Special relativity (overtaking)

In summary, the conversation is discussing a problem involving two trains, A and B, moving at different speeds with respect to an observer C. The observer D, who is in train B, coincides with both events "front of A passing back of B" and "back of A passing front of B". From D's perspective, the speed of A and B are the same, and using velocity addition, the book concludes that the speed of A and B relative to D is c/5. The rest of the problem involves using this information to find the contracted lengths of the trains and solve for the required answer.
  • #1
bigevil
79
0

Homework Statement



This is an example problem although I don't really understand it.

Train A is moving at 4c/5 and train B moves at 3c/5 with respect to observer C.

Let E1 be the event "front of A passing back of B" and E2 be the event "back of A passing the front of B".

An traveller D (who is in train B) walks from the back of B to the front of B and coincides with both E1 and E2. What is the speed he walks at and how long does D measure the overtaking process to be?

2. ?

I just need a little hint here. This is actually an example problem from my book, so I do have the working and answer provided, but I don't actually understand what the significance of the observer D coinciding with both events is.

The book states immediately that as a result of the condition here, and looking at the point of view of D, A moves at speed v wrt D and B moves at speed -v wrt D. Then it uses the velocity addition again to get [tex]\frac{2v}{1+v^2/c^2} = \frac{5c}{13}[/tex]. The book concludes that v = c/5, upon solving the equation given earlier.

Once this is done, the rest is easy: using the velocity of either A or B relative to D, find the contracted length of each train and divide by c/5 to get the required answer, which is [tex]2\sqrt{6}L/c[/tex].

(From an earlier and easier half of the question, velocity addition finds that A moves at 5c/13 with respect to B.)
 
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  • #2
I run into another problem once I go further along this question.

If I solve the equation given here, I actually get v = 2c/5 rather than v = c/5 quoted in the book.
 
  • #3
bigevil said:
I just need a little hint here. This is actually an example problem from my book, so I do have the working and answer provided, but I don't actually understand what the significance of the observer D coinciding with both events is.
Look at things from D's perspective. D observes the length of train A pass him (front to back) in exactly the same amount of time that the length of train B passes him (back to front). That fact allows you to immediately conclude that the speed of A wrt D is the same as the speed of B wrt D (except for the sign, of course). (Assuming the trains have the same rest length L, of course.)

The book states immediately that as a result of the condition here, and looking at the point of view of D, A moves at speed v wrt D and B moves at speed -v wrt D. Then it uses the velocity addition again to get [tex]\frac{2v}{1+v^2/c^2} = \frac{5c}{13}[/tex]. The book concludes that v = c/5, upon solving the equation given earlier.
OK. Given my comment above, does this now make some sense?

bigevil said:
If I solve the equation given here, I actually get v = 2c/5 rather than v = c/5 quoted in the book.
Double check your solution. (Using that same equation, I got the book's answer.)
 
  • #4
Yes, I understand now, thanks Al!
 

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that explains how the laws of physics should be viewed in different inertial frames of reference, particularly when objects are moving at high speeds close to the speed of light.

2. How does special relativity explain overtaking?

Special relativity explains overtaking by taking into account the relative velocities and frames of reference of the objects involved. It shows that the perception of time and distance between two objects can be different for observers in different inertial frames of reference, leading to the phenomenon of overtaking.

3. Can anything travel faster than the speed of light in special relativity?

No, according to special relativity, the speed of light is the fastest speed at which any object can travel. This is because as an object approaches the speed of light, its mass increases infinitely, making it impossible to accelerate any further.

4. How does special relativity affect our everyday lives?

Special relativity has a significant impact on our everyday lives, as it explains many phenomena such as time dilation and length contraction, which are essential for technologies such as GPS. It also provides a better understanding of the nature of space and time.

5. Are there any practical applications of special relativity?

Yes, special relativity has many practical applications, including nuclear energy, particle accelerators, and space travel. It also plays a crucial role in the development of modern technologies such as GPS, which relies on precise timing and measurements influenced by special relativity.

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