SR and reference frames, when does the madness end?

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BoJangles

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<p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">Note I didn&rsquo;t continue this in the other thread I posted which was touching in this subject, as to not interrupt other more advanced ongoing conversations.</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">---</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">Wow special relativity is confusing the hell out of me. </font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">I'm working my way through a Frontiers and controversies in astrophysics lecture by Charles Bailyn (my new favourite lecturer), all is good. This lecture series and others I have worked through are basically introductory courses (as my maths is only just getting up to speed)</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">My goal is to understand relativity though something is just not sticking for me (Note: as I can&rsquo;t ask Charles, I&rsquo;ll ask you guys instead).</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">My understanding of reference frames (simplified) are as follows.</font></p><ul><li><div class="MsoNormal" style="margin-top:0cm;margin-left:0cm;margin-right:0cm"><font face="Calibri" size="3">There is no privileged reference frame (that&rsquo;s an easy one). They are all as good as each other, standing in one, is just as good as observing from another.</font></div></li><li><div class="MsoNormal" style="margin-top:0cm;margin-left:0cm;margin-right:0cm"><font face="Calibri" size="3">A juggler standing on a train doesn&rsquo;t have to compensate for his moving reference frame (I can understand this as well). I.e. juggler on a train juggles just the same as a juggler standing on the solid earth ground (I pulled this example from a </font><span style="font-size:10pt;line-height:115%;font-family:'Arial','sans-serif'">Leonard Susskind's SR lecture)</span><font face="Calibri" size="3">. To me this just means, the momentum given to the juggler is passed into the balls (no need to compensate there).</font></div></li></ul><p>&nbsp;</p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">Say the train is moving at 51% of C (west), and someone on the ground is throwing a ball at 51% of C (East), the person on the train ( in his reference frame) just seen a ball fly away at C + 2%. </font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">Now if old mate juggler went to throw the ball at a speed greater than C, physicists would jump up and down and say can&rsquo;t break the speed of light (maths doesn&rsquo;t allow it). Although no reference frame is privileged, none should be under privileged either. </font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">This kind of says to me, that you actually can&rsquo;t speed anything up to even half the speed of light, just in case someone is looking from a different reference frame going in an equal and opposite direction, or inversely that your own reference frame isn&rsquo;t moving at some massive speed( as proper motion through the galaxy would suggest ).</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">There is something contradicting its self here, and I'm sure it&rsquo;s the basis of why I do not understand this, no matter how many video lectures I sit through.</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal"><font face="Calibri" size="3">In summary of my woes, why do reference frames seem pointless and anti intuitive? Why isn&rsquo;t there are a base reference (the mother of all frames)? Actually you can ignore this one (I don&rsquo;t want to prove physics wrong), just understand it and believe the results.</font></p><p style="margin-top:0cm;margin-left:0cm;margin-right:0cm" class="MsoNormal">&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#808080">-------------- </font></p><p align="center"><font size="1" color="#808080"><em>Let me start out with the standard disclaimer ... I am an idiot, I know almost nothing, I haven’t taken calculus, I don’t work for NASA, and I am one-quarter Bulgarian sheep dog.  With that out of the way, I have several stupid questions... </em></font></p><p align="center"><font size="1" color="#808080"><em>*** A few months blogging can save a few hours in research ***</em></font></p> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Say the train is moving at 51% of C (west), and someone on the ground is throwing a ball at 51% of C (East), the person on the train ( in his reference frame) just seen a ball fly away at C + 2%.<br /> Posted by Manwh0re</DIV></p><p>This is wrong.&nbsp; Only in the frame of the person throwing the ball would 102% C be observed.&nbsp; Separation/closing speeds greater than C witnessed by an outside reference frame are not violations of SR because they observe neither object by themselves with a velocity greater than C. </p><p>In your example above comparing the juggler to the thrown ball, you need to use the SR velocity addition formula which is:</p><p>s = v+u / 1+(v/c)(u/c)&nbsp;</p><p>Where s = velocity between the train and the ball.&nbsp; v = velocity between ball thrower and train and u = velocity between ball thrower and ball.&nbsp; The result is your juggler will see the ball moving at .816 C.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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BoJangles

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>This is wrong.&nbsp; Only in the frame of the person throwing the ball would 102% C be observed.<br />Posted by derekmcd</DIV><br /><br />Hang on a sex, i just said ( my rules ) the person on the ground threw the ball at half C + 1%, which would mean the observer ( as seen through the juggler) seen the ball pass by at 1c + 2%</p><p>:/</p><p>This should be simple maths, the&nbsp;person travelling down X at 1/2C+0.01C, person standing on X0 Throwing a ball -X at -1/2C+ 0.01C</p><p>The person on the train should should see the ball at 2x whatever the combined velocity was?</p><p>Im sure its something simple here Im missing :/...</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#808080">-------------- </font></p><p align="center"><font size="1" color="#808080"><em>Let me start out with the standard disclaimer ... I am an idiot, I know almost nothing, I haven’t taken calculus, I don’t work for NASA, and I am one-quarter Bulgarian sheep dog.  With that out of the way, I have several stupid questions... </em></font></p><p align="center"><font size="1" color="#808080"><em>*** A few months blogging can save a few hours in research ***</em></font></p> </div>
 
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Mee_n_Mac

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Im sure its something simple here Im missing :/... <br />Posted by <strong>Manwh0re</strong></DIV><br /><br />Yes there is.&nbsp; What you're doing is trying to transform a description of motion in 1 reference frame to a description of motion in another reference frame using the wrong rules of transformation. Let me go deeper. Generally we use a set of 3 orthogonal (and thus independant) axes to describe the position of a thing.&nbsp; We say it's some distance in the X dimension, some other distance in the Y dimension and lastly yet another distance in the Z dimension.&nbsp; You can describe motion in this reference frame by adding a dimension of time and noting the position (X,Y,Z) at each instant of time.&nbsp; So far so good.&nbsp; Now say your friend is on a train and he defines his reference frame to have his X be in the direction the train is moving, His Y to be 90 degrees from the X and to the right side of the train and his Z to be up.&nbsp; He's holding the ball in his hands and says in his reference frame it's not moving.&nbsp; In your reference frame it's clearly moving. What's going on ? Well you're both right and there's no mystery, you're both describing the same thing but in different frames. You can come up with a method to transform his description of position and motion in his reference frame into a compatible description in your reference frame. Sparing you the details of the math, you end up using simple addition and perhaps some sines and cosines. You then can transform any description from 1 frame to the other.&nbsp;</p><p>The problem comes (SR-wise) when his frame is moving very fast relative to yours (or visa versa).&nbsp; While you think you can still do the simple math you've always done it turns out this doesn't work anymore.&nbsp; You need a new set of rules to transform position and motion from 1 frame to the other. Those rules are the Lorentz transform. At low relative speeds they give you the same answers you always got or close enough to them that you'd never notice any difference. That you simply can't add velocities like you're used to doing is non-intuitive because in ordinary life the simple approximation to the real truth, the Galilean transformations you have been using, are close enough. It's only at high velocities that you'd begin to notice that they no longer give the right answers.&nbsp; The heart of SR is that in every&nbsp;inertial frame you would/will always measure the speed of light to be C. That has some very&nbsp;non-intuitive consequences.&nbsp; </p><p>OK, enough for now.&nbsp;</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Note I didn&rsquo;t continue this in the other thread I posted which was touching in this subject, as to not interrupt other more advanced ongoing conversations.---Wow special relativity is confusing the hell out of me. I'm working my way through a Frontiers and controversies in astrophysics lecture by Charles Bailyn (my new favourite lecturer), all is good. This lecture series and others I have worked through are basically introductory courses (as my maths is only just getting up to speed)My goal is to understand relativity though something is just not sticking for me (Note: as I can&rsquo;t ask Charles, I&rsquo;ll ask you guys instead).My understanding of reference frames (simplified) are as follows.There is no privileged reference frame (that&rsquo;s an easy one). They are all as good as each other, standing in one, is just as good as observing from another.A juggler standing on a train doesn&rsquo;t have to compensate for his moving reference frame (I can understand this as well). I.e. juggler on a train juggles just the same as a juggler standing on the solid earth ground (I pulled this example from a Leonard Susskind's SR lecture). To me this just means, the momentum given to the juggler is passed into the balls (no need to compensate there).&nbsp;Say the train is moving at 51% of C (west), and someone on the ground is throwing a ball at 51% of C (East), the person on the train ( in his reference frame) just seen a ball fly away at C + 2%. Now if old mate juggler went to throw the ball at a speed greater than C, physicists would jump up and down and say can&rsquo;t break the speed of light (maths doesn&rsquo;t allow it). Although no reference frame is privileged, none should be under privileged either. This kind of says to me, that you actually can&rsquo;t speed anything up to even half the speed of light, just in case someone is looking from a different reference frame going in an equal and opposite direction, or inversely that your own reference frame isn&rsquo;t moving at some massive speed( as proper motion through the galaxy would suggest ).There is something contradicting its self here, and I'm sure it&rsquo;s the basis of why I do not understand this, no matter how many video lectures I sit through.In summary of my woes, why do reference frames seem pointless and anti intuitive? Why isn&rsquo;t there are a base reference (the mother of all frames)? Actually you can ignore this one (I don&rsquo;t want to prove physics wrong), just understand it and believe the results.&nbsp; <br />Posted by Manwh0re</DIV></p><p>&nbsp;Careful, there are privleged reference frams.&nbsp; Spacial relativity is formulated in an inertial reference frame.&nbsp; Inertial reference frames are privleged, and they are the only ones in which the equations of special relativity hold.&nbsp; If you don't pay attention to that restriction you get into trouble with things like the "twin paradox".</p><p>What is true is that one does not prefer one inertial reference frame over anothe inertial reference frame.&nbsp; In particular there is no notion of some frame of "absolute rest".</p><p>Next, and harder, you cannot add velocities linearly between reference frames.&nbsp; You can do that in Newtonian mechanics with what is called the Galilean transformation, but special relativity uses the Lorentz transformation and the formula for addition of velocities is more complicated.&nbsp; </p><dl><dd><img class="tex" src="http://upload.wikimedia.org/math/2/9/1/291bfd8042576ef4c34fb191693e72c0.png" alt="w'=frac{w-v}{1-wv/c^2}." /></dd><dd></dd><dd>http://en.wikipedia.org/wiki/Special_relativity<br /></dd></dl> <div class="Discussion_UserSignature"> </div>
 
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why06

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An Object is approaching you at 102% of C if from an outside reference frame one clocks you heading at 51% of the speed of light towards it and clocks the ball at 51% of the speed of light towards where you were. If the two objects are traveling along the same line of motion. <div class="Discussion_UserSignature"> <div>________________________________________ <br /></div><div><ul><li><font color="#008000"><em>your move...</em></font></li></ul></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>An Object is approaching you at 102% of C if from an outside reference frame one clocks you heading at 51% of the speed of light towards it and clocks the ball at 51% of the speed of light towards where you were. If the two objects are traveling along the same line of motion. <br />Posted by why06</DIV></p><p>I would agree with this statement if you would say that an outside observer sees a closing velocity of 1.02 c.&nbsp; But that is not what you would see if you were the one moving at .51 c relative to that reference frame.</p><p>It is possible to have "phenomena" travel at apparent velocities that exceed c, but not masses and not information-carrying signals.&nbsp; The classic example is to sweep a searchlight against a distant panel.&nbsp; The illuminated spot on that panel can move arbitrarily quickly, faster than c, but of course there is nothing that is actually moving and no information is being transmitted.&nbsp; </p> <div class="Discussion_UserSignature"> </div>
 
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why06

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<p>No, ofcourse not. the ball would actually appear to physically shortened (from what I've read) as it approached the passenger on the train. Space and time are being compressed. However if the ball is heading toward the observer @ v=c and no information, as you said, can travel faster than c what will the train passenger actually see?</p><p>&nbsp;Would light be compressed? will the ball seem brighter</p><p>Time? Will the ball appear to slow down</p><p>Space? Will the amount of space suddenly increase as the two "space-time" wakes approach each other?</p> <div class="Discussion_UserSignature"> <div>________________________________________ <br /></div><div><ul><li><font color="#008000"><em>your move...</em></font></li></ul></div> </div>
 
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Mee_n_Mac

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>No, ofcourse not. the ball would actually appear to physically shortened (from what I've read) as it approached the passenger on the train. Space and time are being compressed. However if the ball is heading toward the observer @ v=c and no information, as you said, can travel faster than c what will the train passenger actually see?&nbsp;Would light be compressed? will the ball seem brighterTime? Will the ball appear to slow downSpace? Will the amount of space suddenly increase as the two "space-time" wakes approach each other? <br />Posted by <strong>why06</strong></DIV><br /><br />The velocity can't equal C but as it gets closer to it your train guy would see a funny ball (short front to back but round otherwise) that is blue in color as it approached and red in color as it departed. <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>No, ofcourse not. the ball would actually appear to physically shortened (from what I've read) as it approached the passenger on the train. Space and time are being compressed. However if the ball is heading toward the observer @ v=c and no information, as you said, can travel faster than c what will the train passenger actually see?&nbsp;Would light be compressed? will the ball seem brighterTime? Will the ball appear to slow downSpace? Will the amount of space suddenly increase as the two "space-time" wakes approach each other? <br />Posted by why06</DIV></p><p>Actually, the ball will not visually appear to be compressed.&nbsp; It will appear to be circular in cross section.&nbsp; But it will not actuall BE circular in cross section.&nbsp; The appearance of circularity arrises from the finite speed of light and the fact that light rays from different areas of the ball travel different distances to your eye.&nbsp; </p> <div class="Discussion_UserSignature"> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Actually, the ball will not visually appear to be compressed.&nbsp; It will appear to be circular in cross section.&nbsp; But it will not actuall BE circular in cross section.&nbsp; The appearance of circularity arrises from the finite speed of light and the fact that light rays from different areas of the ball travel different distances to your eye.&nbsp; <br /> Posted by DrRocket</DIV></p><p>Penrose-Terrell Effect: (the first link is much better)</p><p>http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html</p><p>http://en.wikipedia.org/wiki/Terrell_rotation&nbsp;</p><p>&nbsp;</p><p>Here's a good paper describing different observations including the Penrose-Terrell Effct:</p><p>http://aether.lbl.gov/www/classes/p1...ework/five.pdf</p><p>&nbsp;</p><p>Here's some pretty neat animations, but I can't vouch for how rigorously they applied math to them.</p><p>http://www.anu.edu.au/Physics/Savage/TEE/site/tee/home.html</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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