pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 ... So, now everyone involved agrees on this: • the truck is traveling at 0.6c relative to you, and 0.915c relative to the car. • the car is traveling at 0.7c relative to you, and 0.915c relative to the truck. • you are traveling at 0.6c relative to the truck, and 0.7c relative to the car. ... I'm sure we all (except XYZ) agree with this. Now, could you not add: • you see the car and truck closing at 1.3c • after they all pass, you see the gap between the car and truck growing at 1.3c ...? And given this, is there really such a big issue in ... ... The distance between any observer and any other object can't change at or faster than the speed of light in the inertial frame of reference of the observer and therefore the distance between any two objects can't change at or faster than twice the speed of light in any inertial frame of reference. ... (If that 1.3c closing speed is incorrect, what value would you give it?) Quote
CraigD Posted January 12, 2017 Report Posted January 12, 2017 And given this, is there really such a big issue in ...... The distance between any observer and any other object can't change at or faster than the speed of light in the inertial frame of reference of the observer and therefore the distance between any two objects can't change at or faster than twice the speed of light in any inertial frame of reference. ... I wouldn’t call it a big issue, but a formal technicality. In mathematical physics, “velocity” has a precise definition: change in position divided by change in time. “Position” has a precise meaning that can apply to only a single body in a single coordinate system. So while it’s perfectly OK to define and use “change in distance between body A and body B divided by change in time”, what you’ve defined isn’t technically a velocity. The name-calling mess that is much of this thread shows what can happen when this distinction isn’t carefully made and understood. (If that 1.3c closing speed is incorrect, what value would you give it?)Using the well known, and in this thread, occasionally reviled velocity addition formula according to Special Relativity, the speed at which A measure B to be approaching it, which agrees with at what B measures A to be approaching it, is [math]\frac{0.6+0.7}{1+0.6*0.7} = \frac{130}{142} \,\mbox{c} \, \dot= \, .91549296 \,\mbox{c}[/math] Though the traditional way to appreciate this formula is to derive it from the classical laws of motion and postulates of SR, for people not disposed to deriving stuff, you can get a good feeling about it by playing with it by plugging in various values. Here’s a kinda’ cool bit of play: A >-0.5-> <-0.5-< B, [math]\frac{0.5+0.5}{1+0.5 \cdot 0.5}= \, 0.8[/math], A <-.08-< B (addition) <-.5-< A <-0.8-< B, [math]\frac{0.8-0.5}{1-0.8 \cdot 0.5}= \, 0.5[/math], A <-.05-< B (undoing it with subtraction) A >-1-> <-1-<B, [math]\frac{1-1}{1+1 \cdot 1}= \, 1[/math], A <-1-< B (the constancy of c) And for people who don’t like approximate decimals, here’re some exact, rational number results of repeatedly adding [math]\frac12 \mbox{c}[/math]: [math]\frac{\frac{0}{1}+\frac12}{1+\frac{0}{1} \frac12} = \frac{1}{2}[/math], [math]\frac{\frac{1}{2}+\frac12}{1+\frac{1}{2} \frac12} = \frac{4}{5}[/math], [math]\frac{\frac{4}{5}+\frac12}{1+\frac{4}{5} \frac12} = \frac{13}{14}[/math], [math]\frac{\frac{13}{14}+\frac12}{1+\frac{13}{14} \frac12} = \frac{40}{41}[/math], [math]\frac{\frac{40}{41}+\frac12}{1+\frac{40}{41} \frac12} = \frac{121}{122}[/math], [math]\frac{\frac{121}{122}+\frac12}{1+\frac{121}{122} \frac12} = \frac{364}{365}[/math], [math]\frac{\frac{364}{365}+\frac12}{1+\frac{364}{365} \frac12} = \frac{1093}{1094}[/math], [math]\frac{\frac{1093}{1094}+\frac12}{1+\frac{1093}{1094} \frac12} = \frac{3280}{3281}[/math], [math]\frac{\frac{3280}{3281}+\frac12}{1+\frac{3280}{3281} \frac12} = \frac{9841}{9842}[/math], [math]\frac{\frac{9841}{9842}+\frac12}{1+\frac{9841}{9842} \frac12} = \frac{29524}{29525}[/math], [math]\frac{\frac{29524}{29525}+\frac12}{1+\frac{29524}{29525} \frac12} = \frac{88573}{88574}[/math], [math]\frac{\frac{88573}{88574}+\frac12}{1+\frac{88573}{88574} \frac12} = \frac{265720}{265721}[/math], [math]\frac{\frac{265720}{265721}+\frac12}{1+\frac{265720}{265721} \frac12} = \frac{797161}{797162}[/math], [math]\frac{\frac{797161}{797162}+\frac12}{1+\frac{797161}{797162} \frac12} = \frac{2391484}{2391485}[/math] Ooh, pretty numbers, rendered by LaTeX! :) Quote
pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 (edited) I wouldn’t call it a big issue, but a formal technicality. In mathematical physics, “velocity” has a precise definition: change in position divided by change in time. “Position” has a precise meaning that can apply to only a single body in a single coordinate system. So while it’s perfectly OK to define and use “change in distance between body A and body B divided by change in time”, what you’ve defined isn’t technically a velocity. The name-calling mess that is much of this thread shows what can happen when this distinction isn’t carefully made and understood. Well, yeah, that's why I stuck my oar in. There seems a big to-do between A-wal and OceanBreeze but really it seems mostly nit-picking used to avoid either admitting anything. Using the well known, and in this thread, occasionally reviled Velocity addition formula ... .91549296c ... I don't think anyone (but XYZ) here is arguing against the velocity addition formula. That 0.915c value is already covered in the first two bullet points quoted from Boston University by OceanBreeze (and which I copied without complaint). Are you really claiming the closing speed of the car and truck, as viewed by the person in the middle, is also 0.915c ? (In post #186 OceanBreeze covers the velocity addition formula (which I don't think A-wal has an issue with), but then makes claims about closing speed without giving it nearly as much attention - even though that's the aspect of A-wals posts that he seemed to have the most problem with. I think you've essentially repeated OceanBreezes post). Edit: I really don't think A-wal is saying anything outrageous. Note from wiki (just as an example of common understanding) ... https://en.wikipedia.org/wiki/Speed_of_light#Faster-than-light_observations_and_experiments The rate of change in the distance between two objects in a frame of reference with respect to which both are moving (their closing speed) may have a value in excess of c. However, this does not represent the speed of any single object as measured in a single inertial frame. rate of change ~ A-wal used the word "faster". distance = A-wal used the word distance. This is very normal stuff, I don't understand all the angst. Edited January 12, 2017 by pzkpfw Quote
CraigD Posted January 12, 2017 Report Posted January 12, 2017 Are you really claiming the closing speed of the car and truck, as viewed by the person in the middle, is also 0.915c ?No, the change in distance between A and B as viewed by the person in the middle (let’s call them C) is 1.3 c. The point I’m trying to make is that this rate of change in distance is not, technically, a velocity (or a speed) in the frame of C. Not all rates of changes in distance are speeds, even though they are measured using the same kind of units – units of distance divided by time. I gave a couple of example of rates of changes in distance that aren’t speeds in this post. The example of the sum of speeds of 2 bodies that are moving directly toward or away from one another in the frame not at rest relative to either body is less obviously not a speed than the examples I gave there of the sum of the speeds of every car on Earth (which I estimated is about 300 c) or the rate at which the position marked by a spotlight changes when the spotlight is moved (the lighthouse paradox), but it is nonetheless not technically a speed. Being strict in definition velocity – wikipedia’s “the velocity of an object is the rate of change of its position with respect to a frame of reference” is a typical, and valid, one – is critical in avoiding confusion. There’s nothing wrong with, defining all sorts of mathematical or intuitive quantities with units of D/T, and such constructions can be very useful, but it’s important to understand that they may not be speeds as defined in Galilean/Newtonian or relativistic mechanics, so may not be subject to the laws of those theories. Quote
A-wal Posted January 12, 2017 Report Posted January 12, 2017 (edited) Is that so? This is from Boston University. Scroll down to the paragraph on Relative Velocities. Quoting from that source with a nice picture (just for you) Let's say you now stand on an intergalactic freeway. You see a truck heading in one direction at 0.6c, and a car heading in the opposite direction at 0.7c. What is the velocity of the truck relative to the car? It is not 1.3c, because nothing can travel faster than c. The relative velocity can be found using this equation: In this case, u is the velocity of the car relative to you, v is the velocity of the truck relative to you, and u' is the velocity of the car relative to the truck. Taking the direction the car is traveling to be the positive direction: This in no way validates what you're claiming. The car and the truck are moving at 1.3c relative to each other in your frame of reference. So, now everyone involved agrees on this:• the truck is traveling at 0.6c relative to you, and 0.915c relative to the car. • the car is traveling at 0.7c relative to you, and 0.915c relative to the truck. • you are traveling at 0.6c relative to the truck, and 0.7c relative to the car.• the truck and the car are moving at 1.3c relative to each other from the frame of reference of the observer. (Those are the only velocities that pertain to this situation in Special Relativity. As I said earlier, and I think Craig said the same thing, if you add them together in this way, u' = u + v, you no longer are talking about velocities, but speeds. A velocity has both magnitude and direction. A closing speed has only magnitude, if you assign a direction to it, it is a velocity and cannot be greater than c. How can the third observer (your stickman) assign a direction to the closing speed?)Those are the only relative velocities that the special theory of relative describes but they aren't the only velocities that you can use and speeds of anything below 2c are perfectly valid from the frame of reference of a third object. Speed but not velocity? That sound like BS! The truck and the car are moving at 1.3c relative to each other from the frame of reference of the observer. This is the point that you originally tried to refute."Light moves past every inertial (non accelerating) observer at the same relative velocity, so... A is moving away from a light source and B is moving in the same direction away from the light source and moving at 0.5c relative to A. C moves past A at c and C moves past B also at c. None of them accelerated. This is what's important to understand...Unlike the first example, C passes A at the same speed that C passes B. From A's perspective C moves past themselves at c and still from A's perspective, C moves past B at 0.5c. From B's perspective C moves past A at 1.5c and moves past themselves at c." "The light © passes B at 0.5c in A's frame and passes B at 1c in B's frame. The same light is moving past the same object at different velocities depending on the reference frame. Velocity is a measure of distance in space over distance in time. So in A's frame the light is covering a shorter distance (length contraction) in space over a greater amount of time (time dilation) relative to B than it is in B's frame because it's in A's frame it's overtaking B at half the velocity than it in B's frame (0.5c instead of 1c)." The parts in red are wrong. You cannot use simple addition and subtraction to get these relative velocities. Your result of 1.5 c for the relative velocity of C with respect to A, from B’s perspective should alert you that you are doing something wrong. No relative velocity is ever more than c from any reference frame.As I said before, you do not have even a basic understanding of what SR is all about. Simple addition and subtraction cannot be applied to relative velocities when dealing with relativistic situations. Taking your example: “If an object is moving away from you at 0.75c and anther object is moving away from you at 0.75c in the opposite direction then those two objects are moving away from each other at 1.5c in your reference frame”. Total BOLLOCKS! The relative velocities add this way: Velocity addition: u’ = ( u – v ) / [ 1 – uv/c^2 ]Remember, in SR the maximum possible velocity is c and you will never measure any relative velocity higher than c from any inertial FOR. That is the second postulate of SR. If you were really interested in looking at the science of SR you might correct A-Wal on his velocity addition, that is what a responsible moderator would do, isn't it? He has: The velocity addition formula is applied if you want to work out what the velocity of the two objects moving away from is relative to each other in their frames! In your frame they're both moving away from you at 0.75c so of course they're moving away from each other at 1.5c. Or maybe you think that is right? The two moving objects are moving away from each other at 1.5 c from my frame? Do you realize he is saying that my frame is special in that I can see a relative velocity that they do not see? I have already corrected him two or three times, and showed him the correct math, but you have said nothing. Is that what "science for everyone" means? Everyone can do their own crackpot science?And so on. Count the false statements!Read that last part again, it is important: “The relativistic equation applies to any situation; the one we're used to is a special case that applies only for small velocities”.You're taking that statement out of context. It's only talking about the velocities of objects relative to the observer. It doesn't to apply the relative velocity of two objects in the frame of reference of the observer. The relativistic velocity addition equation is the General solution! We only use u’ = u – v in the special case that applies to small velocities much less than c!My point was that the special theory of relativity only applies to the velocity of objects relative to the observer and you can''t apply those rules to other situations so in this context the relativistic velocity addition formula is nit the general solution. Why are you insisting on use the special case for small velocities and applying it when the velocities are approaching a significant fraction of c?Why the hell would using faster speeds make any difference? :) If an object is moving away from you at 0.9999999999c and another object is moving away from you at 0.9999999999c in the opposite direction then in your frame of reference the objects are moving away from each other at 1.9999999998c. Jesus Christ! Just read the above to see how wrong you are.He can't help you. If I do, you shall remain ignorant and arrogant for the rest of your life. Is that what you want, or do you want to learn something?You listen, you arrogant punk...Face the fact that you are wrong or you will never learn anything and go through life ignorant and arrogant.The remainder of your post is not worth responding to. You can choose to remain ignorant and arrogant, and if so, that is your problem, not mine. Or you can learn something; your choice. At this point I could not care less what you do.Are ignorant and arrogant the new subjective and objective xyz2.0? I wouldn’t call it a big issue, but a formal technicality. In mathematical physics, “velocity” has a precise definition: change in position divided by change in time. “Position” has a precise meaning that can apply to only a single body in a single coordinate system. So while it’s perfectly OK to define and use “change in distance between body A and body B divided by change in time”, what you’ve defined isn’t technically a velocity.No that's not really the issue. As hard as it is to believe OceanBreeze is under the impression that the velocity addition formula applies to the speed that two objects are moving relative to each other from the perspective of a third observer. At least he was, I think he's trying to subtly backpeddle on that now. The name-calling mess that is much of this thread shows what can happen when this distinction isn’t carefully made and understood.So is name calling acceptable here now? Just so I know. No, the change in distance between A and B as viewed by the person in the middle (let’s call them C) is 1.3 c.The point I’m trying to make is that this rate of change in distance is not, technically, a velocity (or a speed) in the frame of C. Not all rates of changes in distance are speeds, even though they are measured using the same kind of units – units of distance divided by time. I gave a couple of example of rates of changes in distance that aren’t speeds in this post. The example of the sum of speeds of 2 bodies that are moving directly toward or away from one another in the frame not at rest relative to either body is less obviously not a speed than the examples I gave there of the sum of the speeds of every car on Earth (which I estimated is about 300 c) or the rate at which the position marked by a spotlight changes when the spotlight is moved (the lighthouse paradox), but it is nonetheless not technically a speed.Being strict in definition velocity – wikipedia’s “the velocity of an object is the rate of change of its position with respect to a frame of reference” is a typical, and valid, one – is critical in avoiding confusion. There’s nothing wrong with, defining all sorts of mathematical or intuitive quantities with units of D/T, and such constructions can be very useful, but it’s important to understand that they may not be speeds as defined in Galilean/Newtonian or relativistic mechanics, so may not be subject to the laws of those theories.If an object is moving away from you at half the speed of light and another object is moving away from you at half the speed of light in the opposite directive to the first object then what speed are they moving away from each other from your perspective? That's not a velocity? Edited January 12, 2017 by A-wal Quote
pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 No, the change in distance between A and B as viewed by the person in the middle (let’s call them C) is 1.3 c. Right, and that's the question I was asking of OceanBreeze and the answer I was expecting. The point I’m trying to make is that this rate of change in distance is not, technically, a velocity ... Sure, but if his use of "velocity" in respect to closing speed is really the crux of what you and OceanBreeze are complaining about, then I'd suggest making it clear that it's just a terminology thing. Rehashing velocity addition again and again is just hiding the point. Quote
pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 ... If an object is moving away from you at half the speed of light and another object is moving away from you at half the speed of light in the opposite directive to the first object then what speed are they moving away from each other from your perspective? That's not a velocity? In short, think of velocity as a thing moving in some direction at some speed (relative to something). Your two objects are moving in different directions relative to the central observer. Yes, their distance is growing at c, but what thing is moving in a direction? Closing speed (and whatever the opposite is called) is not a velocity. Quote
A-wal Posted January 12, 2017 Report Posted January 12, 2017 (edited) In short, think of velocity as a thing moving in some direction at some speed (relative to something). Your two objects are moving in different directions relative to the central observer. Yes, their distance is growing at c, but what thing is moving in a direction?But they can move at away from each other anything under 2c, so to me it seems perfectly valid to say that their velocity relative to each other can be anything under 2c. This is a separate issue anyway. Comments like this... "Taking your example: “If an object is moving away from you at 0.75c and anther object is moving away from you at 0.75c in the opposite direction then those two objects are moving away from each other at 1.5c in your reference frame”. Total BOLLOCKS! The relative velocities add this way: Velocity addition: u’ = ( u – v ) / [ 1 – uv/c^2 ]" ... show that it isn't simply the term 'velocity' that he has issue with. If that's all it was then he'd have just said that. This one is my favourite: "Or maybe you think that is right? The two moving objects are moving away from each other at 1.5 c from my frame? Do you realize he is saying that my frame is special in that I can see a relative velocity that they do not see?" He think it creates a preferred frame of reference. :) Edited January 12, 2017 by A-wal Quote
pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 But they can move at away from each other anything under 2c, so to me it seems perfectly valid to say that their velocity relative to each other can be anything under 2c. No. The velocity of one thing relative to another can not be more than c. So, two things almost at 1c going in opposite directions, can have a distance between them that grows (according to an observer at rest relative to them both) at almost 2c. There's no individual object here here with a velocity that's over 1c. This is a separate issue anyway. Comments like this... "Taking your example: “If an object is moving away from you at 0.75c and anther object is moving away from you at 0.75c in the opposite direction then those two objects are moving away from each other at 1.5c in your reference frame”. Total BOLLOCKS! The relative velocities add this way: Velocity addition: u’ = ( u – v ) / [ 1 – uv/c^2 ]" ... show that it isn't simply the term 'velocity' that he has issue with. If that's all it was then he'd have just said that. This one is my favourite: "Or maybe you think that is right? The two moving objects are moving away from each other at 1.5 c from my frame? Do you realize he is saying that my frame is special in that I can see a relative velocity that they do not see?" He think it creates a preferred frame of reference. :) Yeah, that seems wrong (your quoting is confusing), but it's pretty much what you claimed yourself. You guys keep talking in circles around each other. In alternating posts it seems he says A! and you reply B! Then he says B! and you reply A! Quote
A-wal Posted January 12, 2017 Report Posted January 12, 2017 No. The velocity of one thing relative to another can not be more than c. So, two things almost at 1c going in opposite directions, can have a distance between them that grows (according to an observer at rest relative to them both) at almost 2c. There's no individual object here here with a velocity that's over 1c.No individual can moves because objects can only move relative to something else. In this case the velocity (or closing speed if you prefer) between two objects can be anything under 2c, not 1c. The only difference here is that it requires a third object. That was my original point and that's what he was trying to argue against. Yeah, that seems wrong (your quoting is confusing), but it's pretty much what you claimed yourself.I edited my previous post to make the quotes clearer. I never claimed anything like a preferred frame of reference. He thought that not apply the velocity addition formula to objects that are in motion relative to each from the perspective of a third observer would create a preferred reference frame. Yes really! You guys keep talking in circles around each other. In alternating posts it seems he says A! and you reply B! Then he says B! and you reply A!That's because he keeps trying to argue against a different point (that I never made) to try to evade his earlier misconceptions about relativity. First he says if it's not described in the special theory of relativity then it can't be right so I explain that's it's outside of the framework of that model and the special theory of relativity doesn't apply to this situation and then he argues that my my points don;t apply to the special theory of relativity. That's why it's quite hard to follow the discussion, he keeps putting up strawmen and doing everything he can to distract from his mistakes, that's why I quoted his early comments in post #192, to show what what his original position was and what started this annoying conversation. Quote
OceanBreeze Posted January 12, 2017 Report Posted January 12, 2017 He's quite right, actually. What a strange thing to disagree with. Quote mine much? You cut off the part that that I do say is wrong, leaving only the trivially true part. Nice work! So let me break it down for you. The first sentence is trivially true, and I have never argued against it. Here is the entire quote: There is never a velocity greater than c of an object relative to the observer! It doesn't apply to the velocity of other objects relative to each other. This is painfully obvious and the fact that you can't see it shows that you really have no business attempting to argue about a model that you clearly have no real understanding of. Do you see that part in Bold Red? He is saying the velocity addition formula: Does NOT apply to the velocity of two objects in motion, relative to each other! Do you agree with that? Maybe you do, so that explains your quote mine. How about we have a quick review?This is from Boston University. Scroll down to the paragraph on Relative Velocities. Here we have an observer at rest, standing between the two moving objectsexactly the case that A-wal is having a meltdown over From the Link: Let's say you now stand on an intergalactic freeway. You see a truck heading in one direction at 0.6c, and a car heading in the opposite direction at 0.7c. What is the velocity of the truck relative to the car? It is not 1.3c, because nothing can travel faster than c. The relative velocity can be found using this equation: In this case, u is the velocity of the car relative to you, v is the velocity of the truck relative to you, and u' is the velocity of the car relative to the truck. Taking the direction the car is traveling to be the positive direction: So, now everyone involved agrees on this:• the truck is traveling at 0.6c relative to you, and 0.915c relative to the car. • the car is traveling at 0.7c relative to you, and 0.915c relative to the truck. • you are traveling at 0.6c relative to the truck, and 0.7c relative to the car. Now, here is the really interesting and important part: The relativistic equation for velocity addition shown above can also be used for non-relativistic velocities. We're more used to adding velocities like this : u' = u - v. This is exactly what the relativistic equation reduces to for velocities much less than the speed of light. The relativistic equation applies to any situation; the one we're used to is a special case that applies only for small velocities. I cannot emphasize that last sentence enough! If you are familiar with mathematics, then you may have learned something about General Solutions and Particular Solutions. A General Solution is valid in all circumstances, while a particular solution is valid under certain particular circumstances. The fancy velocity addition formula that you may think should only be used when relativistic velocities are involved, can be used at any time, even for the velocity of a turtle crossing a road. In that case, it will reduce to the approximate particular solution of u' = u – v that we are all familiar with. So, when the velocities involved are <<< c we can use the approximation. When velocities are as significant fraction of c ( about 10% c) we must use the full velocity addition formula, and not the approximation. There is no excuse for adding two relativistic velocities with the approximate formula! A-wal is claiming the exact opposite! He seems to believe that the approximation is just fine as long as there is a third observer present! That is wrong, and the information I posted from Boston U. shows just why it is wrong. What I can’t understand is his resistance to using the correct formula? What everyone needs to realize, this is not just some trick of adding velocities, just to avoid exceeding c! That would be ridiculous, would it not? The reason for the velocity addition formula is because that is the way the Universe works! When objects are in motion, the fabric of space and time changes. We do not notice it at small velocities because the change is infinitesimally small. But at velocities approaching c, the change becomes very large. If you can’t handle the use of a single vector, the use of four-vectors is going to totally boggle your mind! The physics of vector equations, in four dimensions will get to the heart of ‘Einsteinian’ physics, the real physics, the relativistic correct physics. So if you can’t understand now, at this time, what the difference is between a vector , that is velocity, and a scalar, that is speed, you have no hope of ever understanding relativity. The Galilean/Newtonian equations are on the Right, and the Einsteinian/Lorentzian equations on the Left. This is what you are going to be doing if you want to work with Special Relativity. What we have been discussing so far is child’s play, yet some people here cannot handle it, and that is OK, as long as you are willing to learn. When you have been shown to be wrong, start by admitting it, at least to yourself, and then try to learn the correct way. CraigD 1 Quote
sluggo Posted January 12, 2017 Report Posted January 12, 2017 OceanBreeze#182Early in this thread, the topic was the difference in measured speeds of objects depending on which observer was making the measurements. Closing speeds faster than light got introduced along with their exemption since the ftl restriction only applies to objects with mass.Now you refer to Newton's Laws, how to dock a shuttle, and add some demeaning remarks on the side. Do you have training that qualifies you to do psychological evaluations via forums?A basic understanding of physics includes knowing moving masses need a source of energy, but we don't need to reinvent the wheel for every example we use.Both forms (with c=1)V = (v-w)/(1-vw) and V = (v+w)/(1+vw)never exceed 1 if one or both speeds =1. This quote accompanies the 2nd form in the 1905 paper."It follows from this equation that from a composition of two velocities which are less than c, there always results a velocity less than c." When you write "It results in “excessive speed”, What does that even mean? Do you mean a speed greater than c?If that is what you mean, you do not understand SR The graphic explains. In the U-frame A moving at .3c launches a space probe B at the origin. A measures its speed via the radar method with a signal (blue) reflected at R. The round trip A-time is 9.54-4.42 = 5.12. The green lines adjust for A's time dilation which produces the smaller blue light path. The simultaneity convention assigns R to R'. The speed of B as measured by A is 2.56/6.98 = .37. U measures the speed of B using the same method, as .60.If U adds A's speed to the B speed according to A he gets .67. If A could determine his (absolute) speed relative to light the path would be magenta which needs no correction. The right figure is A's perception of events Quote
OceanBreeze Posted January 12, 2017 Report Posted January 12, 2017 Early in this thread, the topic was the difference in measured speeds of objects depending on which observer was making the measurements. Closing speeds faster than light got introduced along with their exemption since the ftl restriction only applies to objects with mass.Now you refer to Newton's Laws, how to dock a shuttle, and add some demeaning remarks on the side. Do you have training that qualifies you to do psychological evaluations via forums?A basic understanding of physics includes knowing moving masses need a source of energy, but we don't need to reinvent the wheel for every example we use. What? I was not the one to introduce the docking scenario, that was you. And you said F=ma did not apply to that situation, because only the distance was closing, or some such nonsense. So yes, I referred to Newton's Laws and linked to a NASA page that explains how those Laws apply to the docking at ISS.Apparently, on this forum we do need to go back to basics for every situation because I see more and more evidence that you people do not understand the basics. In case you forgot, here is your post about the docking. First of all, there are no relativistic velocities involved, and secondly, Newton's Laws are used to perform the docking procedure, as per the link I posted. What about (c±v) Einstein used in the derivation of the coordinate transformations?When docking at the space station, moving 1000's mph, the closing speed is critical at approx 1"/sec. It's important in real world applications, and does not violate any rule in SR. Like all the examples used, it's a changing spatial relation, not moving mass. OceanBreeze#182 A basic understanding of physics includes knowing moving masses need a source of energy, but we don't need to reinvent the wheel for every example we use.Both forms (with c=1)V = (v-w)/(1-vw) and V = (v+w)/(1+vw)never exceed 1 if one or both speeds =1.This quote accompanies the 2nd form in the 1905 paper."It follows from this equation that from a composition of two velocities which are less than c, there always results a velocity less than c." The graphic explains. In the U-frame A moving at .3c launches a space probe B at the origin. A measures its speed via the radar method with a signal (blue) reflected at R. The round trip A-time is 9.54-4.42 = 5.12. The green lines adjust for A's time dilation which produces the smaller blue light path. The simultaneity convention assigns R to R'. The speed of B as measured by A is 2.56/6.98 = .37. U measures the speed of B using the same method, as .60.If U adds A's speed to the B speed according to A he gets .67. If A could determine his (absolute) speed relative to light the path would be magenta which needs no correction. The right figure is A's perception of eventsvelocity add.gif What I see here is the Velocity addition according to Einstein. I am not the one who has a problem with that. You should be explaining it to A-wal, not me. Quote
OceanBreeze Posted January 12, 2017 Report Posted January 12, 2017 Speed but not velocity? That sound like BS! The truck and the car are moving at 1.3c relative to each other from the frame of reference of the observer. This is the point that you originally tried to refute."Light moves past every inertial (non accelerating) observer at the same relative velocity, so... A is moving away from a light source and B is moving in the same direction away from the light source and moving at 0.5c relative to A. C moves past A at c and C moves past B also at c. None of them accelerated. This is what's important to understand...Unlike the first example, C passes A at the same speed that C passes B. From A's perspective C moves past themselves at c and still from A's perspective, C moves past B at 0.5c. From B's perspective C moves past A at 1.5c and moves past themselves at c." "The light © passes B at 0.5c in A's frame and passes B at 1c in B's frame. The same light is moving past the same object at different velocities depending on the reference frame. Velocity is a measure of distance in space over distance in time. So in A's frame the light is covering a shorter distance (length contraction) in space over a greater amount of time (time dilation) relative to B than it is in B's frame because it's in A's frame it's overtaking B at half the velocity than it in B's frame (0.5c instead of 1c)." The parts in red are wrong. You cannot use simple addition and subtraction to get these relative velocities. Your result of 1.5 c for the relative velocity of C with respect to A, from B’s perspective should alert you that you are doing something wrong. No relative velocity is ever more than c from any reference frame. If an object is moving away from you at half the speed of light and another object is moving away from you at half the speed of light in the opposite directive to the first object then what speed are they moving away from each other from your perspective? That's not a velocity? Lets go over this again: Light moves past every inertial (non accelerating) observer at the same relative velocity, so... Good start! A is moving away from a light source and B is moving in the same direction away from the light source and moving at 0.5c relative to A. C moves past A at c and C moves past B also at c. None of them accelerated. Just to be clear, C is moving at c, so C is LIGHT. This is what's important to understand...Unlike the first example, C passes A at the same speed that C passes B. OK still Good. From A's perspective C moves past themselves at c and still from A's perspective, C moves past B at 0.5c. NO! There is your problem! All observers observe the same velocity of Light always and that is c. No observer can ever observe Light moving past another observer at anything other than c. What you are saying violates the constancy of the speed of light. You just said that C is moving past B at c. Now you are saying from A’s perspective, C is moving past B at 0.5c. That is a ridiculous contradiction! Think about it. First of all, how does Observer A manage to observe the velocity of C relative to B?Seriously, how does A do it? The only things he can observe is the velocity of C relative to himself, and the velocity of B relative to himself. There is no way for him to directly observe the relative velocity between C and B. So he observes B is moving at 0.5c relative to himself, and he observes that C is moving at c, relative to himself. To find out the relative velocity between C and B he must use the velocity addition formula: [math]u'\quad =\quad \frac { u\quad -\quad v }{ 1\quad -\quad \frac { uv }{ { c }^{ 2 } } }[/math] let u’ be the relative velocity between B and C. let u = c and v = 0.5c and what do you get for u’? You had better get c because we already know that the velocity of light as seen by B is always c. Do you see how easy and sensible that is? Using your intuitive method, you are saying the relative velocity between C and B is 0.5c even though you are talking about Light, and you know it must be c! Maybe now you see how ridiculous your statement is? It does not matter what frame of reference you are in, the velocity of light relative to any other inertial frame is always c! You will never observe anything different and when you calculate it, it will be c always when you use the General Formula for velocity addition.Your insistence on using the simple addition formula is getting you into ridiculous contradictions! Quote
pzkpfw Posted January 12, 2017 Report Posted January 12, 2017 No individual can moves because objects can only move relative to something else. In this case the velocity (or closing speed if you prefer) between two objects can be anything under 2c, not 1c. The only difference here is that it requires a third object. That was my original point and that's what he was trying to argue against. If you stopped incorrectly insisting on calling that "velocity", this whole mess might go away. Quote
A-wal Posted January 12, 2017 Report Posted January 12, 2017 Quote mine much? You cut off the part that that I do say is wrong, leaving only the trivially true part. Nice work! So let me break it down for you. The first sentence is trivially true, and I have never argued against it. Here is the entire quote: "There is never a velocity greater than c of an object relative to the observer! It doesn't apply to the velocity of other objects relative to each other. This is painfully obvious and the fact that you can't see it shows that you really have no business attempting to argue about a model that you clearly have no real understanding of." Do you see that part in Bold Red? He is saying the velocity addition formula: Does NOT apply to the velocity of two objects in motion, relative to each other! Do you agree with that? Maybe you do, so that explains your quote mine.You know very well that I was saying that the velocity addition formula doesn't apply to the relative motion of two objects from the perspective of a third observer! This is beyond pathetic. How about we have a quick review?This is from Boston University. Scroll down to the paragraph on Relative Velocities. Here we have an observer at rest, standing between the two moving objectsexactly the case that A-wal is having a meltdown over From the Link: Let's say you now stand on an intergalactic freeway. You see a truck heading in one direction at 0.6c, and a car heading in the opposite direction at 0.7c. What is the velocity of the truck relative to the car? It is not 1.3c, because nothing can travel faster than c. The relative velocity can be found using this equation: In this case, u is the velocity of the car relative to you, v is the velocity of the truck relative to you, and u' is the velocity of the car relative to the truck. Taking the direction the car is traveling to be the positive direction: So, now everyone involved agrees on this:• the truck is traveling at 0.6c relative to you, and 0.915c relative to the car. • the car is traveling at 0.7c relative to you, and 0.915c relative to the truck. • you are traveling at 0.6c relative to the truck, and 0.7c relative to the car.• the truck and the car are moving at 1.3c relative to each other in your frame of reference!!! Now, here is the really interesting and important part: The relativistic equation for velocity addition shown above can also be used for non-relativistic velocities. We're more used to adding velocities like this : u' = u - v. This is exactly what the relativistic equation reduces to for velocities much less than the speed of light. The relativistic equation applies to any situation; the one we're used to is a special case that applies only for small velocities. I cannot emphasize that last sentence enough!Obviously the velocity addition formula applies to any velocities, this is really desperate of you. The issue isn't anything to do with how fast they're going, the issue is that you took the velocity addition formula and applied to objects that are motion relative to each other from the perspective of a third observer! That is an incorrect use of the formula by trying to apply to a situation outside of the framework of the model that you took it from. This shows a total lack of comprehension. If you are familiar with mathematics, then you may have learned something about General Solutions and Particular Solutions. A General Solution is valid in all circumstances, while a particular solution is valid under certain particular circumstances. The fancy velocity addition formula that you may think should only be used when relativistic velocities are involved, can be used at any time, even for the velocity of a turtle crossing a road. In that case, it will reduce to the approximate particular solution of u' = u – v that we are all familiar with. So, when the velocities involved are <<< c we can use the approximation. When velocities are as significant fraction of c ( about 10% c) we must use the full velocity addition formula, and not the approximation. There is no excuse for adding two relativistic velocities with the approximate formula!OMFG! :) It's a general solution in the sense that it applies to any speed yes. That doesn't mean that it applies to any situation! In the case of the frame of reference of a third observer it doesn't apply, so in this sense it's not a general solution. A-wal is claiming the exact opposite! He seems to believe that the approximation is just fine as long as there is a third observer present! That is wrong, and the information I posted from Boston U. shows just why it is wrong. What I can’t understand is his resistance to using the correct formula?The information you posted certainly does not show why it's wrong to use the standard non-relativistic velocity addition formula when you're dealing with the relative motion of two objects from the perspective of a third observer, because that is the correct formula in that situation! I'm only opposed to using the relativistic velocity addition formula when it is in fact the incorrect velocity addition formula to use. What everyone needs to realize, this is not just some trick of adding velocities, just to avoid exceeding c! That would be ridiculous, would it not? The reason for the velocity addition formula is because that is the way the Universe works! When objects are in motion, the fabric of space and time changes. We do not notice it at small velocities because the change is infinitesimally small. But at velocities approaching c, the change becomes very large. If you can’t handle the use of a single vector, the use of four-vectors is going to totally boggle your mind! What everyone needs to understand is that this is nothing to do with the discussion and is just a really poor attempt to deflect attention away from your previous fundamental errors about how relativity works! The physics of vector equations, in four dimensions will get to the heart of ‘Einsteinian’ physics, the real physics, the relativistic correct physics. So if you can’t understand now, at this time, what the difference is between a vector , that is velocity, and a scalar, that is speed, you have no hope of ever understanding relativity. The Galilean/Newtonian equations are on the Right, and the Einsteinian/Lorentzian equations on the Left. This is what you are going to be doing if you want to work with Special Relativity.You can't even grasp the basic fundamentals so I'd make sure you fully get to grips with that first. What we have been discussing so far is child’s play, yet some people here cannot handle it, and that is OK, as long as you are willing to learn. When you have been shown to be wrong, start by admitting it, at least to yourself, and then try to learn the correct way.LOL :) You are the last person who should be giving that advice to other people! Lets go over this again: Light moves past every inertial (non accelerating) observer at the same relative velocity, so... Good start! A is moving away from a light source and B is moving in the same direction away from the light source and moving at 0.5c relative to A. C moves past A at c and C moves past B also at c. None of them accelerated. Just to be clear, C is moving at c, so C is LIGHT. This is what's important to understand...Unlike the first example, C passes A at the same speed that C passes B. OK still Good. From A's perspective C moves past themselves at c and still from A's perspective, C moves past B at 0.5c. NO! There is your problem! All observers observe the same velocity of Light always and that is c. No observer can ever observe Light moving past another observer at anything other than c. What you are saying violates the constancy of the speed of light. You just said that C is moving past B at c. Now you are saying from A’s perspective, C is moving past B at 0.5c. That is a ridiculous contradiction!There is no contradiction, you just need to go back to basics and try to understand how difference frames of reference work. All observers observer the same velocity of light relative to themselves but not to other objects. In other words light always moves past YOU at c (providing you're not accelerating) but it will move past other objects at velocities other than c in your frame of reference if those objects are in motion relative to you. Think about it. First of all, how does Observer A manage to observe the velocity of C relative to B?Seriously, how does A do it? The only things he can observe is the velocity of C relative to himself, and the velocity of B relative to himself. There is no way for him to directly observe the relative velocity between C and B. So he observes B is moving at 0.5c relative to himself, and he observes that C is moving at c, relative to himself. To find out the relative velocity between C and B he must use the velocity addition formula: [math]u'\quad =\quad \frac { u\quad -\quad v }{ 1\quad -\quad \frac { uv }{ { c }^{ 2 } } }[/math] let u’ be the relative velocity between B and C. let u = c and v = 0.5c and what do you get for u’? You had better get c because we already know that the velocity of light as seen by B is always c. Do you see how easy and sensible that is? Using your intuitive method, you are saying the relative velocity between C and B is 0.5c even though you are talking about Light, and you know it must be c! Maybe now you see how ridiculous your statement is? It does not matter what frame of reference you are in, the velocity of light relative to any other inertial frame is always c! You will never observe anything different and when you calculate it, it will be c always when you use the General Formula for velocity addition.Your insistence on using the simple addition formula is getting you into ridiculous contradictions!It would be a contradiction if light could move past two objects at the same speed if those objects are in motion relative to each other. For that you have to change the frame of reference. If you stopped incorrectly insisting on calling that "velocity", this whole mess might go away.No that's just a technicality and if that was they issue he originally had he would have simply pointed out that it's not technically velocity. Look, no mention of 'velocity' in the context of closing speed:"Light moves past every inertial (non accelerating) observer at the same relative velocity, so... A is moving away from a light source and B is moving in the same direction away from the light source and moving at 0.5c relative to A. C moves past A at c and C moves past B also at c. None of them accelerated. This is what's important to understand...Unlike the first example, C passes A at the same speed that C passes B. From A's perspective C moves past themselves at c and still from A's perspective, C moves past B at 0.5c. From B's perspective C moves past A at 1.5c and moves past themselves at c." "The light © passes B at 0.5c in A's frame and passes B at 1c in B's frame. The same light is moving past the same object at different velocities depending on the reference frame. Velocity is a measure of distance in space over distance in time. So in A's frame the light is covering a shorter distance (length contraction) in space over a greater amount of time (time dilation) relative to B than it is in B's frame because it's in A's frame it's overtaking B at half the velocity than it in B's frame (0.5c instead of 1c)." The parts in red are wrong. You cannot use simple addition and subtraction to get these relative velocities. Your result of 1.5 c for the relative velocity of C with respect to A, from B’s perspective should alert you that you are doing something wrong. No relative velocity is ever more than c from any reference frame. As I said before, you do not have even a basic understanding of what SR is all about. Simple addition and subtraction cannot be applied to relative velocities when dealing with relativistic situations. Taking your example: “If an object is moving away from you at 0.75c and anther object is moving away from you at 0.75c in the opposite direction then those two objects are moving away from each other at 1.5c in your reference frame”. Total BOLLOCKS! The relative velocities add this way: Velocity addition: u’ = ( u – v ) / [ 1 – uv/c^2 ] Remember, in SR the maximum possible velocity is c and you will never measure any relative velocity higher than c from any inertial FOR. That is the second postulate of SR.If you were really interested in looking at the science of SR you might correct A-Wal on his velocity addition, that is what a responsible moderator would do, isn't it? He has: The velocity addition formula is applied if you want to work out what the velocity of the two objects moving away from is relative to each other in their frames! In your frame they're both moving away from you at 0.75c so of course they're moving away from each other at 1.5c. Or maybe you think that is right? The two moving objects are moving away from each other at 1.5 c from my frame? Do you realize he is saying that my frame is special in that I can see a relative velocity that they do not see? I have already corrected him two or three times, and showed him the correct math, but you have said nothing. Is that what "science for everyone" means? Everyone can do their own crackpot science? He completely misunderstood what the velocity addition formula actually represents and when to apply it and even thought that using it correctly would lead to a preferred frame of reference. He's doing everything he can to try to snake his away out of owning up to his mistakes because he doesn't want to face his limitations so he just keeps arguing points that was never in dispute to try to deflect addition away from his initial false conceptions about what the model his using to copy from actually describes. Don't fall for it! I've dealt with people like this before, really pathetic people with so much of a self-esteem issue that they feel the need to to infest science forums and post what they've just looked up but don't have the capability to grasp any of it. When they're lack of understanding shows itself by taking what they've copied out of context and trying to apply it to situations that it shouldn't be applied to they either react like this (first one to Sluggo):I suggest you should try using Google sometimes Before you post something…..it might save you some embarrassment. You think the distance between the capsule (which has mass) and the ISS (which has more mass) just closes, without any force being involved, because the space between them has no mass? You think Newton’s Laws of Motion, and Force is not required?That is comedy gold. Here, read this (hopefully it is not too high level for you) You listen, you arrogant punk, any discussion about Special Theory of Relativity (SR) must be assumed to be about Einstein’s SR, as that is the only SR that is recognized. Endless arguments and much chaos will ensue if everyone is thinking of their own theory of SR. You can't make up your own crap "theories" in science!This is the problem. You looked up something that you have no real understanding of and dropped yourself in it by taking what the special theory of relativity describes and thinking that it applies to all relative motion when it actually only applies the motion of objects relative to the observer. I'm not making up my own version of relativity by talking about the motion of objects relative to each other from the perspective of a third observer and it certainly doesn't imply a preferred frame of reference. Trying to apply the relativistic velocity addition without changing reference frames is the most fundamental error I've ever seen from somebody pretending to know SR, congratulations. Quote
CraigD Posted January 13, 2017 Report Posted January 13, 2017 So is name calling acceptable here now? Just so I know.No, of course not, name calling is not acceptable. Respect the opinions of others. Don’t be rude and offensive. Etc. Seriously. This thread has some good concepts in it, but they’re buried and obscured by heaps of ad hominem action and reaction. Nobody should call anybody a “punk”, “really pathetic”, or other things in this vein. Everyone, please be polite! If an object is moving away from you at half the speed of light and another object is moving away from you at half the speed of light in the opposite directive to the first object then what speed are they moving away from each other from your perspective? That's not a velocity?Correct, the rate of change in distance between 2 bodies is not a velocity. A velocity consists of a speed and a direction, both of which are relative to a frame of reference. Distances between bodies and their rate of change are very useful (for example, in simulations like the one in this 10-year-old thread), but they’re technically neither speeds nor velocities. Quote
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