Special Relativity- Spacetime Diagrams

In summary, the conversation discusses a problem involving the Lorentz transformation and calculating the time it takes for light to travel from point B to a mirror and back again in both the rest frame and a moving frame. The postulates of the theory are used to derive an expression for the time in the moving frame, and it is noted that the distance between B and M2 in the moving frame is still L. The equations for the time in both frames are found to be 2L = r'c and 2L' = rc, and the four momentum for a massless particle is discussed. Ultimately, the conversation leads to the solution of the problem, which involves using the coordinates (cr',-r'v) in the
  • #1
sebb1e
36
0

Homework Statement



http://www.maths.ox.ac.uk/system/files/private/active/0/b07.2_c7.209.pdf

Q1
(Also in Q2, what is the definition of 4 momentum for a massless particle?)


The Attempt at a Solution



I can do the entire question if I first derive the Lorentz transformation but I suspect that as I can't do the question directly in its intended way I don't really understand what I am doing.

I have done ii) using Bondi's k-factor but not sure this is right as I haven't considered the path of the light from B to M2. I have just drawn a spacetime diagram with R' the inertial frame and drawn a photon leaving R' origin at time t, reaching B at t'=kt (when the light gets back to B from M2) and getting back to R' origin at k^2t=kt' and solving.

I am just getting into a mess trying to do iii) though and absolutlely no idea how to do iv) without Lorentz transformation.

Thanks!
 
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  • #2
Given that this is a past exam, answering your questions should be within the guidelines of Physics Forums. So,

i) What are the two postulates of the theory?
ii) Use these postulates to derive the desired expression. Hint: Pythagorean Theorem.
iii) The result from ii) can be used.
 
  • #3
For both parts ii and iii, calculate the time it takes for the light to go from B to the mirror and back again in the moving frame and compare it to the time τ it takes in the rest frame.
 
  • #4
I'm really not sure how to calculate the time in the moving frame without essentially deriving Bondi's k factor. I think I must be missing something obvious.
 
  • #5
It's just the distance the light travels divided by c.
 
  • #6
Thanks, I get ii) now by using Pythag theorem, wasn't thinking about it in the right way. Am I correct to think that the distance between B and M2 in R' is still L, just that there is time dilation?

iii) Two equations I need are 2L=r'c and 2L'=rc then results follows using ii)

I think the first follows as distance between B and M2 in R' is still L.
I'm not sure how to justify second equation. As there is no triangle to consider as velocity is in the same direction as the light can we say that the light must take r to travel L'?

I get how to do iv) now, we can just fill in the matrix entries from ii) and iii)

Also what is the four momentum for a massless particle on Q2 and do the 3 equations still hold if the particle is massless? I can't find this defined anywhere in my book.

Thank you so much for your help!
 
  • #7
sebb1e said:
Thanks, I get ii) now by using Pythag theorem, wasn't thinking about it in the right way. Am I correct to think that the distance between B and M2 in R' is still L, just that there is time dilation?
M2 is the mirror on the y-axis. Since you can assume y=y', the distance is still L.
iii) Two equations I need are 2L=r'c and 2L'=rc then results follows using ii)

I think the first follows as distance between B and M2 in R' is still L.
I'm not sure how to justify second equation. As there is no triangle to consider as velocity is in the same direction as the light can we say that the light must take r to travel L'?
So you're actually interested in M1. In R', you can't assume the distance between B and M1 is still L. I don't understand your two equations since you seem to be mixing variables from the two frames, like L and r' (which is I assume the round-trip time measured in the R' frame).
Also what is the four momentum for a massless particle on Q2 and do the 3 equations still hold if the particle is massless? I can't find this defined anywhere in my book.
I'm not sure what exactly the problem is looking for in the definition. How does your book define the four-momentum of a massive particle?
 
  • #8
Well the time taken to get from B to M1 in R' is going to be L'/c and in R it takes L/c. How do I move on from here as I can't see how to relate the two equations.


The four momentum is defined as the rest mass multiplied by the four velocity. I don't think the equations will hold for massless particle as m2 appears in the denominator when you are working out gammas.
 
  • #9
sebb1e said:
Well the time taken to get from B to M1 in R' is going to be L'/c and in R it takes L/c. How do I move on from here as I can't see how to relate the two equations.
You're forgetting the motion of the mirrors in R'. It'll take less time than L'/c for the light to go from B to M1 because M1 will be moving toward the light. Similarly, it'll take longer than L'/c to go from M1 back to B because B will be moving away from the light.
The four momentum is defined as the rest mass multiplied by the four velocity. I don't think the equations will hold for massless particle as m2 appears in the denominator when you are working out gammas.
OK, that definition obviously won't work for a massless particle, so perhaps they want you to specify what the components of the four-momentum are for a massless particle.
 
  • #10
In R', a is the time to get from the bulb to M1 and b is the time to get from M1 to the bulb.

ca=L'-av a=L'/(c+v)
cb=L'+bv b=L'/(c-v)Total time to get from bulb to m1 and back again is 2L'c/(c^2-v^2)=(2L'/c)*gamma^2 in R'.

Answer follows as this is equal to r' as the light must return back to the bulb at the same time as that from M2.

In iv) I can get the left column of the matrix entries ok using the co-ordinates (cr',-r'v) in R' is (cr,0) in R. The problem is in getting the two right co-ordinates. I get that (L,L) in R should be equivalent to (cL'/(c+v),cL'/(c+v)) in R' (from a above) but this doesn't yield the right answer, you need to have (cL'/(c-v),cL'/(c-v)) in R'. I can't see what I've done wrong.I've found a definition as E/c(1,e) where e is the unit vector of velocity.
 
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  • #11
Have I got a and b the wrong way round as this would then give me the correct answer for iv)
 
  • #12
sebb1e said:
In iv) I can get the left column of the matrix entries ok using the co-ordinates (cr',-r'v) in R' is (cr,0) in R. The problem is in getting the two right co-ordinates. I get that (L,L) in R should be equivalent to (cL'/(c+v),cL'/(c+v)) in R' (from a above) but this doesn't yield the right answer, you need to have (cL'/(c-v),cL'/(c-v)) in R'. I can't see what I've done wrong.
It worked out for me when I used (cL'/(c+v),cL'/(c+v)) in R' and (L,L) in R.
 

FAQ: Special Relativity- Spacetime Diagrams

What is special relativity?

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

What is a spacetime diagram?

A spacetime diagram is a graphical representation of the relationship between space and time in special relativity. It is a two-dimensional diagram where time is represented on the vertical axis and space is represented on the horizontal axis. Events are plotted as points on the diagram, and lines called worldlines show the path of an object or observer through spacetime.

How do you read a spacetime diagram?

To read a spacetime diagram, you first need to identify the axes. The vertical axis represents time, with the present moment being at the origin (0,0). The horizontal axis represents space, with the positive direction to the right and the negative direction to the left. Events are plotted as points on the diagram, and worldlines show the path of an object or observer through spacetime. The slope of a worldline represents the velocity of the object, and the steeper the slope, the faster the object is moving.

What is the significance of the light cone in a spacetime diagram?

The light cone is the area on a spacetime diagram that represents the possible paths of a light signal. The cone is divided into two parts - the past light cone, which includes all events that could have influenced the current event, and the future light cone, which includes all events that could be influenced by the current event. Anything outside of the light cone is not causally connected to the current event.

How does special relativity affect our understanding of time and space?

Special relativity has had a significant impact on our understanding of time and space. It has shown that time and space are not absolute, but instead are relative to the observer's frame of reference. It has also introduced the concept of time dilation, which states that time passes slower for objects in motion relative to an observer. Additionally, special relativity has led to the famous equation E=mc², which relates energy and mass and has been confirmed by numerous experiments.

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