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