Battery life on VERY fast moving object

[PeterDonis]

The redshift does affect the total energy transferred because of $W=hf$ for each light-pulse ("photon").

This is the relativity forum, not the QM forum. There is no such thing as "photon" here. (A proper QM analysis would not justify the claim you are making either, but that's a matter for a separate thread in the QM forum.)

If by "light pulse" you mean a classical light pulse, the Doppler effect does not mean what you are claiming it means.

Source (see §8):
https://en.wikisource.org/wiki/On_the_Electrodynamics_of_Moving_Bodies_(1920_edition)

This paper does not justify what you are claiming.

 

[Sagittarius A-Star]

If by "light pulse" you mean a classical light pulse,

Yes.

the Doppler effect does not mean what you are claiming it means.
...
This paper does not justify what you are claiming.

Why? It argues classically.

 

[PeterDonis]

Why? It argues classically.

Sure, Einstein's paper is classical, but just saying "classical" doesn't justify the claim you are making. Go read Einstein's paper carefully. What he means by "the energy and frequency of a light complex" (and note that the "frequency" part is not actually classical, he was smuggling in a quantum concept there, but he doesn't use it anywhere else in the paper so it's not really relevant to this discussion) is not what you would need him to mean to justify your claim.

 

[Ibix]

I think the complete analysis is:

In the frame of the rocket, energy $E$ is emitted in time $\Delta t$, consisting of $f\Delta t$ cycles of coherent radiation. The rocket mass is reduced by $E/c^2$.

In the frame of the planet, energy $\gamma E$ is emitted in time $\gamma\Delta t$, consisting of $(f/\gamma)\gamma\Delta t=f\Delta t$ cycles of coherent radiation. The rocket mass is reduced by $\gamma E/c^2$.

Note that the radiation in the planet frame is redshifted compared to the rocket frame, but the power is the same and the beam duration is longer so the total energy is higher.

 

[Sagittarius A-Star]

Sure, Einstein's paper is classical, but just saying "classical" doesn't justify the claim you are making. Go read Einstein's paper carefully. What he means by "the energy and frequency of a light complex" (and note that the "frequency" part is not actually classical, he was smuggling in a quantum concept there, but he doesn't use it anywhere else in the paper so it's not really relevant to this discussion) is not what you would need him to mean to justify your claim.

Unlike in most of today's derivations of the frequency-Doppler formula, he wasn't smuggling in a quantum concept in §7. He used the Lorentz-transformation for the electromagnetic field components.

 

[PAllen]

Unlike in most of today's derivations of the frequency-Doppler formula, he wasn't smuggling in a quantum concept in §7. He used the Lorentz-transformation for the electromagnetic field components.

Yes, this is insufficiently emphasized. Einstein's is one of the few derivations which make no use of any notion frequency being proportional to energy. In doing simple 'kinematic' derivations, I do take this shortcut, but Einstein did not.

 

[PeterDonis]

Unlike in most of today's derivations of the frequency-Doppler formula, he wasn't smuggling in a quantum concept in §7. He used the Lorentz-transformation for the electromagnetic field components.

Yes, but nowhere is there a relation like the Planck relation given between energy and frequency. He just derives the classical relativistic Doppler formula, which is a formula for frequency shift but says nothing about energy.

 

[PAllen]

Yes, but nowhere is there a relation like the Planck relation given between energy and frequency. He just derives the classical relativistic Doppler formula, which is a formula for frequency shift but says nothing about energy.

He shows the energy of a light pulse changes by the same (Doppler) factor as the frequency. He does this by directly analyzing the classical field energy (related to amplitude) nowhere using any quantum relation. He actually expresses surprise at this coincidence - that the same factor applies to both.

 

[Ibix]

A more straightforward case to consider is a rocket in inertial motion in the $+x$ direction. It fires two identical things in the $\pm y$ direction with equal and opposite velocity. One can easily write down the four momentum of the rocket before and after the launch and conserve four momentum, both when the "things" are bullets or laser pulses.

In the frame of the rocket (suppressing the $z$ dimension) the rocket's four momentum is initially $(m,0,0)$. After launch it is $(m-2\delta m,0,0)$ and the laser pulses have four momenta of $(\delta m,0,\pm\delta m)$. Note the conservation of energy and momentum. Lorentz transforming, we have the rocket's initial four momentum $(\gamma m,-\gamma mv,0)$ and final four momentum $(\gamma (m-2\delta m),-\gamma (m-2\delta m)v,0)$ and the laser pulses have four momenta $(\gamma\delta m,-\gamma \delta mv,\delta m)$. Again, energy and momentum are conserved. But there is more energy in the transformed pulses than there was in the original frame.

 

[PeterDonis]

He shows the energy of a light pulse changes by the same (Doppler) factor as the frequency.

He uses the term "energy", but a better term might be "rate of energy delivery" or "radiation pressure" (note that he analyzes the latter in Section 7 as well). The point is that what he means by "energy" is not "the total energy delivered to a receiver by the light pulse". It is something you have to integrate over the pulse time to get the total energy delivered. But the pulse time is longer by exactly the same factor, the Doppler factor, as what Einstein is calling the "energy" is reduced. So the two effects cancel out when we are evaluating the total energy delivered.

 

[Sagittarius A-Star]

In the frame of the rocket (suppressing the $z$ dimension) the rocket's four momentum is initially $(m,0,0)$. After launch it is $(m-2\delta m,0,0)$ and the laser pulses have four momenta of $(\delta m,0,\pm\delta m)$.

That is a transversal Doppler effect in the sender's frame. -> blueshift
It is not a transversal Doppler effect in the receiver's frame -> redshift (reason: aberration).

 

[PAllen]

He uses the term "energy", but a better term might be "rate of energy delivery" or "radiation pressure" (note that he analyzes the latter in Section 7 as well). The point is that what he means by "energy" is not "the total energy delivered to a receiver by the light pulse". It is something you have to integrate over the pulse time to get the total energy delivered. But the pulse time is longer by exactly the same factor, the Doppler factor, as what Einstein is calling the "energy" is reduced. So the two effects cancel out when we are evaluating the total energy delivered.

What I see shown is that the energy per cycle decreases by the same factor that the period increases (in the case of a redshift). That is, E' = D*E, per wave cycle, $\lambda' = \lambda / D$, $f' = D * f$.

Note that a key feature is that the increase in energy per cycle is due to transform of wave amplitude, a purely classical quantity related to energy. In this case, it transforms such the energy changes by the Doppler factor. However, in general, classically, the amplitude is independent of the frequency, so there is no hidden relation between energy and frequency.

 

[Ibix]

That is a transversal Doppler effect in the sender's frame. -> blueshift
It is not a transversal Doppler effect in the receiver's frame -> redshift (reason: aberration).

It shows the pulse energy is larger in the frame where the rocket is moving - that's all.

 

[Sagittarius A-Star]

It shows the pulse energy is larger in the frame where the rocket is moving - that's all.

Yes, that is valid for your "straightforward case" in #39.

It is not valid for what the planet receives in the OP. If the rocket would send-out the laser pulse in transversal direction in it's rest-frame, then it would not hit the planet.

 

[Ibix]

Yes, that is valid for your "straightforward case" in #39.

It is not valid for what the planet receives in the OP. If the rocket would send-out the laser pulse in transversal direction in it's rest-frame, then it would not hit the planet.

Doesn't make a difference - the energy of the rocket still reduces by $\gamma E$ in the planet frame if it reduces by $E$ in the rocket frame. Where does the $(\gamma-1)E$ go if it's not absorbed by the planet? Or are you contending that the rocket energy does not reduce by $\gamma E$ in the planet frame?

 

[Ibix]

Of course, it's worth noting that the rocket can't stay in orbit without some kind of applied force, and that force must be changing as the laser fires to maintain the circular orbit against the recoil of the laser and the changing rocket mass. There may be complications there that I'm not accounting for.

Got to go to bed - work in the morning. Will think more on the train...

 

[Ibix]

Ok, let's have another go. I take @Sagittarius A-Star's point about the aberration.

Work in the rest frame of the planet, which I will call the primed frame for consistency with previous notation. The rocket is instantaneously moving in the -x direction and emits a very short laser pulse towards the planet. The rocket initially has four momentum $(\gamma m,-\gamma mv,0)$, where I'm suppressing the $z$ direction. The emitted pulse has $(E',0,-E')$ and, conserving four momentum, the rocket finally has $(\gamma m-E',-\gamma mv, E')$.

I'm going to pause here. The rocket is no longer in the same orbit, so this little instantaneous laser shot isn't correct for the original experiment. What does the rocket have to do to maintain its speed and direction? Easy solution: transform to the frame where the rocket was initially at rest. In that frame, the rocket has final four momentum $(m-\gamma E',-\gamma vE',E')$.

I'm out of time. But the rocket is going to have to do something complicated. If it's tethered to something (e.g. an identical rocket and laser 180° away in the same orbit) the tether will no longer be straight. Or it's going to have to fire rockets to hold orbit. Either way my initial calculation was naive.

 

[Ibix]

And, critically, I think not all of the energy in the laser is going to come from the battery once we start monkeying around keeping the circular orbit.

Now I really must put the interesting stuff down and do some work...

 

[Sagittarius A-Star]

A more straightforward case to consider is a rocket in inertial motion in the $+x$ direction. It fires two identical things in the $\pm y$ direction with equal and opposite velocity. One can easily write down the four momentum of the rocket before and after the launch and conserve four momentum, both when the "things" are bullets or laser pulses.
...

Let me continue the story of posting #39 in a slightly different way: transversal Doppler shift in the receiver's unprimed frame.

In the unprimed frame (suppressing the $z$ dimension) the rocket's four momentum is initially $(\gamma_1 m,\gamma_1 mv_1,0)$. After launch it is $(\gamma_2(m-2\delta m),\gamma_2 (m-2\delta m)v_2,0)$.

It follows from momentum-conservation in $x$-direction (the rocket becomes faster because of reduced rest-mass):
$\gamma_2 (m-2\delta m)v_2 = \gamma_1 mv_1$
$\Rightarrow m(\gamma_2v_2 - \gamma_1v_1) = 2\gamma_2v_2 \delta m$
$\Rightarrow m = (2\gamma_2v_2 \delta m) / (\gamma_2v_2 - \gamma_1v_1) \ \ \ \ \ (1)$

The laser pulses have each an energy of
$E={1 \over 2}(\gamma_1 m - \gamma_2(m-2\delta m))= \gamma_2\delta m - {m \over 2}(\gamma_2 - \gamma_1)$

Together with equation (1) follows:
$E = \gamma_2\delta m - (\gamma_2 - \gamma_1) \cdot(\gamma_2v_2 \delta m) / (\gamma_2v_2 - \gamma_1v_1)$

$E = \delta m (\gamma_2(\gamma_2v_2 - \gamma_1v_1) - (\gamma_2 - \gamma_1) \cdot(\gamma_2v_2 )) / (\gamma_2v_2 - \gamma_1v_1)$

$E = \delta m (\gamma_2(- \gamma_1v_1) - ( - \gamma_1) \cdot(\gamma_2v_2 )) / (\gamma_2v_2 - \gamma_1v_1)$

$E = \delta m \gamma_2 \gamma_1( - v_1 +v_2) / (\gamma_2v_2 - \gamma_1v_1)$

$E = \delta m { v_2 - v_1 \over v_2/\gamma_1 - v_1/\gamma_2}$

Setting $\Delta v := v_2 - v_1$ :
$E = \delta m \Delta v / ((v_1 + \Delta v)\sqrt{1-v_1^2} - v_1\sqrt{1-(v_1+\Delta v)^2})$

Limit for the case of $\delta m$ is neglectable small compared to $m$, that means the rocket approaches to not become faster after launch:$$lim_{\Delta v \rightarrow 0} E = {1 \over \gamma_1} \delta m$$(=Redshift)

Calculation:
https://www.wolframalpha.com/input?i2d=true&i=Limit[Divide[u,\(40)v+u\(41)Sqrt[1-Power[v,2]]-vSqrt[1-Power[\(40)v+u\(41),2]]],u->0]

Note the conservation of energy and momentum. Again, energy and momentum are conserved. But there is less energy in the transformed pulses than there was in the original frame.

 

[Ibix]

Ok, let's have another go. I take @Sagittarius A-Star's point about the aberration.

Work in the rest frame of the planet, which I will call the primed frame for consistency with previous notation. The rocket is instantaneously moving in the -x direction and emits a very short laser pulse towards the planet. The rocket initially has four momentum $(\gamma m,-\gamma mv,0)$, where I'm suppressing the $z$ direction. The emitted pulse has $(E',0,-E')$ and, conserving four momentum, the rocket finally has $(\gamma m-E',-\gamma mv, E')$.

I'm going to pause here. The rocket is no longer in the same orbit, so this little instantaneous laser shot isn't correct for the original experiment. What does the rocket have to do to maintain its speed and direction? Easy solution: transform to the frame where the rocket was initially at rest. In that frame, the rocket has final four momentum $(m-\gamma E',-\gamma vE',E')$.

I'm out of time. But the rocket is going to have to do something complicated. If it's tethered to something (e.g. an identical rocket and laser 180° away in the same orbit) the tether will no longer be straight. Or it's going to have to fire rockets to hold orbit. Either way my initial calculation was naive.

To continue...

Assuming the rocket isn't tethered, it needs to fire something momentum $(p^t,\gamma vE',-E')$ (where $p^t$ is an unknown) to remain at rest in its initial frame. The energetically cheapest option is another laser, in which case $(p^t)^2=E'^2(1+v^2\gamma^2)=(\gamma E')^2$.

So the rocket needs to fire two laser pulses with, in its frame, energy $\gamma E$ on top of whatever centripetal force is present. The rocket's final momentum in its frame is $(m-2\gamma E',0,0)$ and the two laser pulses have $(\gamma E',\pm\gamma vE',\mp E')$. Transforming back to the planet frame we have the rocket's final four momentum as $(\gamma m-2\gamma^2E',-\gamma mv+2\gamma^2vE',0)$ and the laser pulses having $(E',0,-E')$ and $((2\gamma^2-1)E',-2\gamma^2vE',E')$.

So there's where I went wrong. In order for the rocket to stay in a circular orbit it needs to fire a laser pulse outwards as well. When you include this pulse, the rocket's energy decreases by more in the planet frame than in its rest frame, which is consistent with my original total energy argument. However, most of the energy is carried away outwards in the exhaust plume, and very little towards the planet. That's how @Sagittarius A-Star's point that the pulse is redshifted is consistent with the the energy argument I was making.

 

[davidjoe]

With respect to laser pulses, the effect of red shift, and of the satellite’s pulses emitting toward the fixed receiver from its changing coordinates in orbit, as it moves in orbit at a significant portion of C does explain the distortion and impossibility of a contemporaneous, normal conversation through radio waves, which are experiencing a similar effect to the laser’s pulses.

If the same satellite in orbit was tethered to the receiver tower by a taut, copper cable and the mode of conversation was not radio or light wave transmission, but carried by electrical impulse, one aspect of the original hypothetical that changes is that the satellite’s transmission signal is now confined and does not spread out along the curved path of the satellite, as it travels to the receiver. It’s a straight line transmission.

Does this confinement of electrical current to the wire, which on one end is moving with the satellite at relativistic speed, but on the opposite end is comparatively still, change anything in the analyses of the conversation’s distortion?

The cable if examined in isolation, and assuming it does not experience dilation or aging at different rates along its length, in a sense really does not “know” that it is moving, when only it is examined, again in complete isolation. The given is that a hard limit on the transmission of information, the speed of causality, is C, of course.

 

[PeroK]

The cable if examined in isolation and it does not experience dilation or aging at different rates along its length, in a sense really does not “know” that it is moving, if only it is examined in complete isolation.

The cable will definitely know its moving. In fact, no cable could realistically sustain the centripetal force required to keep an object moving in a circular path at relativistic speed. Even before you get to relativistic speeds, the tension required would exceed what is physically possible.

 

[davidjoe]

The cable will definitely know its moving. In fact, no cable could realistically sustain the centripetal force required to keep an object moving in a circular path at relativistic speed. Even before you get to relativistic speeds, the tension required would exceed what is physically possible.

That is true, but, … theoretically, for relativistic discussion. I’ve seen the spinning puppet in space where all surrounding objects are removed one by one, whose arms do rise. It’s hard to contradict that. But, I don’t know about electricity being impeded by centripetal force…

 

[Sagittarius A-Star]

Compared to a stationary laser equally distant but still, with an identical 10 minute battery, if the planet receives more energy, and/or a longer duration of impulse than the battery is rated to be able to produce, Is this problematic?

For the following answer I assume, that the usable electric energy in the battery is neglectable small compared the the rest-energy of the rocket, so that the laser emission does not significantly change the velocity of the rocket.

The planet receives the laser energy for a duration, that is $\gamma$ times longer than 10 minutes.
But the received power is by the same factor smaller. So the total received energy is not affected by the motion of the battery.

Edit:
But the received energy is by the same factor smaller. So the total received energy is smaller.

Do the words themselves from the ship stretch out into held note long tones with long gaps between them?

Yes.

 

[Ibix]

It’s a straight line transmission.

The wire will be curved, in fact. But time dilation is time dilation and it doesn't care whether you're signalling in a wire or in free space. If the ship signals constantly for one minute while it has a time dilation factor of ten relative to the planet frame, the planet frame will see the transmission taking ten minutes, and it will take ten minutes to receive.

 

[Ibix]

Yes.

Broadly I agree, but I would just repeat what I said before, that what you hear depends a lot on your transmission mode and how resilient your receivers are to the signal timing. Assuming a simple analogue signal it would just sound like playing a cassette tape slow, as you say, although you'd have to retune the receiver because the carrier wave would also redshifted by a factor of ten. Digital could sound quite different depending on how the electronics deals with the slow bit rate and slow on/off transitions, even after you've allowed for the carrier wave frequency change.

 

[davidjoe]

The wire will be curved, in fact. But time dilation is time dilation and it doesn't care whether you're signalling in a wire or in free space. If the ship signals constantly for one minute while it has a time dilation factor of ten relative to the planet frame, the planet frame will see the transmission taking ten minutes, and it will take ten minutes to receive.

I have heard explanations that photons do not “age” or experience time, and why that is, such that a 14B year old photon captured by Hubble “thinks” it just left and further we have probably all seen them slow to either capturable in slow motion or visible-in-real time, in certain mediums. The “speed of light” is extremely slow traveling through those mediums.

I am starting to get that the disparity is - as measured, at two independent locations, but in trying to test this little bit, how that works out, exactly with the complication of a link between them, seems like very fertile subject matter ground.

If I was looking down at that wire from above, similar to how we would look down at the radiation waves, emitting, traversing, and being received, I’m expecting that I’d see, if I could see charge flowing, that it would appear to traverse the wire at a constant rate. Probably true in both directions. Probably similar. I don’t know that this is correct though.

But if I looked down at a train approaching a station and were able to see its horn blast’s waves, I could see them bunch up into what amounts to higher frequency at the station due to the Doppler effect. It makes sense in the space between earth and a satellite. Logic suggests it must happen regardless of medium, so then it must occur in the wire as well as Ibix notes. The problem is the wire is fixed length and unlike the train and station, the source and receiver’s distance remains the same.

It’s harder to envision for me, as a result. It’s like envisioning a garden hose of fixed volume, where water is forced in at a very high rate and pressure, but comes out the other end, extremely slowly, over a very long time, as if it was compressed and with more being stored in the hose than its dimensional capacity permits.

 

[PeroK]

There's nothing like a bit of home-spun physics every now and then!

 

[davidjoe]

There's nothing like a bit of home-spun physics every now and then!

I have read somewhere that the frames must be independent. It was left as a just because, proposition. The lawyer in me is curious if that was because picturing a link, leads to questions, with answers, that lead to more questions, without answers.

I’m a lawyer, though a “but why”, kind of guy, too.

 

[Ibix]

I have heard explanations that photons do not “age” or experience time,

For one, don't talk about photons. They're incredibly subtle things and you will almost inevitably mislead yourself trying to think about them without a solid grounding in the maths of quantum field theory. There's a reason I've been talking about light pulses in this thread - it's all classical physics.

Second, proper time isn't defined along null worldlines, but light can clearly change with time, losing or gaining intensity or reflecting off things. It's overly simplistic to say it doesn't age.

The “speed of light” is extremely slow traveling through those mediums.

There's an important distinction between the constant $c$ and the speed at which light travels. The former is the same everywhere; the latter can change and is only equal to $c$ in vacuum. It's the constant $c$ that is relevant to relativity.

I’m expecting that I’d see, if I could see charge flowing, that it would appear to traverse the wire at a constant rate.

Almost certainly not. One end of the wire is moving at near $c$ in a circle, so the signal can make very little inward progress. Near the center the wire is practically stationary and the inward progress of electrical signals can be as high as normal.

Note that electrical signals are not the same as flows of charge. Electrons in wires move at far less than walking pace, but the light comes on near instantly when you flip the switch.

It’s harder to envision for me, as a result. It’s like envisioning a garden hose of fixed volume, where water is forced in at a very high rate and pressure, but comes out the other end, extremely slowly, over a very long time, as if it was compressed and with more being stored in the hose than its dimensional capacity permits.

Analogy is a bad way to do physics, which is why we use maths. Frankly, you're going to struggle to understand physics at the level you are asking about unless you learn the maths.

If you look back over this thread you will see that I didn't do the maths and got the wrong answer. When I did do the maths I worked out where I'd gone wrong. Even with my experience in relativity theory (I can pass for an expert in a bad light) my intuition misled me - as yours will. And you can't fall back on the maths as I did. You don't really need anything tougher than Pythagoras' theorem, although some calculus helps a lot.