michel123456 Posted February 26, 2011 Posted February 26, 2011 The quick answer is that it's arbitrary. It is a sign convention. Considering your first answer as positive (remember 5,5 light minute = 98931511,25 kilometers). Did you consider my question? Not a second. I would prefer a graph instead of calculations.
Iggy Posted February 26, 2011 Posted February 26, 2011 (edited) Not a second. Correct, the interval is not one second, or a light-second. Measure the spatial distance between events in any frame. Call this [latex]\Delta X[/latex] Measure the time between events in that same frame. Call this [latex]\Delta \tau[/latex] Square (which means "multiply a number by itself") both numbers. Subtract one squared number by the other squared number (either from the other--it doesn't matter which) Whatever the answer is, take the square root of that answer. Whatever you get will be the spacetime interval between those two events. Does this help you with what I asked? Edited February 26, 2011 by Iggy
michel123456 Posted February 26, 2011 Posted February 26, 2011 Correct, the interval is not one second, or a light-second. Measure the spatial distance between events in any frame. Call this [latex]\Delta X[/latex] Measure the time between events in that same frame. Call this [latex]\Delta \tau[/latex] Square (which means "multiply a number by itself") both numbers. Subtract one squared number by the other squared number (either from the other--it doesn't matter which) Whatever the answer is, take the square root of that answer. Whatever you get will be the spacetime interval between those two events. Does this help you with what I asked? I don't care. What i care of is this: simplified spacetime interval considering y=0, z=0 from [math]S^2=x^2+y^2+z^2-c^2t^2[/math] we get [math]S^2=x^2-c^2t^2[/math] [math]x[/math] is the distance, [math]c[/math] is the speed of light, [math]t[/math] is time, [math]S[/math] is the space-time interval. [math]x[/math] is given, it is the distance Earth Sun. [math]t[/math] is the variable. Simplifying more for a given distance of 1 (unities don't matter in this example) we get [math]S^2=1^2-c^2t^2[/math] or [math]S^2=1-c^2t^2[/math] if we put c=1 (can we do that?) we get [math]S^2=1-1^2t^2[/math] or [math]S^2=1-t^2[/math] or [math]S^2+t^2=1[/math] Which is the equation of a circle of radius 1. That's what I care about. That's why I like graphs. Did I make a mistake somewhere? Because if I didn't, it simply shows the importance of the "distance" [math]x[/math] value. But I may be wrong. Please correct me.
Iggy Posted February 26, 2011 Posted February 26, 2011 [math]S^2=x^2-c^2t^2[/math] Correct. [math]x[/math] is the distance, [math]c[/math] is the speed of light, [math]t[/math] is time, [math]S[/math] is the space-time interval. Correct! [math]x[/math] is given, it is the distance Earth Sun. [math]t[/math] is the variable. Not exactly. Both t and x are, in a sense, variables. x is the spatial distance between events and t is the time between events (both expressed in the same frame of reference). I gave both x and t in my question. x is 5 light-minutes and t is 4 minutes. All we have to do is solve for s where I have given you x and t. S is the space time interval. Simplifying more for a given distance of 1 (unities don't matter in this example) we get [math]S^2=1^2-c^2t^2[/math] No, here I think you had the problem. You cannot make x = 1 because x equaled 5 light-minutes in the question that I asked. You cannot simplify x=5 to x=1. [math]S^2=1^2-c^2t^2[/math] No, you should get s2 = x2 - c2t2 where x=5, t=4, and c=1. With that you should be able to solve for s.
michel123456 Posted February 26, 2011 Posted February 26, 2011 I was not answering your question. I am not interested in numerical values. Besides you can change units. It was a small rhetoric presentation of my thoughts on the subject. you didn't answer my question in posts #149 & 151 Positive or negative? Considering your first answer as positive (remember 5,5 light minute = 98931511,25 kilometers).
Iggy Posted February 26, 2011 Posted February 26, 2011 I was not answering your question. I had noticed. I am not interested in numerical values. I didn't realize. I asked for a number. That number. Owl gave his one which you denied. We are waiting for yours. countdown 9. I need just one number. That was fun. I thought you were interested in numerical values.
michel123456 Posted February 26, 2011 Posted February 26, 2011 I had noticed. I didn't realize. I thought you were interested in numerical values. NOW I am interested in numerical value.
Iggy Posted February 26, 2011 Posted February 26, 2011 NOW I am interested in numerical value. The value of S in the diagram?
michel123456 Posted February 26, 2011 Posted February 26, 2011 (edited) No, you should get s2 = x2 - c2t2 where x=5, t=4, and c=1. With that you should be able to solve for s. [math]S^2=25-16=9[/math] [math]S=3[/math] ------------------- edited [math]S=\pm3[/math] Edited February 26, 2011 by michel123456
owl Posted February 26, 2011 Posted February 26, 2011 (edited) The statements contradict each other and you are not seeing why. Get two metal rods that are each 540 million miles long. Lay one to your left, pointing to your left, and another to your right, pointing right. Send someone down the length of each rod at 540 million mph. After one hour each person will be at the end of each rod. Yes? Everyone will agree on this, yes? I put a focused effort into my last post, but you didn't reply to any of it. Then you claim I am contradicting myself yet again with no explanation whasoever, like, for instance, the obvious, how exactly do my statements contradict each other? My original, "what is your point?" was intended to draw out from you what I suspect is your position... namely that added together the two velocities far exceed "C." My point, intended to come after your reply... which never came, was that each can go 540 million mph in opposite directions without violating "C" because they are independent velocities, not cumulative. So I came back and made that point with the two lasers fired in opposite directions, neither ,violating "C"... not a cumulative velocity. But you ignore what you want, most of what I say, and come back with more of the same. It will be a manufacturing nightmare to get two metal rods that long (lame joke), so lets just come back to a little more common sense way of saying it. Each traveler travels 540 million miles in the same hour in opposite directions. Neither is exceeding "C." Still asking, what is your point? And how about answering my previous six replies? Is your time and energy so very much more valuable than mine? I said: "This does not mean that I debate, for instance the Lorentz equations but rather their interpretation. NOW... Now on earth is the same "now" on the sun and everywhere." Lightspeed limit does not change that fact that NOW IS NOW everywhere, just because information takes time to travel from one place to another. Sysyphus replied: "These statements are contradictory. Events which are simultaneous in one reference frame are not simultaneous in any other reference frame. You are not just disputing an interpretation. Whether you realize it or not, you are disputing the demonstrable fact that relativity makes correct predictions." So you say. I say they are not contradictory. What does "length contraction" mean? It means that "Observer A" will see a rod, or sun to earth distance differently than "Observer B." That is the nature of separate frames of reference. It does not mean that the distance between earth and sun actually changes each time it is measued from a different frame of reference, or that a given rod length actually changes for the same reason. Just for once, address what i just said. I aksed, rhetorically: "Are we so human/measurement centered that we no longer credit the cosmos with an existence independent of observational perspective?" Sysyphus replied: "Nobody in this thread is making any such claim." Then explain how sun to earth distance changes with each different frame of reference, and same for a given rod length. Edited to spell Sysyphus name correctly... and more on NOW. Edited February 26, 2011 by owl
Spyman Posted February 26, 2011 Posted February 26, 2011 Edited to spell Sysyphus name correctly... I think you meant: Sisyphus.
Iggy Posted February 27, 2011 Posted February 27, 2011 [math]S^2=25-16=9[/math] [math]S=3[/math] ------------------- edited [math]S=\pm3[/math] I agree. how exactly do my statements contradict each other? That's what I'm trying to explain. I'm using the Socratic method. The first claim I made is that your views contradict themselves. Demonstrating this may take more than a couple posts especially when you are throwing tangents everywhere. I asked two questions in my last post that were not ambiguous at all. If you will please indulge me with two very simple and direct answers.
michel123456 Posted February 27, 2011 Posted February 27, 2011 I agree. (...) My graph gives the correct result. Your post #156 don't give the correct result. I appreciate that you didn't edit it. You should have known that [math]S[/math] cannot be bigger than the radius. That was my point. The radius, which represents distance, don't change, [math]S[/math] does. You may notice that for [math]t=x[/math], [math]S=0[/math], which is the expected result.
Iggy Posted February 27, 2011 Posted February 27, 2011 My graph gives the correct result. Your post #156 don't give the correct result. Only if you use the Euclidean metric. Google Minkowski space. You should have known that [math]S[/math] cannot be bigger than the radius. In a Euclidean metric that is true. Where the metric is s^2 = t^2-x^2 it is not.
DrRocket Posted February 27, 2011 Posted February 27, 2011 Summary: Iggy and Michael are talking past one another. Iggy is correct, but one has to think in the abstract setting of Minkowski space to see why. Such thinking is necessary in special relativity. Michael123456 is trying in good faith, but can't seem to grasp the difference between Minkowski space with the Minkowski metric as opposed to Euclidean space with the Euclidean metric. Michael is out somewhere in left field. He should listen to Iggy. Owl is not even in the ballpark. He is talking to himself.
owl Posted February 28, 2011 Posted February 28, 2011 More old business, and then back to "Iggy's two questions." Dr Rocket: "In general relativity, space is NOT a 3-D volume and time is NOT duration between two chosen events." In the spirit of ontological inquiry, I have been questioning some of the basic assumptions of relativity about the nature of space and time starting with the transition from Euclidean 3-D space to non-Euclidean 4-D spacetime.) If you were to approach this as a respectful dialogue (which you don't) you would not assume the correct status of a "4-dimensional Lorentzian manifold, spacetime." By assuming the premise as true and established, you totally avoid the discussion, sounding like an arrogant, omniscient one, condescending to declare: "Owl is not even in the ballpark*. He is talking to himself" ...because I don't accept the dogma of a "4-dimensional Lorentzian manifold" as established beyond all doubt. (* I am comparing the "ballparks" of Euclid and post Euclid.) I have actually presented the transition from Euclidean to non-Euclidean geometry on these boards in some detail with special focus on how intrinsic curvature in one conceptual manifold is "transformed" into extrinsic curvature in another manifold. In this regard, I quoted at some length from Kelley Ross's essay on The Ontology and Cosmology of Non-Euclidean Geometry. You have sucessfully ignored the whole issue and stuffed me in a category in your mind for one who just doesn't understand relativity. Your first statement above is not an absolute truth just because it is now so well accepted in the relativity community. Here is the most brief summary in contradiction of that dogma: Space is volume described by three axes or dimensions (though, beyond defined borders, there is no end to space... another reference to the title of this topic.). Time can be designated as the duration between two instants, from one "now" to another... or it can be seen as the ongoing "elapsed time" of any/all events . Space and time are not necessarily combined as a new, post Euclidian, Lorentz-ian or Minkowski-an Reality, i.e., the post Euclidean leap is not "proven fact"... a new Reality, as you and most of relativity assume. That'll do for now.
DrRocket Posted February 28, 2011 Posted February 28, 2011 More old business, and then back to "Iggy's two questions." Dr Rocket: "In general relativity, space is NOT a 3-D volume and time is NOT duration between two chosen events." Your first statement above is not an absolute truth just because it is now so well accepted in the relativity community. Blindingly, utterly wrong. What I said is absolutely true in general relativity, precisely as stated. If you have some other view, that view is most certainly not general relativity. Yep, not even in the ballpark. Talking to yourself.
owl Posted March 1, 2011 Posted March 1, 2011 Iggy: "I asked two questions in my last post that were not ambiguous at all. If you will please indulge me with two very simple and direct answers." I presume you are referring to the quote below and that the second question is, "Everyone will agree on this, yes?" "Get two metal rods that are each 540 million miles long. Lay one to your left, pointing to your left, and another to your right, pointing right. Send someone down the length of each rod at 540 million mph. After one hour each person will be at the end of each rod. Yes? Everyone will agree on this, yes?" Ok, so you insist on your metal rods rather than my simplification where each just goes 540 million mph for an hour in opposite directions. (Good grief!) Maybe the rods make the distance traveled more tangible for you. Whatever! Either way, each traveler goes the same 540 million miles in opposite directions. Yes, "After one hour each person will be at the end of each rod." No, I don't think everyone will agree, because some of you believe in "length contraction," by which, in some manner, because of the velocity of each traveler relative to each rod, I think, relativity theorists actually believe that somehow each rod becomes shorter. Please now answer how this is NOT subjective idealism ("subject" being frame of reference): Measurement from frames of reference at high velocity (our travelers) actually changes the reality of what is measured. Another challenge you never answerd but to say, that, no, relativity is not based on subjective idealism... that with no explanation or further comment.) OK, I have now answered your two question and have, I would think, a right to reciprocation vis-a-vis my six replies above. Blindingly, utterly wrong. What I said is absolutely true in general relativity, precisely as stated. If you have some other view, that view is most certainly not general relativity. Yep, not even in the ballpark. Talking to yourself. Forgive me for I have sinned in that I do not accept everything about relativity as absolutely true! You have spoken Ex-Cathedra as a True Believer in the doctrine of relativity. Anyone questioning the dogma is a heretic and a fool. You don't even bother to engage in the conversation enough to address the specifics in my last post. Oh well. I see new developments on the horizon in which good ol' Euclid might find a new respect among relativity theorists, but you will not be among them. Your mind is made up, and you are not about to waste your time with a fool like me. Such fundamentalism right here in science!
Iggy Posted March 1, 2011 Posted March 1, 2011 Either way, each traveler goes the same 540 million miles in opposite directions. Yes, "After one hour each person will be at the end of each rod." Good. Thank you. No, I don't think everyone will agree, because some of you believe in "length contraction," I should have been more specific. If everyone in the thought experiment thinks like you do and believes like you believe then they all agree on the position of everyone at the end of the hour, yes? Let me do a graphic: At t = 0 everyone agrees this is everyone's position as well as the position of the rods, yes? At t = 1 hour everyone agrees that this is everyone and everything's position, yes? Is that correct?
michel123456 Posted March 1, 2011 Posted March 1, 2011 I like Leonardo at the center. You have talent. You are making someone run at 80% of the speed of light from Earth to Jupiter. You should have drawn a flame going out of his bottom. At this speed, t=1hour on the left & right should be deleted. And you must describe the method Leonardo used to measure 540 million miles in the first place, because to an alien it is not 540 million miles*. And it is not 1 hour. It is only for Leonardo. *Unless you accept the statement that Leonardo, who is in the same FOR of its rod, who holds the rod in his hands (wow) measures right and all the others around are wrong.
Spyman Posted March 1, 2011 Posted March 1, 2011 (edited) You are not making any sence at all Michel... In thought experiments like this there is no problem with having fast runners and long rods. Time that has passed can not be deleted, all observers will measure a duration on their clocks. There is no claim that "Leonardo" is unable to make correct measurements in his frame of reference. There is no claim that other observers are unable to make correct measurements in their frames of reference. How different "Aliens" measures distance and durations is irrelevant to the thought experiment. There are three observers, "Leonardo" in the middle, "Giovanni" running left and "Raphael" running right, all three of them have clocks & rulers and are able to make correct measurements. Edited March 1, 2011 by Spyman
michel123456 Posted March 1, 2011 Posted March 1, 2011 Time that has passed can not be deleted, all observers will measure a duration on their clocks. I ment deleted on the graph. With Tipp-Ex. You seem annoyed when I put a doubt on the premises. All observers will make correct measurements in their FOR. But at that speed, they cannot be in the same FOR, especially due to the tremendous acceleration in order to reach that speed. Giovanni & Raphael are decalcomania on their spaceship seats (I can't figure they go by foot). You must know that you cannot extract safe conclusions from a thought experiment based on unphysical situation in the first place. I always wanted to say that. But if you insist, Iggy may proceed.
Iggy Posted March 1, 2011 Posted March 1, 2011 But at that speed, they cannot be in the same FOR I'm using crayons, Tipp-Ex, and stick figures to show that absolute time, length, and simultaneity are inconsistent with a constant speed of light.
Spyman Posted March 1, 2011 Posted March 1, 2011 (edited) Michel, what/which part of my post make you think I am annoyed? They are in different frames due to speed which is part of Iggy's discussion with Owl. But there doesn't have to be any acceleration in the thought experiment, Giovanni & Raphael could have accelerated slowly and safe before and then pass close by Leonardo in opposite directions simultaneous while synchronizing their clocks. "Decalcomania, from the French décalcomanie, is a decorative technique by which engravings and prints may be transferred to pottery or other materials." Since there is no huge accelerations involved both Giovanni & Raphael are safe. If they run or travel with spaceships and what model and color of their spaceships or what engine type they are equipped with and other such technical stuff are totally meaningless and irrelevant to the thought experiment. How they manage to reach that tremendous speed is unimportant to the question, whats important is that the speeds or the situation is NOT unphysical or impossible. Edited March 1, 2011 by Spyman
Iggy Posted March 1, 2011 Posted March 1, 2011 But there doesn't have to be any acceleration in the thought experiment, Giovanni & Raphael could have accelerated slowly and safe before and then pass close by Leonardo in opposite directions simultaneous while synchronizing their clocks. I can affirm that. Michel, by the end of the thought experiment I will have decomposed the frames. I still need to add another element first. Hopefully a quick "yes" from Owl will let me go ahead with that.
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