## EM Drive

Point out news stories, on the net or in mainstream media, related to polywell fusion.

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kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX
I will, but first I must address a misunderstanding of Galilean relativity first, which can be used to describe normal chemical rockets. That is both force and acceleration are invariant under a Galilean transformation, which is fairly easy to show. The transformation is defined in one-dimension as:

v' = v - v_0

For v in the rest frame and v' in the moving one. The acceleration in the rest frame is a=dv/dt, the acceleration in the moving one is a'=dv'/dt. However, dv_0/dt = 0 because these are inertial frames. So, dv'/dt=dv/dt. Mass is invariant in Galilean transformation too, so F = ma = ma' = F'.

You can also understand this by a simple thought experiment. Two people on roller chairs (and a laminate floor) are going past at the same speed. They push on each other such that one comes to rest. The other still accelerates to a faster velocity because of the change of momentum of the one that came to rest. The same applies to rockets. The exhaust can be at rest in your frame but rocket still accelerates. The exhaust can even be moving in the same direction as the rocket. What matters is the relative velocity between the rocket and the exhaust, not you and the exhaust.

Now to SR, I will point to this paper which gives the Lorentz transformation of the force: http://redshift.vif.com/JournalFiles/V1 ... 2N2KHO.pdf. If you transform from the instantaneous rest frame of the rocket to any other frame parallel to the direction of travel, the force is the same. Which is why I say his force (or thrust) transformation does not agree with SR.

Non-parallel transformations do not leave the force the same, however. Also, acceleration and force do not transform the same. While the force may be the same in two frames, the acceleration is not. P = \gamma*m*c*v, which blows up as v approaches c, and so acceleration approaches zero so that v never reaches c. So it is also true that these are not invariant quantities. But since his theory deals with thrusts parallel to velocity only, and derives velocity dependent forces (not accelerations) these effects do not even have comparisons to his theory.
Carter

kcdodd
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Just regular Newtonian physics will do: The same force applied to a moving object for a given time does more work on that object than on a slower moving object over the same time. So if you generate a given amount of thrust, then the thrust does more work on a moving object the faster it goes in that frame.
It is true that force times velocity is the power. But take a moment to look at the total energy change of the system. For simplicity I will call m1 the mass of the rocket and m2 the exhaust (v1 and v2 respectively as well). The total kinetic energy of the system is:

T = 0.5*m1*v1^2 + 0.5*m2*v2^2

The total change of kinetic energy is:

dT = m1*v1*dv1 + m2*v2*dv2

Perforce a Galilean transformation: v' = v - v_0

dv' = dv because dv_0 = 0 for inertial frames. So...

T' = 0.5*m1*v1'^2 + 0.5*m2*v2'^2

dT' = m1*v1'*dv1' + m2*v2'*dv2' = m1*(v1 - v_0)*dv1 + m2*(v2-v_0)*dv2 = m1*v1*dv1 + m2*v2*dv2 - v_0*(m1*dv1 + m2*dv2) = m1*v1*dv1 + m2*v2*dv2

m1*dv1 + m2*dv2 = 0 because of Newtons third law.

So, dT' = dT, and so, the rate of change of total kinetic energy is exactly the same in every reference frame. Which also means the total power output from the rocket does not have to increase to deliver the same force at higher velocities.
Carter

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm
chrismb wrote:
GIThruster wrote:Chris, you've got the example backward. The point is that the higher the velocity of the rocket, the less force its thruster creates--so it's not a situation of "The same force applied. . ."
oh....dear...me....
Chris, Carter got the answers right. You didn't even understand the question.

Just as Carter said, what we came up with was P=T*V.

Carter went still a lot further. And just being honest here, I have participated in these sorts of discussions in several forums, most often with engineers but most enlightening with physicists (who contrary to belief are not always right.) Carter, obviously has the training, clarity of mind and unusually high communications skills to answer a fair question fairly, show how and why and do all that without insult to anyone.

People could learn something here.
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

GIThruster
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Joined: Tue May 25, 2010 8:17 pm
Aero wrote:I know better than to get involved, but I will anyway. . .The fact is, using rocket body centered coordinates, thrust is constant within engine tolerances.
Yes, but the "rocket body centered coordinates" is an accelerating, non-inertial frame, and acceleration, force, velocity and energy are not invariant. This is why you either need to do the calculation in a series of "instantaneous frames of rest" which are then summed, or transform as Carter has. You cannot simply calculate in a non-inertial frame as if it were at rest.

http://en.wikipedia.org/wiki/Non-inerti ... ence_frame

Yes, it's very confusing, but our first year high-school physics teachers did indeed warn us about this. All of us. "Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm
kcdodd wrote:
Just regular Newtonian physics will do: The same force applied to a moving object for a given time does more work on that object than on a slower moving object over the same time. So if you generate a given amount of thrust, then the thrust does more work on a moving object the faster it goes in that frame.
It is true that force times velocity is the power. But take a moment to look at the total energy change of the system. For simplicity I will call m1 the mass of the rocket and m2 the exhaust (v1 and v2 respectively as well). The total kinetic energy of the system is:

T = 0.5*m1*v1^2 + 0.5*m2*v2^2

The total change of kinetic energy is:

dT = m1*v1*dv1 + m2*v2*dv2

Perforce a Galilean transformation: v' = v - v_0

dv' = dv because dv_0 = 0 for inertial frames. So...

T' = 0.5*m1*v1'^2 + 0.5*m2*v2'^2

dT' = m1*v1'*dv1' + m2*v2'*dv2' = m1*(v1 - v_0)*dv1 + m2*(v2-v_0)*dv2 = m1*v1*dv1 + m2*v2*dv2 - v_0*(m1*dv1 + m2*dv2) = m1*v1*dv1 + m2*v2*dv2

m1*dv1 + m2*dv2 = 0 because of Newtons third law.

So, dT' = dT, and so, the rate of change of total kinetic energy is exactly the same in every reference frame. Which also means the total power output from the rocket does not have to increase to deliver the same force at higher velocities.
Carter, would you mind setting aside the Shawyer stuff a moment and clarify something that doesn't make sense to me?

You're saying above what appears to me two mutually exclusive things:

a) "It is true that force times velocity is the power." or what I called P=T*V or Power equals Thrust times Velocity (which seems to say given constant power, thrust is velocity dependent (and inversely proportional) and

b) "the total power output from the rocket does not have to increase to deliver the same force at higher velocities."

Could you clear this up for me?
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX
Well, going back to the kinetic energy, and taking the difference:

dT = m1*v1*dv1 + m2*v2*dv2

and now using conservation of momentum: m1*dv1 + m2*dv2 = 0

dT = (v1 - v2)*m1*dv1

Now, dividing by time difference you get power on the left and force times velocity on the right.

P = (v1 - v2)*F1

And you see it is the relative velocity between the exhaust and the rocket which determines the power, which is the same in every reference same.
Carter

GIThruster
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Joined: Tue May 25, 2010 8:17 pm
Well, I'd prefer not to even look at energy. Let me put this another way.

This issue came up in another forum, because of a blunder I had made several years earlier. I posited to Paul March, that since M-E thrusters ought to be capable of constant force generation, they ought to yield constant acceleration. It ought to be possible given an M-E thruster and sufficient power source, to accelerate constantly at 1 Earth gee, in travels around our planetary system.

As an illustration of the technological possibilities, Paul calculated the travel times to the Moon, Mars (at nearest and farthest distance from Earth, the asteroids, Jupiter, and Saturn (all three at farthest distance from Earth) and got some travel times. The Moon would be about 5 hours, accelerating at 1 gee half way there, and decelerating at 1 gee the second half of the flight.

Now over in another forum, a physicist challenged this. He mistook an engineering and technology illustration as a conservation violation. He didn't use the term, but essentially what he said was P=T*V and constant power does not yield constant acceleration as seen from the "lab", "launch" or "rest" frame. He in fact convinced me I had made the error of thinking very simplistically that F=MA, when in a non-inertial frame, this is really not true unless you refine what you mean by "F".

So, simplest way I can ask--does constant force in a chemical or M-E force generator yield constant acceleration in an inertial frame? Given P=T*V, seems this isn't so and the "one gee solution" is not a workable solution.

This seem true to you?
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

pfrit
Posts: 256
Joined: Thu Aug 28, 2008 5:04 pm
Technically that is true, but in a complex universe, it is not. As the speed of the vehicle increase, and long before the tau factor rears its ugly head, the relative temperature of the ambient space rises which makes the rocket perform less efficiently. Remember that in its most simple form a rockets efficiency is determined by the relative temp of its exhaust compared to the ambient temp. It is much more complicated than that, but the temp difference between exhaust and ambient is always a factor. As the engine increase speed, the relative temp of the environment goes up, so thrust goes down. This is one reason why there is an upper limit to the speed of a Bussard ramjet (Managed to actually bring this around to being somewhat on topic). I will leave it to the readers to figure out why the apparent temp goes up. I leave out the apparent change in the density of the medium that your rocket passes through as it accelerates. I know that this is besides the point, but just felt like joining in.
What is the difference between ignorance and apathy? I don't know and I don't care.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm
kcdodd wrote:So, dT' = dT, and so, the rate of change of total kinetic energy is exactly the same in every reference frame. Which also means the total power output from the rocket does not have to increase to deliver the same force at higher velocities.
Whatever you have done to get there, the conclusion is generally wrong.

If I watch a person standing still then starts walking down the street, let's say his mass is 50 kg and his speed is 2m/s. So he would need to impart 100 J of energy to get his kinetic energy up. But that is in the 'street' frame.

Let's now say we consider the same person in an aeroplane at 200m/s and he starts walking down the aisle in the direction of flight. When stationary in the aisle, he is travelling at 200m/s, which is 1 MJ. Once he is walking down the aisle at 2m/s, he is now going at 202m/s, which is 1020100 J.

That is to say, on the plane he gains 20kJ, but on the ground he gains only 100J. You tell me... do you find it 20 times more difficult to walk forward on a plane than on the ground????

KINETIC ENERGY [and rate of change thereof] IS FRAME DEPENDENT. If you don't understand this then you have no hope of understanding the nature of 'collision energy' and 'centre of mass frame' when it comes to the particle physics you need to comprehend fusion reactions, and if so then this is not really the right forum for you!

I think what this might have moved on to is talking about is the rate of change of kinetic energy of the 'rocket' and 'fuel' as a closed system. This has no connection with how fast the rocket is going to get, though.

GIThruster
Posts: 4686
Joined: Tue May 25, 2010 8:17 pm
I now have four physicists in three different forums arguing over this issue.

Heh! Did someone think this was easy?

Here's the answer to Carter that thus far carries the most clarity and seems to explain a mistake he made (I am not yet taking sides):
-----------------------------
Okay, here’s why Carter(?) got it wrong:

First, a simple example.

Suppose I am sitting here in my chair, fixed, inertial (but not inert), and I see a 100kg body going past at 100m/sec in the +X direction. From 1/2mV^2, I calculate its kinetic energy at 5E5 Joules. Now, suppose an observer, Andrew, is travelling along with the body, also at 100 m/sec in the +X direction, but still not accelerating, just moving with constant velocity. Since for him the velocity of the body is zero, he calculates the kinetic energy of the body (T, following Carter’s notation, which is standard for physics), as zero.

Now, suppose a 1000N force is imposed on the body for one second. [Just to ease the worried minds of anyone who might worry about HOW I exerted the force on the body, let’s pretend that it’s got a really big electric charge on it, it has just passed though a coarse-mesh grid, and I then charge the grid, so that its resulting electric field repels the body. It’s sorta like a really big electron gun. It doesn’t really matter. But, feel free to think of it like that if it worries you.]

F=ma

From my standpoint, I see a 1000N force acting on the body for 1 second, 1000N = 100kg * a -à a = 10 m/sec^2. So I see the velocity increase by 10 m/sec in one second. Velocity at the end of the one second of thrust is 110 m/sec as I measure it. Therefore from T = 1/2mV^2, I calculate T= 6.05E5 J. Delta T is 1.05 E5 J, so in that one second (Delta T)/t = 1.05E5/1 = 1.05 E5 W of power was expended.

From Andrew’s standpoint, that same 1000N of force is exerted. He also sees the body accelerate at 10 m/sec for 1 second, so he also sees the body gain 10 m/sec. From 1/2mV^2 he calculates the new kinetic energy of the body, at the end of the one second of acceleration, as 5000 J. So, he calculates Delta T as 5000-0 = 5000 J. Therefore, from his perspective, P= (Delta T)/t = 5000/1 = 5000 W.

Delat T does not equal Delta T’, to use Carter’s notation.

QED

Where did Carter make his mistake?

In diagnosing the difficulty, notice that what made the difference, above, was the difference in starting velocity at the beginning of the acceleration. That’s v_0 in
Carter’s notation. He gets rid of v_0 in his derivation by assuming m1dv1 + m2dv2 is zero by Newton’s Third Law. Apparently that’s not right.

Newton’s Third Law is most commonly expressed as, “For every action, there is an equal and opposite reaction.” Normally, of course, we think of “action” as being synonymous with “force”.

So what he’s really saying, or attempting to say, is that the force on the rocket is the opposite of the force on the propellant. Sure enough. The problem is that in the formulation that he presents, he has lost the sign information. The form of the equation that he gives is only valid if you use the vector representation, which he is not doing. That automatically takes care of the signs. Carter is not keeping track of the signs.

Note that you lose all sign information in the original equation for T and T’ because you take the square of the velocity, which is positive definite.

Carter takes the derivative, I prefer to keep the dt in the denominator because it helps me to think about what I’m doing, but he cancels it out. Whatever. Notice, there is no sign information in his dT equation [dT' = m1*v1'*dv1' + m2*v2'*dv2'] about which way v1 is going or which way v2 is going or whether dv1 or dv2 are positive or negative. It could be any mixture of signs, really, for all that we know. But we know that we are interested in the increase in the kinetic energy of the rocket plus the increase in kinetic energy of the propellant, so even though they are accelerating in opposite directions and travelling in opposite directions ( well, if the rocket is going slow enough originally), we are taking all the signs as positive.

[ Maybe the rocket, right before it started thrusting, was travelling in the +X direction at 5000 m/sec. The propellant then too was travelling in the +X direction at 5000 m/sec then, of course. Let’s say the rocket engine has an exhaust velocity, relative to the engine, of 3000 m/sec. A pretty good engine. After the thrusting, the propellant has a velocity of 2000 m/sec in the +X direction, and the rocket has a velocity of more than 5000 m/sec, also in the +X direction. You see that in Carter’s T’ frame, v1’ and v2’ are both positive, as is dv1’, but dv2’ is negative. That information is not in his dT’ equation.]

For Carter to do his derivation correctly, he must make it the absolute value of velocity that he works with. Then, what he did will give a correct result, right down until he says m1dv1 + m2dv2 is zero by Newton’s Third Law. To make that correct with absolute value, you have to say, “abs(m1dv1) - abs( m2dv2) = zero by Newton’s 3rd.” That’s different from what he’s got, [dT' = m1*v1*dv1 + m2*v2*dv2 - v_0*(m1*dv1 + m2*dv2) ] so he can’t really eliminate that v_0 term, so he can’t really concluded dT=dT’.

QED
"Courage is not just a virtue, but the form of every virtue at the testing point." C. S. Lewis

kcdodd
Posts: 722
Joined: Tue Jun 03, 2008 3:36 am
Location: Austin, TX
The problem between my answer and your answer is that I am considering the total change of energy, while both of you are concerned with the change in energy of the rocket only, which is definatly frame dependant and I have not argued against.

P1 = v1*F1 != v1*F1 - v_0*F1

I get v_0 (frame dependant term) because I don't include anything else. However, the power used to accelerate the exhaust is also.

P2 = v2*F2 != v2*F2 - v_0*F2

However, conservation of momentum mandates that F1 = -F2, so that when I calculate P = P1 + P2 I get

P = v1*F1 - v_0*F1 + v2*F2 - v_0*F2 = v1*F1 - v_0*F1 - v2*F1 + v_0*F1 = (v1 - v2)*F1

Which is precicely what I already got the other way. The total power is what is being supplied by the rocket engine. Not just P1 or just P2. And you can see that P1 or P2 could even be negative depending on which frame you're in, which would not even make sense if you claimed that the power output of the rocket was then negative. However, P1+P2 is always positive and the same value. Ignoring SR of course. I know this may be counter intuitive but thats sometimes physics for you.
Last edited by kcdodd on Tue Mar 08, 2011 9:41 pm, edited 1 time in total.
Carter

ltgbrown
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Alright, I am sure I am in way over my head, but isn't kinetic energy not constant? Isn't chemical energy (in the case of the rocket) or electric energy(from whatever source) (in the case of the ME thruster) being converted into kinetic energy and therfore changing total system kinetic energy and momentum?
Famous last words, "Hey, watch this!"

kcdodd
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Location: Austin, TX
Yes, exactly. But we are talking not of whether energy is constant, but whether the rate of change of the energy is the same in every reference frame. So, if chemical energy is being converted into kinetic energy at a rate of 10J/s in one frame, it has to be 10J/s in every other reference frame too. It just doesn't make sense otherwise. Again, not including SR.
Carter

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm
GIThruster wrote:Where did Carter make his mistake?
He didn't make one. What he was talking about was the total kinetic energy change of two bodies undergoing some sort of re-ordering of their momentum.

What he said (in prose) was that for a given Centre of Mass system - like a rocket and its fuel - the rate of change of kinetic energy for changes in that CoM system are frame invariant. And this is correct. But it just doesn't tell you anything about the misunderstanding that GIT posed a page back, which is the claim that thrust drops off with velocity.

chrismb
Posts: 3161
Joined: Sat Dec 13, 2008 6:00 pm
ltgbrown wrote:Alright, I am sure I am in way over my head, but isn't kinetic energy not constant? Isn't chemical energy (in the case of the rocket) or electric energy(from whatever source) (in the case of the ME thruster) being converted into kinetic energy and therfore changing total system kinetic energy and momentum?
KE varies according to the observer. But if the KE of an observed system changes, then it changes equally for all. An isolated system can only change its KE if its incident parts react against each other. To do so requires energy. The total change of that energy is a function of the energy involved in that process, so it is clear that it is the same amount of energy.

For example; if you have a projectile composed of two parts and some explosive bolt in the middle goes off, causing the thing to split open, then clearly the energy involved in that explosion will go into the kinetic energy of the parts [to be precise, a given fraction of it]. Now, as the chemical energy of that explosive is frame invariant, so the change of kinetic energy must also be frame invariant.

But, clearly, kinetic energy of some object with respect to different observers varies. If you drive down the road at 30mph, you could argue that you've just imparted to planet earth a 30mph speed - but clearly you have not expended that much energy to give it to the earth to travel at 30mph!!!