Rosnet
07-30-05, 12:45 PM
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<Font face= "times new roman" size= "4"><B>Length Contraction - I</B></Font><Br>
This is the second of my posts introducing Special Relativity. The first one sought to introduce the postulates, and use them to arrive at time dilation. This one deals with Length Contraction, mainly. I also make a small analogy of space-time with the idea of rotations in a plane.<Br>
The Special theory of Relativity denies the notions of absolute Space and absolute Time. Both of these are relative, depending on who makes the observation. But they are tied up with each other. If one of them changes, the other also changes. In other words, Relativity 'merges' Space and Time into Spacetime.<Br>
Here, I'll demonstrate how the idea of 'length', is tied up with that of time. There are different ways to arrive at this result. First, I'll do it my favorite way. This is one in which length contraction is shown to 'cancel out' time dilation. The thought behind is so simple that it makes length conraction appear as an obvious result of time dilation.<Br>
Imagine that Einstein is standing near a railway track, while James Watt goes by in a train moving at uniform speed " v ". The same picture, seen through the eyes of Watt, will be different. Watt will see the ground, and Einstein who is standing on the ground, move past him (and the train) with a constant speed " v ". In Watt's frame he and the train are stationary. It is the ground that is moving.<Br>
For the past few days, Einstein has been hearing Quantum Mechanists all around him yelling that nothing exists until it is measured. "Poor James", he thinks, "If I don't do something quick and measure him, he will cease to exist". So he decides to measure the length of Watt's train. He remembers that he's left his ruler at home. Moreover how could he possibly take the length of a speeding train with a tiny ruler, even if he had one? Fast thinker that he is, he whips out his timer, and clocks the time that the length of the train takes to pass by him.<Br>
<Img src= "http://www.geocities.com/virusmakermad/einstein.gif"></Img><Br>
In the diagram, the first section shows the moment t<Sub>0</Sub> = 0 on Einstein's timer, when the train's front is directly in front of him. The train is moving with a speed " v ". The second section shows the moment t<Sub>0</Sub> = t<Sub>0</Sub>, when the back end of the train is in front of Einstein. So the length of the train is given by " l = vt<Sub>0</Sub> ". The subscript "0" attached to a quantity (time, here) means that it pertains to the measurement of (or by) an object which is at rest in the frame under consideration. Here, the time " t<Sub>0</Sub> " is measured by a timer which is at rest in this frame. But " l " is the length of the moving train, and hence does not have the subscript.<Br>
<Img src= "http://www.geocities.com/virusmakermad/watt.gif"></Img><Br>
This diagram shows the situation as Watt sees it. As noted earlier, Watt will deem himself and the train stationary, and will see Einstein, who is holding the timer, moving backwards (to the right, in the picture) with a speed " v ". This means that in Watt's frame, Einstein's timer is dilated. Therefore, by the time the timer registers a time interval of t<Sub>0</Sub>, Watt's clock, which is ticking faster than the timer, will have registered a time " t = γt<Sub>0</Sub> ". The dilation factor γ (gamma) is given by γ = (1 - v<Sup>2</Sup>/c<Sup>2</Sup>)<Sup>-1/2</Sup>. Remember that in this frame, " t " is the actual time that elapses according to Watt. You have to be careful not to inconsistently mix frames. So this gives the length of the train as seen by Watt. Since the train is at rest in this frame, we'll call the length " l<Sub>0</Sub> ". And since, according to Watt, Einstein (who, in this frame, is moving with speed " v ") took a time " t " to cross the length of the train, we have the relation " l<Sub>0</Sub> = vt ".<Br>
In summary, we have the following relations:<Br>
l = vt<Sub>0</Sub><Br>l<Sub>0</Sub> = vt<Br>
Dividing the first equation by the second,<Br>
l/l<Sub>0</Sub> = t<Sub>0</Sub>/t<Br>l = l<Sub>0</Sub>t<Sub>0</Sub>/t<Br>
<B>l = l<Sub>0</Sub>/γ </B><Br>
The length of an object which is moving with uniform speed is contracted in the direction of motion, by the same factor (γ ) by which its clock slows down. As its speed approaches light speed, γ approaches infinity. Which implies that its length approaches zero, and its clock takes longer and longer time to tick. We can see here how length contraction is inevitably connected to time dilation.
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<Table width= "100%" height= "100%" cellpadding= "15"><Tr>
<Td bgcolor= "#F0F0FF" valign= "top" align= "left">
<Font face= "times new roman" size= "4"><B>Length Contraction - I</B></Font><Br>
This is the second of my posts introducing Special Relativity. The first one sought to introduce the postulates, and use them to arrive at time dilation. This one deals with Length Contraction, mainly. I also make a small analogy of space-time with the idea of rotations in a plane.<Br>
The Special theory of Relativity denies the notions of absolute Space and absolute Time. Both of these are relative, depending on who makes the observation. But they are tied up with each other. If one of them changes, the other also changes. In other words, Relativity 'merges' Space and Time into Spacetime.<Br>
Here, I'll demonstrate how the idea of 'length', is tied up with that of time. There are different ways to arrive at this result. First, I'll do it my favorite way. This is one in which length contraction is shown to 'cancel out' time dilation. The thought behind is so simple that it makes length conraction appear as an obvious result of time dilation.<Br>
Imagine that Einstein is standing near a railway track, while James Watt goes by in a train moving at uniform speed " v ". The same picture, seen through the eyes of Watt, will be different. Watt will see the ground, and Einstein who is standing on the ground, move past him (and the train) with a constant speed " v ". In Watt's frame he and the train are stationary. It is the ground that is moving.<Br>
For the past few days, Einstein has been hearing Quantum Mechanists all around him yelling that nothing exists until it is measured. "Poor James", he thinks, "If I don't do something quick and measure him, he will cease to exist". So he decides to measure the length of Watt's train. He remembers that he's left his ruler at home. Moreover how could he possibly take the length of a speeding train with a tiny ruler, even if he had one? Fast thinker that he is, he whips out his timer, and clocks the time that the length of the train takes to pass by him.<Br>
<Img src= "http://www.geocities.com/virusmakermad/einstein.gif"></Img><Br>
In the diagram, the first section shows the moment t<Sub>0</Sub> = 0 on Einstein's timer, when the train's front is directly in front of him. The train is moving with a speed " v ". The second section shows the moment t<Sub>0</Sub> = t<Sub>0</Sub>, when the back end of the train is in front of Einstein. So the length of the train is given by " l = vt<Sub>0</Sub> ". The subscript "0" attached to a quantity (time, here) means that it pertains to the measurement of (or by) an object which is at rest in the frame under consideration. Here, the time " t<Sub>0</Sub> " is measured by a timer which is at rest in this frame. But " l " is the length of the moving train, and hence does not have the subscript.<Br>
<Img src= "http://www.geocities.com/virusmakermad/watt.gif"></Img><Br>
This diagram shows the situation as Watt sees it. As noted earlier, Watt will deem himself and the train stationary, and will see Einstein, who is holding the timer, moving backwards (to the right, in the picture) with a speed " v ". This means that in Watt's frame, Einstein's timer is dilated. Therefore, by the time the timer registers a time interval of t<Sub>0</Sub>, Watt's clock, which is ticking faster than the timer, will have registered a time " t = γt<Sub>0</Sub> ". The dilation factor γ (gamma) is given by γ = (1 - v<Sup>2</Sup>/c<Sup>2</Sup>)<Sup>-1/2</Sup>. Remember that in this frame, " t " is the actual time that elapses according to Watt. You have to be careful not to inconsistently mix frames. So this gives the length of the train as seen by Watt. Since the train is at rest in this frame, we'll call the length " l<Sub>0</Sub> ". And since, according to Watt, Einstein (who, in this frame, is moving with speed " v ") took a time " t " to cross the length of the train, we have the relation " l<Sub>0</Sub> = vt ".<Br>
In summary, we have the following relations:<Br>
l = vt<Sub>0</Sub><Br>l<Sub>0</Sub> = vt<Br>
Dividing the first equation by the second,<Br>
l/l<Sub>0</Sub> = t<Sub>0</Sub>/t<Br>l = l<Sub>0</Sub>t<Sub>0</Sub>/t<Br>
<B>l = l<Sub>0</Sub>/γ </B><Br>
The length of an object which is moving with uniform speed is contracted in the direction of motion, by the same factor (γ ) by which its clock slows down. As its speed approaches light speed, γ approaches infinity. Which implies that its length approaches zero, and its clock takes longer and longer time to tick. We can see here how length contraction is inevitably connected to time dilation.
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