Can magnetic field expand faster than light?

AI Thread Summary
The discussion centers around the possibility of a magnetic field expanding faster than the speed of light, particularly in the context of electromagnetic oscillators and their frequencies. Participants emphasize that changes in magnetic fields propagate at the speed of light, meaning that even if a magnetic field's polarity is altered, distant points in the field will not reflect this change instantaneously. The conversation also touches on the relationship between electric and magnetic fields, noting that a time-varying magnetic field generates an electric field and vice versa. Additionally, the effects of a rotating permanent magnet are discussed, with the conclusion that it does emit electromagnetic waves, which can transfer momentum. Overall, the thread explores the fundamental principles of electromagnetism and the constraints imposed by the speed of light.
  • #51
Tominator said:
But force produced by electromagnets is far more greater, than a force produced by beam of light of lightbulb, or laser. ... the magnetic force here is far superior to force caused by laser or light of a lightbulb.
That is certainly true, but not relevant for propulsion in space. You can generate large forces that will cause a lot of stress in the structure of your device, but the only way to actually propel your device in space is by throwing momentum off the back. All that matters is how much momentum you can toss off and how tightly you can collimate it towards the back. For EM the momentum is given by E/c and so the only remaining question is the focus, which would be far better for a laser than for your electromagnet.
Tominator said:
The question is, wheather the changes in a field at GHz frequencys can be considered not only as EM waves, but also as a propagating fields.
EM waves are the same thing as propagating fields.
 
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  • #52
DaleSpam said:
That is certainly true, but not relevant for propulsion in space. You can generate large forces that will cause a lot of stress in the structure of your device, but the only way to actually propel your device in space is by throwing momentum off the back. All that matters is how much momentum you can toss off and how tightly you can collimate it towards the back. For EM the momentum is given by E/c and so the only remaining question is the focus, which would be far better for a laser than for your electromagnet.
If I set frequencys and distance between the two electromagnets properly, I should be able to get focused force instead of the "lot of stress". Because as you have said:
DaleSpam said:
Your electromagnet system is different from the lightbulb system in the sense that it doesn't use reflection to direct the EM momentum. But they are the same in the sense that the propulsion is due to conservation of momentum between the device and the EM fields.
DaleSpam said:
EM waves are the same thing as propagating fields.
I want the electromagnet to bounce off EM waves, but as long as the EM waves can be considered a propagating field, the electromagnet should gain the same momentum, as if it bounced off the field of the generator itself in the same point. Shouldn't it?
It can even bounce off the field of the generator, but till the change caused by this bouncing off gets to the generator, the generator has to be turned off (for this, GHz frequencys are needed). So there will be no change in momentum of the generator, but there will be significant change in momentum of the electromagnet.
 
  • #53
Tominator, you have a really annoying habit of repeatedly asking the same question over and over and over and somehow thinking that the answer will change.
Tominator said:
If I set frequencys and distance between the two electromagnets properly, I should be able to get focused force instead of the "lot of stress". ... So there will be no change in momentum of the generator, but there will be significant change in momentum of the electromagnet.
What do you think dynamic stress is besides a change in momentum of one part of a structure and not another?

Go ahead and do the math. I have already told you the conclusion, but you obviously need to work it out for yourself. Again, it is not force that you need to focus, it is momentum.
 
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  • #54
DaleSpam said:
Tominator, you have a really annoying habit of repeatedly asking the same question over and over and over and somehow thinking that the answer will change.What do you think dynamic stress is besides a change in momentum of one part of a structure and not another?
Sorry Dalespam, but it is because I still have some doubts about your conclusion that it will be capable only of acceleration simmilar to acceleration gained by laser.
Because if the "primary" electromagnet bounces off the field of the "secondary" one (in close proximity), the force is far greater, than force produced by laser. And because the "secondary" electromagnet is switched off before the "primary" could apply a force on it (the "primary" will actually bounce off "secondary's" field, not knowing, that the "secondary" electromagnet was turned off), it will produce acceleration in one direction. Am I wrong? If yes, where exactly?
What is dynamic stress?

DaleSpam said:
Go ahead and do the math. I have already told you the conclusion, but you obviously need to work it out for yourself. Again, it is not force that you need to focus, it is momentum.

Well, I am not so good at maths, instead of doing math, I try to visualize it, that is why I like physics much more.
If I apply a force on something in free space, it will accelerate and also gain momentum.
I do not need to care about momentum, because in this case it will be conserved thanks to the EM fields.
 
  • #55
Tominator said:
Sorry Dalespam, but it is because I still have some doubts about your conclusion that it will be capable only of acceleration simmilar to acceleration gained by laser. ... Well, I am not so good at maths
That's fine. There is no reason to trust me, but you cannot have it both ways. You need to make the (several years of) mental effort to learn this stuff (including the math) yourself. Otherwise your ideas are nothing more than wishful thinking or daydreaming.

Tominator said:
What is dynamic stress?
Stress (actually strain) is a deformation of a structure from its "resting" shape. Dynamic strain is simply strain that is not constant in time. So, if we have a spring it starts out unstressed, if we suddenly squeeze it with a certain force then it will start changing shape over time (dynamic strain), eventually it will settle down to a new shorter length that is constant over time (static strain, or just strain). If all parts of a structure move at the same velocity then there is no change in the structure's shape (no dynamic strain), but if different parts of a structure are moving at different velocities then the structure's shape is changing (dynamic strain).
 
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  • #56
DaleSpam said:
That's fine. There is no reason to trust me, but you cannot have it both ways. You need to make the (several years of) mental effort to learn this stuff (including the math) yourself. Otherwise your ideas are nothing more than wishful thinking or daydreaming.

Stress (actually strain) is a deformation of a structure from its "resting" shape. Dynamic strain is simply strain that is not constant in time. So, if we have a spring it starts out unstressed, if we suddenly squeeze it with a certain force then it will start changing shape over time (dynamic strain), eventually it will settle down to a new shorter length that is constant over time (static strain, or just strain). If all parts of a structure move at the same velocity then there is no change in the structure's shape (no dynamic strain), but if different parts of a structure are moving at different velocities then the structure's shape is changing (dynamic strain).

Well trusting you is the easyer and probably also wiser way. And after thinking for a while, I realized that you were right.
Or I would produce small acceleration by using current pulses, or dynamic stress by using normal AC currrent. But still I want to learn a bit more about these things.
If an EM wave interfere with oscilating mag field of EM oscilator on the same frequency, in the way that they would be canceled, could it affect the EM oscilator? (its momentum)
Or is the oscilating field in close proximity from the EM oscilator considered EM wave?
 
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  • #57
dalespam. I've got news for you. Your statement
"The magnitude of the momentum of light is given by p = E/c, so it is independent of the frequency"
Wrong, wrong ! E = hf. Energy of a photon is directly proportional to frequency.
 
  • #58
sorry to alter the tone of this topic, but i have had problems posting a new topic.

Could someone explain what electromagnetic waves are? what they are made of, as it appears to me they are made of nothing, but obviously that cannot be right. It is what the actual wave itself is made of, that i am curious to understand.
 
  • #59
azzkika said:
sorry to alter the tone of this topic, but i have had problems posting a new topic.

Could someone explain what electromagnetic waves are? what they are made of, as it appears to me they are made of nothing, but obviously that cannot be right. It is what the actual wave itself is made of, that i am curious to understand.

Never mind I have been asking similar question only months ago.
EM waves are waves caused by alternation of magnetic and electric fields. When there is change in magnetic field there is also change in electric field and vice versa.
If the change is fast enough, it produces a wave capable of self propagating - EM wave.
 
  • #60
yes i understand how waves are made, but not what they actually are made of. eg, a wave in the sea is made of water, or water is what propagates the wave energy. i suppose what I'm getting at, is what is the physical property of EM waves, or are they made of nothing?
 
  • #61
Tominator said:
If the change is fast enough, it produces a wave capable of self propagating - EM wave.

There is no speed requirement. Any change in electric or magnetic field, no matter how slow, propagates as an EM signal with speed c (with a very long wavelength).
 
  • #62
azzkika said:
yes i understand how waves are made, but not what they actually are made of. eg, a wave in the sea is made of water, or water is what propagates the wave energy. i suppose what I'm getting at, is what is the physical property of EM waves, or are they made of nothing?

The nature of every wave is energy. The EM waves are made of energy as well. The water waves needs water as environment to propagate, but again, they are nothing else than energy. It is said that EM waves does not need any environment to propagate. But on the first page of this forum DaveC said that electric and magnetic field extend to infinity. So then the enviroment, in which EM waves propagate would be everywhere. But EM waves are not like the water waves, they are composed of photons. Photon is something between wave and particle. But I still do not understand it so well to teach you, there are better explanations of EM waves all around the forum, you only need to do a bit of searching.

Mapes said:
There is no speed requirement. Any change in electric or magnetic field, no matter how slow, propagates as an EM signal with speed c (with a very long wavelength).

I have heard, that EM waves with frequency under 50Hz are considered oscilations of the field.
 
  • #63
Never heard of a cutoff at 50 Hz. Can you provide a reference so's I might learn more?
 
  • #64
Mapes said:
Never heard of a cutoff at 50 Hz. Can you provide a reference so's I might learn more?

By " I have heard " I really meant heard, so sorry but I can't provide you with any references. I meant it more as a question, because I was not and I am not sure about it. But it is something I have heard at school, so I expected it to be right. Maybe my memory messed it up somehow with the equation of progressive Em wave. You might post a new thread with this question if you want.
 
  • #65
map19 said:
dalespam. I've got news for you. Your statement
"The magnitude of the momentum of light is given by p = E/c, so it is independent of the frequency"
Wrong, wrong ! E = hf. Energy of a photon is directly proportional to frequency.
I am not wrong, but I guess I should have been more clear. I was thinking specifically in terms of a space propulsion system. So if you have a vehicle that carries a finite quantity of fuel then your E is fixed. Assuming that you could convert your fuel to light of two different frequencies, then for the lower frequency you would generate more photons of lower energy each than for the higher frequency. The total momentum of all the photons would be the same at both frequencies.
 
  • #66
Hmmm
But isn't the light propulsion extremely inefficient? Because if you compare the energy (power) used to "throw off" light, to the momentum gained, it looks to me that it is extremely inefficient. If you use the same energy to push off a magnet by electromagnet, the momentum gained is far more greater.
 
  • #67
I already answered that several times Tominator.
 
  • #68
DaleSpam said:
I already answered that several times Tominator.

Yes sorry, I only wanted to take it step by step, as you have suggested.
Well, if something is inefficient, but working, there is always possibility to make it work as efficiently, as technology will let us. Because waves are actually energy, the light propulsion is gaining momentum directly by "throwing off" energy. That seemingly violates the law of conservation of momentum, but as far as I know, law of conservation of energy is superior to law of conservation of momentum.
doesnt the law of conservation of momentum apply only on closed systems? (no energy out or in)
If my assumptions are correct, than it is not a physical non-sense to run around the law of conservation of momentum, by using non-closed system and law of conservation of energy. Thus creating propulsion system, capable of creating force in one direction directly from energy. (?)
 
  • #69
Tominator said:
Hmmm
But isn't the light propulsion extremely inefficient? Because if you compare the energy (power) used to "throw off" light, to the momentum gained, it looks to me that it is extremely inefficient. If you use the same energy to push off a magnet by electromagnet, the momentum gained is far more greater.
Actually light is the by far most efficient thing to throw off.

First you know the formula E=mc^2, right? It means that energy=mass.
Then you also got from special relativity that mass is just energy with its momentum vector along the time axis, and that energy is the length of these momentum vectors, and that no matter the direction the object always move in it with the speed of c. Of course we use the general coordinates of ct being the time axis.

Then you see that you get the most momentum per energy by using pure light, since it got no mass which gives 0 momentum per energy and everything in between light and mass is just a mix.

Anyway the reason we don't propagate things using light is because we aren't that good at transforming mass to energy.


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.

If you want to compare anything you should compare the magnetic field with the electric field, and then the magnetic field is weaker unless you move at the speed of light in which case they are equal. The only reason you notice magnetic fields more is because a wire can have charges move relative to each other while still have a total charge of 0 meaning that you get magnetic fields without electric fields, however if you would isolate just a few grams of purely charged matter you would get extreme forces.
 
  • #70
Tominator said:
Because waves are actually energy, the light propulsion is gaining momentum directly by "throwing off" energy. That seemingly violates the law of conservation of momentum, but as far as I know, law of conservation of energy is superior to law of conservation of momentum.
No, conservation of energy and conservation of momentum are on equal footing. They are the timelike and spacelike parts of the http://en.wikipedia.org/wiki/Four-momentum#Conservation_of_four-momentum" respectively. This is what Klockan3 is referring to in his second paragraph.

Tominator said:
doesnt the law of conservation of momentum apply only on closed systems? (no energy out or in)
Yes, same with the conservation of energy.

Tominator said:
If my assumptions are correct, than it is not a physical non-sense to run around the law of conservation of momentum, by using non-closed system and law of conservation of energy. Thus creating propulsion system, capable of creating force in one direction directly from energy. (?)
You cannot run around either conservation law. All you can do is understand them and use them to accomplish your design goal. TANSTAAFL
 
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  • #71
Klockan3 said:
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?

DaleSpam said:
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?
 
  • #72
Tominator said:
Does it mean that conservation of momentum would not work in not closed system?
Yes.
Tominator said:
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.
 
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  • #73
Tominator said:
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.
 
  • #74
Thanks for your explanations guys. I have not ever taught about it this way.

DaleSpam said:
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
Klockan3 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?
 
  • #75
Tominator said:
On the other hand momentum of the system will not be conserved.
The momentum of the system (including the fields) will be conserved.
 
  • #76
DaleSpam said:
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)?
 
  • #77
Did you read the link on 4-momentum that I posted earlier?
 
  • #78
DaleSpam said:
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.
 
  • #79
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.
 
  • #80
DaleSpam said:
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.
DaleSpam said:
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 :
DaleSpam said:
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.
 
  • #81
Tominator said:
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.
 
  • #82
DaleSpam said:
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.
 
  • #83
The radius of the Earth is 46 Hz, but who's counting.
 
  • #84
Tominator said:
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.
 
  • #85
DaleSpam said:
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:

Tominator said:
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.
 
  • #86
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.
 
  • #87
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.
 
  • #88
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".
 
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  • #89
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?
 
  • #90
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.
 
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  • #91
DaleSpam said:
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.
 
  • #92
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.
 
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  • #93
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.
 
  • #94
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).
 
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  • #95
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?
 
  • #96
Tominator said:
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.
Tominator said:
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.
Tominator said:
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.
 
  • #97
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.
 
  • #98
Hi DaleSpam,

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

DaleSpam said:
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.
 
  • #99
Tominator said:
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.

Tominator said:
If the power is fixed, will the force decrease with time?
Yes.

Tominator said:
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.
 
  • #100
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)?
 

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