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#11
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On May 5, 2:44Â*pm, " wrote:
On 5 mayo, 17:32, rbwinn wrote: On May 5, 1:44�pm, YBM wrote: rbwinn a �crit : You neglect the fact that if x or x' is negative in the equations x=ct, x'=ct', then either the velocity of light has to be negative or time has to be negative. � You're right that in the case of x=ct, when t in negative, so is x.... So what ? The equations you gave would apply only to positive values of x and x', Where did you get this ? I used the equation of propagation of light x=ct, which applies either when t or x are negative or positive, and the LT which apply for any x,t, positive or negative. Well, for example, light is emitted at the origins of S and S' when they coincide. Â*According to Einstein, the light would propagate in S as a sphere with a radius of ct, and Â*in S' as a sphere with a radius of ct', except that the sphere in S' is an oblate sphere because of the distance contraction. Â*So we consider a photon proceeding from the origins of S and S' at t=t'=0 in the Â*-x direction. Â*When a time of t has transpired in S, a time of t' has transpired in S'. Â*The photon is at the coordinate x in S and at the coordinate x' in S'. Â*Both x and x' are negative. Â*Both t and t' are positive. Â*The velocity of the photon is -c, not c as you insist it would be. Â*The Lorentz equations themselves show that x=(-c)t. meaning that a photon is traveling in the +x direction relative to the x axis. � x=ct means that light is traveling in the (Ox) (call this +x if you want) direction and that x=0 at t=0. This equation of movement applies as well for x,t being negative or positive. The Lorentz equation works because it is showing velocity of light, not speed of light as scientists say it does. Â*If the equations were using speed of light, you would be able to reduce them down by the rules of algebra, and they would still work. Â*They will not work if you reduce them down past Â* Â* Â* Â* Â* Â*t'=(t-vx/c^2)/sqrt(1-v^2/c^2) with the equations x=ct, x'=ct', because if you do, the velocity of a photon is wrong. Â*Why not reduce the numerator to t(1-v/c)? Â* Â*If you did, you would have to put a -c into the equation for c every time x was negative. Â*So if you reflect light from a mirror, the velocity of a photon changes, and you have to change from c to -c. The Lorentz equations do this automatically with the value of x, however, they do so at the price of a distance contraction. Â* Â* Â*So what is your theory about how light exists if nothing can be accellerated to the speed of light? Robert B. Winn We all agree with you. So in the interest of science, and since you have the uttermost knowledge about this subject, please disconnect yourself right now from your Internet and start writing a book or paper about this new science. We, in the mean time will contact the editors of some prestigious Journals, such as Science and Nature, and let them know that you are about to provide to the scientific community with the most important discovery of this century, so they can be ready when your work is ready. Miguel Rios- Hide quoted text - What is that supposed to mean? Do I seem like a Party member to you? I will post the equations here. If you do not like them, wander around in darkness with the other scientists of our time. x'=x-vt y'=y z'=z t'=t x=wt x'=wn' w=velocity of light wn'=wt-vt n'=t(1-v/c) OK, it is all there. So what did that change? Do you see any scientists saying that "scientific time" can be called something other than t'? Robert B. Winn |
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#12
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rbwinn wrote:
Well, for example, light is emitted at the origins of S and S' when they coincide. According to Einstein, the light would propagate in S as a sphere with a radius of ct, and in S' as a sphere with a radius of ct', except that the sphere in S' is an oblate sphere because of the distance contraction. It's a sphere, but it is oblate ? Can't you realize you contradict yourself ? They are spheres in both frames. So we consider a photon proceeding from the origins of S and S' at t=t'=0 in the -x direction. So ? Then the equation of movement is x=-ct. speed is c, velocity is (-c,0,0), LT (with speed c as a parameter) will show that in S' the equation of movement is x'=-ct'. When a time of t has transpired in S, a time of t' has transpired in S'. The photon is at the coordinate x in S and at the coordinate x' in S'. Both x and x' are negative. Both t and t' are positive. The velocity of the photon is -c, not c as you insist it would be. The Lorentz equations themselves show that x=(-c)t. I've never said that velocity (which is a vector anyway) is -c or c. Speed is c anyway, and speed is the value to use in LT, not velocity. Then the (real) LT would show that if x=-ct then x'=-ct, just plug this in my demo above, you'll see. What are you trying to prove ? That by using some kind of signed speed (what you improperly call "velocity") in the LT instead of the speed of light they are supposed to use (and are DEFINED as such) you obtain wrong results ? So what ? Just use the real LT, you'll get the right ones... This is just plain ridiculous... The Lorentz equation works because it is showing velocity of light, not speed of light as scientists say it does. I just used speed of light (not velocity) in the LT, and show you by trivial algebra that they "works". I suggest you to consider you failed to grasp many basic points about what are coordinates, events, transformation, equation of movements, etc. This is quite obvious... If the equations were using speed of light, you would be able to reduce them down by the rules of algebra, and they would still work. They will not work if you reduce them down past t'=(t-vx/c^2)/sqrt(1-v^2/c^2) with the equations x=ct, x'=ct', I did. It took 3 lines of elementary algebra. because if you do, the velocity of a photon is wrong. As you can see, this is not true. Why not reduce the numerator to t(1-v/c)? If you did, you would have to put a -c into the equation for c every time x was negative. let's try : x'/t' = (x-vt)/(t-vx/c2) = (ct-vt)/(t-vct/c2) = t(c-v)/( t (1 - v/c) ) = c(1 - v/c) / (1 - v/c) = c = x'=ct' I wonder how you could think that by using different, but valid, algebraic transformations, one could obtain different results ? Especially on such trivial ones... So if you reflect light from a mirror, the velocity of a photon changes, and you have to change from c to -c. The Lorentz equations do this automatically with the value of x, however, they do so at the price of a distance contraction. Complete nonsense... So what is your theory about how light exists if nothing can be accellerated to the speed of light? Unrelated nonsense. |
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#13
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On May 5, 3:48�pm, YBM wrote:
rbwinn wrote: Well, for example, light is emitted at the origins of S and S' when they coincide. �According to Einstein, the light would propagate in S as a sphere with a radius of ct, and �in S' as a sphere with a radius of ct', except that the sphere in S' is an oblate sphere because of the distance contraction. � It's a sphere, but it is oblate ? Can't you realize you contradict yourself ? They are spheres in both frames. So we consider a photon proceeding from the origins of S and S' at t=t'=0 in the �-x direction. So ? Then the equation of movement is x=-ct. speed is c, velocity is (-c,0,0), LT (with speed c as a parameter) will show that in S' the equation of movement is x'=-ct'. When a time of t has transpired in S, a time of t' has transpired in S'. �The photon is at the coordinate x in S and at the coordinate x' in S'. �Both x and x' are negative. �Both t and t' are positive. �The velocity of the photon is -c, not c as you insist it would be. �The Lorentz equations themselves show that x=(-c)t. I've never said that velocity (which is a vector anyway) is -c or c. Speed is c anyway, and speed is the value to use in LT, not velocity. Then the (real) LT would show that if x=-ct then x'=-ct, just plug this in my demo above, you'll see. The speed c is squared everywhere it appears in the Lorentz equations. Have you ever wondered why? Well, there is a reason. (- c)^2 = c^2 So with regard to whether the Lorentz equations are using speed of light or velocity of light can only be determined by x and x', since Einstein says x=ct, x'=ct', and I say that the equations should show velocity of light. So if x is a negative number in the Lorentz equation for t', is c positive or negative in Einstein's little equation? Well, it has to be negative if t is positive, so the Lorentz equations are using velocity of light, just as I said. What are you trying to prove ? That by using some kind of signed speed (what you improperly call "velocity") in the LT instead of the speed of light they are supposed to use (and are DEFINED as such) you obtain wrong results ? So what ? Just use the real LT, you'll get the right ones... The Lorentz equations and their accompanying distance contraction are completely unnecessary. Use the correct velocities for photons, and you can use the Galilean transformation equations. This is just plain ridiculous... The Lorentz equation works because it is showing velocity of light, not speed of light as scientists say it does. I just used speed of light (not velocity) in the LT, and show you by trivial algebra that they "works". I suggest you to consider you failed to grasp many basic points about what are coordinates, events, transformation, equation of movements, etc. This is quite obvious... Well, what is quite obvious to me is that the velocity of a photon traveling in the -x direction has a velocity of -c, something no scientist will admit. So here we are. You think I am in doubt about who is right and who is wrong? If the equations were using speed of light, you would be able to reduce them down by the rules of algebra, and they would still work. �They will not work if you reduce them down past � � � � � �t'=(t-vx/c^2)/sqrt(1-v^2/c^2) with the equations x=ct, x'=ct', I did. It took 3 lines of elementary algebra. because if you do, the velocity of a photon is wrong. � As you can see, this is not true. Why not reduce the numerator to t(1-v/c)? � �If you did, you would have to put a -c into the equation for c every time x was negative. � let's try : � x'/t' = (x-vt)/(t-vx/c2) = (ct-vt)/(t-vct/c2) � � � � = t(c-v)/( t (1 - v/c) ) � � � � = c(1 - v/c) / (1 - v/c) � � � � = c = x'=ct' I wonder how you could think that by using different, but valid, algebraic transformations, one could obtain different results ? Especially on such trivial ones... Well, this is not trivial. Your claim is that a photon traveling on the x' axis in the -x' direction has a velocity of c. I say it has a velocity of -c. We cannot both be correct. Sorry, but you are the one who is wrong. So if you reflect light from a mirror, the velocity of a photon changes, and you have to change from c to -c. The Lorentz equations do this automatically with the value of x, however, they do so at the price of a distance contraction. Complete nonsense... So now you are going to explain why a photon traveling with a velocity of c has a velocity of c after it is reflected back in the direction it came from with a mirror. We are all eager to hear your explanation. � � �So what is your theory about how light exists if nothing can be accellerated to the speed of light? Unrelated nonsense. What a surprise that you would not want to explain your ideas. Robert B. Winn |
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#14
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rbwinn wrote:
The speed c is squared everywhere it appears in the Lorentz equations. Have you ever wondered why? Well, there is a reason. (- c)^2 = c^2 Just read any derivation of LT. Another argument is that any kind of transformation cannot depend on the direction of a specific velocity because they are arbitrate given the system of coordinate you choose. So with regard to whether the Lorentz equations are using speed of light or velocity of light can only be determined by x and x', since Einstein says x=ct, x'=ct', and I say that the equations should show velocity of light. You confuse speed and velocity, this is hopeless... Anyway let's try to clarify the issue : Velocity is a 3-D vector, in the case of a uniform movement this is quite simple : if a body is moving at velocity (vx,vy,vz) then its equation of movement is : (vector OM)=V.t + OA where M is the position of the body at time t, A is the coordinate of the body at time t=0. When you are considering bodies moving along the x axis you can omit y and z coordinate, hence identifying OM with x. In the case of light moving in "+x" direction : V=(c,0,0) x=ct In the case of light moving in "-x" direction : V=(-c,0,0) x=-ct Speed in the norm of the velocity vector : here in both case : c. So if x is a negative number in the Lorentz equation for t', is c positive or negative in Einstein's little equation? Well, it has to be negative if t is positive, so the Lorentz equations are using velocity of light, just as I said. It doesn't make sense... How could someone be confused on such elementary issues ? Your problem is not about LT, but about what a equation of movement is, about what speed and velocity are... No doubt you'll have some trouble with Galilean transformations as well : it's all about the same things ! [...] Well, what is quite obvious to me is that the velocity of a photon traveling in the -x direction has a velocity of -c, something no scientist will admit. So here we are. You think I am in doubt about who is right If you limit yourself to movments on the x axis (both direction), you are perfectly right in calling (c,0,0) as c and (-c,0,0) as -c, both are velocities, not speed. Everyone would admin that -c or c can represent a velocity on such case. [...] Well, this is not trivial. Your claim is that a photon traveling on the x' axis in the -x' direction has a velocity of c. I say it has a velocity of -c. We cannot both be correct. Sorry, but you are the one who is wrong. I've never made such a claim... Please write less and read more. P.S. your newsreader has problems with tabulations in messages, you'd better switch to some decent software. |
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#15
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On May 5, 5:24�pm, YBM wrote:
rbwinn wrote: The speed c is squared everywhere it appears in the Lorentz equations. �Have you ever wondered why? �Well, there is a reason. �(- c)^2 = c^2 Just read any derivation of LT. Another argument is that any kind of transformation cannot depend on the direction of a specific velocity because they are arbitrate given the system of coordinate you choose. So with regard to whether the Lorentz equations are using speed of light or velocity of light can only be determined by x and x', since Einstein says x=ct, x'=ct', and I say that the equations should show velocity of light. � You confuse speed and velocity, this is hopeless... Anyway let's try to clarify the issue : Velocity is a 3-D vector, in the case of a uniform movement this is quite simple : if a body is moving at velocity (vx,vy,vz) then its equation of movement is : (vector OM)=V.t + OA where M is the position of the body at time t, A is the coordinate of the body at time t=0. When you are considering bodies moving along the x axis you can omit y and z coordinate, hence identifying OM with x. In the case of light moving in "+x" direction : V=(c,0,0) x=ct In the case of light moving in "-x" direction : V=(-c,0,0) x=-ct Speed in the norm of the velocity vector : here in both case : c. So if x is a negative number in the Lorentz equation for t', is c positive or negative in Einstein's little equation? �Well, it has to be negative if t is positive, so the Lorentz equations are using velocity of light, just as I said. It doesn't make sense... How could someone be confused on such elementary issues ? Your problem is not about LT, but about what a equation of movement is, about what speed and velocity are... No doubt you'll have some trouble with Galilean transformations as well : it's all about the same things ! [...] Well, what is quite obvious to me is that the velocity of a photon traveling in the -x direction has a velocity of -c, something no scientist will admit. �So here we are. �You think I am in doubt about who is right If you limit yourself to movments on the x axis (both direction), you are perfectly right in calling (c,0,0) as c and (-c,0,0) as -c, both are velocities, not speed. Everyone would admin that -c or c can represent a velocity on such case. [...] Well, this is not trivial. �Your claim is that a photon traveling on the x' axis in the -x' direction has a velocity of c. �I say it has a velocity of -c. �We cannot both be correct. �Sorry, but you are the one who is wrong. I've never made such a claim... Please write less and read more. P.S. your newsreader has problems with tabulations in messages, you'd better switch to some decent software. Good to see you back down. I will just have to live with whatever Google does with tabulations. I am not subsidized by the government the way scientists are. Here is what I do with the Galilean transformation equations. x'=x-vt y'=y z'=z t'=t x=wt x'=wn' w=velocity of light wn'=wt-vt n'=t(1-v/w) n' is time on a cesium clock in S'. There is no distance contraction. Robert B. Winn |
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#16
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On May 5, 4:30 pm, rbwinn wrote:
On May 5, 12:07 pm, YBM wrote: You neglect the fact that if x or x' is negative in the equations x=ct, x'=ct', then either the velocity of light has to be negative or time has to be negative. Nu, so what is wrong with the time being negative? I asked the same question days ago. |
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#17
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rbwinn wrote:
[lot's of stuff he didn't even read] Good to see you back down. I will just have to live with whatever Google does with tabulations. I am not subsidized by the government the way scientists are. Mozilla Thunderbird is free. Here is what I do with the Galilean transformation equations. [snip] I don't care. You have problem with basic issues which concern GT as well as LT. It's not LT you don't undersand, it's x=ct or x=-ct. Evading outside of the scope of what you pretended before is a useless strategy : I don't care beating you, I'd like to understand at least a bit on basic physics. Please read carefully what *you* said at the initial post of this thread, and as crefully what I responded then. |
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#18
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rbwinn wrote:
[lot's of stuff he didn't even read] Good to see you back down. I will just have to live with whatever Google does with tabulations. I am not subsidized by the government the way scientists are. Mozilla Thunderbird is free. Here is what I do with the Galilean transformation equations. [snip] I don't care. You have problem with basic issues which concern GT as well as LT. It's not LT you don't undersand, it's x=ct or x=-ct. Evading outside of the scope of what you pretended before is a useless strategy : I don't care beating you, I'd like see you understand at least a bit on basic physics. Please read carefully what *you* said at the initial post of this thread, and as crefully what I responded then. |
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#19
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On 5 mayo, 18:40, rbwinn wrote:
On May 5, 2:44 pm, " wrote: On 5 mayo, 17:32, rbwinn wrote: On May 5, 1:44�pm, YBM wrote: rbwinn a �crit : You neglect the fact that if x or x' is negative in the equations x=ct, x'=ct', then either the velocity of light has to be negative or time has to be negative. � You're right that in the case of x=ct, when t in negative, so is x.... So what ? The equations you gave would apply only to positive values of x and x', Where did you get this ? I used the equation of propagation of light x=ct, which applies either when t or x are negative or positive, and the LT which apply for any x,t, positive or negative. Well, for example, light is emitted at the origins of S and S' when they coincide. According to Einstein, the light would propagate in S as a sphere with a radius of ct, and in S' as a sphere with a radius of ct', except that the sphere in S' is an oblate sphere because of the distance contraction. So we consider a photon proceeding from the origins of S and S' at t=t'=0 in the -x direction. When a time of t has transpired in S, a time of t' has transpired in S'. The photon is at the coordinate x in S and at the coordinate x' in S'. Both x and x' are negative. Both t and t' are positive. The velocity of the photon is -c, not c as you insist it would be. The Lorentz equations themselves show that x=(-c)t. meaning that a photon is traveling in the +x direction relative to the x axis. � x=ct means that light is traveling in the (Ox) (call this +x if you want) direction and that x=0 at t=0. This equation of movement applies as well for x,t being negative or positive. The Lorentz equation works because it is showing velocity of light, not speed of light as scientists say it does. If the equations were using speed of light, you would be able to reduce them down by the rules of algebra, and they would still work. They will not work if you reduce them down past t'=(t-vx/c^2)/sqrt(1-v^2/c^2) with the equations x=ct, x'=ct', because if you do, the velocity of a photon is wrong. Why not reduce the numerator to t(1-v/c)? If you did, you would have to put a -c into the equation for c every time x was negative. So if you reflect light from a mirror, the velocity of a photon changes, and you have to change from c to -c. The Lorentz equations do this automatically with the value of x, however, they do so at the price of a distance contraction. So what is your theory about how light exists if nothing can be accellerated to the speed of light? Robert B. Winn We all agree with you. So in the interest of science, and since you have the uttermost knowledge about this subject, please disconnect yourself right now from your Internet and start writing a book or paper about this new science. We, in the mean time will contact the editors of some prestigious Journals, such as Science and Nature, and let them know that you are about to provide to the scientific community with the most important discovery of this century, so they can be ready when your work is ready. Miguel Rios- Hide quoted text - What is that supposed to mean? Do I seem like a Party member to you? I will post the equations here. If you do not like them, wander around in darkness with the other scientists of our time. x'=x-vt y'=y z'=z t'=t x=wt x'=wn' w=velocity of light wn'=wt-vt n'=t(1-v/c) OK, it is all there. So what did that change? Do you see any scientists saying that "scientific time" can be called something other than t'? Robert B. Winn Who are you talking to then?. Are you a scientist? Do you see any scientist posting here? Why are you wasting your time with the wrong people? Are you trying to prove something to yourself? Miguel Rios |
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#20
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On May 5, 5:42�pm, YBM wrote:
rbwinn wrote: [lot's of stuff he didn't even read] Good to see you back down. �I will just have to live with whatever Google does with tabulations. �I am not subsidized by the government the way scientists are. Mozilla Thunderbird is free. Here is what I do with the Galilean transformation equations. [snip] I don't care. You have problem with basic issues which concern GT as well as LT. It's not LT you don't undersand, it's x=ct or x=-ct. Evading outside of the scope of what you pretended before is a useless strategy : I don't care beating you, I'd like to understand at least a bit on basic physics. Please read carefully what *you* said at the initial post of this thread, and as crefully what I responded then. Well, you say there are issues with the Galilean transformation equations. I don't see any. t'=t. You have another clock running slower than a t' clock. Convert the rate of time to t'. Everything works. What is the problem you see? Robert B. Winn |
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