Pendulum on a Relativistic Train

In summary, the pendulum Period is not a proper time interval for observer T, observer G or both T and G.
  • #1
Jefffff
25
0

Homework Statement



In a thought experiment, a train is moving at a speed of 0.95c relative to the ground. A pendulum attached to the ceiling of the train is set into oscillation. An observer T on the train and an observer G on the ground measure the period of oscillation of the pendulum. State and explain whether the pendulum period is a proper time interval for observer T, observer G or both T and G.

Homework Equations


[/B]
Lorentz transformations and special relativity equation?

The Attempt at a Solution



I believe that T experiences the proper time period for the pendulum because we know that for points A and B that are static in space, a person stationary to either A or B measures the proper length and the person traveling from A to B at a speed of a fraction of light measures the proper time. However, since in this situation A and B are not stationary to G, they are stationary to T, therefore T measures the proper time for the oscillation period of the pendulum.

Any help would be greatly appreciated. Thanks!
 
Physics news on Phys.org
  • #2
So, you're thinking that I, as an observer watching the pendulum as the train passes through, will see a pendulum that is not swinging properly.

Let's make this pendulum a simple sphere swinging from a line in a vacuum - as viewed from the train. Clearly, I will not see a sphere - because it will be foreshortened by Lorentz contraction. But the shape of that sphere doesn't matter. Also, I will see a sphere of much higher mass. But the mass of the sphere will also not affect it's period. Finally, I will see a sphere that is swinging slower than the train passengers see. So, at first it would appear that both the T's and the G's should calculate the same pendulum rate - and at least one of them should end up wondering why the pendulum isn't behaving properly.

So your question is this: What is the difference between T's and G's calculation of the pendulum rate of swing?

I will give you this hint: You're thinking from the "G" point of view. You need to consider what the world looks like from the "T" point of view.
 
  • #3
According to T, the pendulum appears to be swinging like a normal pendulum. T is also stationary relative to the equilibrium of the pendulum's oscillation so it should measure proper time for the period. T also measures the proper length. To G, the pendulum would appear to swing faster from left to right (assuming the train travels from left to right) and slower from right to left.
 
  • #4
Jefffff said:
According to T, the pendulum appears to be swinging like a normal pendulum. T is also stationary relative to the equilibrium of the pendulum's oscillation so it should measure proper time for the period. T also measures the proper length. To G, the pendulum would appear to swing faster from left to right (assuming the train travels from left to right) and slower from right to left.
You need more hints.
First, let's make this simple. Have the train moving from west to east - and the pendulum swinging north and south.

Let's make the length of the pendulum 2.3 meters. Since that length is perpendicular to motion, it will be 2.3 meters for both T and G.
Here is a pendulum calculator:
https://www.easycalculation.com/physics/classical-physics/simple-pendulum.php

Compute the period as seen by T and G based on that calculator.
Then, compute the ratio of the pendulum frequencies (T/G) based on the relativistic velocity.
If the results do not agree, explain why.
If you get to that point and it still hasn't struck you what is going on, I'll give you one more hint.
 
  • #5
You're right. Only the T sees proper time - because proper time is from the view of the pendulum.

I was over-working the problem. The force of gravity in the proper time reference frame is much higher than for G, so the pendulum will be swinging much faster for T than for G. But that has nothing to do with proper time.
 
  • #6
Okay, thanks for your help!
 

FAQ: Pendulum on a Relativistic Train

What is a pendulum on a relativistic train?

A pendulum on a relativistic train is a thought experiment used to illustrate the effects of special relativity on a moving system. It involves a pendulum swinging back and forth on a train moving at a constant velocity close to the speed of light.

What is the significance of this thought experiment?

This experiment helps to demonstrate the concepts of time dilation and length contraction, which are fundamental principles of special relativity. It also highlights the limitations of classical mechanics in explaining the behavior of objects in extreme conditions.

How does the speed of the train affect the pendulum's motion?

As the train approaches the speed of light, the pendulum's period (time for one swing) increases due to time dilation. This means that the pendulum will appear to move slower to an observer on the train compared to an observer on the ground. The length of the pendulum will also appear shorter due to length contraction.

What is the relationship between the train's velocity and the pendulum's period?

The period of the pendulum is directly proportional to the train's velocity. This means that as the train's speed increases, the pendulum's period also increases. This is because time is passing slower for the pendulum on the train compared to the observer on the ground.

Can this thought experiment be applied to real-world situations?

While this thought experiment may seem abstract, the principles it illustrates have been observed in various experiments and are applicable in real-world situations. For example, GPS satellites use special relativity to account for the time dilation experienced by the satellites due to their high speeds in orbit.

Similar threads

Replies
14
Views
902
Replies
9
Views
1K
Replies
7
Views
3K
Replies
13
Views
1K
Replies
18
Views
2K
Replies
1
Views
2K
Replies
3
Views
7K
Back
Top