Time Dilation to Length Contraction

In summary: Ha's frame and (2)...the end of the motion of the broom in Ha's frame. So the time dilation equation would apply to each of those events separately.
  • #1
Biest
67
0
So I have been trying to wrap my head around time dilation and length contraction... It is all good until i come to the point to derive length contraction from time dilation...

So we have

[tex] \gamma \Delta \tau = \Delta t [/tex]

[tex] \gamma L = L_0[/tex]

So now from my lecture notes i have an observer Ha riding a broom at speed v, and a stationary observer He (so yeah if anybody has the book it is Hartle's Gravity book chapter 4 problem 19). In Ha's frame [itex] L_0 = v \Delta \tau [/itex] and in He's frame [itex] L = v \Delta t [/itex], so now when i substitute that into the equation for time dilation I will not get the result for length contraction. Am I misinterpreting the question or something... Ironically, the gravitational red-shift seems a easier from Hartle's explanation... Maybe I am just missing a [itex] \gamma [/itex] somwhere.

Thanks for any help.

Cheers,

Biest
 
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  • #2
Biest said:
In Ha's frame [itex] L_0 = v \Delta \tau [/itex] and in He's frame [itex] L = v \Delta t [/itex], so now when i substitute that into the equation for time dilation I will not get the result for length contraction.
Why not? Hint: Eliminate v in the second equation.
 
  • #3
[tex] L = \Delta t[/tex]?

What I did in detail was:

[tex] \gamma \frac{L_0}{v} = \frac{L}{v} [/tex]

which gets me

[tex] \gamma L_0 = L [/tex]

instead of having

[tex] L_0 = \gamma L [/tex]
 
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  • #4
Biest said:
[tex] L = \Delta t[/tex]?
No. Rearranging [itex] L_0 = v \Delta \tau [/itex], you get [itex]v = L_0/\Delta \tau[/itex]. Plug that into [itex] L = v \Delta t [/itex].
 
  • #5
Careful: In this problem, the single "moving clock" is in He's frame. So the time dilation formula should read: [tex] \Delta \tau = \gamma \Delta t [/tex]
 
  • #6
Doc Al said:
No. Rearranging [itex] L_0 = v \Delta \tau [/itex], you get [itex]v = L_0/\Delta \tau[/itex]. Plug that into [itex] L = v \Delta t [/itex].



[itex] L = v \Delta t [/itex]
becomes

[itex] L = \frac{L_0}{\Delta \tau} \Delta t [/itex]

where

[itex] \frac{\Delta t}{\Delta \tau} = \gamma [/itex]

so we get

[itex] L = L_0 \gamma [/itex]
 
  • #7
Doc Al said:
Careful: In this problem, the single "moving clock" is in He's frame. So the time dilation formula should read: [tex] \Delta \tau = \gamma \Delta t [/tex]

Well that explains it... I was just completely confused by that... Thanks for the patience...

Just to summarize...in this case we consider He even though standing still as moving and Ha as stationary. such that the proper time [tex] \Delta \tau [/tex] become the [tex] \Delta t [/tex] in the equation in the op
 
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  • #8
Do you understand why it's that way?
 
  • #9
Doc Al said:
Do you understand why it's that way?

Yes, cause we are considering He to be moving with respect to Ha. If we consider Ha as stationary, the time required to pass He, the time measures by Ha, will longer then what He measures. In the now stationary frame the length stays constant while in Hes frame it will be contracted
 
  • #10
Biest said:
Yes, cause we are considering He to be moving with respect to Ha. If we consider Ha as stationary, the time required to pass He, the time measures by Ha, will longer then what He measures. In the now stationary frame the length stays constant while in Hes frame it will be contracted
But they are in relative motion--who's stationary is arbitrary. So that reasoning would work both ways.

But Ha is the one carrying the broom. When he measures his time, he's actually using two clocks: one at the front of the broom; one at the rear. He's measuring the time that it takes He to move from one end of the broom to the other. The time dilation formula applies to the time as measured on a single moving clock (i.e., "moving clocks run slow"). So, Ha can apply the formula to He's clock reading--but He cannot apply it to Ha, since Ha's time involves multiple clocks and synchronization.

Make sense?
 
  • #11
Sry to drag this on. So basically what happens is that the moving frame in case of a rod, broom whatever, uses two clocks to measure its time, while the stationary frame is using one. So we have 1 vs. 2 clocks, where we consider the frame with one clock as the frame in which proper time is measured
 
  • #12
Right. Proper time is always measured by a single clock. In this problem, the non-broom frame is measuring the time for the broom to pass using a single clock, so we can use the time dilation formula to relate that time to that measured in the other frame (using multiple clocks).
 
  • #13
This goes back to the concept of the two ends being "events" in the moving frame while in the stationary frame they are one
 
  • #14
The ends of the broom are just things, not events.

I'd say that the events in question are: (1) observer He passes the front end of the broom; (2) observer He passes the back end of the broom.

In the He frame, these events happen at the same place. Not so in the Ha frame.
 
  • #15
Sorry about the confusing with the events. Hartle calls the ends "two simultaneous events." Thank you very much for the help.
 
  • #16
Biest said:
Sorry about the confusing with the events. Hartle calls the ends "two simultaneous events." Thank you very much for the help.
That's sloppy talk. (I'll have to look up what he says exactly.) What he should say is that when measuring the length of a rod, one measures two simultaneous events: (1) Left end passes by clock A which reads time t_1; (2) Right end passes by clock B which also reads t_1. Those event are simultaneous in the frame doing the measurement of the length of the moving rod. You need to measure the positions of both ends at the same time. (Of course, in the frame of the rod, those events are not simultaneous.)
 
  • #17
Doc Al said:
That's sloppy talk. (I'll have to look up what he says exactly.) What he should say is that when measuring the length of a rod, one measures two simultaneous events: (1) Left end passes by clock A which reads time t_1; (2) Right end passes by clock B which also reads t_1. Those event are simultaneous in the frame doing the measurement of the length of the moving rod. You need to measure the positions of both ends at the same time. (Of course, in the frame of the rod, those events are not simultaneous.)

The sentences goes:

"The length of the rod is the distance between two simultaneous events at its ends. But the notion of simultaneity is different in different inertial frames."

I understand that simultaneity is relative. it was just confusing as to how it is done in lecture versus the book cause my prof. jumped from the time dilation straight to length contraction without putting any steps between.
 
  • #18
Biest said:
The sentences goes:

"The length of the rod is the distance between two simultaneous events at its ends. But the notion of simultaneity is different in different inertial frames."
That's perfectly stated. (I would expect no less from Hartle.)
 
  • #19
Doc Al said:
That's perfectly stated. (I would expect no less from Hartle.)

Yeah i was just confused by how my prof. approached it, which Hartle does in the problem (chapter 4 problem 17) and how he does it using an analogue of the problem. It is also cleared in the problem cause the clocks are drawn in. So you can see where the "singular" clock is positioned. I find it clearer in the book. also how he defines dilation... but what can you do if you are a string theorist and "haven't done this in years"
 
  • #20
Biest said:
but what can you do if you are a string theorist and "haven't done this in years"
uh oh... :rolleyes: My sympathies! :smile:
 
  • #21
Doc Al said:
uh oh... :rolleyes: My sympathies! :smile:

At least i get to learn how to brood force my way through finding geodesics of a 2-sphere... as if the christoffel symbol never existed
 
  • #22
Yikes.
 
  • #23
Doc Al said:
Yikes.

Involves only couple derivatives. a couple hundred sign errors and 3 hours later you ge t an answer... ohh that homework was so enjoyable esp. deriving the christoffel symbol from scratch :smile:
 

Related to Time Dilation to Length Contraction

1. What is time dilation and length contraction?

Time dilation and length contraction are two concepts in the theory of relativity that describe how time and space are perceived differently by different observers. Time dilation refers to the slowing down of time for an object that is moving at high speeds, while length contraction refers to the shortening of an object's length in the direction of its motion.

2. How does time dilation affect the measurement of time?

Time dilation means that time appears to pass slower for objects moving at high speeds. This effect is only noticeable at speeds close to the speed of light and is caused by the stretching of space-time. This means that the measurement of time for an object in motion will appear slower compared to an object at rest.

3. Can time dilation and length contraction be observed in everyday life?

No, time dilation and length contraction are only significant at extremely high speeds, close to the speed of light. In our everyday lives, we do not experience these effects because we are not moving at such speeds. However, these effects have been observed and confirmed through experiments involving high-speed particles and atomic clocks.

4. What is the equation for time dilation and length contraction?

The equation for time dilation is t = t0 / √(1-v2/c2), where t is the measured time, t0 is the time at rest, v is the velocity of the object, and c is the speed of light. The equation for length contraction is L = L0 * √(1-v2/c2), where L is the measured length, L0 is the length at rest, v is the velocity of the object, and c is the speed of light.

5. What is the relationship between time dilation and length contraction?

Time dilation and length contraction are two sides of the same coin. They are both consequences of the theory of relativity and are related to each other through the Lorentz factor. As an object's speed increases, time dilation increases while length contraction decreases, and vice versa. This shows that the perception of time and space are relative and depend on the observer's frame of reference.

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