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#1120 by Tellmeagain on 29 Sep, 2016 21:55
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Ok, I understand the difference. You're saying it takes an increasing power (electric) to maintain a constant force. If I'm accelerating an object with a rail gun that would seem to be the case but I'm providing a force in a fixed frame. But there are ways of applying a fixed force at a fixed power in an accelerating frame. Photon beam propulsion does provide a fixed force at a fixed power up to relativistic speeds or at least relativistic effects limit things. And rockets certainly do provide a constant thrust for a constant power, for a limited time. So that disproves your asserting that such a constant force at constant power is outside the bounds of classical mechanics. The confusion is because the rocket applies it's thrust in a constantly accelerating reference frame and normal simpler mechanical systems apply a force from a fixed reference frame. That makes all the difference.
...SNIPPED
This understanding is not correct, "rocket applies it's thrust in a constantly accelerating reference frame". Actually rocket applies it's thrust in an inertial reference frame. In the "constantly accelerating reference frame" (I assume you are talking about the rocket itself as a frame) there is no thrust, because there is no acceleration. In an inertial reference frame, the rocket equation can be derived as
F(external force)=Ma+(v(of rocket)-v(of exhaust))dM/dt. This equation is good for any inertial reference frame. For rocket that does not experience external force (air drag, gravity, etc), the equation becomes Ma=(v(of rocket)-v(of exhaust))(-dM/dt). This right side is just the thrust, and it is with any inertial reference frame, and not with the constantly accelerating reference frame. For details you can see my discussion in the pdf file ( Woodward_updated.pdf , in section III) downloadable in this post,
http://forum.nasaspaceflight.com/index.php?topic=40959.msg1589319#msg1589319
Professor Woodward derived the rocket equation from F=dp/dt=d(Mv)=Mdv/dt+vdM/dt, where he implicitly talked about the v in Mdv/dt as relative to an inertial frame but the v in vdM/dt as relative to the rocket. This kind of derivation is incorrect. What dp/dt=Mdv/dt+vdM/dt means physically needs careful analysis. We can not mindlessly apply chain rule of derivative to dp/dt without knowing what it really means. They way I used in my pdf file causes no confusion.
I didn't ask this in my previous comment on your report. But can you point out where Woodward actually does a derivation of the Rocket Equation, because in the paper all he does is state the Rocket Equation as you have done. Then talk about the differences between the force represented by the Ma term which you have rewritten as Mdv/dt and the vdM/dt term. This is specifically done to call out the situation where if the mass change of a vehicle is perpendicular to the motion of that vehicle does it affect the velocity of that vehicle. I don't see any derivation being done here more an interpretation of what the terms in the equations mean physically. with the question about the effect of motion on the vehicle being left to the reader to answer. Which I said seems to have been done because in a MET there is a component whose mass is actively being changed, as the Mach Effect Transient mass fluctuation is triggered. Since this is explained as an interaction with the rest of the mass of the far universe, via a gravitational version of absorber theory. I do not think it a stretch that he also believes the mass change in a MET also occurs perpendicular to the motion of the vehicle a MET is propelling.
The derivation is this part: dp/dt=Mdv/dt+vdM/dt (the first part of his equation (16)). He explicitly stated that this was due to Newton's second law (which is F=dp/dt). How he reached from Newton's second law to the rocket equation is called a derivation. I did not rewritte Ma as Mdv/dt+vdM/dt but he did. See his equation 16). You have not define what is a "mass change of a vehicle is perpendicular to the motion of that vehicle". -
#1121 by spupeng7 on 30 Sep, 2016 00:10
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Wondering now, if it is worth saying this again...
The proportional relationship between energy of motion and the square of its velocity is a relative one. It cannot be otherwise or it would impede our freedom of movement. When an emdrive is tested in space that relationship will have no relevance beyond getting folk over the hump of credulity which it invokes.
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#1122 by meberbs on 30 Sep, 2016 04:13
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E = γmc² - the definition of relativistic energy
Not sure where you found that equation, but the definition of relativistic energy is E2 = (m*c2)2 + (p*c)2, where p = γ*m0*v
In the limit v << c for special relativity you get classical mechanics, so your result won't change in that respect, but the derivative will be more annoying.
Some things can't be saved by any amount of context. It doesn't matter that there are correct ways to get to the conclusion, the statements in the paper are completely wrong, even if the conclusion is right....
...With that said, I think any criticism of the third part of Woodward's Paper first has to explain why the Mach Effect transient mass fluctuation, as described by Woodward, either doesn't exist; or could not have the behavior described.
You are expecting criticism of the paper to answer things not said in the paper. The paper does not make the argument that F = k*P does not apply to the drive, and it does not describe any of the aspects of the Mach effect thruster that make it an open system, nor does it even mention transient mass fluctuations, let alone explain how those could resolve this issue as they are categorically different than normal expelling exhaust.
...
** Special relativity modifies this slightly, allowing the constant k to be up to 1/c due to 0 rest mass particles carrying momentum and energy.
I agree that none of those things are mentioned. However, it is the lack of mentioning those things that I believe has lead to a lot of the criticisms of that said paper. Context is very important.
...However, once you understand the idea Woodward is proposing with Mach Effect Theory. Then I would strongly argue that the first 3/4 of the paper makes sense. the last 1/4 is still a little problematic for me personally but that's because I don't know if the description he provides at the end is enabled by the mass fluctuations taking place in the MET. I suspect that is the case; since the kinetic energy of the internals of the MET device would be very different from the rest of the ship it is propelling.
The last 1/4 is the worst, but that is a close call. Earlier, he goes through all of the correct steps to do a proof by contradiction that if a closed system obeys the figure of merit equation, then CoE is broken. While he is ambiguous about what went wrong, he is clear that he thinks that because you arrived at a contradiction you can't learn anything from that line of logic, and the logic must be broken. As I said, this is a proof by contradiction. His statements demonstrate a lack of understanding of basic logic.
The remaining section about rockets and cars with sand falling out of them is simply irrelevant. The mass leaving the system of those examples is completely different than the Mach effect where it is a mass fluctuation that only produces a propulsive effect due to theories that are not discussed in the paper at all. -
#1123 by meberbs on 30 Sep, 2016 06:01
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Ok, I understand the difference. You're saying it takes an increasing power (electric) to maintain a constant force. If I'm accelerating an object with a rail gun that would seem to be the case but I'm providing a force in a fixed frame. But there are ways of applying a fixed force at a fixed power in an accelerating frame. Photon beam propulsion does provide a fixed force at a fixed power up to relativistic speeds or at least relativistic effects limit things. And rockets certainly do provide a constant thrust for a constant power, for a limited time. So that disproves your asserting that such a constant force at constant power is outside the bounds of classical mechanics. The confusion is because the rocket applies it's thrust in a constantly accelerating reference frame and normal simpler mechanical systems apply a force from a fixed reference frame. That makes all the difference.
First, while Shawyer may have said the EMDrive is an open system, none of his theory supports that statement at all. He claims no new physics needed, and does not describe anything that the device pushes against, or any type of exhaust from the device.
You also make a point about a 'closed' system. Shawyer claims the EmDrive is open, as does Woodward about his device.
Would you agree with this: If an EmDrive works at all, making a constant force in some frame, then it would keep accelerating as long as power were applied. Agree? If not you should at least admit an EmDrive up to k=2/c. But I've already shown photon recycling can break the 2/c limit. Thanks.
I am not sure where you got 2/c, but the limit is 1/c from special relativity. I am guessing you may have been thinking about an external laser being reflected by the ship, but that is not a case of constant force, due to the doppler effect as the device accelerates. Photon recycling is completely different, because then you have another body involved on the other end, and you have to account for that in your calculations.
Rockets do not provide constant thrust for constant power. First, you have to define your system. If you take the whole rocket at the beginning (including fuel) then you will find that center of mass never moves, in that sense force is 0.
For a non-trivial example, Let's instead draw the box only around the payload (here I am including tanks, etc. as part of the payload), and exclude the fuel.
definitions:
mp = mass of payload
mf = initial mass of fuel
me = rate that mass is expelled (take this as a positive number)
ve = velocity that fuel is expelled at (ship frame, again positive number)
Using the rocket equation:
Force on payload as a function of time is: F = mp*ve*me / ( mp + mf - me*t)
While this is an increasing function of time, the power applied to the payload to cause the force is increasing faster:
dEp/dt = v*mp*ve*me / ( mp + mf - me*t)
where v = ve * ln( (mp+mf) / (mp + mf - me*t) )
The ratio of force to power is therefore 1/v, which is a function of time, not a constant.
The reason the increasing power is possible is because while the chemical energy (or electric for ion thrusters) per unit mass expelled remains constant, that expelled mass had gained kinetic energy from the previously expelled mass, some of which is transferred to the payload as the fuel is sent out the back.
I could post the rest of the energy balance equations, but equations are a pain to type and I think I have made my point.
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#1124 by Star One on 30 Sep, 2016 06:40
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As this is an important clarification by them I have posted it in both relevant threads.
CUBESAT MISSION CLARIFICATIONQuoteThere has been a lot of erroneous information in media articles regarding Cannae’s upcoming launch of a cubesat mission into LEO. To clarify our previous post and press release: Cannae is not using an EmDrive thruster in our upcoming launch. Cannae is using it’s own proprietary thruster technology which requires no on-board propellant to generate thrust. In addition, this project is being done as a private venture. Cannae is only working with our private commercial partners on the upcoming mission.
http://cannae.com/cubesat-mission-clarification/ -
#1125 by RERT on 30 Sep, 2016 09:29
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E = γmc² - the definition of relativistic energy
Not sure where you found that equation, but the definition of relativistic energy is E2 = (m*c2)2 + (p*c)2, where p = γ*m0*v
At least some questions I can answer. J.G.Taylor, 'Special Relativity', Clarendon Press, 1975, equation 5.19.
E as defined is the time component of the energy-momentum 4-vector, with the momentum you mention as the space component. As a component of a 4-vector, it is not a Lorentz invariant.
In his equation 5.20, he notes the algebraic identity
(E²/c²)-p²=m²c² where m is the rest mass, manifestly a Lorentz invariant
If you plug in E = γmc² and p=γmv, you will find that this is always true because of the definition of γ.
The other way round, if you plug in your own definition of p and the regular definition of γ, you will find this equation reduces to my definition of E.
If you work through E = γmc² for small velocities, you will find that to first order in (v²/c²), E = mc²+½mv², ie. rest energy plus kinetic energy.
Bottom line is that I'm pretty sure I'm right on this narrow point.
I'm still scratching my head as to why dE/dt always has to be zero in the rest frame. -
#1126 by RERT on 30 Sep, 2016 09:41
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Quote from: Star One
As this is an important clarification by them I have posted it in both relevant threads.
CUBESAT MISSION CLARIFICATION
(snip)
I read this as saying 'We are not testing Shawyer's stuff, we are testing our own version of a resonant cavity thruster'. Do you suspect something else?
(modify to remove typo) -
#1127 by Star One on 30 Sep, 2016 10:22
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Quote from: Star One
As this is an important clarification by them I have posted it in both relevant threads.
CUBESAT MISSION CLARIFICATION
(snip)
I read this as saying 'We are not testing Shawyer's stuff, we are testing our own version of a resonant cavity thruster'. Do you suspect something else?
(modify to remove typo)
It's probably a propriety thing that have to be clear on this. -
#1128 by Notsosureofit on 30 Sep, 2016 13:29
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#1129 by M.LeBel on 30 Sep, 2016 14:35
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Simple comment here.
Whether you consider V < C or V<<C or V <<<< C does not matter. Relativity still holds. There is no set point where relativity does not hold or stops!
The engineering way is to round off the numbers... But you can't round off/away the principle. Whatever effects you expect from relativity still happens at V <<<<<<< C It is just smaller, not gone.
Remember this when trying to figure out how the universe works; does it work according to principles or does it work as we like to round it off?
Marcel,
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#1130 by meberbs on 30 Sep, 2016 15:02
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You are right, I made a stupid mistake when I went through the algebra to check that equation.E = γmc² - the definition of relativistic energy
Not sure where you found that equation, but the definition of relativistic energy is E2 = (m*c2)2 + (p*c)2, where p = γ*m0*v
At least some questions I can answer. J.G.Taylor, 'Special Relativity', Clarendon Press, 1975, equation 5.19.
E as defined is the time component of the energy-momentum 4-vector, with the momentum you mention as the space component. As a component of a 4-vector, it is not a Lorentz invariant.
In his equation 5.20, he notes the algebraic identity
(E²/c²)-p²=m²c² where m is the rest mass, manifestly a Lorentz invariant
If you plug in E = γmc² and p=γmv, you will find that this is always true because of the definition of γ.
The other way round, if you plug in your own definition of p and the regular definition of γ, you will find this equation reduces to my definition of E.
If you work through E = γmc² for small velocities, you will find that to first order in (v²/c²), E = mc²+½mv², ie. rest energy plus kinetic energy.
Bottom line is that I'm pretty sure I'm right on this narrow point.
I'm still scratching my head as to why dE/dt always has to be zero in the rest frame.
The dE/dt is only instantaneously 0. If you allow any finite time to pass, it is non-zero. One way to think about it is you are taking an infinitesimal increase in velocity dv, so you are still in effectively the rest frame. The kinetic energy in the rest frame is always 0 by definition. Also, when considering infinitesimals, the momentum is m*dv, but the energy is 0.5*m*dv^2, and squared infinitesimals are infinitely smaller than infinitely small, so they are 0 as long as there are any first order infinitesimals around.*
* mathematicians would not like my phrasing here, but I am trying to explain this conceptually, so I am skipping the rigor. -
#1131 by Bob012345 on 30 Sep, 2016 16:56
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Ok, I understand the difference. You're saying it takes an increasing power (electric) to maintain a constant force. If I'm accelerating an object with a rail gun that would seem to be the case but I'm providing a force in a fixed frame. But there are ways of applying a fixed force at a fixed power in an accelerating frame. Photon beam propulsion does provide a fixed force at a fixed power up to relativistic speeds or at least relativistic effects limit things. And rockets certainly do provide a constant thrust for a constant power, for a limited time. So that disproves your asserting that such a constant force at constant power is outside the bounds of classical mechanics. The confusion is because the rocket applies it's thrust in a constantly accelerating reference frame and normal simpler mechanical systems apply a force from a fixed reference frame. That makes all the difference.
...SNIPPED
This understanding is not correct, "rocket applies it's thrust in a constantly accelerating reference frame". Actually rocket applies it's thrust in an inertial reference frame. In the "constantly accelerating reference frame" (I assume you are talking about the rocket itself as a frame) there is no thrust, because there is no acceleration. In an inertial reference frame, the rocket equation can be derived as
F(external force)=Ma+(v(of rocket)-v(of exhaust))dM/dt. This equation is good for any inertial reference frame. For rocket that does not experience external force (air drag, gravity, etc), the equation becomes Ma=(v(of rocket)-v(of exhaust))(-dM/dt). This right side is just the thrust, and it is with any inertial reference frame, and not with the constantly accelerating reference frame. For details you can see my discussion in the pdf file ( Woodward_updated.pdf , in section III) downloadable in this post,
http://forum.nasaspaceflight.com/index.php?topic=40959.msg1589319#msg1589319
Ok, but you're just nitpicking. The rocket frame always has a co-moving instantaneous rest frame. -
#1132 by Bob012345 on 30 Sep, 2016 17:05
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Photon beam propulsion does provide a fixed force at a fixed power up to relativistic speeds or at least relativistic effects limit things.
Another limiting factor for a photon rocket is the final mass. A photon rocket converts mass into the energy of photons. Sooner or later it will run out of "fuel" mass, likely well before relativistic effects come into play. The formula can be found here: https://en.wikipedia.org/wiki/Photon_rocket
A photon rocket is anything that releases EM radiation as a means of propulsion. It could be a laser shooting out the back of a rocket. In that case it would provide the same thrust (laser power/c) as long as electrical power were provided. Perhaps you are thinking of an antimatter-matter rocket which is a special case. -
#1133 by wicoe on 30 Sep, 2016 17:13
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A photon rocket is anything that releases EM radiation as a means of propulsion. It could be a laser shooting out the back of a rocket. In that case it would provide the same thrust (laser power/c) as long as electrical power were provided. Perhaps you are thinking of an antimatter-matter rocket which is a special case.
If I understand it correctly, ANY photon rocket has to lose mass since otherwise it would violate CoE. Essentially, with every photon emitted, the system loses a bit of mass (corresponding to the energy of the photon emitted). The exact mechanism is irrelevant. In other words, an autonomously-powered laser also loses mass as it emits photons. -
#1134 by Bob012345 on 30 Sep, 2016 17:37
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Ok, I understand the difference. You're saying it takes an increasing power (electric) to maintain a constant force. If I'm accelerating an object with a rail gun that would seem to be the case but I'm providing a force in a fixed frame. But there are ways of applying a fixed force at a fixed power in an accelerating frame. Photon beam propulsion does provide a fixed force at a fixed power up to relativistic speeds or at least relativistic effects limit things. And rockets certainly do provide a constant thrust for a constant power, for a limited time. So that disproves your asserting that such a constant force at constant power is outside the bounds of classical mechanics. The confusion is because the rocket applies it's thrust in a constantly accelerating reference frame and normal simpler mechanical systems apply a force from a fixed reference frame. That makes all the difference.
First, while Shawyer may have said the EMDrive is an open system, none of his theory supports that statement at all. He claims no new physics needed, and does not describe anything that the device pushes against, or any type of exhaust from the device.
You also make a point about a 'closed' system. Shawyer claims the EmDrive is open, as does Woodward about his device.
Would you agree with this: If an EmDrive works at all, making a constant force in some frame, then it would keep accelerating as long as power were applied. Agree? If not you should at least admit an EmDrive up to k=2/c. But I've already shown photon recycling can break the 2/c limit. Thanks.
I am not sure where you got 2/c, but the limit is 1/c from special relativity. I am guessing you may have been thinking about an external laser being reflected by the ship, but that is not a case of constant force, due to the doppler effect as the device accelerates. Photon recycling is completely different, because then you have another body involved on the other end, and you have to account for that in your calculations.
Rockets do not provide constant thrust for constant power. First, you have to define your system. If you take the whole rocket at the beginning (including fuel) then you will find that center of mass never moves, in that sense force is 0.
For a non-trivial example, Let's instead draw the box only around the payload (here I am including tanks, etc. as part of the payload), and exclude the fuel.
definitions:
mp = mass of payload
mf = initial mass of fuel
me = rate that mass is expelled (take this as a positive number)
ve = velocity that fuel is expelled at (ship frame, again positive number)
Using the rocket equation:
Force on payload as a function of time is: F = mp*ve*me / ( mp + mf - me*t)
While this is an increasing function of time, the power applied to the payload to cause the force is increasing faster:
dEp/dt = v*mp*ve*me / ( mp + mf - me*t)
where v = ve * ln( (mp+mf) / (mp + mf - me*t) )
The ratio of force to power is therefore 1/v, which is a function of time, not a constant.
The reason the increasing power is possible is because while the chemical energy (or electric for ion thrusters) per unit mass expelled remains constant, that expelled mass had gained kinetic energy from the previously expelled mass, some of which is transferred to the payload as the fuel is sent out the back.
I could post the rest of the energy balance equations, but equations are a pain to type and I think I have made my point.
Sorry but a rocket absolutely can provide a constant thrust for constant power, they do it all the time. I thought it was plainly obvious that I am referring to a flight regime where the mass change is trivial compared to the overall mass. Then it can be at or near a constant for a while. I really am not debating about the rocket equation over the whole trip. The main point is to compare it to the EmDrive which purports to have a constant thrust at a constant power. I'm not saying it must or it's always exactly thus but that it is possible. So, for my purposes, yes, the rocket has a constant thrust for a given power over some regime of flight.
Yes, you can do twice as well with an external beam as propulsion than if you have the beam emitting from the rocket because you have twice the momentum change. No, the Doppler shift does happen but it's effect is very small until very high speeds are reached and that is not the main point we are debating. That's a second order issue.
Photon recycling does involve another body but not another energy source. So you can do better than 2/c over the whole trip. If fact if you recycle at an effective momentum of 4 * power/c you run into 'over unity' past c/2.
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#1135 by Bob012345 on 30 Sep, 2016 17:53
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A photon rocket is anything that releases EM radiation as a means of propulsion. It could be a laser shooting out the back of a rocket. In that case it would provide the same thrust (laser power/c) as long as electrical power were provided. Perhaps you are thinking of an antimatter-matter rocket which is a special case.
If I understand it correctly, ANY photon rocket has to lose mass since otherwise it would violate CoE. Essentially, with every photon emitted, the system loses a bit of mass (corresponding to the energy of the photon emitted). The exact mechanism is irrelevant. In other words, an autonomously-powered laser also loses mass as it emits photons.
Yes but it's very small, total energy/c^2 which would be something like a kg over a trip to the nearest star. -
#1136 by Slyver on 30 Sep, 2016 20:18
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The dE/dt is only instantaneously 0. If you allow any finite time to pass, it is non-zero. One way to think about it is you are taking an infinitesimal increase in velocity dv, so you are still in effectively the rest frame. The kinetic energy in the rest frame is always 0 by definition. Also, when considering infinitesimals, the momentum is m*dv, but the energy is 0.5*m*dv^2, and squared infinitesimals are infinitely smaller than infinitely small, so they are 0 as long as there are any first order infinitesimals around.*
* mathematicians would not like my phrasing here, but I am trying to explain this conceptually, so I am skipping the rigor.
My calc and de teacher (he was completely amazing (and ridiculously difficult), so I took him for 3 semesters) just rolled over in his grave, or he would, if he were dead. -
#1137 by spupeng7 on 30 Sep, 2016 21:08
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Grave rolling aside, if the energy I generate by accelerating to 50m/s on my motorcycle is relative to the centre of our galaxy then I would need a megawatt power station to drive it.
Conservation of energy is broken, of course it is. Conservation of momentum is absolute and should remain valid for continuous accelerations in free space, but energy remains relative. JMN..
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#1138 by wicoe on 30 Sep, 2016 21:17
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Grave rolling aside, if the energy I generate by accelerating to 50m/s on my motorcycle is relative to the centre of our galaxy then I would need a megawatt power station to drive it.
Conservation of energy is broken, of course it is. Conservation of momentum is absolute and should remain valid for continuous accelerations in free space, but energy remains relative. JMN..
It's not though - you always get the same increase/decrease in the energy of a system, regardless of the ref frame (must be inertial though). When you accelerate on a motorcycle, you push the earth backwards a little bit. The total energy of the system (motorcycle + earth) increases by exactly the same amount in any inertial ref frame, even one moving at hundreds km/sec relative to us. This amount corresponds to the power you "spent" while accelerating. -
#1139 by spupeng7 on 30 Sep, 2016 21:29
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Grave rolling aside, if the energy I generate by accelerating to 50m/s on my motorcycle is relative to the centre of our galaxy then I would need a megawatt power station to drive it.
Conservation of energy is broken, of course it is. Conservation of momentum is absolute and should remain valid for continuous accelerations in free space, but energy remains relative. JMN..
It's not though - you always get the same increase/decrease in the energy of a system, regardless of the ref frame (must be inertial though). When you accelerate on a motorcycle, you push the earth backwards a little bit. The total energy of the system (motorcycle + earth) increases by exactly the same amount in any inertial ref frame, even one moving at hundreds km/sec relative to us. This amount corresponds to the power you "spent" while accelerating.
Yes, and for the emdrive the system for which the energy is total, must include the rest of the universe. The connection between the craft and the universe may be difficult to visualise but it gets a lot easier if you appreciate it as the mechanism of inertia itself.
Mach inspired Einstein, maybe Einstein could have learnt even more from him if there were an emdrive to consider at that time
