Schwarzschild Extension Coordinate Transformation Algebra

In summary: Hopefully with this information someone may find it easier to spot why I am getting a negative - been llooking for a long time.Thanks
  • #1
binbagsss
1,259
11
So I have the metric as ##ds^{2}=-(1-\frac{2m}{r})dt^{2}+(1-\frac{2m}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}##*

I have transformed to coordinate system ##u,r,\phi, \theta ##, where ##u=t-r*##(2),
where ##r*=r+2m In(\frac{r}{2m}-1)##
and to the coordinate system ##v,r,\phi, \theta ##,
where ##v=t+r*##,(1)

From (1) and (2) I see that ##dt=dv-\frac{dr}{(1-\frac{2m}{r})}## and ##dt=du+\frac{dr}{(1-\frac{2m}{r})}##
(On a side note, what is the proper name of these types of derivative expressions?)

Substituting these into * in turn it is easy enough to get the metrics:(which I believe are correct?).

Question:

I am now want to get the metric using both \(v\) and \(u\) in favour of \(r\) and \(t\).
To do this I make use of:
##\frac{1}{2}(v-u)=r+2M In(\frac{r}{2M}-1) ##
therefore ##\frac{1}{2}(dv-du)(1-\frac{2m}{r})=dr##

and I sub this into either (1) or (2),
say (1) , I then get:
##ds^{2}= - (1-\frac{2M}{r}) dudv+r^{2}d\Omega^{2}##

And the first term is a minus sign out.
(in accord to source sean m carroll lecture notes on general relativity eq.7.73.)

I have no idea why I am a sign out,

Thanks,your assistance is greatly appreciated !
 
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  • #2
Bump.
 
  • #3
binbagsss said:
Substituting these into * in turn it is easy enough to get the metrics:(which I believe are correct?).

I don't see any metrics here. Did you leave them out by mistake?
 
  • #4
Apologies!
metrics are:
##ds^{2}=-(1-\frac{2M}{r})dv^{2}+2dvdr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2} ##
##ds^{2}=-(1-\frac{2M}{r})du^{2}-2dudr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2} ##
 
  • #5
binbagsss said:
metrics are

Ok, those look correct.

binbagsss said:
I have no idea why I am a sign out

I don't think you are. I think you are just using an opposite sign convention from Carroll for the definition of ##r## in terms of ##u## and ##v##. If you look at Carroll's equation 7.74, it has ##u - v## where you have ##v - u##. That's why he has an opposite sign in 7.73 from yours.

(Carroll also has a factor of 1/2 in front of 7.73, whereas you do not. That's because he also scales ##u## and ##v## differently than you are. None of these differences affect the physics; they're just different conventions for the math.)
 
  • #6
PeterDonis said:
Ok, those look correct.
I don't think you are. I think you are just using an opposite sign convention from Carroll for the definition of ##r## in terms of ##u## and ##v##. If you look at Carroll's equation 7.74, it has ##u - v## where you have ##v - u##. That's why he has an opposite sign in 7.73 from yours.

(Carroll also has a factor of 1/2 in front of 7.73, whereas you do not. That's because he also scales ##u## and ##v## differently than you are. None of these differences affect the physics; they're just different conventions for the math.)

I used a different notation than Carrol ## u bar=v##.
I cancelld the ##1/2## using ##dudv=dvdu##
 
  • #7
binbagsss said:
I used a different notation than Carrol ubar=v u bar=v.
I cancelld the 1/21/2 using dudv=dvdu

Hm, yes, I see. So much for that theory. :oops:

I'm not sure what's going on. I get the minus sign in the ##du dv## term the same way you do. Physically, the minus sign makes sense: a line element with ##du## and ##dv## both the same sign should lie inside one of the light cones (future for ##du## and ##dv## both positive, past for ##du## and ##dv## both negative), and so should be timelike and have a negative squared length. Unless Carroll is using a different sign convention somewhere else that I haven't spotted, the only other thing I can think of is that his equation 7.73 has a typo.
 
  • #8
Looking at the next metric, equation 7.77, it has a negative sign. (And I believe this to be corect as it agrees with http://www.damtp.cam.ac.uk/user/hsr1000/black_holes_lectures_2014.pdf , page 33, eq 2.35). Looking at how u' and v' are defined as functions of u,v, I am getting a sign change when going from the metric in u,v to u'v'. (whilst the cambridge notes, equation 2.32, defines u' with a neg sign compared to Carroll, either definition gives arise to a neg sign) and so it appears that there should be a sign change and so equation 7.73 in carroll should have a positive.

Hopefully with this information someone may find it easier to spot why I am getting a negative - been llooking for a long time.

Thanks

(Cambridge ntes and Carroll have defined their u and v, u' and v' the other way around).
 
Last edited:
  • #9
Sorry just to add both sources are using the same metric signature - (-,+,+,+)).
 

Related to Schwarzschild Extension Coordinate Transformation Algebra

1. What is the Schwarzschild Extension Coordinate Transformation Algebra?

The Schwarzschild Extension Coordinate Transformation Algebra is a mathematical framework used in general relativity to describe the transformation of coordinates between two reference frames in the vicinity of a spherically symmetric, non-rotating mass. It is named after the German physicist Karl Schwarzschild, who first derived it in 1916.

2. Why is the Schwarzschild Extension Coordinate Transformation Algebra important?

The Schwarzschild Extension Coordinate Transformation Algebra is important because it allows us to describe the effects of gravity on space and time in the vicinity of a massive object. It is essential for understanding the behavior of objects near black holes, and it is also used in other areas of physics such as cosmology and astrophysics.

3. How does the Schwarzschild Extension Coordinate Transformation Algebra work?

The algebra involves a set of equations that relate the coordinates of an observer in one reference frame to those of an observer in a different reference frame. It takes into account the curvature of spacetime caused by the presence of a massive object and allows us to calculate the effects of gravity on the movement of objects and the flow of time.

4. What are some applications of the Schwarzschild Extension Coordinate Transformation Algebra?

One of the most significant applications of this algebra is in the study of black holes. It allows us to calculate the event horizon of a black hole, which is the point of no return for any object that falls into it. It is also used in the study of gravitational waves, which are ripples in the fabric of spacetime caused by massive objects accelerating.

5. Are there any limitations to the Schwarzschild Extension Coordinate Transformation Algebra?

While the algebra is an essential tool in general relativity, it has limitations. It is only applicable to spherically symmetric, non-rotating masses, and it does not take into account the effects of other forces, such as electromagnetism. Additionally, it breaks down at the singularity of a black hole, where the equations become undefined.

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