GR: "Proper Distance" Meaning and Usage

[Passionflower]

Now if I set r0 = infinite in your equation, it reduces to:

<br /> \frac{dr}{dt} = \sqrt{\frac{2m}{r}(1-v_0^2) +2mv_0^2} <br />

which does not agree with either the mathpages coordinate velocity or proper velocity:

You might want to double check that.

I get:

<br /> \sqrt {2m \left( {r}^{-1}-{\frac { \left( -r+1 \right) {{\it <br /> v_0}}^{2}}{r}} \right) }<br />

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v_0^2) +v_0^2} <br />

This formula is identical to mine if we take r0 = infinity.

This is identical:

<br /> \sqrt {{v_0}^{2}+{\frac {1-{v_0}^{2}}{r}}}=\sqrt {{r}^{-1}-{\frac {<br /> \left( -r+1 \right) {v_0}^{2}}{r}}}<br />

<br /> \frac{dr single-quote}{dt single-quote} = \sqrt{\frac{2m(r^{-1} - r_0^{-1})}{(1-2m/r_0)}} <br />

It looks like you are right on that one.

 

[yuiop]

Now if I set r0 = infinite in your equation, it reduces to:

<br /> \frac{dr}{dt} = \sqrt{\frac{2m}{r}(1-v_0^2) +2mv_0^2} <br />

You might want to double check that.

I get:

<br /> \sqrt {2m \left( {r}^{-1}-{\frac { \left( -r+1 \right) {{\it <br /> v_0}}^{2}}{r}} \right) }<br />

Starting with your equation:

<br /> \sqrt {2m \left( {r}^{-1}-{\frac { \left( -r+1 \right) {{\it <br /> v_0}}^{2}}{r}} \right) }<br />

\Rightarrow \sqrt {\frac{2m}{r}-\frac { 2m\left( -r+1 \right) <br /> v_0^2}{r}} <br />

\Rightarrow \sqrt {\frac{2m}{r} +2mv_0^2-\frac { 2mv_0^2}{r}} <br />

\Rightarrow \sqrt {\frac{2m}{r}(1-v_0^2) +2mv_0^2}<br />

So we both get the same result for the reduction of your equation when r0 = infinite, and this does not agree exactly with the equation I gave in #89:

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2} <br />

Do you now at least agree that the equations I gave in #89 are correct?

I note that sometimes you are explicitly using rs = 2m and other times using rs = 1 which might lead to errors.
Your generic equation is close for the r0 = infinity case if we only use rs =1 but it is miles out for the v0 = 0 and r0<infinite case.

 

[Passionflower]

Ok, let's take a look at the formulas you presented.

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2} <br />

When I enter m=1/2, v0=1/2 and r=1 I get a v>1 this does not seem right. In fact v reaches 1 already at r=4/3 using your formula. Unless I am mistaken v=1 should only happen at the EH. My formula gives v=1 at r=1.

I note that sometimes you explicitly using rs = 2m and other times using rs = 1 which might lead to errors.

Yes, I apologize for that, I changed many formulas around to express them in terms of m instead of rs but some have not been updated.

 

[yuiop]

Ok, let's take a look at the formulas you presented.

The local velocity of a falling particle with initial velocity v0 at infinity, according to an observer at r, was derived in #79 and is:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2} <br />

When I enter m=1/2, v0=1/2 and r=1 I get a v>1 this does not seem right. In fact v reaches 1 already at r=4/3 using your formula. Unless I am mistaken v=1 should only happen at the EH. My formula gives v=1 at r=1.

I am not sure how you get that result. I get:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2} = \sqrt{\frac{1}{1}(1-(0.5)^2) +(0.5)^2} = \sqrt{1-0.25 + 0.25} = 1<br />

Try your formula:

<br /> <br /> \frac(dr&#039;}{dt&#039;} = \sqrt {2m \left( {r}^{-1}-{\frac { \left( -r+1 \right) {{\it <br /> v_0}}^{2}}{r}} \right) }<br /> <br />

using m=7, r=2m and v0 =1/2 and you will see where the problem is. (My (mathpages) equation works for any arbitrary value of m.)

 

[Passionflower]

I am not sure how you get that result. I get:

<br /> \frac{dr&#039;}{dt&#039;} = \sqrt{\frac{2m}{r}(1-v0^2) +v0^2} = \sqrt{\frac{1}{1}(1-(0.5)^2) +(0.5)^2} = \sqrt{1-0.25 + 0.25} = 1<br />

Try your formula:

<br /> <br /> \sqrt {2m \left( {r}^{-1}-{\frac { \left( -r+1 \right) {{\it <br /> v_0}}^{2}}{r}} \right) }<br /> <br />

using m=7, r=2m and v0 =1/2 and you will see where the problem is. (My equation works for any arbitrary value of m.)

Sorry yuiop I am definitely not having my day, must be my jet lag due to moving from China to the USA.

Ok, now I think your formulas look good, I am going to play with them a little. Would it be possible to get a generic "Fermi distance formula"? I am going to take your formulas to get a generic distance formula as well.

Ok, here are the plots for the formulas you presented, and they look good. One thing is very different, the 'dropped from a r value' observers. In my plot the distance decrease is definitely not linear while using your formulas it appears linear. But I think yours are correct and mine were wrong.

In the last graph the stationary observers are obviously not in motion so each point on the plot represents a different observer at a different r value.

 

[pervect]

The paramaterized equations for a general spacelike geodesic will be given by the functions r(s) and t(s) that satisfy

<br /> \frac{dt}{ds} = \frac{C}{1-2m/r}<br />

<br /> \left(\frac{dr}{ds}\right)^2 = 1 + C^2 - 2m/r<br />

where C is an arbitrary constant. To get the fermi-normal distance associated with the worldline of some particular time-like observer, you'll have to choose the correct value of C so that your geodesic is orthogonal to the worldline of your time-like obsever at the point where they cross.

 

[Roshanjha14]

Can anyone suggest me any book to start with general theory of relativity and cosmology?

 

[Roshanjha14]

Can anyone describe me simplest way that what the general theory relativity acutually tell.