Product of tangent vectors \pm1 affine parameter

  • Thread starter binbagsss
  • Start date
  • #1
1,211
9
If ##\sigma## is an affine paramter, then the only freedom of choice we have to specify another affine parameter is ##a\sigma+b##, a,b constants. [1]

For the tangent vector, ##\xi^{a}=dx^{a}/du##, along some curve parameterized by ##u##

My book says that ' if ##\xi^{a}\xi_{a}\neq 0##, then by suitable affine parameterization we can arrange such that ##\xi^{a}\xi_{a}=\pm1##,

Question:

What does it mean by some suitable affine parameterization? so say if ##\sigma## is a affine parameter and we do not have ##\xi^{a}\xi_{a}=\pm1##, is it saying that we can use [1] and carefully choose ##a## and ##b## such that this is the case?

I've often seen proper time used such that ##\xi^{a}\xi_{a}=\pm1## is the case.
Why is this?

Or Is this part of the definition of proper time, are there any other 'known' parameters for which ##\xi^{a}\xi_{a}=\pm1## or is the affine parameter for which this holds unique?

Thanks in advance.
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
Yes, the point is that if ##\xi^a \xi_b \neq 0##, then you can pick ##a## such that ##\xi^a \xi_a = \pm 1## (##b## does not really enter into it, it is just a translation along the curve). The point is that this parametrises the curve using the curve length as parameter. In Minkowski space, for time-like curves, this means parametrising the curve with the proper time.
 
  • #3
1,211
9
Yes, the point is that if ##\xi^a \xi_b \neq 0##, then you can pick ##a## such that ##\xi^a \xi_a = \pm 1## (##b## does not really enter into it, it is just a translation along the curve). The point is that this parametrises the curve using the curve length as parameter. In Minkowski space, for time-like curves, this means parametrising the curve with the proper time.
Ok. So if i use an affine parameter ##\tau+b## I Still achieve ##\xi^a \xi_a =\pm 1##?

For a null geodesic we can't use ##\tau## as it is always zero.
However am I correct in thinking that the paramter ##\tau+b## would be plausible?

Thanks.
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
16,829
6,643
Ok. So if i use an affine parameter ##\tau+b## I Still achieve ##\xi^a \xi_a =\pm 1##?

For a null geodesic we can't use ##\tau## as it is always zero.
Yes, if you have ##\tau' = \tau + b##, you would get ##d\tau'/d\tau = 1## and thus ##dx^\mu/d\tau = dx^\mu/d\tau'##. It is just a change of what you call "proper time equals zero".


However am I correct in thinking that the paramter ##\tau+b## would be plausible?
No, it would not be. Just by the same argumentation as above. You would still be trying to parameterise with the proper time, just with a different definition of proper time equal to zero. It does not work for light-like curves. You can still find an affine parameter, but it cannot be based on curve length (i.e., proper time).
 
  • #5
1,211
9
Yes, if you have ##\tau' = \tau + b##, you would get ##d\tau'/d\tau = 1## and thus ##dx^\mu/d\tau = dx^\mu/d\tau'##. It is just a change of what you call "proper time equals zero".




No, it would not be. Just by the same argumentation as above. You would still be trying to parameterise with the proper time, just with a different definition of proper time equal to zero. It does not work for light-like curves. You can still find an affine parameter, but it cannot be based on curve length (i.e., proper time).
Is a re-scale of prper time ok, so ##\lambda=a\tau##?
 

Related Threads on Product of tangent vectors \pm1 affine parameter

  • Last Post
Replies
7
Views
5K
  • Last Post
Replies
8
Views
3K
Replies
7
Views
1K
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
Replies
16
Views
3K
Replies
10
Views
2K
Replies
20
Views
328
Replies
3
Views
765
  • Last Post
3
Replies
70
Views
1K
Top