studiot

Perhaps @MigL knows the answer to this but I would be very suprised if any useful rocket was designed for constant mass exhaust. Different accelerations and velocities will also be required at different stages of the flight. This implies that the force applied to the rocket will vary over the flight time. These are parameters that can be worked out for a desired flight history.

It is the experience of all my family and friends that curry, casserole and cottage pie all taste better the second time around. So my question is Is there any way to skip the first day ?

I feel this thread is wandering further and further off topic which is (not was) the formal logical soundness of a point particle. The Cosmologist has not answered my question to him about quantum numbers. In my oldfashioned day we had two quantum numbers concerned with 'angular momentum. There were 4 in all labelled, n, l m and s respectively. The first 3 had a classical interpretation, although they mark stationary point solutions to the Schrodinger equation. These describe the energy, position size and shape of the obitals. The final one, s, has only a quantum interpretation although it is common to refer to it as the spin or quantum spin of the electron about its own axis, the actual behaviour is not exactly that of a mechanically spinning anything. The second one, the azimuthal quantum number, l, can be associated with classical angular momentum of the electron as a point particle rotating about the nucleus. So where are we in relation to point particles ? Incidentally as regards the so called infinite limit that also seems to be causing trouble, We have no problem with density being a point function limit of mass over volume, nor of pressure being the force over zero area in the limit. I respectfully suggest crossing swords with a real organic chemist about chirality would be difficult. Chirality embraces a host of phenomena of which handedness is only one, but I'm sure exchemist would run rings round any of us on this subject.

Surely you dont need a course in basic calculus ? Definition momentum = mass x velocity p = mv Therefore if both m and v are variable [math]\frac{{dp}}{{dt}} = m\frac{{\partial v}}{{\partial t}} + v\frac{{\partial m}}{{\partial t}}[/math] Where both m and v are functions of t. To work this out you would need to have equations for both m and v as functions of t, unless you could devise an equation connecting m and v. Do you have such equations ? Some treatments specify that the mass flows out at a constant rate, that is [math]\frac{{\partial m}}{{\partial t}} = C[/math] Alternatively you can calculate the changing mass flow required to produce constant acceleration.

You said To which I replied Yes if the lifetime is 1 second* the muon expeience 1 second of 'life'. But the rest of what I said means that the lab observer sees (measures, experiences) the moving muon taking much longer than this to die, whilst the muon in his hand take exactly 1 second to die. *(Actually the time is about 2 microseconds, not 1 second)

Just saying it doesn't make it so. I don't understand why you have this problem as an electrical engineer you must be familiar with the equation Instantaneous power = instantaneous voltage x instantaneous current. The analogy is exact Instantaneous force = instanteous mass x instantaneous voltage. So spit out your objection fully please.

That was a particularly high handed dismissal of my request for you to describe what you see as 'suspicious' You made the claim, not I, so it is up to you to justify it.

By the way. There is nothing in classical EM theory to connect charge directly to energy. If there were it could be like heat and we would observe convertion as well as exhaustion of the source. In that sense charge is more like gravity. Both are inexhaustible supplies of their particular effect.

I don't think so. That would be true if and only if 'space' was embedded in a higher dimensional manifold. Current theory is that this is not the case and that the physical manifold we call space is not embedded in anything. That is possible because of Gauss famous little theorem - The theorema egregium - which distinguishes between intrinsic and extrinsic curvature.

Why should that be suspicious ? All the equation tells us is that force is the product of two independent quantities. Independent means that each or both can be varied without reference to the other. Setting one or both constant just makes for easier maths.

We have certainly not exhausted or explored all the rooms in 'dimension house' yet. But we have found some phenomena that run counter to 'common sense' or our ordinary experience. Nor have we arrived at a consistent definition of dimension, suitable for all purposes. We have also has a couple of threads of discussion here at SF about the subject.

It must be because the thing you and I call space does not accelerate, or even move.

I understand what you are getting at but Trying to play the smartass with everybody has led you to post some ridiculous English. Nothing 'happens to points of view'. They do not change. However the muon experiment is really good in that it allows comparison of effects in two different frames and introduces of the second fundamental point of relativity. Of course it does. That is exactly your point and it is correct as far as it goes. But it is not complete because there are two different frames involved. The ground or laboratory observer measures different times in his laboratory for the same effect to occur when the muons with him in the laboratory, and when they are approaching it at speed.

So as you have been told several times f = ma is used but only in the appropriate place and is not the beginning of the dynamical analysis of a rocket. Only in a simple analysis.

At least we can agree on this Wry smile. What do you make of Physical Similarity and Dimensional Analysis Duncan or Dimension Theory Hurewicz and Wallman ?

But I didn't say that and it isn't true all the time. I said At the start the momentum (note not the rate of change of momentum) is exactly zero. That is not the same thing. dp/dt comes later in the analysis when we can say that it starts off at zero and is conserved so the total momentum of the whole system is always zero. That fact lets us form an equation between parts of the system that do not have zero momentum by themselves. So we can then say that the upward momentum - the downward momentum must equal zero Which is a very useful condition. This equality leads us to consider the third law forces between the parts by way of the second law. The logic of all this is inexorable if you follow it correctly. But it is easy to leave a step or two out when you are so familiar with it you become bored with it.

+1 But we do not deal in unsupported assumptions. We (like to think) we have sound reasons for these assumptions It does and it compresses. The compression of rocks at depth is enormous.

But Sears started, where I and swansont suggested you start - with conservation of mementum. Sears also derives the conditions where f = ma is applicable first, before he uses it also as I indicated in my last post. Scienceforums offers you the facility to view enlarged vrsions of the attachments, and I specifically uploaded large enough versions for this. At full enlargement it seems just fine to me. Note yet again I am saying ( as are others) you need both versions for the rocket. You still haven't acknowledged what I said about simultaneous equations. As an electrical engineer try this. V = IR and P = IV Are two different equations about different physical quantities. Because they have some common simultaneous variables you can substitute to find an equation for say power in terms of either voltage or current and resistance. P = V2/r = I2R These can be extended further to find energy and other properties. The rocket is an equivalent situation in mechanics.

This is not maths or physics. The constants can't be zero. E = e2/r is not dimensionally consistent. It may be of interest to you as a new member to discuss ways of presenting mathematics to best suit yourself. Scienceforums offers particularly wide ranging possibilities - better than any other I know of. Although the Tex / markup is not as good as some. There is the sandbox https://www.scienceforums.net/forum/99-the-sandbox/ for testing and also there are several tutorials specific to this site . Don't hesisitate to ask how to do some of this when you need it.

always ?? I'm fed up with members that don't bother to read what's offered, just argue apparantly for the sake of it. Here is a simple treatment which explains why using f = ma makes life more difficult. Because m is not necessarily constant. Professor Sears had a particularly clear way of putting matters.

I am neither fond nor not fond of you. However you specifically offered So I asked to take you up on your offer. So far you have expended several posts avoiding doing so.

Which is exactly why I asked @martillo to start at the beginning. When the rocket is standing ready to launch it has exactly zero momentum. So we can write a conservation of momentum equation right there. This is usually taken for granted and also what I am referring to when I asked about what are they hiding ? Of course but what I am saying is that in order to solve or analyse this problem we start with a complete rocket and fuel as 'the system' and then add further equations coming from the application of mechanics to parts of that system. I further point out that all of the equations are simultaneous so that one can substitute from one into another Again a tried and tested technique. I am also say that there is nothing revolutionary in this, indeed the technique is well used in many simpler systems.

So no mathematics then ? Just like your hand wavy references.

I'm willing to be educated, but the rules here are quite clear. I should not need to go offsite to read the material fundamental to the issue. So fire away with your mathematics please.

I asked you what you know. You haven't quoted a single number. So tell me just how you know what v is ? And you are telling me that i should use f = ma instead of some other equation. What exactly will this give me and where will I get the numbers to put into this equation ? And where exactly does your f = ma fit into your claim that you know v and -ve ? Why do you think I specified a rocket ready to launch ? If you don't want to calculate anything why are we going through all this rigmarole ?