

Yes. Gron derives in his book the following equation (4.50):
$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$

 

[jartsa]

At the bow charge a battery by energy $E=UC$. (Volts times Coulombs)

Then lower the energy and the battery to the stern, getting out of the elecric energy this much mechanical energy: $mgh = gh*(E/c^2)$Then get the all the remaining electric energy back trough the electric wire. You should now have the original energy E, so the elecric energy you got back must be: $E - gh*(E/c^2)$.Using E=UC and realising that the Coulombs can not change, we come to the conclusion that the voltage U must have changed, by factor of $1-(gh*c^2)$. When voltage changes by that factor, then power changes by that factor squared.

By voltage changing by some factor I mean that voltages measured at the bow and at the stern differ by that factor.

 

[Ibix]

I see no reason, why this should not be also the exact answer (at least on front of the Rindler horizon), reason:

I don't agree. The factor $1+gl/c^2$ doesn't go to infinity as the stern approaches the horizon, which the exact factor must do.

I did the complete maths. Imagine that the bow emits a light pulse received at the stern. Let us work in an inertial frame where the origin is the focus of the Rindler hyperbolae and the stern received the light pulse at the moment it is instantaneously at rest. In this frame, the factor $f$ is the kinematic Doppler effect due to the velocity of the bow at the time it emitted the pulse. We can write the worldline of the bow$$x^2-c^2t^2=\frac{c^2}g$$and the light pulse$$x=\left(\frac{c^2}g-l\right)-ct$$and note by differentiating the former that the velocity, $v=dx/dt$, of the bow satisfies $vx=c^2t$. This system solves to give us the velocity at emission as $$v=cgl\frac{gl-2c^2}{(gl-2c^2)^2-2c^4}$$Note that this velocity is negative. Thus to get a blueshift we want to plug it into the Doppler shift expression $$\begin{eqnarray*}f&=&\sqrt{\frac{c-v}{c+v}}\\&=&\frac 1{1-gl/c^2}\end{eqnarray*}$$This does indeed behave as $1+gl/c^2$ for small $l$, but also has the expected infinity at $l=c^2/g$ when the rocket's stern is at the Rindler horizon.

There seem to be some typos in what @jartsa wrote, but assuming that he intends his factor to be $1-gh/c^2$ (he uses $h$ instead of $l$) and a divisor rather than a multiplier we agree.

 

[Ibix]

...and I've just noticed that @pervect specified the length of the rocket as $d$ in the OP. Oh well!

 

[Sagittarius A-Star]

$$\begin{eqnarray*}f&=&\sqrt{\frac{c-v}{c+v}}\\&=&\frac 1{1-gl/c^2}\end{eqnarray*}$$This does indeed behave as $1+gl/c^2$ for small $l$, but also has the expected infinity at $l=c^2/g$ when the rocket's stern is at the Rindler horizon.

Yes, I was incorrect. An easier deviation is possible in the accelerated frame (with $x=0$ for the bow, which has the porper acceleration $g$) from:

Yes. Gron derives in his book the following equation (4.50):
$$d\tau = dt\sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{v^2}{c^2}}$$

The stern has a negative $x$-coordinate and the velocity $v=0$, and you get:
$$dt = d\tau / \sqrt {(1+ \frac{g\,x} {c^2})^2 - \frac{0}{c^2}} = d\tau / (1- \frac{g\,d} {c^2})$$
This becomes close to infinite, if the stern (with finite proper time $\tau$) is close to the Rindler horizon.
$$f = dt/d\tau$$.

 

[pervect]

I agree with the $1/(1-gd/c^2)$ answer. It's also compatible with a longer analysis I did for the voltage in a transmission line with zero current using Maxwell's equations - the transmission line with zero current being replaced with a parallel plate capacitor as the inductance doesn't matter.

 

[PeterDonis]

The factor $1+gl/c^2$ doesn't go to infinity as the stern approaches the horizon, which the exact factor must do.

The factor you give goes to zero as the stern approaches the horizon, since "approaches the horizon" means $l \rightarrow - c^2 / g$. (The bow is at $l = 0$ in these coordinates.) This is to be expected since the factor you give is the time dilation factor for the observer at the stern, relative to the observer at the bow--it is, heuristically, the rate at which the bow observer calculates the stern observer's clock to be running, relative to his own.

The reciprocal of this factor is what is needed to solve the problem posed in the OP. With proper attention to signs, this agrees with the formula $1 / \left( 1 - g d / c^2 \right)$.

 

[jartsa]

There seem to be some typos in what @jartsa wrote

Yes typos. Can't edit the post anymore.

As there seems to exist a redshift effect of DC-voltage, does a moving tram observe a changed voltage of the overhead wire, because of the motion?

 

[Sagittarius A-Star]

While I believe I have an answer to this problem
...
The received power at the bow of the space-ship is 1 watt, steady-state direct current, with no time variations, as measured locally by an observer at the bow.

What is the transmitted power drawn from the battery at the stern of the spaceship as measured by a local observer?

I agree with the $1/(1-gd/c^2)$ answer.

That is not the answer, we gave to your question. The given answers were the square of it as factor for the power:

Thus, if I'm not mistaken, the power received is $f^2$ times the power emitted.

I agree. Reason: $P=I^2R$, with $I=dq/d\tau$.

$1-(gh*c^2)$. When voltage changes by that factor, then power changes by that factor squared.

Do you agree to the "square-answers"?

 

[jartsa]

That is not the answer, we gave to your question. The given answers were the square of it as factor for the power:
Do you agree to the "square-answers"?

Yes I agree.

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How does an inertial observer explain all of this stuff? The battery does some work on the spaceship increasing its kinetic energy? And as it takes an increasing amount of time to transmit a parcel of energy through the line, the energy stored in ithe transmission line increases. So then the rate at which the parcels of energy arrive at the bow is decreased, and the usable electric energy of each parcel is decreased?

 

[Sagittarius A-Star]

How does an inertial observer explain all of this stuff? The battery does some work on the spaceship increasing its kinetic energy? And as it takes an increasing amount of time to transmit a parcel of energy through the line, the energy stored in ithe transmission line increases.

This depends on, which inertial frame you choose, and at which time you valuate the quantities. If you choose for example a "co-moving" inertial frame, in which the center-of-momentum of the spaceship is at rest at a certain point in time, then at this point in time the kinetic energy of the spaceship is zero. Before this point in time, the spaceship moved into the opposite direction as in which it accelerates and had therefore some kinetic energy.

So then the rate at which the parcels of energy arrive at the bow is decreased, and the usable electric energy of each parcel is decreased?

Yes. Frequency (=rate) and electromagnetic energy in vacuum transform each according to the Doppler formula (in an inertial frame). For the usable electric power via a constant current, it should be overall the same effect (Doppler formula squared in an inertial frame).