Special relativity - measure of a rod and simultaneity

In summary, the conversation discusses a special relativity problem involving two inertial frames of reference and a rod with fixed length in one of the frames. The first method of finding the length contraction is through Lorentz transformations, while the second method involves finding the difference in time between two events. The conversation also mentions the concept of relativity of simultaneity. Finally, an alternative method to derive length contraction is suggested.
  • #1
Gom
2
3
Homework Statement
The problem requests to find Lorentz contraction by using two different methods, the first one being Lorentz transformations, and the second, by following some steps described below.
Relevant Equations
Lorentz transformations, Lorentz contraction
Hi, I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, ##O## and ##O'##, the last one moving with relative velocity ##v## in the ##x## direction. There's a rod with length ##L'## fixed to frame ##O'##, such that front end ##A## is at ##x'=0##, and back end ##B## is at ##x'=L'##. Clocks are synchronized at time ##t=t'=0##, when position is ##x=x'=0##. An observer from ##O## measures the rod, and the result is ##L##. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as ##x'=L'##, we have ##L'=\gamma L##, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$

Now I need to find the result ##L'=\gamma L##, but with another method. First, the problem requests to find the ##\Delta t'## (from the point of view of ##O'##) since front end ##A## of the rod is measured by ##O##, until back end ##B## is measured by ##O##. Of course, both events are simultaneous for ##O##, then ##\Delta t=t_B-t_A=0##, so ##t_B=t_A=t##. This is what I did (I'm not sure if it's correct)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$
$$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that back end ##B## is measured before front end ##A##.

After that, the problem requests to find the position of coordinate origin at ##O##, at the moment when back end ##B## is measured by ##O##, as seen by ##O'##; and also the position of coordinate origin at ##O##, at the moment when front end ##A## is measured by ##O##, as seen by ##O'##. Finally, with this and ##\Delta t'=-\gamma vL/c^2##, I should find again the difference of length between ##L## and ##L'##.

Please let me know if I wasn't clear. Thanks.
 
  • Like
Likes berkeman
Physics news on Phys.org
  • #2
Gom said:
Homework Statement:: The problem requests to find Lorentz contraction by using two different methods, the first one being Lorentz transformations, and the second, by following some steps described below.
Relevant Equations:: Lorentz transformations, Lorentz contraction

Hi, I´m trying to solve a special relativity problem, and I think I need some help. There are two inertial frames of reference, ##O## and ##O'##, the last one moving with relative velocity ##v## in the ##x## direction. There's a rod with length ##L'## fixed to frame ##O'##, such that front end ##A## is at ##x'=0##, and back end ##B## is at ##x'=L'##. Clocks are synchronized at time ##t=t'=0##, when position is ##x=x'=0##. An observer from ##O## measures the rod, and the result is ##L##. Now, from Lorentz transformations, we know that

$$x'=\gamma(x-vt)=\gamma(L-v.0)=\gamma L$$

And, as ##x'=L'##, we have ##L'=\gamma L##, with

$$\gamma=\frac{1}{\sqrt(1-(v/c)^2)}$$
Okay, so you transformed the event ##(0, L)##, which is the end of the rod at time ##t = 0## to get the x-coordinate of the rod in ##O'##. And, as the rod is stationary in ##O'## that end remains at ##\gamma L## independent of time.

Gom said:
Now I need to find the result ##L'=\gamma L##, but with another method. First, the problem requests to find the ##\Delta t'## (from the point of view of ##O'##) since front end ##A## of the rod is measured by ##O##, until back end ##B## is measured by ##O##. Of course, both events are simultaneous for ##O##, then ##\Delta t=t_B-t_A=0##, so ##t_B=t_A=t##. This is what I did (I'm not sure if it's correct)

$$t'_A=\gamma(t_A-vx_A/c^2)=\gamma(t-v.0/c^2)=\gamma t$$
$$t'_B=\gamma(t_B-vx_B/c^2)=\gamma(t-vL/c^2)=\gamma t-\gamma vL/c^2$$

Therefore,

$$\Delta t'=t'_B-t'_A=-\gamma vL/c^2$$

Which means that back end ##B## is measured before front end ##A##.
First of all, what the problem calls the "back end" seems to be the front end to me. It's ahead in the direction of motion. The rod could be moving backwards, of course.

In any case, what you've found is that the event ##(0, L)## in ##O## transforms to the event ##(-\frac{\gamma vL}{c^2}, x')## in ##O'##.

In other words, a clock at the end of the rod in ##O'## would read ##t' = -\frac{\gamma vL}{c^2}## as that end of the rod passes some marker at the point ##x = L## in ##O## and a local clock in ##O## reads ##t = 0##.

That's one example of the relativity of simultaneity.
Gom said:
After that, the problem requests to find the position of coordinate origin at ##O##, at the moment when back end ##B## is measured by ##O##, as seen by ##O'##; and also the position of coordinate origin at ##O##, at the moment when front end ##A## is measured by ##O##, as seen by ##O'##. Finally, with this and ##\Delta t'=-\gamma vL/c^2##, I should find again the difference of length between ##L## and ##L'##.

Please let me know if I wasn't clear. Thanks.
I'm not sure why the problem asks you to do this. It seems a little muddled to me.

An alternative method to derive length contraction is as follows:

1) Calculate the trajectory of each end of the rod in frame ##O## using the data from frame ##O'##,

2) Calculate the length of the rod in ##O## using distance between where the front and back of the rod are at the same time in frame ##O##.
 
  • Like
Likes Gom
  • #3
Thanks PeroK. The excercise doesn't allow to use another method, so I will keep looking for it. However, I would like to know about the method you suggested. Is there any bibliography where I can find it explained?
 

Related to Special relativity - measure of a rod and simultaneity

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that explains the relationship between space and time. It states that the laws of physics are the same for all observers in uniform motion, and the speed of light is constant in all inertial frames of reference.

2. How is the length of a rod measured in special relativity?

In special relativity, the length of a rod is measured differently for observers in different frames of reference. The length of an object appears shorter when it is moving at high speeds relative to an observer, due to the effects of time dilation and length contraction.

3. What is simultaneity in special relativity?

Simultaneity refers to events that occur at the same time according to an observer. In special relativity, simultaneity is relative and can differ between observers in different frames of reference due to the effects of time dilation and the finite speed of light.

4. How does special relativity affect our understanding of time?

Special relativity states that time is not absolute and can be perceived differently by observers in different frames of reference. Time can appear to pass slower for objects in motion, and the concept of simultaneity is relative. This challenges our traditional understanding of time as a universal constant.

5. What are some practical applications of special relativity?

Special relativity has many practical applications, including GPS technology, particle accelerators, and nuclear power plants. It also plays a crucial role in modern physics and our understanding of the universe, helping us explain phenomena such as black holes and the expanding universe.

Similar threads

  • Introductory Physics Homework Help
2
Replies
36
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
657
  • Introductory Physics Homework Help
Replies
3
Views
847
  • Introductory Physics Homework Help
Replies
2
Views
855
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
951
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
983
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top