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Basic question about TD and LC

by mananvpanchal
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mananvpanchal
#1
Mar10-12, 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?
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Doc Al
#2
Mar12-12, 09:51 AM
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Quote Quote by mananvpanchal View Post
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],
OK, you've chosen events that take place at different times but at the same location in A. An example of this could be two time readings on a clock at rest in A.

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]
Yes. This is the normal time dilation formula--moving clocks are measured to run slowly. So B will measure a greater time between those events than would A.

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]?
I don't understand this comment. Why would you think this? (Note that the simple time dilation formula only applies to time intervals between events taking place at the same location in the moving frame, where Δxa = 0.)

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].
Now you have chosen events that take place at the same time, but at different positions in A. Could be simultaneous measurements of the ends of a rod in A.

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]
OK. This should make sense. After all, in B's frame the rod is moving. So the positions of those events will be further apart.

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?
In order for B to make a measurement of the length of a rod that is at rest in A, B must measure the positions at the same time. When he does that, then he'll see the usual length contraction. Note that here the measurements of the endpoints are not made at the same time according to B.
mananvpanchal
#3
Mar14-12, 01:18 AM
P: 215
Hello Doc Al

http://www.astro.ucla.edu/~wright/Al...rian-anim.html

Click image for larger version

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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].
Quote Quote by Doc Al
Yes. This is the normal time dilation formula--moving clocks are measured to run slowly. So B will measure a greater time between those events than would A.
Yes, you are right. We can easily see in the above link and the image that if we put the two points of [itex]t_b[/itex] on parallel lines to [itex]x_a[/itex]. The time-component of B frame (perpendicular distance between this two parallel lines) is still bigger than time-component of A frame. So, B feels more time between two events, and so B feels A's clock running slowly


Quote Quote by Doc Al
OK. This should make sense. After all, in B's frame the rod is moving. So the positions of those events will be further apart.
Quote Quote by Doc Al
In order for B to make a measurement of the length of a rod that is at rest in A, B must measure the positions at the same time. When he does that, then he'll see the usual length contraction. Note that here the measurements of the endpoints are not made at the same time according to B.
If we put two points of [itex]x_b[/itex] on parallel lines to [itex]t_a[/itex]. We can still see that space component (perpendicular distance between this two parallel lines) is still bigger in B frame than space component of A frame. So, B feels length between two events is expanded. I cannot understand this. The rod is moving in B frame. So length between two events should be contracted, not expanded.

Doc Al
#4
Mar14-12, 04:48 AM
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Basic question about TD and LC

Quote Quote by mananvpanchal View Post
If we put two points of [itex]x_b[/itex] on parallel lines to [itex]t_a[/itex]. We can still see that space component (perpendicular distance between this two parallel lines) is still bigger in B frame than space component of A frame. So, B feels length between two events is expanded. I cannot understand this. The rod is moving in B frame. So length between two events should be contracted, not expanded.
Only if those events marked the ends of the rods at the same time according to B. If the events had the same time coordinates in frame A they would not have the same time coordinates in frame B. (Simultaneity is frame dependent.)

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.
mananvpanchal
#5
Mar14-12, 05:15 AM
P: 215
Quote Quote by Doc Al View Post
Only if those events marked the ends of the rods at the same time according to B. If the events had the same time coordinates in frame A they would not have the same time coordinates in frame B. (Simultaneity is frame dependent.)

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.
http://en.wikipedia.org/wiki/Length_...representation

Click image for larger version

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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]?
Doc Al
#6
Mar14-12, 06:02 AM
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Quote Quote by mananvpanchal View Post
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.
No. If the two events happen at the same location according to B's frame, then of course A will measure a greater time interval than B. This is equivalent to there being a clock at rest in B, so B measures the shortest time between those events. Realize that since the events happen at different locations in A there are multiple clocks involved. So you cannot simply apply the time dilation formula to that time interval.

Nonetheless, both frames see the other's clocks as running slow (and out of sync along the direction of motion).
mananvpanchal
#7
Mar14-12, 07:20 AM
P: 215
Quote Quote by Doc Al View Post
No. If the two events happen at the same location according to B's frame, then of course A will measure a greater time interval than B. This is equivalent to there being a clock at rest in B, so B measures the shortest time between those events. Realize that since the events happen at different locations in A there are multiple clocks involved. So you cannot simply apply the time dilation formula to that time interval.

Nonetheless, both frames see the other's clocks as running slow (and out of sync along the direction of motion).
Please, try to get my point the events is not happened at same location in B's frame. The events is happened at same location in A's frame.

Please, look at the image below.

Click image for larger version

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Please, look at left hand side of image. AB is time duration between two events in A's frame happened at same location and at different time. Where AC is time duration between the two events in B's frame. If we want to know how much time elapsed in B's frame between those two events, we want to take parallel line to [itex]x_a[/itex]. From this we get AC' as time elapsed in B's frame. We can say AC' > AB, so we can say that A's clock running slower than B's clock.

On the right hand side of image. The two events A and B happened at same time and at different location in A's frame. When we transform the events co-ordinates into B's frame. We get A and C point. The two points do not have same time component. So, B' frame cannot find length contraction appropriately. So, the condition is time component of B's frame should be same for events. So we are taking a parallel line to [itex]t_b[/itex] and we get AB as contracted length. Is this not a length measured in A's frame? Can we not take parallel line to [itex]t_a[/itex] rather than [itex]t_b[/itex] as we have done to get time dilation (taking parallel line to [itex]x_a[/itex] rather than [itex]x_b[/itex])?

EDIT: If we take parallel line to [itex]t_b[/itex], we get back the same point from which we transformed. Which doesn't give contracted length. It gives length measured in A's frame at rest. Which should be longest relative to any other frame.
mananvpanchal
#8
Mar15-12, 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].
rksingh
#9
Mar15-12, 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
mananvpanchal
#10
Mar16-12, 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.
ghwellsjr
#11
Mar16-12, 09:35 AM
PF Gold
P: 4,745
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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