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Basic question about TD and LCby mananvpanchal
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#1
Mar1012, 10:33 PM

P: 215

Hello, I have basic confusion about Time Dilation and Length Contraction. I have struggled much, but I haven't succeed. Please. help me to clear it.
In A frame I have [itex][t_{a1}, x_{a1}][/itex] and [itex][t_{a2}, x_{a2}][/itex]. If I assume c=1 and [itex]x_{a1}=x_{a2}[/itex], and if I transform it to B frame which is moving with v speed relative to A frame. I get this. [itex]t_{b1}=\gamma (t_{a1}vx_{a1})[/itex] [itex]t_{b2}=\gamma (t_{a2}vx_{a2})[/itex] If I simply do [itex]t_{b2}t_{b1}[/itex], I get [itex]t_{b2}t_{b1} = \gamma (t_{a2}t_{a1})[/itex] [itex]\Delta t_b = \gamma \Delta t_a[/itex] [itex]\Delta t_b > \Delta t_a[/itex] If As I understand [itex]\Delta t_a[/itex] is elapsed time in A frame and [itex]\Delta t_b[/itex] is elapsed time in B frame then it would be [itex]\Delta t_b < \Delta t_a[/itex], then why [itex]\Delta t_b > \Delta t_a[/itex]? Now, If I have [itex][t_{a1}, x_{a1}][/itex] and [itex][t_{a2}, x_{a2}][/itex] where c=1 and [itex]t_{a1}=t_{a2}[/itex]. then, [itex]x_{b1}=\gamma (x_{a1}vt_{a1})[/itex] [itex]x_{b2}=\gamma (x_{a2}vt_{a2})[/itex] If I simply do [itex]x_{b2}x_{b1}[/itex], I get [itex]x_{b2}x_{b1} = \gamma (x_{a2}x_{a1})[/itex] [itex]\Delta x_b = \gamma \Delta x_a[/itex] [itex]\Delta x_b > \Delta x_a[/itex] If As I understand [itex]\Delta x_a[/itex] is length in A frame and [itex]\Delta x_b[/itex] is length in B frame then it would be [itex]\Delta x_b < \Delta x_a[/itex], then why [itex]\Delta x_b > \Delta x_a[/itex]? What am I thinking wrong here? 


#2
Mar1212, 09:51 AM

Mentor
P: 41,311




#3
Mar1412, 01:18 AM

P: 215

Hello Doc Al
http://www.astro.ucla.edu/~wright/Al...riananim.html Please, look at above image when we transform the points on [itex]t_a[/itex]. We get points on [itex]t_b[/itex]. And when we transform the points on [itex]x_a[/itex]. We get points on [itex]x_b[/itex]. 


#4
Mar1412, 04:48 AM

Mentor
P: 41,311

Basic question about TD and LC
In order for B to measure the length of a moving rod, he must measure the position of the ends of the rod at the same time according to him. If he does, then he'll measure the usual length contraction. 


#5
Mar1412, 05:15 AM

P: 215

Ok, now see above image and the link. As you said if we want length contraction, then we should take the two events of [itex]x_b[/itex] on two parallel lines to [itex]t_b[/itex]. So as wiki article says OB > OA, we can achieve length contraction in our case. But, the same thing we can apply to time component too. As shown in image. We can take the two events of [itex]t_b[/itex] on two parallel lines to [itex]x_b[/itex]. And now B sees A's clock is running faster than his own's clock. Using parallel lines to [itex]x_a[/itex] we can explain time dilation. Using parallel lines to [itex]t_b[/itex] we can explain length contraction. Why can we not explain time dilation using parallel lines to [itex]x_b[/itex]? Why can we not explain length contraction using parallel lines to [itex]t_a[/itex]? 


#6
Mar1412, 06:02 AM

Mentor
P: 41,311

Nonetheless, both frames see the other's clocks as running slow (and out of sync along the direction of motion). 


#7
Mar1412, 07:20 AM

#8
Mar1512, 12:56 AM

P: 215

Please, look at left hand side of post #7's image.
The two events A and B happens at same location in A's frame. A measures time duration between this two events as AB in his own clock. When we transform the two events into B's frame. We get A and C. If we take C on parallel line to [itex]x_a[/itex], we get C' point on [itex]t_a[/itex]. So we can conclude that B's frame measures more time duration AC' between this two events in his own clock. So B can conclude that A's clock running slowly than his own clock. The same as above, two events A and B happens at same time in A's frame (right side of image). A measures length between two events as AB with his own ruler. When we transform the two events into B's frame. We get A and C. If we take C on parallel line to [itex]t_a[/itex], we get C' point on [itex]x_a[/itex]. So we can conclude that B's frame measures more length AC' between this two events with his own ruler. So B can conclude that A's ruler is expanded than his own ruler. If we take parallel line to [itex]t_b[/itex] to get length contraction then there are two problem with that. 1. AC is not length measured by B's frame at same time, so if we want to solve this by taking parallel line to [itex]t_b[/itex], we get same point B on [itex]x_a[/itex]. Where AB is distance measured by A's frame not distance measured by B's frame at same time. 2. If we want length contraction and we want to solve this by taking parallel line to [itex]t_b[/itex] then we have to take parallel line to [itex]x_b[/itex] to get time dilation, but we cannot get time dilation by this method. If we want time dilation then we have to take parallel line to [itex]x_a[/itex]. 


#9
Mar1512, 06:49 AM

P: 1

the basic thing about time dilation and length contraction is that both are reciprocal in nature. to understand more about the meaning of reciprocal character you try find the 'twin paradox' and the 'barn and ladder paradox' respectively.
though both paradoxes can't be resolved within the realm of str as they involve accelrated frames but they do give a meaning of reciprocity 


#10
Mar1612, 09:12 AM

P: 215

Hello, Doc Al
My post #8 is the best description of my understanding. Please, flash some light on it. Thanks. 


#11
Mar1612, 09:35 AM

PF Gold
P: 4,684

I thought you came to an understanding on the length contraction part of your questions in your other thread, Length Contraction rearrangement. Was I wrong?
BTW, this is why you shouldn't ask the same question in multiple threads. 


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