dragonstar57 Posted February 28, 2011 Share Posted February 28, 2011 (edited) if object A is moving at X and object B is moving in the opposite direction also at X than either object is moving away from the other at 2x right? but why does this break down when the statement of "x=C" is added to the situation? also how does dilation effect the trajectory of an object? if object x and y are moving at point Z and X is moving at C and is 5000 kilometers away from point z. object y is 400 kilometers away from point z and is moving at 5000 KMS object Y's trajectory is on a right angle to x's. how will time dilation effect when one gets there relative to the other and at what speed are the objects moving relative to one another? ps. sorry if this is difficult to understand or if i left out an important value. Edited February 28, 2011 by dragonstar57 Link to comment Share on other sites More sharing options...
ydoaPs Posted February 28, 2011 Share Posted February 28, 2011 (edited) Velocities do not add linearly. http://en.wikipedia.org/wiki/Velocity-addition_formula [math]S=\frac{v+u}{1+\frac{vu}{c^2}}[/math] Where S is the sum of the two velocities, v is the velocity of an object, and u is the velocity of the other. Edited February 28, 2011 by ydoaPs Link to comment Share on other sites More sharing options...
dragonstar57 Posted February 28, 2011 Author Share Posted February 28, 2011 Velocities do not add linearly. could you expand on this a little please? are you saying that you don't add the velocities of 2 objects to find out the velocity of one object if one of them is assumed to be stationary? Link to comment Share on other sites More sharing options...
swansont Posted February 28, 2011 Share Posted February 28, 2011 could you expand on this a little please? are you saying that you don't add the velocities of 2 objects to find out the velocity of one object if one of them is assumed to be stationary? Precisely. The Galilean world where you would simply use v + u is only an approximation, valid at low speeds. Link to comment Share on other sites More sharing options...
IM Egdall Posted February 28, 2011 Share Posted February 28, 2011 Say object A is moving to the left at 30,000 miles an hour relative to you. And say object B is moving to the right at 60,000 miles an hour relative to you. Now at what speed is object B moving away from object A? Newton would say 30,000 plus 60,000 equals 90,000 miles an hour. But this is incorrect. Per the relativistic formula given above, the true answer is 89,999.99964 miles an hour. (If you set w = 30,000 and v = 60,000 and c = 670 million and solve the equation; this is what you get.) So for small speeds relative to the speed of light, Newton's simple formula is an excellent approximation. But at speeds approaching the speed of light, the asnwers are quite different. For example, say object A is moving at 90 percent the speed of light and object B is moving at 80 percent the speed of light in the opposite direction. At what speed is B moving away from A? Newton says 0.90 plus 0.80 equals 1.7. This is 1.7 times the speed of light. But if we set w = 0.90 and v = 0.80 and c = 1 in Einstein's equation, we get the answer of 0.988. This is 0.988 times the speed of light. In fact, per Einstein's formula, no matter what the speeds of the two objects, the resultant speed is never greater than the speed of light. Link to comment Share on other sites More sharing options...
zapatos Posted February 28, 2011 Share Posted February 28, 2011 Is this true from everyone's (A, B, and you) perspective? Link to comment Share on other sites More sharing options...
dragonstar57 Posted February 28, 2011 Author Share Posted February 28, 2011 so in "if object x and y are moving at point Z and X is moving at C and is 5000 kilometers away from point z. object y is 400 kilometers away from point z and is moving at 5000 KMS object Y's trajectory is on a right angle to x's." how how does that work? there sems to be some kind of paradox there... Link to comment Share on other sites More sharing options...
IM Egdall Posted March 2, 2011 Share Posted March 2, 2011 Is this true from everyone's (A, B, and you) perspective? The speed you measure depends on your perspective (more formally, your frame of reference). But per special relativity, no matter what your frame of refernce, the speed you measure for any object cannot exceed the speed of light. More precicely, no object with mass can reach the speed of light. And light itself (or any particle with zero mass) always measures as traveling at the speed of light, no matter what speed you are traveling at. Link to comment Share on other sites More sharing options...
dragonstar57 Posted March 2, 2011 Author Share Posted March 2, 2011 The speed you measure depends on your perspective (more formally, your frame of reference). But per special relativity, no matter what your frame of refernce, the speed you measure for any object cannot exceed the speed of light. More precicely, no object with mass can reach the speed of light. And light itself (or any particle with zero mass) always measures as traveling at the speed of light, no matter what speed you are traveling at. but that seems to suggest that things can be in more than one place at one time Link to comment Share on other sites More sharing options...
IM Egdall Posted March 3, 2011 Share Posted March 3, 2011 but that seems to suggest that things can be in more than one place at one time Huh? Please give me an example. Link to comment Share on other sites More sharing options...
dragonstar57 Posted March 3, 2011 Author Share Posted March 3, 2011 Huh? Please give me an example. i didn't calculate it but if they were to intersect at point z would they not have met before their speeds would otherwise imply? Link to comment Share on other sites More sharing options...
IM Egdall Posted March 4, 2011 Share Posted March 4, 2011 i didn't calculate it but if they were to intersect at point z would they not have met before their speeds would otherwise imply? No. Everything makes sense. I need a more detailed description of the example you have in mind to explain this better. Are "they" two bodies moving relative to each other? And are they moving towards each other so that they intersect at some point in time? Link to comment Share on other sites More sharing options...
md65536 Posted March 4, 2011 Share Posted March 4, 2011 i didn't calculate it but if they were to intersect at point z would they not have met before their speeds would otherwise imply? I recently read in Carl Sagan's Cosmos that some of Einstein's early thought experiments leading to special relativity involved imagining such collisions. See: http://www.american-buddha.com/journeys.space.time.htm and search for "cart" to skip to the relavent bit. The paradoxes and impossible situations happen only without special relativity. Link to comment Share on other sites More sharing options...
dragonstar57 Posted March 11, 2011 Author Share Posted March 11, 2011 No. Everything makes sense. I need a more detailed description of the example you have in mind to explain this better. Are "they" two bodies moving relative to each other? And are they moving towards each other so that they intersect at some point in time? yes they are two bodies and they are moving at a point i labeled as point z (i believe that it is in the op) if you were to draw a line from the bodies to the point the angle that that would form would be 90 degrees Link to comment Share on other sites More sharing options...
michel123456 Posted March 11, 2011 Share Posted March 11, 2011 Precisely. The Galilean world where you would simply use v + u is only an approximation, valid at low speeds. I thought relativistic composition of velocities was valid only between FOR. It seems to me your answers differ from thread to thread. Link to comment Share on other sites More sharing options...
swansont Posted March 11, 2011 Share Posted March 11, 2011 I thought relativistic composition of velocities was valid only between FOR. Galilean refers to a transformation between reference frames. The Lorentz transformation reduces to Galilean at low speeds: if v and u are small, vu/c^2 can be ignored. It seems to me your answers differ from thread to thread. Or you could view it as the answers are consistent, and this is indicative of you having a misunderstanding. Link to comment Share on other sites More sharing options...
Spyman Posted March 11, 2011 Share Posted March 11, 2011 I thought relativistic composition of velocities was valid only between FOR. Yes, but clear distinctions between frames and with respect to what objects have velocities are necessary. If A is moving at speed X and B is moving in the opposite direction also at speed X, then either: 1. they are both separating from each other at X, or 2. they are each separating from a central point at X In case 1: A and B will measure the other to travel at X, but if you want to know how fast a third observer in a central point will measure their separation you need to switch frame. In case 2: The central observer will measure their separation to be 2X, but if you want to know how fast A or B measures the separation from the other then you need to switch frame. Link to comment Share on other sites More sharing options...
michel123456 Posted March 11, 2011 Share Posted March 11, 2011 (edited) Galilean refers to a transformation between reference frames. The Lorentz transformation reduces to Galilean at low speeds: if v and u are small, vu/c^2 can be ignored. Or you could view it as the answers are consistent, and this is indicative of you having a misunderstanding. There is no doubt I have a misunderstanding. But statements like "The Galilean world where you would simply use v + u is only an approximation, valid at low speeds" are there to increase the misunderstanding. In another thread, all of you experts show that in certain case, one MUST use the Galilean addition. That was about the case of closing speed, where you made me admit , IIRC, that relativistic addition is WRONG. Edited March 11, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted March 11, 2011 Share Posted March 11, 2011 In another thread, all of you experts show that in certain case, one MUST use the Galilean addition. That was about the case of closing speed, where you made me admit , IIRC, that relativistic addition is WRONG. That's because the two situations are different. The Lorentz transform reduces to the Galilean transform when the speeds are small. Closing speed is a completely different situation, and the velocity addition formula doesn't apply — you aren't doing a transform to a different frame. The v+u for closing speed isn't the result of using the velocity addition formula, it's from using the definition of velocity. Link to comment Share on other sites More sharing options...
mooeypoo Posted March 12, 2011 Share Posted March 12, 2011 But statements like "The Galilean world where you would simply use v + u is only an approximation, valid at low speeds" are there to increase the misunderstanding. That's quite rude, michel123456. If you don't understand, or if you think the statement was unclear, you can ask and participate in a civil debate. A backdoor ad hominem attack against someone who's trying to help is still a fallacy, and is still against etiquette. ! Moderator Note Argument from ignorance doesn't justify ad hominem.* *The above statement was meant to increase the effectiveness of this thread. Link to comment Share on other sites More sharing options...
michel123456 Posted March 12, 2011 Share Posted March 12, 2011 (edited) It was not intended to be "ad hominem". Statements like "use the Galilean velocity addition at low speed" and "use the relativistic composition of velocities at high speed" are a source of misunderstanding for everyone. Iggy, supported by Janus, Sisyphus, Spyman & Swansont (in alphabetical order, I hope I didn't miss anyone) have showned an example (closing speed) where Galilean velocity addition can theoretically reach 2c. I think it is a valuable information. And there is a difference between ignorance and non-understanding At this moment I know there is something I don't understand, although I understand the explanation. ......... Is that understandable? Edited March 12, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2011 Share Posted March 12, 2011 It was not intended to be "ad hominem". Statements like "use the Galilean velocity addition at low speed" and "use the relativistic composition of velocities at high speed" are a source of misunderstanding for everyone. But the statement was "statements like … are there to increase the misunderstanding," implying intent. You didn't merely say that you were confused, or misunderstood. You blamed me for intentionally causing the confusion. Iggy, supported by Janus, Sisyphus, Spyman & Swansont (in alphabetical order, I hope I didn't miss anyone) have showned an example (closing speed) where Galilean velocity addition can theoretically reach 2c. Until you brought it up, there was no mention of closing speed in this thread. The topic was the velocity addition formula, used when one of the objects is at rest, using values from a different frame of reference, i.e. you are transforming the values from one frame into another. My answer was spot-on: the relativistic velocity addition formula reduces to the classical one at low speeds, IOW, the classical equation does not hold at high speeds. This, of course, assumes it is a situation where you should be using the equation. If you want to talk further about closing speeds, you should do it in the closing speed thread. Link to comment Share on other sites More sharing options...
michel123456 Posted March 12, 2011 Share Posted March 12, 2011 (edited) edited & (deleted) Edited March 12, 2011 by michel123456 Link to comment Share on other sites More sharing options...
swansont Posted March 12, 2011 Share Posted March 12, 2011 ! Moderator Note Posts on the level of discussion of science have been moved http://www.scienceforums.net/topic/55653-the-level-of-discussion/ Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now