Can magnetic field expand faster than light?

[Tominator]

Anyway, your thing about electomagnets, the problem with your analysis is that when you look at electro magnets strength you look at stable fields. Stable fields never give out any momentum, its just oscillating fields. Oscillating magnet fields becomes electro magnetic radiation in the same way oscillating electric fields creates electro magnetic radiation, they are the same phenomenon.

Well, I meant to turn on electromagnet, which will push off permanent magnet, so both of them will gain momentum. Part of the energy in form of the current will be consumed. If we use the same energy to produce a light beam, to give us some momentum, the momentum gained will be far smaller than the one gained by electromagnet, using the same energy. This does not make sense to me, because you are telling me that light is the most efficient thing to throw off. So how is that possible?

Yes, same with the conservation of energy.

You cannot run around either conservation law. All you can do is understand them and use them to accomplish your design goal. TANSTAAFL

Does it mean that conservation of momentum would not work in not closed system? Is it enough to put energy to the system, to make it not closed?

 

[Dale]

Does it mean that conservation of momentum would not work in not closed system?

Yes.

Is it enough to put energy to the system, to make it not closed?

Any external force makes the system non-isolated. It does not have to "put energy". For example, the energy to accelerate an automobile comes from the gasoline, not the road, but it is the force of the road that makes the car a non-isolated system whose momentum can change. If the car were in space its momentum could not change regardless of how much gasoline its engine used.

 

[Klockan3]

Well, I meant to turn on electromagnet, which will push off permanent magnet, so both of them will gain momentum. Part of the energy in form of the current will be consumed. If we use the same energy to produce a light beam, to give us some momentum, the momentum gained will be far smaller than the one gained by electromagnet, using the same energy. This does not make sense to me, because you are telling me that light is the most efficient thing to throw off. So how is that possible?

You forget to include the energy contained in the permanent magnet, since all mass you shoot away basically means you wasted E=mc^2 energy.

You can think like this for simplicity's sake, there are 2 kinds of particles in the world; Particles which move and particles which stands still. Both are made up of energy. Photons are made up of just particles that move, and resting mass are made up of just particles which stand still, while moving mass are a mix between moving particles and particles which stand still.

Now these particles can interact with each other, irl its a lot more complex but you can say that if you add two particles moving in the opposite direction together you can get an immobile particle, in the same way you can split an immobile particle into two moving which moves in opposite directions.

Anyway, everything you shoot away from your spaceship means that you must split immobile particles into moving, giving you one which moves in positive direction with the ship, this one you want to keep, then you also gets one moving in the other direction, this is the one we leave behind. Now, pure light is to just send that moving particle away, anything else you are also dumping a lot of immobile particles with that moving one and thus wasting a ton of extra energy.

Due to this spaceship fuel should ideally go in an as fast velocity as possible since then you use the least amount of immobile per mobile, nothing else really matters since energy is energy and no type of transforming energy is more powerful than anything else.

And yes a closed system is not closed if you add energy, since energy and momentum are really the same thing.

 

[Tominator]

Thanks for your explanations guys. I have not ever taught about it this way.

Any external force makes the system non-isolated. It does not have to "put energy". For example, the energy to accelerate an automobile comes from the gasoline, not the road, but it is the force of the road that makes the car a non-isolated system whose momentum can change. If the car were in space its momentum could not change regardless of how much gasoline its engine used.

So generally, momentum is always conserved. But what if we use, what Klockan3 has said

And yes a closed system is not closed if you add energy, since energy and momentum are really the same thing.

Lets consider a system with moving magnet and stationary coil. If light-weight magnet approaches heavy coil, a current will be induced in the coil. Part of the momentum of the magnet will be given to those charged particles. Current can be considered a form of energy as well as moving charged particles. So if we charge an acumulator by these charged particles, law of conservation of energy will be conserved. On the other hand momentum of the system will not be conserved.
How much of the momentum can be transformed to electric energy this way?
If we put current to a coil, to push off a magnet, would the coil gain the same momentum as the magnet?

 

[Dale]

On the other hand momentum of the system will not be conserved.

The momentum of the system (including the fields) will be conserved.

 

[Tominator]

The momentum of the system (including the fields) will be conserved.

Since, according to Klockan3, the energy and momentum is the same thing, wouldn't it be direct voiolation of conservation of energy, if we could draw off energy from the system, and also conserve momentum?
On the other hand, if we want to use it for space propulsion purposes, it does not metter, the only thing that matters is that the momentum of magnet and coil (not considering fields) would not conserve.
How big could the change in momentum of the coil and magnet system be (not considering the fields)?

 

[Dale]

Did you read the link on 4-momentum that I posted earlier?

 

[Tominator]

Did you read the link on 4-momentum that I posted earlier?

Well, I have clicked on the link, but those patterns scared me off. After your latest post, I have tried to read it again, but I did not understand it well.
From what I understood, the example was about subatomic particles, and it was mentioned there that "it is useful in relativistic calculations", so I do not see a connection to the case described previously. Coil or magnet can not gain the invariant mass, because they can not be considered particles. Or does it apply also on objects which are moving nowhere near the speed of light?
On the other hand, if the accumulator is charged by induced current, the system will become not closed, meaning that momentum does not have to be conserved.

 

[Dale]

Momentum is conserved for both subatomic particles and larger systems. It applies at all velocities. Anyway, the point is that, as Klockan3 and I have said energy is momentum in the time direction. Momentum is a vector quantity, so its conservation means that each component is individually conserved.

Here is a brief example for your scenario. Let's say that your vehicle (including fuel and payload) masses 10 metric tons (10000 kg). At rest, that corresponds to a four-momentum of (2998,0,0,0) GNs. If you convert 10 kg of fuel completely to light and throw it in the -x direction then that is a four-momentum of (3,-3,0,0) GNs. So by conservation of the four-momentum we know that the four-momentum of the remaining vehicle fuel and payload is (2995,3,0,0) GNs, which is a speed of 300 km/s.

 

[Tominator]

Here is a brief example for your scenario. Let's say that your vehicle (including fuel and payload) masses 10 metric tons (10000 kg). At rest, that corresponds to a four-momentum of (2998,0,0,0) GNs. If you convert 10 kg of fuel completely to light and throw it in the -x direction then that is a four-momentum of (3,-3,0,0) GNs. So by conservation of the four-momentum we know that the four-momentum of the remaining vehicle fuel and payload is (2995,3,0,0) GNs, which is a speed of 300 km/s.

Wow, that is a speed. The only bad thing about it is that our technology is not there yet.

Momentum is conserved for both subatomic particles and larger systems. It applies at all velocities. Anyway, the point is that, as Klockan3 and I have said energy is momentum in the time direction. Momentum is a vector quantity, so its conservation means that each component is individually conserved.

That is far more better explanation, than on wikipedia. But I still do not understand the post nr.75 :

The momentum of the system (including the fields) will be conserved.

Did you mean the EM wave, generated when EM field is formed arround the coil, and the electrical field of condensator=accumulator?
But then, if I understand the 4-momentum correctly, the change in mechanical momentum of the system would have to be equal to the momentum of the EM wave.

 

[Dale]

Did you mean the EM wave, generated when EM field is formed arround the coil, and the electrical field of condensator=accumulator?
But then, if I understand the 4-momentum correctly, the change in mechanical momentum of the system would have to be equal to the momentum of the EM wave.

The change of the momentum of the vehicle will be equal and opposite to the momentum of the EM wave so that the total momentum of the system (vehicle + EM) is conserved.

 

[Tominator]

The change of the momentum of the vehicle will be equal and opposite to the momentum of the EM wave so that the total momentum of the system (vehicle + EM) is conserved.

You have already told me that. This time I was not asking the same question again. The post nr. 75 of yours, which I did not understand well, referred to the post nr.74.
In this post I asked about momentum of a system composed of a coil and magnet, in which magnet is approaching the coil.

 

[Phrak]

The radius of the Earth is 46 Hz, but who's counting.

 

[Dale]

You have already told me that. This time I was not asking the same question again. The post nr. 75 of yours, which I did not understand well, referred to the post nr.74.
In this post I asked about momentum of a system composed of a coil and magnet, in which magnet is approaching the coil.

OK Tominator, get a blackboard and write 100 times: "Momentum is conserved for any isolated system". The details of the system are irrelevant.

 

[Tominator]

OK Tominator, get a blackboard and write 100 times: "Momentum is conserved for any isolated system". The details of the system are irrelevant.

I have already understood that. I have not said it is not conserved, only in the post 74 but that was because I considered the system not closed. But when you introduced me the
4-momentum, I realized it should be conserved, even thoug I previously taught it is not closed system. But I was not sure, so I wrote this:

Did you mean the EM wave, generated when EM field is formed arround the coil, and the electrical field of condensator=accumulator?
But then, if I understand the 4-momentum correctly, the change in mechanical momentum of the system would have to be equal to the momentum of the EM wave.

I was only asking how is the momentum conserved, not wheather it is conserved or not.
On the other hand it still seems to me weird that if I turn the momentum in closed system to energy, the momentum will stay conserved.

 

[Dale]

OK, I think I understand the question now. You don't "turn momentum to energy". Energy already is momentum in the time direction. So it is not a question of converting one into the other at all. Remember, the conservation of a vector quantity means that each component is independently conserved.

If I still missed the question perhaps you can rephrase.

 

[Tominator]

You have cleared that one up for me, thanks.
Although, there are some cases, where I really do not know, how is the momentum conserved there. For example:
There is a ship and atom. The ship exerts a strong force, in tiniest fraction of a second, on the atom. Assuming, the atom is able to accelerate that fast, it will be as if the ship bounced of a wall, wouldn't it? Because the atom can not go faster than light, there is a light speed "wall".
The Faster the bounc off, the stronger the inertial effect. So at some point, even if I bounce off a ball, it will be as if I bounced off a wall.
Well, momentum is always conserved, so I am not certain wheather are my assumptions with the wall and ball correct.

 

[Dale]

Let's say that the ship masses 10,000 kg and is traveling at 300 km/s, so it has a momentum of 3E6 kg km/s. Let's say the atom is a really heavy one like uranium, it masses 4E-25 kg, and let's say it is traveling at -300 km/s, so it has a momentum of -1E-22 kg km/s. By conservation of momentum after they collide the ship's momentum is still 3E6 kg km/s (to 28 digits of precision) so it is still traveling at 300 km/s.

So conservation of momentum does not imply that it is "as if the ship bounced off a wall".

 

[Tominator]

Thanks for answer, on the other hand we probably would not need to care about the momentum of the sip, because there will be no ship left.
But what if the ship and atom do not colide but the ship applies force on the atom to push it off and thus gain momentum. Since the weight of atom is small, the change in its momentum is enormous, even with several N of force. If I am correct, than energy used for longer period of time to create lower force, is equal to energy used to create massive force in a fraction of a second.
The Faster the bounce off, the stronger the inertial effect. So at some point, even if I bounce off a ball, it will be as if I bounced off a wall.
If big force is applied on the atom in a split of a second, the change in momentum of the atom (p=mv) would be tremendous and it would happen in a split of a second. The atom can not accelerate beyond the speed of light, so for the ship, it will be as if it bounced of a wall. Correct?

 

[Dale]

I still don't get your "wall" comments. Perhaps you can work out a concrete example.

Btw, relativistic momentum is a non-linear function of velocity: p=mv/sqrt(1-v²/c²) which increases without bound as the velocity approaches c.

 

[Tominator]

I still don't get your "wall" comments. Perhaps you can work out a concrete example.

Btw, relativistic momentum is a non-linear function of velocity: p=mv/sqrt(1-v²/c²) which increases without bound as the velocity approaches c.

Yes, I was totaly wrong with the "wall" comments, cos I did not realize that I can not bounce of, or push off something traveling almost at the speed of light in the oposite direction. On the other hand, isn't it more efficient to push off something in fraction of a second rather than slowly?
Recently, I have done some calculations. I wanted to know, how heavy could the ship, powered by 150MW nuclear reactor be, to be able to lift of and accelerate at 15m/s.sqrt (at 5m/s.sqrt, if we count the gravitational acceleration). I found out, that even with engine efficiency at 50%, the ship can weight around 670tuns. Am I correct?
If yes, why arent we using it, the ion engine shoul be in principal able to do the job.

 

[Dale]

Hi Tominator,

You may want to read the https://www.physicsforums.com/showthread.php?t=199087" by D_H, one of our resident rocket scientists.

Acceleration of a rocket is not just a function of engine power and vehicle mass, it also depends on the exhaust velocity and exhaust mass flow rate. Specifically, if you look at the last two equations on D_H's page you can combine them to determine the required exhaust velocity and the mass flow rate of the exhaust given a specific power, acceleration, and vehicle mass.

For your scenario, with a 75 MW power, 15 m/s² acceleration, and a 670 tonne vehicle, you find that the required exhaust velocity is 15 m/s and the required mass flow rate is 673 tonne/s. That exhaust velocity is unrealistically low and the mass flow rate is unrealistically high (the rocket burns all its fuel in less than 1s).

For a more realistic exhaust velocity of 4000 m/s, with the same 75 MW power plant, and the same 15 m/s² acceleration, your maximum vehicle mass is 2.5 tonne and your ehxaust mass flow rate is 9 kg/s. Even that mass flow rate will use up the entire load of fuel in less than 5 minutes.

 

[Tominator]

Hi DaleSpam,

Thanks for the link, but I did not mean any particular type of engine, such as rocket engine that is why I wrote ship. The nuclear reactor's output is elecricity, so there would be an engine running on electricity required. Rockets are very inefficient, because the exhaust gases are heated, thus major part of the energy is consumed by heating the gases.

I counted the possible weight of the ship with 150MW reactor from these equations: W=F.s, P=W/t and s=a.tsqrt/2 (15m/s.sqrt is the acceleration of the ship)
from these, considering time=1s, you got m=2W/a.sqrt

You might say, that the ship needs to exhaust something to gain momentum but till it is in the atmosphere, it can use the air around to repell off, above the atmosphere it might use ion engines.

The exhaust velocity of ion engine is around 40 000m/s in vacum. The energy here is used to push off the ions, so it should be much more efficient than conventional rocket engines.

 

[Dale]

So using an exhaust velocity of 40 km/s, a power of 75 MW, and an acceleration of 15 m/s² we get that the vehicle mass is 250 kg and that it consumes fuel at a rate of 94 g/s.

So, the point is that an engine with a higher exhaust velocity provides less thrust for a given power than an engine with a lower exhaust velocity. It is more efficient in terms of delta-v/kg of fuel (the exhausted fuel has more momentum per kg of fuel), but produces less thrust in terms of N/W (more of the energy is in the exhaust).

 

[Tominator]

That is strange, because I have heard about concept called NERVA, which is actually nuclear powered rocket. It never flew, because of the radioactive pollution it would cause. But it was certainly, according to what I know, able to fly. But the amount of fuel was almost like with normal rocket.
On the other hand, I still do not know, why are those equations, I have used, incorrect. I have read the link, you have sent me, but I still can not figure out, why am I getting different answer by the euation, I have mentioned in previous post.
If one object applies force on another object the reaction has to be equivalent, no matter, what the weight ratio is between the two objects. If the force is the same, the momentum has to be the same too. So how can the possible weight of the ship depend on the weight of the fuel exhausted per second?

 

[Dale]

I still do not know, why are those equations, I have used, incorrect. I have read the link, you have sent me, but I still can not figure out, why am I getting different answer by the euation, I have mentioned in previous post.

The first clue that something is wrong in that equation is that you just randomly specified t=1s. That should clue you in that your result depends on time and is only valid at one time point.

If one object applies force on another object the reaction has to be equivalent, no matter, what the weight ratio is between the two objects. If the force is the same, the momentum has to be the same too.

Yes, that is correct. That is what Newton's 3rd law says.

So how can the possible weight of the ship depend on the weight of the fuel exhausted per second?

Because you specified a power, not a force. For a given fixed power the thrust force is a function of the exhaust velocity. Once you have fixed the power and the exhaust velocity then the exhaust mass flow rate and the thrust are uniquely determined by conservation of momentum (derived by D_H in the tutorial). Then if you further fix the acceleration that uniquely determines the vehicle mass by Newton's 2nd law.

 

[Dale]

Hi Tominator,

I hope my previous post was not too terse. Anyway, I wanted to explain a little more. I really recommend that you read the rocket tutorial in depth and ask me any specific questions that you have.

I mentioned that your formula was only valid at one time point:
P = W/t = (F.s)/t = ((m a).(1/2 a t²))/t = ma²t/2

What this means is that if you apply a constant force on an object (starting at rest at the origin) then the power increases linearly with time.

 

[Tominator]

Hi DaleSpam,

Thanks for your answers and recomendations, you wrote it clearly, but it took me some time to process it.

Because you specified a power, not a force. For a given fixed power the thrust force is a function of the exhaust velocity. Once you have fixed the power and the exhaust velocity then the exhaust mass flow rate and the thrust are uniquely determined by conservation of momentum (derived by D_H in the tutorial). Then if you further fix the acceleration that uniquely determines the vehicle mass by Newton's 2nd law.

Well, I can not argue about that. From what I have read, the equation is for a rocket burning its fuel till it reaches desired change in velocity. But there are work-arounds, for example by using more stages the rocket will be more efficient.

I mentioned that your formula was only valid at one time point:
P = W/t = (F.s)/t = ((m a).(1/2 a t²))/t = ma²t/2

What this means is that if you apply a constant force on an object (starting at rest at the origin) then the power increases linearly with time.

If the power is fixed, will the force decrease with time?
But what if we use the energy in pulses? Can the efficiency be enhanced by pulses? (Thanks to the inertia of the exhausted fuel)
What I mean is a kind of a pulse drive
With the ion engine, it is possible, (or in recent futurre it will be possible), to generate thrust by pulses. By using more ion engines, the pulses can be synchronized like in a combustion engine.

 

[Dale]

Well, I can not argue about that. From what I have read, the equation is for a rocket burning its fuel till it reaches desired change in velocity. But there are work-arounds, for example by using more stages the rocket will be more efficient.

Yes. In that case the equations on the tutorial page apply to each stage individually. Reducing the mass of the vehicle at each stage helps with efficiency.

If the power is fixed, will the force decrease with time?

Yes.

But what if we use the energy in pulses? Can the efficiency be enhanced by pulses? (Thanks to the inertia of the exhausted fuel)
What I mean is a kind of a pulse drive
With the ion engine, it is possible, (or in recent futurre it will be possible), to generate thrust by pulses. By using more ion engines, the pulses can be synchronized like in a combustion engine.

You can certainly use the energy in pulses, and in fact it is generally optimal to do so. However, the substitution s = 1/2 a t² is only valid for uniform acceleration starting from the origin, so your equation would not apply. You would need to use the more general equations on the tutorial page.

 

[Tominator]

Hm, I was really surprised, why we do not use nuclear power to get a ship to orbit, now I understand.
Anyway, recently I have got an idea: if it is possible to generate plasma, by using high frequency EM field, it should be possible to generate plasma also by using high frequency one-way current pulses. Would this have a propulsion effect(in the atmosphere)?

 

[Dale]

Yes, this is the principle used by the ill-fated sharper image "ionic breeze". You can use that principle to build a little hovercraft called an http://en.wikipedia.org/wiki/Ionocraft" . The thrust that these generate is currently not enough to lift the high voltage power supply that they need, but I think that is probably a practical limitation rather than a fundamental one. They make good little science fair projects.

 

[Tominator]

I have heard about the Ionocrafts, but I assume, they do not use plasma. I was thinking about creating plasma by the EM field and thus making a kind of EM jet engine. Because the plasma will be generated only by using electricity and of course the air, there will be no need for fuel while in the atmosphere. On the other hand, this concept would be even more inefficient than rockets.

Is there an ionizator used in the ionocrafts?

I thought, that if ionized gas runnig around in the nozzle in the direction from the air-intake to the part, where air is exhausted was used there, than the ionized gas can make the air accelerate. The ionized gas would not be exhausted, it would only circle around grabbing some air with and accelerating it in the right direction. Would this be more efficient than today's ionocrafts?

 

[Dale]

Well, the word "plasma" isn't well defined to my knowledge. A plasma is a partially ionized gas, and an ionocraft does partially ionize the air, but the degree of ionization is so low that it doesn't have much of the "collective" behavior typical of more completely ionized plasmas.

I don't know about the efficiency of the circulating nozzle idea. You could certainly keep the ions in the nozzle longer and so perhaps get more ionization for the same ionization current, but I am not clear about how you would turn the circulation into thrust.

 

[Tominator]

Well, the word "plasma" isn't well defined to my knowledge. A plasma is a partially ionized gas, and an ionocraft does partially ionize the air, but the degree of ionization is so low that it doesn't have much of the "collective" behavior typical of more completely ionized plasmas.

I don't know about the efficiency of the circulating nozzle idea. You could certainly keep the ions in the nozzle longer and so perhaps get more ionization for the same ionization current, but I am not clear about how you would turn the circulation into thrust.

I was counting with air as a propellant. The air will be sucked in, thanks to circulation of ionized gas, then compressed and accelerated out of the nozzle. (also by the circulation) So no energy would be wasted on creating plasma. It should work like a propeller made out of gas. Would it work?

 

[Dale]

I don't know. I am having a hard time visualizing how the circulating ions would act like a propeller. I am afraid that you will have to work this one out yourself, but I can also tell you that plasma mechanics is an incredibly advanced subject. To really learn it will literally take years of serious effort.