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Lorentz Transformation

  • Thread starter Kunhee
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  • #1
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Homework Statement


A rocket is traveling toward a galaxy with speed v.
a) If NASA says that distance from earth to the galaxy is d, what is the distance d' from earth to the galaxy according to the astronauts?
b) The astronauts experience a travel time to the galaxy t' and NASA records the travel time as t. For NASA, d = vt. Show that the astronauts also find that distance = speed x time.

Homework Equations



c = 3 x 10^8
Lorentz Factor = (1-(v/c)^2)^(-1/2)
x' = 1/LF ( x - vt)
t' = 1/LF ( t - vx / c^2)

The Attempt at a Solution



Currently I only learned the equations above. I do not know the x', x, t', or t. Is this an intuitive question?
 

Answers and Replies

  • #2
PeroK
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The Attempt at a Solution



Currently I only learned the equations above. I do not know the x', x, t', or t. Is this an intuitive question?
If you are asking that sort of question, you have a long way to go. What physics have you previously studied?
 
  • #3
Simon Bridge
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This is not an intuitive question.
You need to look up "introduction to special relativity" someplace (hint: google) and get back to us.
Since this is homework, you must be doing some sort of course that will have lessons, notes, and a text book... have you tried those?
You may be able to figure out what is intended by noticing that there are primed and unprimed variables in the question though.
 
  • #4
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a)
d' = d - vt?

b)
x' = 1/LF (x - vt)
x' = 1/(1-(v/c)^2)^(-1/2) (x - vt)
x' = 1/(1) ( x- vt)
x' = x- vt
same as d' = d - vt

So if these are correct, could you explain to me why the notion of "time travel" is required to make this problem that of special relativity? Or does it simply mean that by making t' equal to t, we are dealing with change of time = 0.
 
  • #5
PeroK
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a)
d' = d - vt?

b)
x' = 1/LF (x - vt)
x' = 1/(1-(v/c)^2)^(-1/2) (x - vt)
x' = 1/(1) ( x- vt)
x' = x- vt
same as d' = d - vt

So if these are correct, could you explain to me why the notion of "time travel" is required to make this problem that of special relativity? Or does it simply mean that by making t' equal to t, we are dealing with change of time = 0.
None of that makes any sense to me. Are you a high school student, learning SR on your own or a university student?
 
  • #6
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My apologies. I am learning on my own.
 
  • #7
PeroK
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My apologies. I am learning on my own.
What physics do you already know?
 
  • #8
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I only learned high school physics.
 
  • #9
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I am using Kleppner and Kolenkow's textbook Intro to Mechanics.
 
  • #10
PeroK
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I am using Kleppner and Kolenkow's textbook Intro to Mechanics.
They cover length contraction. Does that seem relevant here?
 
  • #11
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Yeah this is a length contraction question.

x' = 1/LF (x - vt)
x' = 1/LF (x - v(0)) because change in t = 0
x' = 1/LF x

So here, x' is d' that astronauts measure from earth to the galaxy
and x is the d that NASA measures from earth to the galaxy.
Am I correct so far?
 
  • #12
PeroK
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Yeah this is a length contraction question.

x' = 1/LF (x - vt)
x' = 1/LF (x - v(0)) because change in t = 0
x' = 1/LF x

So here, x' is d' that astronauts measure from earth to the galaxy
and x is the d that NASA measures from earth to the galaxy.
Am I correct so far?
Personally, I would solve this problem using length contraction and time dilation, as the information is given as lengths and time intervals. The problem is stated in those terms and not in terms where you can immediately use the Lorentz Transformation.

If you want to use Lorentz, then you are going to have to restate the problem in terms of position and time coordinates, to which you can apply Lorentz.

Saying something like x' = d' is sloppy. To use Lorentz, you need more precision.

I'm not sure if that makes any sense tto you.
 
  • #13
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I see, the question is under the category "Lorentz Transformation" so I did not look ahead yet at Length Contraction and Time Dilation. I will look at those sections and try again.
 
  • #14
PeroK
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I see, the question is under the category "Lorentz Transformation" so I did not look ahead yet at Length Contraction and Time Dilation. I will look at those sections and try again.
I think K&K is excellent, but I hadn't looked at their section on SR. Just taking a quick look at it I can't find your problem (I have the second edition). Also, it seems a bit brief for learning this on your own. There is a free pdf available here:

http://www.lightandmatter.com/sr/

Another book I would recommend is:

https://www.amazon.co.uk/Special-Relativity-Thomas-M-Helliwell/dp/1891389610

Both of these are entire texts on SR - rather than just a couple of chapters - and may be more appropraite for learning on your own.
 
  • #15
Simon Bridge
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Accessible - but a bit of a crash course:
http://www.physicsguy.com/ftl/html/FTL_intro.html
... while you are working on trying other texts.

Basically we don;t give full physics lessons here, though we can help with lessons and resources and give you a hand over the tricky parts. ie.

Links for learning special relativity:
http://physicsdatabase.com/2013/11/18/10-great-links-for-learning-special-relativity/

... these are college level but you don't need more than Newtonian physics and quadratic equations, basic algebra and the ability to draw graphs. If you can do kinematics using velocity-time diagrams you can learn special relativity. Understanding takes practise.
 
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  • #16
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Thank you Simon!
 
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