Alias Moniker Posted July 23, 2014 Posted July 23, 2014 (edited) If I'm traveling 299,792,457 m/s, then relative to me, $c$ would only be +1 m/s. But according to special relativity, $c$ should be a constant velocity relative to me, no matter what velocity I'm traveling, as long as my velocity is not $c$. From Wiki:The Principle of Invariant Light Speed – "... light is always propagated in empty space with a definite velocity [speed] c which is independent of the state of motion of the emitting body." (from the preface).[1] That is, light in vacuum propagates with the speed c (a fixed constant, independent of direction) in at least one system of inertial coordinates (the "stationary system"), regardless of the state of motion of the light source.Indicating that C is not a constant value TO me, but that it is a constant value RELATIVE to me. That no matter what speed I'm traveling, C is always a constant value (+298,792,458) RELATIVE to my velocity. As long as my velocity is not $c$.So that would mean, that even if I'm traveling at 299,792,457 m/s, relative to me, $c$ should still be a value that is constantly greater to me if I'm traveling 299,792,457 m/s as it is to me when I'm traveling at 0 m/s.C = V + 299,792,458 m/s. Edited July 23, 2014 by Alias Moniker -2
ACG52 Posted July 23, 2014 Posted July 23, 2014 So that would mean, that even if I'm traveling at 299,792,457 m/s, the speed of light, relative to me, should still be 299,792,457 m/s, instead of 1 m/s. No, it's 299,792,458, relative to everything. Including you.
xyzt Posted July 23, 2014 Posted July 23, 2014 (edited) If I'm traveling 299,792,457 m/s, then relative to me, $c$ would only be +1 m/s. Why do you think that? Turns out that , in SR speeds do not add like in Newtonian mechanics, so, if you are traveling at speed v wrt a reference frame and if you watch a light beam, that light beam travels at..... [math]\frac{c-v}{1-\frac{cv}{c^2}}=c![/math] with respect to you. From your post(s) you seem to believe that light is traveling at [math]c-v[/math], which would give you 1ms but your "belief" is trumped by the physical reality. It is clear that you have a great difficulty accepting the physical reality. Edited July 23, 2014 by xyzt
Alias Moniker Posted July 23, 2014 Author Posted July 23, 2014 (edited) Why do you think that? I'm traveling 1 m/s. You're traveling 2 m/s. Relative to me, your speed is +1 m/s. The speed of light is always constant in a vacuum, regardless of the velocity of the observer. I'm traveling at 1 m/s C is traveling at C Relative to me, C is traveling at C. I'm traveling 299,792,457 m/s. C is traveling at C Relative to me C is traveling at C. That is what a "constant speed, regardless of the velocity of the observer" would look like. If you assign a number to speed, then what you see is, I'm traveling at 0 m/s C is traveling at 299,792,458 Relative to me, C is traveling at 299,792,458 But I'm traveling at 299,792,457 C is traveling at 299,792,458 Relative to me, C = 1 That is not a "constant speed, regardless of the velocity of the observer". C is not a "speed". Edited July 23, 2014 by Alias Moniker
xiaojun Posted July 23, 2014 Posted July 23, 2014 Why do you think that? Turns out that , in SR speeds do not add like in Newtonian mechanics, so, if you are traveling at speed v wrt a reference frame and if you watch a light beam, that light beam travels at..... [math]\frac{c-v}{1-\frac{cv}{c^2}}=c![/math] with respect to you. From your post(s) you seem to believe that light is traveling at [math]c-v[/math], which would give you 1ms but your "belief" is trumped by the physical reality. It is clear that you have a great difficulty accepting the physical reality. Calculated by this formula is the speed of light is not equivalent velocity, the speed of light 1
Alias Moniker Posted July 23, 2014 Author Posted July 23, 2014 Why do you think that? Turns out that , in SR speeds do not add like in Newtonian mechanics, so, if you are traveling at speed v wrt a reference frame and if you watch a light beam, that light beam travels at..... [math]\frac{c-v}{1-\frac{cv}{c^2}}=c![/math] with respect to you. From your post(s) you seem to believe that light is traveling at [math]c-v[/math], which would give you 1ms but your "belief" is trumped by the physical reality. It is clear that you have a great difficulty accepting the physical reality. Turns out, you think the ability to do something mathematically means that it is ALWAYS POSSIBLE in reality.
xyzt Posted July 23, 2014 Posted July 23, 2014 Turns out, you think the ability to do something mathematically means that it is ALWAYS POSSIBLE in reality. Turns out that this is exactly the way things happen in reality. I am sorry it offends your "beliefs". I'm traveling 1 m/s. You're traveling 2 m/s. Relative to me, your speed is +1 m/s. You are confusing relative speed with closing speed. If you travel at [math]v_1[/math] and I travel at [math]v_2[/math] wrt the same reference frame , coming towards me, then our closing speed is [math]v_2+v_1[/math] If I travel at [math]v_1[/math] wrt a reference frame F and you travel at [math]v_2[/math] wrt me, then you are traveling at [math]\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}[/math] wrt the frame F. Tough, this is the reality. Calculated by this formula is the speed of light is not equivalent velocity, the speed of light This is not a coherent sentence.
Alias Moniker Posted July 23, 2014 Author Posted July 23, 2014 Turns out that this is exactly the way things happen in reality. I am sorry it offends your "beliefs". You are confusing relative speed with closing speed. If you travel at [math]v_1[/math] and I travel at [math]v_2[/math] wrt the same reference frame , coming towards me, then our closing speed is [math]v_2+v_1[/math] If I travel at [math]v_1[/math] wrt a reference frame F and you travel at [math]v_2[/math] wrt me, then you are traveling at [math]\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}[/math] wrt the frame F. Tough, this is the reality. This is not a coherent sentence. What frame is C in again?
xyzt Posted July 23, 2014 Posted July 23, 2014 What frame is C in again? All of them. You need to come to grips with this FACT.
Alias Moniker Posted July 23, 2014 Author Posted July 23, 2014 (edited) I mean, can you substitute C for V1 or V2? Can you do this math for "C", or do you have to use 299,792,458 m/s? Edited July 23, 2014 by Alias Moniker
xyzt Posted July 23, 2014 Posted July 23, 2014 I mean, can you substitute C for V1 or V2 if it is not a number? Yes, you can. When you do that, something magic happens. The resultant speed is ....c!
Alias Moniker Posted July 23, 2014 Author Posted July 23, 2014 (edited) Yes, you can. When you do that, something magic happens. The resultant speed is ....c! And is that the same result that you get when you use 299,792,458 m/s instead of C? Edited July 23, 2014 by Alias Moniker
xyzt Posted July 23, 2014 Posted July 23, 2014 And is that the same result that you get when you use 299,792,458 m/s instead of C? Yes. Have you had any algebra classes? Because you seem not to know even the basics. How old are you? 1
hypervalent_iodine Posted July 23, 2014 Posted July 23, 2014 ! Moderator Note Seriously? Quit the spamming, stick to your already existing thread on this topic or we will ban you.
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