EM Drive

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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.

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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 coarsemesh 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 50000 = 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
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 coarsemesh 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 50000 = 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

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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.
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

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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!"

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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
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