  # Pete

Senior Members

367

## Everything posted by Pete

1. The work done on the body is zero since the force on the body (only force acting here is grafvity) is perpendicular to the displacement (F*dr = 0 where "*" = dot product). Think of sliding a body over a surface of zero friction - no work is done there either. The runner does no work agains friction either (for the most part anyway). However there is work being done inside the body in terms of biochemical reactions (muscle contractions etc). That's why we get tired when we run and even when all we do is press our hands together or stand. If there is a non-zero amount of work done on a then the velocity of the body changes. The work-energy theorem tells us that the work done on a body by conservative forces equals the change in the kinetic energy of the body. Positive work acclerates the body increasing the bodies kinetic energy. Negative work slows the body down thus decreasing the bodies kinetic energy. That is correct. If you're asking if work is done to move the body across the room then the answer is no. If you're asking if work is done ti life the body then the answer is yes. Those other texts are talking about the entire process of lifting and moving. The net result is a change in the bodies gravitational potential and any change in kinetic energy that may arise depending on how the body is moved (e.g. from a state of rest to a state of rest or from a state of rest to a state of motion etc.)
2. QM also says that there are systems/states which are certain even when they aren't being observed. Such states are called eigenstates.
3. I believe that swansont is refering to what he believes are philosophical/conceptual difficulties, not physical ones. Merged post follows: Consecutive posts merged He's referring to particle's with a finite/non-zero proper mass. Light has zero proper mass so what he's saying does not apply to light.
4. I'm assuming that the 500 seconds you stated above is an approximation to the actual travel time. Typically one has to state with respect to what frame of reference one is measuring time so that 500 s is with respect to a particular frame of reference. The precise travel time is only a very small difference from that in the absense of gravitational effects. This means that approximately its 500 seconds from all frames of reference which are moving slowing with respect to the sun. The 186,000 miles per second is the speed of light in flat spacetime as measured from an inertial frame of reference. The speed of light is a function of the gravitational potential and thus varies with position. Yes. To be precise that does not prove GR. It merely comnfirms a prediction made by GR. Space can be curved by gravity. the presence of time dilation does not gaurentee spatial curvature.
5. No. Its not meaningful to speak of something being bent in time.
6. What do you mean here when you say that the uncertainty is inherent in the system? The value calculated for the uncertainty is uniquely determined by the quantum state that the system is in. E.g. $\Psi$ determines $\Delta X$. Is that what you meant?
7. What does it mean for a universe to be inside out??
8. By definition the term freefall means that the only force acting on a body is the force of gravity. Thus one body orbiting another is in freefall (as in falling freely). In such cases one can construct a locally inertial frame of reference with the body in freefall at the origin. The term weight refers to the force that supports a body in a gravitational field. If there is no supporting force then the body is weightless.
9. You can learn about it in Black Holes and Time Warps: Einstein's Outrageous Legacy, by Kip S. Thorne and Exploring Black Holes by Taylor and Wheeler - See http://www.eftaylor.com/pub/chapter2.pdf
10. That's a pretty lame response. When you post comments as you keep insisting on doing which are intended to provoke (or don't you realize that what you are posting are provoking comments?) then you should be prepared to accept the responses that you get when you take such action and not constantly whine about them as you keep doing. Are you so ignorant that you couldn't figure that out and instead responed with a totally irrelavant comment as you have in your last response?
11. Prove it. swansont - Given Ophiolite's misuse of the term in this thread I now see that I'll have to agree with you after all.
12. Actually I was more interested in concrete examples rather than a yes/no response.
14. No. Inertia is the property of matter which resists changes in the motion of the body.
15. Infraction regarding what? Asking why someone posted something the way they did or for objecting to certain assertions that were made? I'm not the one who isn't focusing on the physics.
16. Its annoying when a moderator asserts that someone is being melodramatic when they ask a question to someone who doesn't seem to be taking a derivation/proof seriously. Such comments have an effect which is opposite to the desired one. Have you ever considered using Pm for such comments?
17. Of course I understand why someone would ask a poster to define a term they used since, as you say, that may very well be the case and that's perfectly understandable. Recall that I provided a definition when one was requested. Unfortunately that was not enough since people kept making invalid assumptions regarding my comments. E.g. you and Mr Skeptic assumed that I didn't understand why somone might ask for a definiton. Can we now drop this definition thing and stay with the original question. I'm not interested in discussing the definition myself. Thanks. That's nonsense. I hope you weren't serious since you are clearly using the term incorrectly. Didn't you read swansont's definition? In this case being closed minded would mean that I would not care about the reaons for posting a definition. That is clearly not the case. I was interested in the reasons. I considered them and in the end I disagreed with them. I disagree. I was giving an example of this term being understood without needing a dictionary. When the crackpots show up they should be corrected on the fly since they don't care what the actual definition is. Being crackpots they are going to disagree with the definition regardless of where it comes from.
18. I'm not saying that a definition is not neccesary. I'm saying that the definition is that what one normally means by "open minded." I'm sure you've heard this term, haven't you? Have you ever known there to be two different meanings to it? This is a commonly used term and already has a well-known definition and I was asking why swansont thinks it needs to be defined. There are many terms which really do need to be defined. For example; were you aware that when the author of a QM text uses the term "momentum" he doesn't mean what everyone else does (i.e. mechanical momentum). He is referring to canonical momentum. That is something rarely mentioned in QM texts but its important to keep in mind, especially since they are spelled exactly the same. Lost my chance at what?
19. The later is not what being open-minded means. Only crackpots use the term in that sense and I don't tend to define terms with crackpots in mind. I'm curious was to why you think it needs to be defined. Have you never heard or used this term before? One merely uses common sense when answering this question. One does not need a dictionary to have a conversation in most cases and I believe that this is one of those cases where none is needed. I E-mailed two physicists friends of mine and each of them knew exactly what one means when speaking of being open minded. Many books refers to the idea that scientists should be open minded and I don't recall one of them needing to define it. And Aristotle's comment does not contradict what one normally considers to be open minded.
20. That was my point and he should merely state that and then he should look the derivation up in a text like Rindler's or Tolman's. This is not new physics or something I created/imagined. When people do things like this (put terms like "proof" in quotes) it tends to indicate that they're not taking the subject seriously. melodrama ? What makes this a melodrama? Is it because I wanted to know why Severian has this attitude before he starts attempting to follow a derivation? It will just make it harder for him to follow. Its not wise assuming something is in error before you start to follow it. If Severian figures out what a box is and why rigidity has nothing to do with the topic/derivation then we can move forward. Its not as if boxes (or box-like objects such as a cylinder) haven't been used in relativity thought experiments before. Einstein was the first to do so in fact in what has come to be known as Einstein's famous "photon in a box" experiment. Severian - If you have Rindler's SR text then I suggest that you read the chapter on continuum mechanics and learn this material from there or perhaps even from Tolman's SR/GR text. Do you have either of these books?
21. Yes. I'm very aware of that fact. But so what? Why do you mention it? Nothing in what I've said so far has anything to do with the problems of rigidity. A box with a gas in it is in dynamic equilibrium. Do you understand what Rindler means when he speaks of the problems of the rigidity of an object? He is not referring to any problem with a box having rigid walls and thus can contain a gas and maintain the shape of a box. He's referring to the fact that it a disturbance on an object does not propagate instantaneously throughout the body. Rindler is a supurb author and relativist. His books are very good in that they are quite comprehensive. They address subtleties that most other texts ignore. By rigid I'm refering to the fact that the walls of the box are sturdy (i.d. they are not made out of rubber). Now that I've answered your questions perhaps you'll answer mine. Why do you keep putting proof in quotes?
22. As opposed to what? Why do you keep putting proof in quotes? You're telling me that you don't know what a box is? When following a derivation one starts with basic knowledge (like what a box is. If you can't figure out what a box is then look the term up in a dictionary) and goes from there. If, during the process, the construction of the box becomes important then one becomes more precise at that time. I'll give you a hint - What I mean by "box" is identical to what the author of this paper means The mass of a gas of massless photons, H. Kolbenstvedt, Am. J. Phys. 63(1), January 1995 I can upload this onto my website if you are unable to figure out what a box is. If you have Schutz's new book Gravity from the Ground Up then crack it open since he means the exact same thing. A box is a box is a box. A box is rigid by definition. In any case the example I gave was about a rod, not a box. And the expression for the energy-momentum tensor can easily be ontained merely by transorming the tensor from a frame of rest to a moving frame using a Lorentz transformation
23. The existance of red shift and blue shift does not require relativity. The existance of red/blue shift can easily be understood in terms of the Doppler effect. Are you familiar with it.
24. As I've already stated, your counter example is flawed. Is there a reason why you ignored the explanation I gave as to why? I don't follow. When exactly did you make this assumption? Was it before or after I gave a proof? And why would you make such an assumption? You asked for proof of my assertion so I provided a proof by giving a counter example which clearly stated that there were external forces acting on the body. I don't follow. How is that relevant to the example I gave? Why do you think that "it is not surprising"? From your comments it seems to me that we're not talking about the same thing. I had stated that one cannot define proper mass for a body in all possible cases. To prove such an assertion one merely has to provide an example where a proper mass can't be defined and that is exactly what I did. Its the stress that is non-zero. The total force on the body could very well be zero though. I neglected to mention that fact. I.e. that even when the total external forces on the object is zero the example I gave still holds. Let me be more concrete example: Consider an inertial frame of reference S in which there is a rod lying at rest along the x-axis. Let there be applied a force to each end of the rod (these forces are parallel to the x-axis) which are equal and opposite. The total force is zero but now there is stress in the rod. Now transform to the inertial frame S' which is in standard configuration with S. The rod is moving in this frame with speed v. The inertial mass of a body is defined as m = p/v. The rest mass is defined in the limit as v -> 0. If the stressed rod were lying parallel to the y-axis rather than the x-axis then m would have a different value. It is important to keep in mind here that in such a case the relation E = mc2 no longer holds. If E is the total inertial energy of the rod (defined as the kinetic energy plus kinetic energy) and p is the total momentum of the rod then the quantity (E,p) will not be a 4-vector and, contrary to your assumption, the magnitude will not be invariant and thus no proper mass can be defined. I.e. there does not exist a 4-momentum for the object in such a case. Since there is no 4-momentum one can't define a magnitude for it and thus one cannot define a proper mass. A similar case is to consider the mass of a gas which is contained in a box. The inertial mass of the gas will be a function of the pressure of the gas. To be precise the inertial mass density of the gas is given by Inertial mass density = $\rho + p/c^2$ This is an example when the 4-momentum won't give you the inertial mass of quantity and one must use the stress-energy-momentum tensor. The frame is always implied. There is little use in stating "As measured in frame S" since it is implied that one is refering to a particular frame of reference. This is normally true whenever any frame dependant quantity is being discussed. In case anyone is wondering whether this all merely academic or not, consider the fact that when one calculates mass density in such applications such as calculating active gravitational mass. In cosmology there are objects known as a vacuum domain wall and the better known cosmic strings. A domain wall has tension in it (negative pressure) which acts to oppose the positive mass-energy and gives a net result of a negative active gravitational mass. This means that a vacuum domain wall gravitational repels objects away from it (these objects are described in Peebles book on cosmology).
×