Is Time Slower or Faster Inside a Moving Train?

In summary: the light signal will reach the rear of the train before the rear clock has had a chance to reach the zero reading.
  • #1
matdoya
6
0
Dear forum,

when I first read about special relativity I came up with this paradox.

I went to the teacher who gave me the booklet but he couldn't answer my question.

I'm quite confident that this forum is the place to answer my question:

Does time goes slower or faster inside the train?

www.trainparadox.tk

grtz,

S.

ps: you can't react on the site itself but this forum will do just fine of course...
 
Physics news on Phys.org
  • #2
matdoya said:
Dear forum,

when I first read about special relativity I came up with this paradox.

I went to the teacher who gave me the booklet but he couldn't answer my question.

I'm quite confident that this forum is the place to answer my question:

Does time goes slower or faster inside the train?
That's a meaningless question. One of the basic concepts of "relativity" is that any statement or question about properties of an even must include "relative to ..."
From a frame of reference motionless relative to the Earth and moving relative to the train, time goes slower in the train. From a frame of reference motionless relative to the train and moving relative to the earth, time has not changed in the train but time is goes slower outside the train.

www.trainparadox.tk

grtz,

S.

ps: you can't react on the site itself but this forum will do just fine of course...
Also you state at your site:
"If you speed is s1 and the speed of the train is s2 you move at speed s1 + s2 when you are stepping in the same direction and s1 - s2 (or s2 - s1) in the other case.
Einstein stated that this is not the case for the speed of light."

That is not true. The same formula applies to any speeds: if your speed is s1 (relative to the train) and you are on a train with speed s2 (relative to the ground) then your speed, relative to the ground, is
[tex]\frac{s1+ s2}{1+ \frac{s1s2}{c^2}}[/tex]
And if either s1 or s2 is equal to c, the result will be just c.
 
Last edited by a moderator:
  • #3
matdoya said:
ps: you can't react on the site itself but this forum will do just fine of course...
In addition to what HallsofIvy points out regarding addition of velocities, your analysis ignores both length contraction and--more importantly--the relativity of simultaneity. If you want more detailed comments, post your analysis here.
 
  • #4
Doc Al said:
In addition to what HallsofIvy points out regarding addition of velocities, your analysis ignores both length contraction and--more importantly--the relativity of simultaneity. If you want more detailed comments, post your analysis here.

First of all:
congrats about the truly professional way of welcoming me here...
right to the case and stuff,
I'm quite impressed.

Secondly:
I must admit that I'm relatively new to relativity, especially special relativity so try to bear with me...

And now the analysis:

I gave the article on wikipedia about the ladder paradox an attempt, but I'm not sure I get it quite well, so once again: bare with me.

As I stated on the site (I hope you don't expect me to repeat the whole page here, so I quote the site):
site said:
For simplicity I didn't put the Lorentz contraction in.
Because the train is moving at a speed (d.c)/l the train has an actual length of l*sqrt(1-d²)
but this doesn't solve the paradox and would only make the formulas longer.

So my best guess until now is that the relativity of simultaneity will do the trick...

I quickly glanced at the formulas of relativity of simultaneity and in my example I would have to substitute t2, v and x by something complicated so I'm going to do the math later on when I find the courage for it...

Sorry for the useless post.

S.
 
  • #5
matdoya said:
So my best guess until now is that the relativity of simultaneity will do the trick...

I quickly glanced at the formulas of relativity of simultaneity and in my example I would have to substitute t2, v and x by something complicated so I'm going to do the math later on when I find the courage for it...

First, I am going to be unprofessional and say welcome to PF matdoya :smile:

Second, you are right that "relativity of simultaneity" will do the trick, but it is not complicated. Basically, what appears simultaneous to one observer will not appear simultaneous to another observer with relative motion. The equation for relativity of simultaneity is simply -Lv/c^2 where L is the length of the train as measured by an obvserver onboard the train. For example if the train has a rest length of 1.0 lightsecond and a velocity relative to the track of 0.8c then when the clocks on the train are synchronised according to Einstein's method, the clock at the rear of the train will be ahead of the clock at the front of the train by 0.8 seconds. If a light signal is sent from the front of the train when the front clock is reading zero seconds, the rear clock will already have a head start of 0.8 seconds showing on it at the moment the light signal started out. When the light signal reaches the rear, the rear clock will have advanced by 0.2 seconds and will be reading 1 second when the signal arrives. On the other hand when a signal starts out from the rear of the train at time zero according to the rear clock the front clock will be reading -0.8 seconds and the front clock advances by 1.8 seconds in the time it takes the light signal from the back to catch up with the front and the front clock will also be reading 1 second when the signal arrives. In both cases the amount the clocks advance during the signal travel times, is less than elapsed time measured by observers at rest with the track. Given that at a relative velocity of 0.8c the time dilation factor is 1.66666 and length contraction factor is 0.6, you should have enough information to work out the ellapsed times measured by the track observers.
 
  • #6
kev said:
First, I am going to be unprofessional and say welcome to PF matdoya :smile:

tsss ;)

kev said:
Second, you are right that "relativity of simultaneity" will do the trick, but it is not complicated. Basically, what appears simultaneous to one observer will not appear simultaneous to another observer with relative motion. The equation for relativity of simultaneity is simply -Lv/c^2 where L is the length of the train as measured by an obvserver onboard the train. For example if the train has a rest length of 1.0 lightsecond and a velocity relative to the track of 0.8c then when the clocks on the train are synchronised according to Einstein's method, the clock at the rear of the train will be ahead of the clock at the front of the train by 0.8 seconds. If a light signal is sent from the front of the train when the front clock is reading zero seconds, the rear clock will already have a head start of 0.8 seconds showing on it at the moment the light signal started out. When the light signal reaches the rear, the rear clock will have advanced by 0.2 seconds and will be reading 1 second when the signal arrives. On the other hand when a signal starts out from the rear of the train at time zero according to the rear clock the front clock will be reading -0.8 seconds and the front clock advances by 1.8 seconds in the time it takes the light signal from the back to catch up with the front and the front clock will also be reading 1 second when the signal arrives. In both cases the amount the clocks advance during the signal travel times, is less than elapsed time measured by observers at rest with the track. Given that at a relative velocity of 0.8c the time dilation factor is 1.66666 and length contraction factor is 0.6, you should have enough information to work out the ellapsed times measured by the track observers.

Yeah,
well,
I kinda hold on to the idea until I can do the whole calculation and the whole equation is solved on my own special way, so expect either the aha-post or, I hope, a calculation that needs further investigation...

Anyway, I know already know that an important element was left out,
chances are that the site will be off-line soon...

best regards,

Sander
 

1. What is the "An other train paradox?"

The "An other train paradox" is a thought experiment that explores the concept of relativity and time dilation. It involves two trains traveling at different speeds and raises questions about how time is experienced by observers in different frames of reference.

2. Who came up with the "An other train paradox?"

The "An other train paradox" was first proposed by Albert Einstein in his theory of special relativity in 1905. However, the paradox has been expanded upon and studied by many scientists and philosophers since then.

3. How does the "An other train paradox" challenge our understanding of time?

The paradox challenges our understanding of time by showing that it is relative and can be experienced differently by observers in different frames of reference. It also highlights the concept of time dilation, where time appears to pass slower for objects in motion compared to those at rest.

4. Is the "An other train paradox" a physical or thought experiment?

The "An other train paradox" is primarily a thought experiment, as it is used to illustrate and explore theories of relativity and time. However, it can also be considered a physical experiment, as it has been tested and observed in real-world scenarios, such as the famous Hafele-Keating experiment in 1971.

5. What are the real-life implications of the "An other train paradox?"

The "An other train paradox" has significant implications for our understanding of time, space, and the nature of reality. It has also led to advancements in technology, such as the development of atomic clocks and GPS systems that take into account the effects of time dilation.

Similar threads

  • Special and General Relativity
2
Replies
44
Views
3K
  • Special and General Relativity
Replies
29
Views
1K
  • Special and General Relativity
3
Replies
71
Views
4K
  • Special and General Relativity
Replies
20
Views
1K
Replies
35
Views
1K
  • Special and General Relativity
Replies
9
Views
868
Replies
9
Views
1K
  • Special and General Relativity
Replies
27
Views
4K
Replies
12
Views
1K
  • Special and General Relativity
Replies
6
Views
2K
Back
Top