swansont Posted January 20, 2010 Posted January 20, 2010 There are two components to translating time in LT. t' = ( t - vx/c²)λ x is not the distance traveled of the object, it is a position in the stationary coordinates. So, I do not know what you are driving at here. If x is is a constant, then v must be zero. If v is nonzero, then x is a variable. The frequency of the clock is changed because that is how time is measured. So the frequency of the satellite is programmed in orbit so that it will register at a frequency of 10.23 MHz. Five sources of relativistic effects contribute in Figure 2. The effects are emphasized for several different orbit radii of particular interest. For a low earth orbiter such as the Space Shuttle, the velocity is so great that slowing due to time dilation is the dominant effect, while for a GPS satellite clock, the gravitational blueshift is greater http://relativity.livingreviews.org/Articles/lrr-2003-1/ Now, I see t' = tλ, where t is the satellite clock, so that t = t'/λ in the equations, but I do not see what you are proposing. time ≠ frequency e.g. if you look at eqn 34,you see a frequency shift due to dilation, integrated over the path to get the time dilation. It doesn't look like your equation because Ashby is using spherical coordinates, as one should for such an exercise. But if you do the math right, you should come up with the same answer. Merged post follows: Consecutive posts mergedI found a paper today that sees the same thing I see with GPS. (2) the speed of light is dependent on the receiver’s tranlational motion relative to the ECI frame if the source is stationary (although simultaneity is the key of GPS operations and the relativity of simultaneity of Special Relativity disagrees with the basic operational principle of GPS) The term -vxλ/c² is the relativity of simultaneity. Then his major conclusions is: From the calculations given before, we find that the speed of light in a reference frame moving relative to the ECI frame is c – v or c + v. http://web.stcloudstate.edu/ruwang/IAINpaper2000.pdf The problem with this conclusion he implies the speed of light is not a constant. He should have said the constant speed of light is measured c – v or c + v. This is similar to the error in SR. The speed of light is a constant but the measurement technique fails. So, light speed is not c + v and c - v. That represents our failure to measure it correctly. You found someone with the same misconceptions perhaps (this was a conference proceeding, i.e. not a peer-reviewed paper — do you have a record of how many attendees objected to the conclusions?) or you have misinterpreted it. What about the thousands of papers and scientists that disagree with you? But you can't use GR in multiple frames to show that c is not constant, because different GR frames will measure different values due to the additional effects, as I have explained before. This is getting tedious.
vuquta Posted January 20, 2010 Author Posted January 20, 2010 If x is is a constant, then v must be zero. If v is nonzero, then x is a variable. You cannot possibly be asserting any type of reasonable logic based on this statement. You said it, prove it. time ≠ frequency e.g. if you look at eqn 34,you see a frequency shift due to dilation, integrated over the path to get the time dilation. It doesn't look like your equation because Ashby is using spherical coordinates, as one should for such an exercise. But if you do the math right, you should come up with the same answer. Merged post follows: Consecutive posts merged You should not have ever made this statement. Yes, he is doing a path integral for time since all paths are spherical. This is correct. But, since you now adhere to this logic, where is your simultaneity term? It is not in there and now you are forced to agree with my conclusions. Thanks. You found someone with the same misconceptions perhaps (this was a conference proceeding, i.e. not a peer-reviewed paper — do you have a record of how many attendees objected to the conclusions?) or you have misinterpreted it. What about the thousands of papers and scientists that disagree with you? But you can't use GR in multiple frames to show that c is not constant, because different GR frames will measure different values due to the additional effects, as I have explained before. This is getting tedious. I was very clear that c is a constant and cut the author for not stating this. But, it is tedious since all this evidence is proving that the SR method of measuring light speed fails. Light is c, but SR cannot measure it. I can help you with this. Light moves through space at c. Now, a frame wants to measure light. Does this frame move through space. I will give you your answer, yes it does. So, while light moves through space, what is the frame doing? You have no answer.
swansont Posted January 20, 2010 Posted January 20, 2010 You cannot possibly be asserting any type of reasonable logic based on this statement. You said it, prove it. By definition. v = dx/dt You should not have ever made this statement. Yes, he is doing a path integral for time since all paths are spherical. This is correct. But, since you now adhere to this logic, where is your simultaneity term? It is not in there and now you are forced to agree with my conclusions. Thanks. Near as I can tell, this "simultaneity term" is a requirement made up by you. i.e having that specific form of a term show up in the equation. What you are looking for is the time dilation correction, and it's there. If you do the math in spherical coordinates, it will look different. And you get what is in the Ashby article. It corrects the time for the two coordinate systems. You don't need any extra term — that covers it all, if you do the analysis correctly, which is where you run onto trouble. I was very clear that c is a constant and cut the author for not stating this. But, it is tedious since all this evidence is proving that the SR method of measuring light speed fails. Light is c, but SR cannot measure it. It quite obviously fails when you try and apply it under conditions where the postulates do not apply, that is, under non-inertial conditions. I can help you with this. Light moves through space at c. Now, a frame wants to measure light. Does this frame move through space. I will give you your answer, yes it does. So, while light moves through space, what is the frame doing? You have no answer. The frame is at rest, by definition. Insisting otherwise is one reason why you can't do the analysis properly.
vuquta Posted January 20, 2010 Author Posted January 20, 2010 By definition. v = dx/dt I agree, but what are you doing with this? I am not following it. Near as I can tell, this "simultaneity term" is a requirement made up by you. i.e having that specific form of a term show up in the equation. What you are looking for is the time dilation correction, and it's there. If you do the math in spherical coordinates, it will look different. And you get what is in the Ashby article. It corrects the time for the two coordinate systems. You don't need any extra term — that covers it all, if you do the analysis correctly, which is where you run onto trouble. Well, there is time dilation and then there is the simultaneity term. So, for co-local events, time dilation applies. But, for events ie light events, t' = ( t - vx/c² )λ. Since an event is light travel, t = x/c, when lookin only at the x-axis point. t' = ( x/c - (v/c)(x/c)λ. t' = x/c(1 - (v/c))λ. = x/c(c/c - (v/c))λ = x/c√[( c - v )/(c + v)] So, t' = x/c√[( c - v )/(c + v)]. This implies, the further light travels down the x-axis, the less synchronous the moving clock and the stationary clock become per second. This is an extra term compare to time dilation. Time dilation desyncronizes at a constant rate per second. On the other hand with LT, the clocks in the moving frame desyncronize at an ever increasing rate per second compared to the corresponing stationary clock at x as x increases, which is the distance from the origin in the stationary frame. That does not show up in GPS. It quite obviously fails when you try and apply it under conditions where the postulates do not apply, that is, under non-inertial conditions. No, because all the other adjustments for path and GR are included in GPS. Further, the thought experiment I did was for 3.9 km which is a straight line comparison in both frames and LT did not apply to this experiment but is supposed to. The frame is at rest, by definition. Insisting otherwise is one reason why you can't do the analysis properly. Is the frame at some absolute rest? No. It is a simple thought process. Light moves through space at c. Does the light receiver move through space also in some unknown way? Naturally, I can claim the frame is at rest in space but is that realistic?
swansont Posted January 20, 2010 Posted January 20, 2010 I agree, but what are you doing with this? I am not following it. Go back and reread then. Well, there is time dilation and then there is the simultaneity term. No this is an artificial distinction you are making. The time difference between two frames is time dilation. Period. You are trying to break it into two parts — a phase (time) shift, which you are calling time dilation, and a "simultaneity term" but that's incorrect. You are misapplying/misinterpreting the Lorentz transformation. The first term is there in case you haven't synchronized the clocks, and they have a time difference at the beginning of the motion. If you have synchronized them when they are co-located (in a 1-D problem), you set t to zero. So, for co-local events, time dilation applies. But, for events ie light events, t' = ( t - vx/c² )λ. Since an event is light travel, t = x/c, when lookin only at the x-axis point. t' = ( x/c - (v/c)(x/c)λ. t' = x/c(1 - (v/c))λ. = x/c(c/c - (v/c))λ = x/c√[( c - v )/(c + v)] So, t' = x/c√[( c - v )/(c + v)]. This implies, the further light travels down the x-axis, the less synchronous the moving clock and the stationary clock become per second. This is an extra term compare to time dilation. Time dilation desyncronizes at a constant rate per second. On the other hand with LT, the clocks in the moving frame desyncronize at an ever increasing rate per second compared to the corresponing stationary clock at x as x increases, which is the distance from the origin in the stationary frame. The equation is linear in x, so this last statement is clearly false. That does not show up in GPS. <Sigh> Yes it does. It's just done in a different coordinate system with a different formulation, but it's there. There is clearly a kinematic time dilation term. Is the frame at some absolute rest? No. It is a simple thought process. Light moves through space at c. Does the light receiver move through space also in some unknown way? Naturally, I can claim the frame is at rest in space but is that realistic? Yes it is realistic, and it has experimental confirmation of being realistic.
vuquta Posted January 20, 2010 Author Posted January 20, 2010 Go back and reread then. Ok, this does not mean anything. No this is an artificial distinction you are making. The time difference between two frames is time dilation. Period. You are trying to break it into two parts — a phase (time) shift, which you are calling time dilation, and a "simultaneity term" but that's incorrect. You are misapplying/misinterpreting the Lorentz transformation. hmmmmmm If I set x = vtλ/(1+λ), I find that t'=t using t' = ( t - vx/c² )λ Where is the time dilation? So, when you try to connect frames with time dilation exclusively, this point breaks that argument. The first term is there in case you haven't synchronized the clocks, and they have a time difference at the beginning of the motion. If you have synchronized them when they are co-located (in a 1-D problem), you set t to zero. t' = ( t - vx/c² )λ OK, let's assume they are co-located. So, you set t = 0. t' = ( t - vx/c² )λ = ( 0 - vx/c² )λ So, t' = -vxλ/c² Therefore, for x > 0, since v > 0, then t < 0. That means we can go back into the past. Yes it is realistic, and it has experimental confirmation of being realistic. No, there are no one way speed of light timing experiments done with clocks. At one mile, these timing experiments would need to function at the pico-second level. Anyway, you claim light moves through space at one speed c. So do I. A frame is just a coordinate system and is not space. So, how is your coordinate system moving through space while light does?
swansont Posted January 20, 2010 Posted January 20, 2010 If I set x = vtλ/(1+λ), I find that t'=t using t' = ( t - vx/c² )λ Where is the time dilation? So, when you try to connect frames with time dilation exclusively, this point breaks that argument. Why would you use x = vtλ/(1+λ)? Other than to fudge a result, that is. x = vt by definition. There are no primed values in the equation. You can't use some other arbitrary equation. t' = ( t - vx/c² )λ OK, let's assume they are co-located. So, you set t = 0. t' = ( t - vx/c² )λ = ( 0 - vx/c² )λ So, t' = -vxλ/c² Therefore, for x > 0, since v > 0, then t < 0. That means we can go back into the past. Are you sure of the sign convention? No, there are no one way speed of light timing experiments done with clocks. At one mile, these timing experiments would need to function at the pico-second level. Anyway, you claim light moves through space at one speed c. So do I. A frame is just a coordinate system and is not space. So, how is your coordinate system moving through space while light does? What is space?
vuquta Posted January 21, 2010 Author Posted January 21, 2010 Why would you use x = vtλ/(1+λ)? Other than to fudge a result, that is. x = vt by definition. There are no primed values in the equation. You can't use some other arbitrary equation. No, Using LT by Einstein, t' = ( t - vx/c² )λ Then I set x = vtλ/(1+λ) in which case t' = t. This is all legal. Then, there is no time dilation. Now what? Are you claiming Einstein was wrong? Me too. Originally Posted by vuquta t' = ( t - vx/c² )λ OK, let's assume they are co-located. So, you set t = 0. t' = ( t - vx/c² )λ = ( 0 - vx/c² )λ So, t' = -vxλ/c² Therefore, for x > 0, since v > 0, then t < 0. That means we can go back into the past. Are you sure of the sign convention? Well, this is Einstein's junk, so according to him, that is the result. What is space? Yea, that which light moves. Oh, so do frames in some unknown way. Actually, though off this track, I don't know. I really don't know what space is. I do know this though. While light moves through space, so does the frame. Therefore, SR's claim that t = d/c is universal for measuring light speed is false because this claims the "at rest" frame is at absolute rest. Merged post follows: Consecutive posts mergedThis was my original post..perhaps I posted in the wrong Forum but I never stated that I had all the answers, The introduction to the Forum was for all levels of Knowledge of Physics, not for Mathmaticians only.Math is not the 'Holy Grail' and dose not hold all the answers, although I respect your experties in Math and cannot begin to be as good, I do have a good knowledge of Physics and life knowledge I just stated what I think and did not wish to encroach on your held beliefs. Math is a means to an end and Equations can be used to prove or disprove any theory, but although I believe that Math is useful, it is not always the answer to the question, that is why I named my thread as a 'Holistic' view, meaning an all round view of how things work. I do not mind your ridicule or your condecending attitude if that is what makes you feel superior and I do not visit the Forum every day but I will continue to post if I think that my thoughts are valid. So try not to confuse me in Math as it wont work.. I can only give you an opinion. For some reason, this place/universe operates according to the rules of math. There would be no current technology if this were false. So, when describing events and behaviors in the universe, math has been shown to be the best model for its description. Don't ask me why because I do not know.
mooeypoo Posted January 21, 2010 Posted January 21, 2010 (edited) since x=vt by definition, by setting x = vtλ/(1+λ) you essentially said that your t is equal to tλ/(1+λ) which is not what you wrote above. If x=vt, how is it legal to arbitrarily choose a value for x and then be all surprised when the equation fails? Merged post follows: Consecutive posts merged So, when describing events and behaviors in the universe, math has been shown to be the best model for its description. Don't ask me why because I do not know. Because it's very clear, very objective, you can't really interpret anything twice (assuming you do things right and you're well-defining everything) and it has powerful predictability values. That's why. Edited January 21, 2010 by mooeypoo Consecutive posts merged.
insane_alien Posted January 21, 2010 Posted January 21, 2010 if it can't be measured then it does not affect reality and is therefore irrelevant. if it has an effect on reality then it is measurable and therefore is relavant. it cannot be both.
vuquta Posted January 21, 2010 Author Posted January 21, 2010 since x=vt by definition, by setting x = vtλ/(1+λ) you essentially said that your t is equal to tλ/(1+λ) which is not what you wrote above. Let's see the math. You are wrong. I said set x = vtλ/(1+λ)\, then t'=t. Because it's very clear, very objective, you can't really interpret anything twice (assuming you do things right and you're well-defining everything) and it has powerful predictability values. That's why. Yes, but I do not know why math is the best language for the universe. Just because it works is not an answer as to why. Merged post follows: Consecutive posts mergedif it can't be measured then it does not affect reality and is therefore irrelevant. if it has an effect on reality then it is measurable and therefore is relavant. it cannot be both. This is geocentric thinking similar to SR saying an "rest frame". We abandoned this thinking years ago, sans SR. Measurability does not create reality. I will give an example. Take photosynthesis. Energy transfer rates between pigments are very rapid, and charge separation in reaction centers occurs in 3-30 picoseconds http://photoscience.la.asu.edu/photosyn/education/photointro.html Plants operate in the pico second range. Yet, our ability to measure this only became real in the last decade. However, plants have been around 4 billion years. Now, did photosynthesis become real in the last decade or was it real for 4 billion years? Your logic is similar to much of quantum theory which is absurd in this sense as I showed.
mooeypoo Posted January 21, 2010 Posted January 21, 2010 Let's see the math. You are wrong. I said set x = vtλ/(1+λ)\, then t'=t. You are not making sense. The definition of x=vt. That's where you start from, no matter what. That's its definition. If you want to set x=vtλ/(1+λ)\ then it means that you changed either t or v to equal something other than their original selves. You can't have BOTH t=t' AND x=vtλ/(1+λ)\ If x=vtλ/(1+λ)\ and therefore t=tλ/(1+λ)\ Let's take a more 'basic' equation for example. F=ma. I can, mathematicall, claim that F=bsm/x but that would make my a=bs/x by definition, which would then lead you to some very awkward problems with the definition of a(acceleration = [math]dx^2/d^2 t[/math]) You can't just arbitrarily change the formulas that were derived mathematically and then exclaim you found an error. Yes, but I do not know why math is the best language for the universe. Just because it works is not an answer as to why.
vuquta Posted January 21, 2010 Author Posted January 21, 2010 You are not making sense. The definition of x=vt. That's where you start from, no matter what. That's its definition. If you want to set x=vtλ/(1+λ)\ then it means that you changed either t or v to equal something other than their original selves. You can't have BOTH t=t' AND x=vtλ/(1+λ)\ If x=vtλ/(1+λ)\ and therefore t=tλ/(1+λ)\ Let's take a more 'basic' equation for example. F=ma. I can, mathematicall, claim that F=bsm/x but that would make my a=bs/x by definition, which would then lead you to some very awkward problems with the definition of a(acceleration = [math]dx^2/d^2 t[/math]) You can't just arbitrarily change the formulas that were derived mathematically and then exclaim you found an error. Yes, but I do not know why math is the best language for the universe. Just because it works is not an answer as to why. We are talking about SR. Do you know SR? If you do not, I can show you some things.
mooeypoo Posted January 21, 2010 Posted January 21, 2010 Yes, but I do not know why math is the best language for the universe. Just because it works is not an answer as to why. You either have a language problem, vuquta, and you didn't UNDERSTAND what I wrote, or you have an attitude problem and you ignored what I wrote. Either way, this isn't the first time. You need to stop dismissing valid claims just because you feel like it. Measurability does not create reality. It does in science. I will give an example. Take photosynthesis. Energy transfer rates between pigments are very rapid, and charge separation in reaction centers occurs in 3-30 picoseconds http://photoscience.la.asu.edu/photosyn/education/photointro.html And you can measure it with equipment, and you can measure its direct and indirect results. It affects reality. You can measure its effects on reality, hence, it's measurable. Plants operate in the pico second range. But the general effect is accumulative and is detected. Case in point: we detect (and know quite well on the operations of) photosynthesis. Your point just made the opposite case of what you claim. Yet, our ability to measure this only became real in the last decade. Yes, until we were able to measure this, it was uncertain that the phenomena exists. We are not claiming that measurements *MAKE* reality, we are claiming that measurements make what we KNOW OF reality. If you claim something totally new, you need to propose not only a way to observe it, but to observe its effects on other things. We've never seen or observed a Black Hole either, but we can detect its effects on other objects and explain it. The problem with your assertion is not that there's no observation to prove it, it's that there are observations and mathematical data to disprove it. You're just insisting on ignoring them. Your logic is similar to much of quantum theory which is absurd in this sense as I showed. You didn't show anything, you only insist on ignoring what people explain to you, give faulty mathematics, claim you know all and take us on a circular path of redefined definitions. This thread is going in circles; you had ajb explain it to you in a geometric perspective, Klaynos from an experimental perspective and swansont from a theoretical perspective, and that was after I spent half a thread with you trying to convince you to stick to the proper definitions for the sake of a clear debate. It's time you stop lecturing and take a breather to actually read what people write to you. And don't you dare say that your questions were unanswered. They were. Multiple times, you just ignore them or dismiss them and then wait enough time to hope we forgot, so you can ask them again and claim we have no answer. How many more people do you want to prove to you that you are wrong (and from how many angles) before you start considering the possibility? If the answer is "infinite" or "never" then you're in the wrong forum, and the wrong profession. ~moo
Cap'n Refsmmat Posted January 21, 2010 Posted January 21, 2010 Let's see the math. You are wrong. I said set x = vtλ/(1+λ)\, then t'=t. Why do you set x = vtλ/(1+λ) rather than just x=vt?
Chriton Posted January 21, 2010 Posted January 21, 2010 Math is the language of physics. You will never be able to grasp the more complicated parts of physics without it. Don't worry, you're not alone. Many of us here, me included, will never have a complete understanding of physics, but we have learned not to dismiss what we don't fully grasp. You really need to stop referring to science as a "held belief". The scientific method ensures that evidence takes precedence over personal wishes and hopes. This is wrong. If a theory is valid, math can't be used to disprove it. Maths used in physics is not like statistics; you can't manipulate the numbers to give you the answer you want if it's not the right answer. Science wouldn't be the accurate methodology it is if one person's answers were just as valid as everyone else's. It's not so much about right and wrong as it is about the best working explanation of reality. You might try reading the work of the thousands of people who have worked and studied all their lives to produce answers their peers judge to be the best working explanation of reality before you start rewriting their work. No ridicule or condescension intended, EVER. Yes! dont get me wrong, I like Maths, I am not up to your standard of equations but Physics is about the Physical universe around us and not all explanations are mathematical. Over the centuries Phyicists and Scientists have came up with mathmatical explanations of how things work, and others have disproved their Math so Physics has been an arguments of how the Math works in physics, nothing is set in stone, even in this Thread there is disagreements about the speed of light and how you measure it, I do not knock the people who dedicate their lives to understanding the meaning of existence but some are blinkered in how they percieve what they are actualy trying to understand, I try to be open minded and although I am not a Mathmatition I try to use Logic and Math to think about things.
vuquta Posted January 21, 2010 Author Posted January 21, 2010 (edited) You either have a language problem, vuquta, and you didn't UNDERSTAND what I wrote, or you have an attitude problem and you ignored what I wrote. Either way, this isn't the first time. You need to stop dismissing valid claims just because you feel like it. I do not think so. I gave the point x = vtλ /(1+λ ) Then you plug this into t' = ( t - vx/c²)λ When you do you get t'=t. Do you need help with the language of math? I can do this proof for you. It is not hard. But, math facts are math facts. And you can measure it with equipment, and you can measure its direct and indirect results. It affects reality. You can measure its effects on reality, hence, it's measurable. But the general effect is accumulative and is detected. Case in point: we detect (and know quite well on the operations of) photosynthesis. Your point just made the opposite case of what you claim. You are saying measuring changes reality. We were on a different subject. The poster was claiming measuring creates reality. It this your contention also? I have a refutation to this absurd argument. Yes, until we were able to measure this, it was uncertain that the phenomena exists. We are not claiming that measurements *MAKE* reality, we are claiming that measurements make what we KNOW OF reality. This is good and getting there. The problem with your assertion is not that there's no observation to prove it, it's that there are observations and mathematical data to disprove it. You're just insisting on ignoring them. No, I am saying measurability does not determine reality. Where did you get this? You didn't show anything, you only insist on ignoring what people explain to you, give faulty mathematics, claim you know all and take us on a circular path of redefined definitions. This thread is going in circles; you had ajb explain it to you in a geometric perspective, Klaynos from an experimental perspective and swansont from a theoretical perspective, and that was after I spent half a thread with you trying to convince you to stick to definitions for the sake of a clear debate. Here is the problem. I have given a point that breaks Sr. You cannot refute it. In addition, I have made the realistic statement that while light moves through space so does the frame in some unknown way. Instead of leacturing, prove my statements are false. I am being very specific. If I am going to debate someone and I give clear math, which I have, I expect clear math in return. Do you have this? Merged post follows: Consecutive posts mergedWhy do you set x = vtλ/(1+λ) rather than just x=vt? Lt is the equation for the light sphere to map 4-D events to 4-D events. You are thinking x-axis only. This point requires a y-component or v > c. Or, z and y can trade off such that (ct)² = x² + y² + z². So, x = vtλ/(1+λ) requires a y value at least. Given that fact, two frames are forced into simultaneity. Edited January 21, 2010 by vuquta Consecutive posts merged.
mooeypoo Posted January 21, 2010 Posted January 21, 2010 I do not think so. I gave the point x = vtλ /(1+λ ) Then you plug this into t' = ( t - vx/c²)λ When you do you get t'=t. Do you need help with the language of math? I can do this proof for you. It is not hard. It's not hard if you ignore the REST of what I said, as you did. Look at my "F=ma" example. I won't answer anything else until you stop answering partial questions and nitpicking what you're comfortable with answering. I'm done going in circles with you. We're going to do this point by point and relate to the *full* points made, or not at all. Make your choice. ~moo
vuquta Posted January 21, 2010 Author Posted January 21, 2010 It's not hard if you ignore the REST of what I said, as you did. Look at my "F=ma" example. I won't answer anything else until you stop answering partial questions and nitpicking what you're comfortable with answering. I'm done going in circles with you. We're going to do this point by point and relate to the *full* points made, or not at all. Make your choice. ~moo Let's hold on here. This point I offered is valid. You refuse to deal with it. You are operating in your comfort zone. You are in circles. Since you know what you are doing, why not operate on it? What is your F=ma example? How is that relevent?
mooeypoo Posted January 21, 2010 Posted January 21, 2010 It's not valid. I explained why. swansont explained why. ajb explained why. Klaynos explained why. Until you deal with our ACTUAL explanations and not your partial-description of our straw-manned explanations, we have nothing to discuss further.
vuquta Posted January 21, 2010 Author Posted January 21, 2010 It's not valid. I explained why. swansont explained why. ajb explained why. Klaynos explained why. Until you deal with our ACTUAL explanations and not your partial-description of our straw-manned explanations, we have nothing to discuss further. I am sorry, but they did nothing and you either. I gave a specific point under LT. Normally folks run and cry when presenting. I don't. But, this point I have is not a strawman argument. So, if you are unable to handle this specific mathematical point I gave just simply say you give up instead of all this. Say it is beyond you. Why attack me? So here it is again, x = vtλ/(1+λ) This point causes simultaneity between the frames at any time t which of course forces SR into submission. Why not deal in this extremely specific subject?
mooeypoo Posted January 21, 2010 Posted January 21, 2010 You say the operation is legal. You were told it's not. swansont explained the problem. I gave you an example. If you have a problem with my example, you're welcome to ask, but at the very LEAST you should go over it. The formulas used in SR have been derived, they were not just spewed onto a piece of paper randomly. The formulas have meaning. you can't just inject whatever you want into them, be inconsistent in yuor definitions, and claim that disproves the math. You seem to either need to go over the definition of a formula in mathematics, or some basic mathematical rules. ~moo
vuquta Posted January 21, 2010 Author Posted January 21, 2010 You say the operation is legal. You were told it's not. swansont explained the problem. I gave you an example. If you have a problem with my example, you're welcome to ask, but at the very LEAST you should go over it. The formulas used in SR have been derived, they were not just spewed onto a piece of paper randomly. The formulas have meaning. you can't just inject whatever you want into them, be inconsistent in yuor definitions, and claim that disproves the math. You seem to either need to go over the definition of a formula in mathematics, or some basic mathematical rules. ~moo No sorry the point works. You can look at the full math here with the light sphere. So, if your friends think they have proved it wrong, they are mistaken and are welcome to bring their math forward and I will show them why they are wrong.
mooeypoo Posted January 21, 2010 Posted January 21, 2010 It doesn't work. Are we going to spend our time arguing like children? "yes it does! no it doesn't!" Stomping your feet on the ground, covering your ears and singing "LALALA" loudly will not make your ideas correct. You STILL haven't read the (multiple) explanations. They *have brought* the math forward. You're just ignoring it. Merged post follows: Consecutive posts mergedThis thread is temporarily closed for review. Vuquta, I recommend you take the time to go over the posts that were made for you and consider them. Whether we open the thread or not, it will serve as a good exercise either to continue refuting the claims or - perhaps! - learn from them. Thread closed, for now, until further review.
Cap'n Refsmmat Posted January 21, 2010 Posted January 21, 2010 This is staying closed. The discussion was going nowhere.
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