Mach Effect progress

[chrismb]

this trouble only comes up when engineers pretend to be physicists and ignore the need to do transforms that is so obvious to the physicist.

The same laws of physics are observed by engineers and physicists. There is no difference in how they deal with creating macroscopic energy from nothing....It doesn't happen in any known way.

Just stick with one singular non-accelerating frame, and the maths of the engineer, physicist or high school student will all end up with the same answer so long as they do the maths right. And that is ALWAYS deltaKE<inputEnergy .

If the statement above is in doubt then it needs to be demonstrated algebraically that it is wrong. The time for hand-waving about it is over.

[GIThruster]

IIRC GIThruster said that the math disproving GoatGuy's math was posted in detail on a mailing list.

Jim specifically asked me not to get involved in this sort of discussion because he rightly notes that those indulging in this do not have the tools nor the training to provide a useful discourse. He's right. And just as Heidi has noted recently on Jim's mail list, when someone who has the proper skills wants to write a paper showing the kinds of troubles engineers pretending to be physicists so often offer, she'll answer them in the peer reviewed journals. Until that time this is not worth her time, nor Jim's.

Maybe this is all easier for me to stomach since I don't have even the skills of a trained engineer, and so I'm much more comfortable recognizing my limitations.

If you have no training in GR and in using Lorentz transforms, you are best suited to leave this topic alone and focus on issues you can rightfully expect to understand.

[GIThruster]

The same laws of physics are observed by engineers and physicists.

chris, engineers are taught from day one, just as I was in high school and college in the lower level courses in physics I've taken; that one needs to be especially careful with non-inertial frames of reference.

All thrusters put out constant stationary thrust. That thrust is not constant or invariant when the thruster is accelerating.

You do not have the proper tools to do this calculation. You are in fact ignoring that acceleration is a serious issue. You're doing what your high school teachers told you not to do.

If you don't believe me, do the calculation you propose with a chemical thruster on an arm supplied with all the fuel it needs and you'll see, that chemical thruster likewise will go overunity. What this means is, your method is wrong.

This is becasue force is not invariant:

http://en.wikipedia.org/wiki/Invariant_(physics)

[Barry Kirk]

Even if the thrust is reduced when the M-E thruster is accelerating, which basically works out to the energy drain I was talking about earlier. It is still possible that the M-E thruster could be significantly more efficient about turning energy into delta-V.

A good example is a Jet engine is far more efficient at turning fuel mass into thrust than a rocket engine. The "ISP" for a jet engine can be far higher than any chemical rocket engine.

For example, the GE-CF6 on a B-747 at subsonic speeds has an ISP of about 6000... Well over 10 times the best Hydrolox engine in existance.

If a M-E thruster can achieve a decent thrust to weight ratio and get an equivalent ISP, getting off this bloody rock gets really really cheap.

I said equivalent ISP... what I'm getting at is the mass of fuel burned to generate the energy to generate the thrust...

At that point payload mass fractions get very high and you can build heavier more robust vehicles.

Even if you can only get an ISP of 1000 with the same thrust to weight ratio of a hydrolox engine... Re-usable space planes become affordable. And by affordable I mean that you might have to mortgage your house to afford a ticket to orbit.

Equivalent to the relative costs of people purchasing a 3rd class ticket on a steamer 100 years ago to immigrate to the USA.

If you don't have to carry your fuel, as an example was the 200 mile long launch cable I mentioned a couple of posts back costs could go down even more. I envisioned the 200 mile long launch cable as a booster. You get your energy from the cable for the first 200 miles and than switch to onboard power sources.

If nukes become acceptable or you can use beamed power, than you don't need a launch cable.

[Barry Kirk]

this trouble only comes up when engineers pretend to be physicists and ignore the need to do transforms that is so obvious to the physicist.

The same laws of physics are observed by engineers and physicists. There is no difference in how they deal with creating macroscopic energy from nothing....It doesn't happen in any known way.

Just stick with one singular non-accelerating frame, and the maths of the engineer, physicist or high school student will all end up with the same answer so long as they do the maths right. And that is ALWAYS deltaKE<inputEnergy .

If the statement above is in doubt then it needs to be demonstrated algebraically that it is wrong. The time for hand-waving about it is over.

Chris,

You can't have a non-accelerating M-E thruster that generates a non-zero force. By definition, if the M-E thruster is generating a force, than a=F/m... Now the acceleration could be really small because the M-E thruster is bolted to the Earth. Effectively in this case the acceleration is zero. Therefore, the distance that the M-E thruster applies it's force through is zero, and since work or energy is force time distance and distance is zero, the energy is zero.

Jim's table top experiments where the M-E thruster is effectively bolted to the ground is not going to generate any energy because it's applying a force through zero distance.

If the M-E thruster is trying to accelerate just itself and a reasonable payload, well by definition it is accelerating and that is a different domain.

[Barry Kirk]

Just looking back at my own post and trying to understand what I wrote...

Hope other people can understand it better than I can...

[GIThruster]

That makes perfect sense, Barry; and is in fact why from Jim's earliest writings on M-E thrusters, he has always stipulated they generate stationary force, just like any chemical, Hall or Ion thruster. It's also why when Paul and I worked together to flesh out the "one gee solution" used in the WarpStar travel times around our planetary system, we noted that we did not do any relativistic corrections to the math. In order to do the calculations correctly, one needs Lorentz transforms which neither of us have been trained in.

Point in fact, no baseline thrust or thrust efficiency will be constant under acceleration as viewed from a launch frame. If the thruster is permitted to accelerate, it's thrust will vary as a function, I believe, of velocity (an inverse quadratic function). This goes back to the observation that acceleration is linear with force, but kinetic energy is quadratic with velocity.

Basically, in order to avoid a conservation violation, one needs to do the math with transforms--the tools of GRT.

[CaptainBeowulf]

chris, engineers are taught from day one, just as I was in high school and college in the lower level courses in physics I've taken; that one needs to be especially careful with non-inertial frames of reference.

All thrusters put out constant stationary thrust. That thrust is not constant or invariant when the thruster is accelerating.

You do not have the proper tools to do this calculation. You are in fact ignoring that acceleration is a serious issue. You're doing what your high school teachers told you not to do.

Chris - I understand your concerns but what GIT says above rings a bell in my mind. This is a very tricky business. If you want to try doing the math yourself and posting it I think some of us will try to critique it as best we're able, but I don't know if anyone will be able to discuss it properly.

I think we're better off watching M-E experimentation and waiting to see if the results cause a more extensive literature to develop around it. If there are two people arguing the math from both sides of this issue who know exactly what they're doing, we might be able to follow along, but without the physics training we're just going to wander off onto tangents and go in circles.

[chrismb]

Jim specifically asked me not to get involved in this sort of discussion because he rightly notes that those indulging in this do not have the tools nor the training to provide a useful discourse. He's right.

blah blah blah...

So let it be properly understood - someone comes up with a theory in which the conclusions appear to contradict the laws of physics, and as soon as anyone wants to lay out a few basic calculations to see if it is physically justified the proponents merely have to claim that others do not have the skills to question it because only the proponents understand their own theory?

Yeah! Right!

.. and even worse, there are people prepared to nod willingly and say 'yes, master, we do not understand. Only the anointed ones can know!'

If that is truly Woodward's complaint, he should seek to find people to educate so that they understand.

This forum is willing and able.

As GIT has portrayed it, this is a perfect example of pathological science.

It is staggering that there are people here who are prepared to nod along with such claims, without asking 'show us some numbers'.

It HAS to be physical and justifiable at the macroscopic level of thermodynamics. Hand-flappy arguments are over. Forever repeating the claim that any rocket system will hit over-unity does not make it true. Show up the colour of those equations.

[Barry Kirk]

I'm pretty sure that Jim's experimental M-E thruster never moved or only moved sub millimeter distances.

Could somebody here please verify that.

This means we are talking about a statically generated force.

I can take two metal plates held at a rigid distance from each other and give them both a positive charge. Is there a force between them? Yes. Is any energy created? No... Because they aren't moving.

The same applies to a static M-E thruster. No movement means no energy.

I don't know if Jim has addressed the issue of what happens when the M-E thruster starts to move because.

1. The thrust is higher.
2. It's not bolted to the Earth.

If those two metal plates discussed above are allowed to move, well that physics is pretty well understood and does not violate any conservation laws.

[GIThruster]

So let it be properly understood - someone comes up with a theory in which the conclusions appear to contradict the laws of physics. . .

It only appears that way to people who don't understand GRT. There are many physicists who have looked at this over the years, including the physics staff at Fullerton when Jim first began to publish and all those who read his peer reviewed works since. There are several physicists on Jim's reading list, including one, Jack Sarfatti; who have objections to Jim's work; but they never make the kinds of objections you're making because they have real training in GRT and they know better.

[GIThruster]

Could somebody here please verify that.

This means we are talking about a statically generated force.

Yes. All thrusters produce a stationary force that when allowed to accelerate, varies. There is nothing unique to M-E thrusters in this regard. That's why it is completely appropriate to look at the case of a chemical thruster and you'll see it "appears" to violate standard physics in just the same sort of way that GoatGuy, Andrew and chris have all focused on.

Not to put too fine a point on it, but Paul March is an extremely accomplished engineer, and he knows better than to make these sorts of mistakes as well as to try to do real calculations without transforms.

[chrismb]

Is there a force between them? Yes. Is any energy created? No... Because they aren't moving..

No KE. But the electric field has an energy. The plates will have had energy added to them by the addition of the charge. Work will have been done to push the charge on to them.

Two charged objects will [attempt to] move apart/move together so as to minimise the energy of the electric field between them.

There is never any force, except from a field that 'is trying to' minimise its energy. It is the rate of change of energy of the field that is the force.

[chrismb]

the case of a chemical thruster and you'll see it "appears" to violate standard physics in just the same sort of way that GoatGuy, Andrew and chris have all focused on.

WHHEERRREEE is this crazy maths GIT keeps saying exists???

Just show the forum the equations. No-one here believes that there is any 'violation' when calculating input energy versus output KE with conventional chemical thrusters. Only GIT beleives he's read it somewhere, once, and tells everyone they don't understand it so he's not going to show them.

PLEEEEEEASE show the forum what it is that is claimed to 'violate' standard physics.

JUST SHOW THE CALCS!!!

[GIThruster]

Chris - I understand your concerns but what GIT says above rings a bell in my mind. This is a very tricky business.

Just as a whimsical aside, about a year or so ago, one of the physicists on Jim's list took an interest in the conservation issue. He's an extremely accomplished PhD physicist who works for NASA at JSC. His training however is primarily in the Standard Model, not in GRT.

He took up this issue and championed the conservation violation argument for about a week of very lengthy posts, often more than a couple screen pages long. He related personal stories about how confusing dealing with non-inertial frames is. He then came to the wrong conclusion, and Jim very gently and politely pointed him to the proper solution. M-E thrusters act just as any other thrusters when looking at the conservation issue.

There was also a physicist, an MS not a PhD, over at NBF who I engaged on this issue and I will own that he had me befuddled. He was maintaining that if one looks at the "launch frame" which an M-E spacecraft accelerates in, one avoids all the need for transforms. He was wrong, because the stationary thrust of any thruster is only constant when it is at rest. He was doing just as GoatGuy, Andrew and chris are doing and failing to note that thrust or better "force" is only constant when the thruster is stationary.

It is very easy to screw this stuff up. That's why you not only need to have access to the proper tools in GRT, but you need to be proficient in their use, in order to make sense of this issue. You need to not only understand which qualities are variant and invariant, but you need a deep understanding of why they are so. This only comes from real proficiency in GRT.