What is the relationship between speed and coordinates in general relativity?

[DrGreg]

Yeah, and we give a relation of the form

$$g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$$ (1)

where in the $$\gamma_v$$, the appearing v is the local 3-velocity

$$v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau).$$

Take the sides of the last equation in (1) to the power 2 and write $$(v^0)^2=V^2-v^2$$ where $$V^{\alpha}$$ denotes the 4-velocity of the particle. Hence

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and with $$u^2=(u^0)^2$$ we obtain finally

$$V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

If your method doesn't lead to this equation, then it doesn't work!

AB

In a private message, kev asked me to comment on this post. But I can't work out what

$$(v^0)^2=V^2-v^2$$

is supposed to mean.

I'm not sure where this is going anyway. It seems to me that the original question was answered by my post #9 (& comments in #11).

 

[yuiop]

Yeah, and we give a relation of the form

$$g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$$ (1)

where in the $$\gamma_v$$, the appearing v is the local 3-velocity

$$v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau).$$

Take the sides of the last equation in (1) to the power 2 and write $$(v^0)^2=V^2-v^2$$ where $$V^{\alpha}$$ denotes the 4-velocity of the particle. Hence

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and with $$u^2=(u^0)^2$$ we obtain finally

$$V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

If your method doesn't lead to this equation, then it doesn't work!

Hi Altabeh,

I would like to understand your notation better as I would like to know if my method outlined in #18 agrees with your method and because I think I might learn something from you. It might help people that refer to this thread in the future too. You never know, the OP might even take a look at this thread.

Could you state the exact location and state of motion of the observer for each velocity measurement $u^0, v^0, u^i$ and $v^i$ and if you are using proper time in each case.

Could you also confirm if all these velocities are 3 velocities (i.e. everything other than V)?

 

[Altabeh]

In a private message, kev asked me to comment on this post. But I can't work out what $$(v^0)^2=V^2-v^2$$ is supposed to mean.

Here v=(v^1,v^2,v^3) is the proper 3-velocity and $$V$$ represents the magnitude of the 4-velocity $$v^{\alpha}$$.

I'm not sure where this is going anyway. It seems to me that the original question was answered by my post #9 (& comments in #11).

This doesn't have anything to do with the original question. I actually answered by this to a question brought up by kev in post #18.

Hi Altabeh,

Hi

I would like to understand your notation better as I would like to know if my method outlined in #18 agrees with your method and because I think I might learn something from you. It might help people that refer to this thread in the future too. You never know, the OP might even take a look at this thread.

In the setup I use to denote the 4-velocities of observer, $$u^{\alpha}$$, and particle, $$v^{\alpha}$$, I assume the observer is instantaneously at rest at the event the proper 4-velocity $$v^{\alpha}$$ is observed. Thus we have

$$u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)$$

at the event of observation. From now on it is easy to calculate the result already obtained by DrGreg, i.e. $$d\tau = dt \sqrt{1 - 2GM/rc^2}{\sqrt{1-v^2/c^2}}$$.

But about your method I think you should first elaborate the steps you take to for example get the magnitude of the proper 3-velocity, $$V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } .$$ And then explain what do you mean by $$W = c = \gamma_v(c, u_x, u_y, u_z) = \gamma_v\sqrt{ c^2 - u_x^2 - u_y^2 - u_z^2 }$$? Here I see some misleading arguments like equality of a 4-vector with a scalar value! After doing so, I'd be able to get the idea of your calculations!

Could you also confirm if all these velocities are 3 velocities (i.e. everything other than V)?

The only 3-velocity, which is basically a coordinate velocity, is the velocity we use in the Lorentz factor $$\gamma_v=\frac{1}{\sqrt{1-v^2/c^2}}.$$ Other than that, all velocities used in calcualtions are proper 4-velocities!

AB

 

[yuiop]

Hi Altabeh,

Take the sides of the last equation in (1) to the power 2 and write $$(v^0)^2=V^2-v^2$$ where $$V^{\alpha}$$ denotes the 4-velocity of the particle.

Here v=(v^1,v^2,v^3) is the proper 3-velocity and $$V$$ represents the magnitude of the 4-velocity $$v^{\alpha}$$.

OK, $$(v^0)^2$$ is a 4 velocity so it has the value c^2 and $$V^{2}$$ is also a 4 velocity so it also has the value c^2 by definition so:

$(v^0)^2=V^2-v^2$ means $c^2 = c^2-v^2$ which is only true when v=0. Is that what you intended?

In the setup I use to denote the 4-velocities of observer, $$u^{\alpha}$$, and particle, $$v^{\alpha}$$, I assume the observer is instantaneously at rest at the event the proper 4-velocity $$v^{\alpha}$$ is observed. Thus we have

$$u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)$$

at the event of observation.

I take "observer is instantaneously at rest at the event" to mean that the measurement of the particle's 4 velocity $(v^{0})^2$ is made by an observer with 4 velocity $(v^{0})^2$ who is at rest in the Schwarzschild coordinates. (i.e this observer is an accelerating hovering observer). Can I assume that $(v^{i})^2$ is the 4 velocity of the particle according to an observer at infinity and and $(u^{i})^2$ is the 4 velocity of the hovering observer at r according to the observer at infinity?

But about your method I think you should first elaborate the steps you take to for example get the magnitude of the proper 3-velocity, $$V = \sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } .$$

$$\sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } .$$ was meant to be the magnitude of the coordinate 3 velocity according to an observer at infinity in Schwarzschild coordinates. It is not intended to be a proper 3 velocity.

It is obtained by starting from the local coordinate 3 velocity of the particle

$$\sqrt{( u_x^2 + u_y^2 + u_z^2 )}$$ = $$\sqrt{ (dr '/dt ')^2 + (dy '/dt ')^2 + {dz '/dt ')^2 }$$
as measured by an observer that is stationary at r with respect to the Schwarzschild coordinates. In other words this local observer has $dr = d(\Omega) =0$. The primed coordinates are meant to indicate local measurements made using local rulers and clocks. Note that the local geometry can be described by Minkowski space and this local static observer does not need to concern himself with the radius or angular measurements.

Now the Schwarzschild metric in units of G=c=1 is:

$$dS = (1-2m/r) dt^2 - dr^2/(1-2m/r) - r^2d\theta^2 - r^2 sin^2(\theta) d\phi^2 }$$

can be expressed in a slightly simpler form as:

$$dS = (1-2m/r) dt^2 - dx^2/(1-2m/r) - dy^2 - dz^2 }$$

if we align the equator and the principle meridian with the location of the particle for convenience, a bit like we often align the boost direction in a Lorentx transformation in Minkowski space with the x coordinate.

Now we know when transforming from the coordinates of the particle (dt,dx,dy,dz) as measured by a observer at infinity to the coordinates of the same particle (dt',dx',dy',dz') as measured by a stationary observer at r in Schwarzschild coordinates that:

$$dt ' = dt\sqrt{1-2m/r}$$
$$dx ' = dx/\sqrt{1-2m/r}$$
$$dy ' = dy$$
$$dz ' = dz$$

using the above relations, it is straightforward to work out that for a local coordinate 3 velocity of

$$\sqrt{ (dx '/dt ')^2 + (dy '/dt ')^2 + {dz '/dt ')^2 }$$

that the coordinate 3 velocity according to an observer at infinity :

$$\sqrt{ (dx /dt)^2 + (dy /dt)^2 + (dz /dt)^2 }$$

is equal to:

$$\sqrt{ (dx '/dt ')^2(1-2m/r)^2 + (dy '/dt ')^2(1-2m/r) + (dz '/dt ')^2(1-2m/r) }$$

which is basically the expression I gave earlier for the coordinate 3 velocity for an observer at infinity:

$$\sqrt{ g^2 u_x^2 + g u_y^2 + g u_z^2 } .$$

and because it is a coordinate 3 velocity that does not involve proper times, the components are completely independent of each other unlike the components of a 4 velocity.

And then explain what do you mean by $$W = c = \gamma_v(c, u_x, u_y, u_z) = \gamma_v\sqrt{ c^2 - u_x^2 - u_y^2 - u_z^2 }$$? Here I see some misleading arguments like equality of a 4-vector with a scalar value! After doing so, I'd be able to get the idea of your calculations!

OK, I think I should have used c^2 rather than c for the 4 velocity so I will restate it as:

$$W^2 = c^2 = (\gamma_v)^2(c, u_x, u_y, u_z) = (\gamma_v)^2( c^2 - u_x^2 - u_y^2 - u_z^2 )$$

Now $$\gamma_v$$ is $$1/\sqrt{1-v^2/c^2}$$ (where v is the local coordinate 3 velocity) and $$\gamma_v = dt '/d\tau$$ so:

$$W^2 = c^2 = ( c^2dt '^2/d\tau^2 - u_x '^2 dt '^2/d\tau^2 - u_y '^2 dt '^2/d\tau^2 - u_z '^2 dt '^2/d\tau ^2 ) =$$

$$( c^2dt '^2/d\tau^2 - d_x '^2/d\tau^2 - d_y '^2 /d\tau ^2 - d_z '^2 /d\tau^2 )$$

which is the more familiar way of expressing a 4 velocity.

The only 3-velocity, which is basically a coordinate velocity, is the velocity we use in the Lorentz factor $$\gamma_v=\frac{1}{\sqrt{1-v^2/c^2}}.$$ Other than that, all velocities used in calculations are proper 4-velocities!

OK, I thought you were talking about 3 velocities because you defined $$v^{i}$$ as

$$v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau)$$

which only has 3 components.

Hopefully we are getting nearer to understanding each other's notations and calculations and most of the confusion is probably due to me because I am not familiar with the more formal notation style.

[P.S.] In #20 you also said

$$g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$$ (1)

which seems to imply $$g_{ii}u^iv^i = 0$$.

Have I read that right?

 

[Altabeh]

Hi Altabeh,
OK, $$(v^0)^2$$ is a 4 velocity so it has the value c^2 and $$V^{2}$$ is also a 4 velocity so it also has the value c^2 by definition so:

$(v^0)^2=V^2-v^2$ means $c^2 = c^2-v^2$ which is only true when v=0. Is that what you intended?

Actually $$v=v^i:=(v^1,v^2,v^3)$$ and $$V$$ is the magnitude of the four-velocity $$v^{\alpha}$$ and is by definition equal to $$\sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=c.$$

I take "observer is instantaneously at rest at the event" to mean that the measurement of the particle's 4 velocity $(v^{0})^2$ is made by an observer with 4 velocity $(v^{0})^2$

Your notations are not correct. The particle's 4-velocity is $$v^{\alpha}$$ not $$(v^{0})^2$$ and the observer's 4-velocity is $$u^{\alpha}$$ not $(v^{0})^2.$

(who is at rest in the Schwarzschild coordinates. i.e this observer is an accelerating hovering observer).

A hovering observer is not at rest unless it is said that the observer is hovering at rest which I think is what you meant.

Can I assume that $(v^{\alpha})$ is the 4 velocity of the particle according to an observer at infinity and and $(u^{\alpha})$ is the 4 velocity of the hovering observer at r according to the observer at infinity?

I edited the quote above (note that the Latin symbols mostly run over the values 1,2,3 and the Greek ones take the values 0,1,..,3.) You can assume that the $(u^{\alpha})$ is the 4 velocity of the radial hovering observer at infinity and consequently $(v^{\alpha})$ is the 4 velocity of the particle according to him.

It is obtained by starting from the local coordinate 3 velocity of the particle

$$\sqrt{( u_x^2 + u_y^2 + u_z^2 )}$$ = $$\sqrt{ (dr '/dt ')^2 + (dy '/dt ')^2 + {dz '/dt ')^2 }$$

Now you can suppose that $$\sqrt{( u_x^2 + u_y^2 + u_z^2 )}$$ is measured by the observer at infnitiy.

Note that the local geometry can be described by Minkowski space and this local static observer does not need to concern himself with the radius or angular measurements.

This is even possible for the observer who is at rest at infinity and for a Schwarzschild metric the outside observer being at rest at an infnitely enough distance from the gravitating body spacetime can be taken to be Minkowski. So I don't know how you are going to define a transformation between a local stationary observer at r and an observer at rest at infinity. Let's first clear this up and then go on!

AB

 

[yuiop]

Actually $$v=v^i:=(v^1,v^2,v^3)$$ and $$V$$ is the magnitude of the four-velocity $$v^{\alpha}$$ and is by definition equal to $$\sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=c.$$

When I asked you to define your velocity symbols $u^0, v^0, u^i$ and $v^i$ in #37, you replied that other than v used in the Lorentz factor, "all velocities used in calculations are proper 4-velocities", but here you seem to be making it clear that $v^i$ is a 3 velocity and from #38 you said "Here v=(v^1,v^2,v^3) is the proper 3-velocity" I think we can establish that by $v^i$ you mean the proper 3 velocity or what is sometimes called the spatial component of the 4 velocity. Since we are talking about a 3 velocity it is not invariant and we need to make clear who makes measures $v^i$, but I think from the context of your earlier posts it is the measurement made by a stationary observer at radial coordinate r.

At least we are agreed that $V^2$ has a magnitude of $c^2$ and so we can say:

$$(u^o)^2 = c^2 -v^2$$

Now if we assume by $$v^2$$ you mean $$( v^1(\tau), v^2(\tau), v^3(\tau) )$$ then the quantity on the right of the above equation is not invariant and nor is it invariant if we assume you mean $$(v^1(t), v^2(t), v^3(t) )$$.

You still have not explicitly stated what you mean by the symbols $u^0, v^0, u^i$.

Your notations are not correct. The particle's 4-velocity is $$v^{\alpha}$$ not $$(v^{0})^2$$ and the observer's 4-velocity is $$u^{\alpha}$$ not $(v^{0})^2.$

The particle's 4-velocity is $$v^{\alpha}$$ (as measured by who?) not $$(v^{0})^2$$ and the observer's 4-velocity is $$u^{\alpha}$$ (as measured by who?) not $(u^{0})^2.$. Thank you for telling me what $$(v^{0})^2$$ and $(u^{0})^2.$ do not mean, but that still does not tell me exactly what you do mean by the symbols $u^0, v^0, u^i$.

I edited the quote above (note that the Latin symbols mostly run over the values 1,2,3 and the Greek ones take the values 0,1,..,3.) You can assume that the $(u^{\alpha})$ is the 4 velocity of the radial hovering observer at infinity and consequently $(v^{\alpha})$ is the 4 velocity of the particle according to him.

In #37 you said

In the setup I use to denote the 4-velocities of observer, $$u^{\alpha}$$, and particle, $$v^{\alpha}$$, I assume the observer is instantaneously at rest at the event the proper 4-velocity $$v^{\alpha}$$ is observed. Thus we have

$$u^{\alpha}=(dt/ d\tau,0,0,0)=(1,0,0,0)$$

at the event of observation.

which appears to imply by the observer "at rest at the event", that $$u^{\alpha}$$, and particle, $$v^{\alpha}$$ are measured by a local observer at r rather than at infinty as you state in comment above. Anyway, this still does not clarify what you mean by your symbols $u^0, v^0, u^i$.

I have plenty more to say and I will put that in my next post, but for now I do not want anything to distract you from clearly defining what you mean by your symbols $u^0, v^0, u^i$ and $v^i$ that you used in the your equation $g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$ back in #18.

 

[yuiop]

This is even possible for the observer who is at rest at infinity and for a Schwarzschild metric the outside observer being at rest at an infnitely enough distance from the gravitating body spacetime can be taken to be Minkowski. So I don't know how you are going to define a transformation between a local stationary observer at r and an observer at rest at infinity. Let's first clear this up and then go on!

In units of G=M=c=1, if we had a non rotating gravitational body of radius r=4M and I told you the vertical velocity of a particle on the surface of this body is zero and the horizontal velocity is 0.9c according to the measurements in coordinate time of a local stationary observer on the surface of that body, then I am sure you could tell me what the coordinate 3 velocity of the particle is according to an observer at infinity, even though I have not specified the local horizontal velocity in terms of r and $$d\theta$$ or $$d\phi$$.

We are agreed that any observer can treat the local geometry as Minkowski spacetime but if an observer at infinity is making measurements of distant events at r then that is non-local and he can not use Minkowski coordinates and in this case he hase to use Schwarzschild coordinates.

If it helps please assume that when I used ux, uy and uz to describe velocity components I mean:

$$ux = dr/dt$$
$$uy = r d\theta/dt$$
$$uz = r sin(\theta) d\phi/dt$$

and by dx, dy, dz I mean:

$$dx = dr$$
$$dy = r d\theta$$
$$dz = r sin(\theta) d\phi$$

By making those substitutions my equations are perfectly conventional and my use of shorthand notation of ux, uy, uz for velocity components is not conceptionally different to your notation of $$v^1, v^2, v^3$$ for the velocity components.

 

[Altabeh]

When I asked you to define your velocity symbols $u^0, v^0, u^i$ and $v^i$ in #37, you replied that other than v used in the Lorentz factor, "all velocities used in calculations are proper 4-velocities", but here you seem to be making it clear that $v^i$ is a 3 velocity and from #38 you said "Here v=(v^1,v^2,v^3) is the proper 3-velocity" I think we can establish that by $v^i$ you mean the proper 3 velocity or what is sometimes called the spatial component of the 4 velocity. Since we are talking about a 3 velocity it is not invariant and we need to make clear who makes measures $v^i$, but I think from the context of your earlier posts it is the measurement made by a stationary observer at radial coordinate r.

You seem to be understanding me badly. I have not yet entered the Schwarzschild metric and my notations are so clear. I told you that the quantities with only Latin indices all refer to the 3-quantities whereas quantities with Greek indices include one more component, the null time component so when encountering them you should bear in mind that they are all 4-quantities. For example, $$v^i$$ is a 3-velocity while $$v^{\alpha}$$ is a 4-velocity (which is measured according to a comoving observer).

At least we are agreed that $V^2$ has a magnitude of $c^2$ and so we can say:

$$(u^o)^2 = c^2 -v^2$$

This is not true unless I accept that $$u^0$$ is just a typo and it must have been $$v^0$$.

Now if we assume by $$v^2$$ you mean $$( v^1(\tau), v^2(\tau), v^3(\tau) )$$ then the quantity on the right of the above equation is not invariant and nor is it invariant if we assume you mean $$(v^1(t), v^2(t), v^3(t) )$$.

By $$v^2$$ I mean $$(v^1)^2+(v^2)^2+(v^3)^2.$$

You still have not explicitly stated what you mean by the symbols $u^0, v^0, u^i$.

I ignore my own observer and rather go right into the setup you are looking for. (Please use this notation to keep being in agreement with the previous notations used in this thread!) The coordinate 4-velocity $$u^{\alpha}=(u^0,u^i)=(1,0,0,0)$$ belongs to the local stationay observer at $$r$$ i.e. observer A. And the 4-velocity $$v^{\alpha}=(v^0,v^i)=(1,v^i)$$ is for a particle moving in the Schwartzschild spacetime measured by the observer A. The 4-velocity $$v'^{\alpha}=(v'^0,v'^i)=(1,v'^i)$$ is measured by an observer at infinity, i.e. observer B.

The particle's 4-velocity is $$v^{\alpha}$$ (as measured by who?) not $$(v^{0})^2$$ and the observer's 4-velocity is $$u^{\alpha}$$ (as measured by who?) not $(u^{0})^2.$.

Do you think we need to clarify who is measuring the coordinate velocity of an observer!?

Now take the linearized isotropic Schwartzschild metric:

$$ds^2 = (1-2m/r) dt'^2 - (1+2m/r) (dx'^2+ dy'^2 +dz'^2 ).$$

So that the measurements of observer A are related to those of B by

$$dt= \sqrt{1-2m/r}dt',$$
$$dx= \sqrt{1+2m/r}dx',$$
$$dy= \sqrt{1+2m/r}dy',$$
$$dz= \sqrt{1+2m/r}dz'.$$

One can check the validity of these formulae by knowing that $$2m<r$$ and thus $$\sqrt{1-2m/r}<1,$$ $$\sqrt{1+2m/r}>1.$$ meaning that the time dilation and length contraction occur correctly.

By looking at the invariant proper velocity formula,

$$V=\sqrt{g_{\mu\nu}v^{\mu}v^{\nu}}=\sqrt{(1-2m/r)-(1+2m/r)(v'^i)^2}=\sqrt{1-(v'^i)^2-(2m/r)(1+(v'^i)^2)}.$$

and neglecting the terms being of the order 2 in $$2m/r$$ we get

$$V=\sqrt{\frac{(1-4m/r)-(1+4m/r)(v^i)^2}{1-2m/r}}.$$

Assuming that the 3-velocity $$v^i$$ is small, this approximately is

$$V=\sqrt{\frac{1-4m/r}{1-2m/r}}+O((v^i)^2),$$

implying that the circumferences under which observer B measures the particle's velocity has really tiny effects on the measurements due to being at an infinitely large distance from the particle. Also you can see that if $$m=0$$, one would get the formula

$$(v^0)^2=V^2+(v^i)^2.$$

AB

 

[yuiop]

Hi Altabeh,
Thanks for clearing some things up, but I now think the main reason your post in #20 (quoted below) has been confusing the heck out of me is that you made a mistake in the middle of your derivation.

Yeah, and we give a relation of the form

$$g_{00}u^0v^0+g_{ii}u^iv^i=g_{00}u^0v^0=\gamma_vc^2$$ (1)

where in the $$\gamma_v$$, the appearing v is the local 3-velocity

$$v^{i}=(dx^1/d\tau,dx^2/d\tau,dx^3/d\tau).$$

Take the sides of the last equation in (1) to the power 2 and write $$(v^0)^2=V^2-v^2$$ where $$V^{\alpha}$$ denotes the 4-velocity of the particle. Hence

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

and with $$u^2=(u^0)^2$$ we obtain finally

$$V^2={\frac {-{v}^{4}{g_{{00}}}^{2}{u}^{2}+{c}^{6}+{v}^{2}{g_{{00}}}^{2}{u}^{ 2}{c}^{2}}{ \left( {c}^{2}-{v}^{2} \right) {g_{{00}}}^{2}{u}^{2}}}.$$

If your method doesn't lead to this equation, then it doesn't work!

AB

When Dr Greg asked:

...
I can't work out what $$(v^0)^2=V^2-v^2$$ is supposed to mean.

you replied:

Here v=(v^1,v^2,v^3) is the proper 3-velocity and $$V$$ represents the magnitude of the 4-velocity $$v^{\alpha}$$.

So we establish that by v^2 in your equation $$(v^0)^2=V^2-v^2$$ you mean $$v^2 = (v^i)^2 = (dx^1/d\tau)^2 + (dx^2/d\tau)^2 + (dx^3/d\tau)^2$$.
Taking that into account, what you should have said is:

$$(v^0)^2= V^2 + v^2$$

where the (+) sign rather than the (-) sign that you used in your orginal equation is the key correction.

This is the proof that you should have used a (+) sign:

$$c^2 = (v^0)^2 - (v^i)^2$$

$$(v^i)^2 = [ (v^0)^2 - c^2 ]$$

Substitute the above equation into the equation below:

$$(v^0)^2 = V^2 + (v^i)^2$$

and you obtain:

$$(v^0)^2 = V^2 + [ (v^0)^2 - c^2 ]$$

Now we have agreed $$V^2 = c^2$$ so the equation above is now correct.

Later on in this equation:

$$g^2_{00}(u^0)^2(V^2-v^2)=\frac{c^4}{1-v^2/c^2},$$

you use $$v^2$$ to mean proper 3 velocity $$(v^i)^2$$ on the left of the equation and also use $$v^2$$ to mean coordinate 3 velocity on the right of the equation, so you are using the same symbol to mean two different things in the same equation which leads to errors.

This is not true unless I accept that $$u^0$$ is just a typo and it must have been $$v^0$$.

Yes I made a typo. I meant to quote your equation $$(v^0)^2=V^2-v^2$$.

By $$v^2$$ I mean $$(v^1)^2+(v^2)^2+(v^3)^2.$$

I was trying to clarify whether the $$v^2$$ you used in the $$(v^0)^2=V^2-v^2$$ was in terms of coordinate time or proper time and this answer does not make it make it clear but from the context of other posts I am now convinced you mean $$v^2$$ in terms of proper time in that particular equation.

Now take the linearized isotropic Schwartzschild metric:

$$ds^2 = (1-2m/r) dt'^2 - (1+2m/r) (dx'^2+ dy'^2 +dz'^2 ).$$

Please not the isotropic metric! I know Eddington came up with these sorts of metrics but they really are a horror and they mean the local speed of light is neither constant nor isotropic or the local observer has to use different length rulers for vertical and horizontal measurements. In this forum it often stressed that the local speed of light is always c and introducing metrics where that is not true is going to confuse things. I think this is worthy of further discussion but we should probably start a new thread for that to avoid going too far off topic.

 

[Altabeh]

Hi Altabeh,
... is that you made a mistake in the middle of your derivation.

Yes, the sign was incorrect and thank you for noticing it. I also corrected this in my recent post.

you use $$v^2$$ to mean proper 3 velocity $$(v^i)^2$$ on the left of the equation and also use $$v^2$$ to mean coordinate 3 velocity on the right of the equation, so you are using the same symbol to mean two different things in the same equation which leads to errors.

I believe that my notation is flawed in this case and I'd be glad if you could open a new thread and write up in a clear way all these things we talked about in like 10 or more posts!

AB

 

[DrGreg]

I believe that my notation is flawed in this case and I'd be glad if you could open a new thread and write up in a clear way all these things we talked about in like 10 or more posts!

I was getting very confused, so I think it's an excellent idea to start again, but we don't need a new thread if we're still on the same subject.

To avoid confusion I propose to adopt the convention of using upper-case for 4-vectors. We should also take care not to use the same letter with different meanings and whenever we refer to a 3-vector, or components of a 4-vector, we make it very clear which coordinate system we are talking about. I hope also everyone is happy to work with the convention that G = 1 = c.

To set the ball rolling and establish some notation I'll rewrite my previous solution using the method George Jones first mentioned. Recall the original question was

Let's say we use the Schwarzschild metric.
Now, I can parametrize the wordline of a certain object by giving a
parametrisation $(t(\tau),r(\tau),\theta(\tau),\phi(\tau))$, where $\tau$ is the proper time of
that object. Now, I would like to know how to measure the speed of the object;
that is, the magnitude of its 3-velocity. I can find the 4-velocity, but can I deduce in a meaningful way the speed, i.e. as observed by an observer 'at rest' in the frame
(of course, 'at rest' is a relative concept - I'm referring to the observer at infinity, where the Schwarzschild metric is more or less flat)?

Let V be the 4-velocity of an object to be measured, and U be the 4-velocity of any hovering local observer "at rest" at the event where the measurement is required.

In the locally-Minkowski coordinates of a free-falling and momentarily-at-rest local observer at that event, $\textbf{U} = (1, 0, 0, 0)$ and $\textbf{V} = (\gamma, \gamma w, 0, 0)$, where w is the magnitude of the 3-velocity that you want to measure (let's assume it's along the x-axis) and $\gamma = (1-w^2)^{-1/2}$. Therefore

$$g(\textbf{U}, \textbf{V}) = g_{\alpha\beta}U^{\alpha}V^{\beta} = \gamma$$​

But g(U,V) is invariant, i.e. the same in all coordinate systems, so you can also calculate it in Schwarzschild coordinates, and equate the two values.

Now, in Schwarzschild coordinates $(t, r, \theta, \phi)$, $g_{00} = (1 - 2M/r)$ and the other components of the metric don't matter in this case. We have

$$\textbf{U} = \left(\frac{1}{\sqrt{1 - 2M/r}}, 0, 0, 0\right)$$​

(The last three components must be zero and the first is chosen to ensure that g(U,U) = 1.) And

$$\textbf{V} = \left(\frac{dt}{d\tau}, \frac{dr}{d\tau}, \frac{d\theta}{d\tau}, \frac{d\phi}{d\tau}\right)$$​

And so

$$\frac{1}{\sqrt{1-w^2}}= g_{\mu\nu}V^{\mu}U^{\nu} = \frac{dt}{d\tau}\sqrt{1 - 2M/r}$$​

That tells you how to calculate w from $dt/d\tau$ and r for any particle.

If the question had specified that the path was parameterised by Schwarzschild-coordinate-time instead of proper time, the final step would be to note that

$$\textbf{V} = \frac{dt}{d\tau} \left(1, \frac{dr}{dt}, \frac{d\theta}{dt}, \frac{d\phi}{dt}\right)$$​

and as g(V,V) = 1 this gives $dt/d\tau$ in terms of $r, \theta, dr/dt, d\theta/dt, d\phi/dt$.

Now kev, if you want to discuss an alternative method we can do so.

 

[Altabeh]

This is a decent notation and it sets the stage for kev's further insights on the subject to be given systematically.

AB

 

[yuiop]

Hi Altabeh and Dr Greg,

I accept the challenge but I am I am a little short on time right now. I would also like to agree some further conventions first.

Can we assume that when talking about 4 velocities or components of 4 velocities that the velocity components are in terms of proper time and when talking about 3 velocities we should explicitly state whether we are talking about proper velocity (the 3 spatial components of a 4 velocity) or coordinate velocity or agree a notation to make the difference clear.

There is also the issue of primed variables. Should primed velocity components indicate the measurements of a local observer at r in Schwarzschild coordinates or the measurements of an observer at infinity or should we just reley on the context of the surrounding text to make the difference clear?

 

[DrGreg]

Can we assume that when talking about 4 velocities or components of 4 velocities that the velocity components are in terms of proper time

4-velocities are always differentiated with respect to proper time, so yes.

...and when talking about 3 velocities we should explicitly state whether we are talking about proper velocity (the 3 spatial components of a 4 velocity) or coordinate velocity or agree a notation to make the difference clear.

There is also the issue of primed variables. Should primed velocity components indicate the measurements of a local observer at r in Schwarzschild coordinates or the measurements of an observer at infinity or should we just reley on the context of the surrounding text to make the difference clear?

My suggestion is to always spell out explicitly what you mean, to avoid the confusions that previously arose in this thread.

 

[yuiop]

NOTE: After comments by Dr Greg and Altabeh it has become clear to me that I my conclusions in this post are not correct and I will be posting corrections in a later post.

This is a decent notation and it sets the stage for kev's further insights on the subject to be given systematically.

In this thread I am not offering insights, but just trying to come to a better understanding of 3 and 4 velocity in the context of the Schwarzschild geometry. If my learning curve helps someone else in the process then all well and good.

... always spell out explicitly what you mean, to avoid the confusions that previously arose in this thread.

In the interests of being explicit I would like to adopt the primed convention suggested by Altabeh in an earlier post of indicatiing the measurements made by an observer at infinity in Schwarzschild coordinates by primed symbols and local measurements by a an observer at r (or measurements by an observer in Minkowski spacetime) by unprimed symbols. Using this notation the classic Schwarzschild metric would be written as:

$$d\tau^2 = (1-2M/r) {dt '}^2 - \frac{1}{(1-2M/r)}{dr '}^2 - r^2{d\theta '}^2 - r^2 sin^2(\theta) {d\phi '}^2 }$$

Using this convention, the 4 velocity of a stationary observer at r, according to the observer at r is:

$$\textbf{U} = \left(1, 0, 0, 0\right)$$​

and the magnitude of the same 4 velocity is:

$$||\textbf{U}|| = 1$$​

The 4 velocity of the stationary observer at r, according to the observer at infinity is:

$$\textbf{U} ' = \left(\frac{1}{\sqrt{1 - 2M/r}}, 0, 0, 0\right)$$​

The magnitude of U' according to the observer at infinity is:

$$||\textbf{U} '|| = \frac{1}{\sqrt{1 - 2M/r}}$$​

The significant observation I make here is that while the magnitude of a 4 velocity is always 1 in Minkowski coordinates, the same is not true in Schwarzschild coordinates. Agree or dissagree?

Now let us say the 3 velocity of a vertically falling test particle is w, then the 4 velocity according to a local stationary observer at r is

$$\textbf{V} = (\gamma, \gamma w, 0, 0)$$​

where $\gamma = (1-w^2)^{-1/2}$, then the 4 velocity of the same test particle according to the observer at infinity is:

$$\textbf{V} ' = \left(\frac{\gamma}{(1-2M/r)}, \frac{\gamma w}{(1-2M/r)}, 0, 0\right)$$​

and the magnitude of the 4 velocity of the falling particle, according to the observer at infinity is:

$$||\textbf{V} '|| = \frac{1}{\sqrt{1 - 2M/r}}$$​

which is at least in agreement with the magnitude given earlier for the 4 velocity of the stationary observer at r, according to the observer at infinity.

This means that the equations I gave for 4 velocity in Schwarzschild coordinates in #18 of this thread are wrong because I was assuming the mgnitude of a 4 velocity is always c. (I think the equations I gave for coordinate 3 velocity transformations in Schwarzschild geometry in the same post are still good.)

...

$$\frac{1}{\sqrt{1-w^2}}= g_{\mu\nu}V^{\mu}U^{\nu} = \frac{dt}{d\tau}\sqrt{1 - 2M/r}$$​

That tells you how to calculate w from $dt/d\tau$ and r for any particle.

Using primed notation for explicitness again, I can write the equation given above by Dr Greg as:

$$\frac{1}{\sqrt{1-w^2}}= \frac{dt '}{d\tau}\sqrt{1 - 2M/r}$$​

making clear that w is a local 3 velocity according to a stationary observer at r and (dt'/dtau) is the time dilation according to an observer at infinity.

The OP asked:

From special relativity, I recall that you could write the 4-velocity
as (gamma,gamma*v) where gamma is the usual gamma factor and v is the three-velocity.
I'm looking for something analogous - if that's meaningful.

If my reasoning above is correct, then the answer is yes there is something anaogous in Schewarzschild coordinates, if we understand gamma to mean 1/(1-2M/r)*1/(1-v^2) and not just 1/(1-v^2) as in the Special Relativity context. I have changed v to lower case in the above quote, in line with our agreed notation for 3 velocity.)

 

[DrGreg]

In the interests of being explicit I would like to adopt the primed convention suggested by Altabeh in an earlier post of indicatiing the measurements made by an observer at infinity in Schwarzschild coordinates by primed symbols and local measurements by a an observer at r (or measurements by an observer in Minkowski spacetime) by unprimed symbols. Using this notation the classic Schwarzschild metric would be written as:

$$d\tau^2 = (1-2m/r) {dt '}^2 - \frac{1}{(1-2M/r)}{dr '}^2 - r^2{d\theta '}^2 - r^2 sin^2(\theta) {d\phi '}^2 }$$

To be consistent, you need to prime all the symbols:
$$d\tau^2 = (1-2m/r') {dt '}^2 - \frac{1}{(1-2M/r')}{dr '}^2 - r'^2{d\theta '}^2 - r'^2 sin^2(\theta') {d\phi '}^2 }$$

Using this convention, the 4 velocity of a stationary observer at r, according to the observer at r is:

$$\textbf{U} = \left(1, 0, 0, 0\right)$$​

and the magnitude of the same 4 velocity is:

$$||\textbf{U}|| = 1$$​

Well, you haven't explicitly spelled out exactly which coordinates this observer is using.

The 4 velocity of the stationary observer at r, according to the observer at infinity is:

$$\textbf{U} ' = \left(\frac{1}{\sqrt{1 - 2M/r}}, 0, 0, 0\right)$$​

The magnitude of U' according to the observer at infinity is:

$$||\textbf{U} '|| = \frac{1}{\sqrt{1 - 2M/r}}$$​

The significant observation I make here is that while the magnitude of a 4 velocity is always 1 in Minkowski coordinates, the same is not true in Schwarzschild coordinates. Agree or dissagree?

First of all, 4-vectors are concepts that exist independently of coordinate systems. So really U and U' are the same 4-vector, but you are expressing the vector's components relative to two different coordinate systems. Loosely speaking, the "physicist's definition" of a 4-vector is something whose components transform under the rule

$$U'^{\alpha} = \frac{\partial x'^{\alpha}}{\partial x^{\beta}} U^{\beta}$$​

and this is true for 4-velocities.

Second, the "length" of a 4-vector is calculated via the metric

$$||\textbf{U}|| = \sqrt{|g_{\alpha\beta}U^{\alpha}U^{\beta}|}$$​

so, in your primed coordinates you have to use the Schwartzschild metric components and you still get an answer of 1.

The magnitude of a 4-velocity is always 1, no matter what coordinates you use. This follows from the two equations

$$||\textbf{U}||^2 = \left|g_{\alpha\beta}\frac{dx^{\alpha}}{d\tau}\frac{dx^ ^{\beta}}{d\tau}\right|$$
$$d\tau^2 = |g_{\alpha\beta}dx^{\alpha}dx^{\beta}|$$​

Now let us say the 3 velocity of a vertically falling test particle is and w, then the 4 velocity according to a local stationary observer at r is

$$\textbf{V} = (\gamma, \gamma w, 0, 0)$$​

where $\gamma = (1-w^2)^{-1/2}$, then the 4 velocity of the same test particle according to the observer at infinity is:

$$\textbf{V} ' = \left(\frac{\gamma}{(1-2M/r)}, \frac{\gamma w}{(1-2M/r)}, 0, 0\right)$$​

You will need to explain why you think that is true (which will involve clarifying what coordinates you are using). I won't comment any further on what you said after that.

 

[Altabeh]

The magnitude of U' according to the observer at infinity is:

$$||\textbf{U} '|| = \frac{1}{\sqrt{1 - 2M/r}}$$​

You can't use the locally Minkowski metric where you're applying the Schwartzschild metric to the components of the 4-velocity $$U'$$ at the same time. The magnitude of $$U'$$ is

$$||\textbf{U} '|| = \sqrt{g_{\mu\nu}U^{\mu}U{\nu}}=\sqrt{(1 - 2M/r)\frac{1}{1 - 2M/r}}=1.$$​

Agree or dissagree?

Disagreed for the reason given above.

where $\gamma = (1-w^2)^{-1/2}$, then the 4 velocity of the same test particle according to the observer at infinity is:

$$\textbf{V} ' = \left(\frac{\gamma}{(1-2M/r)}, \frac{\gamma w}{(1-2M/r)}, 0, 0\right)$$​

The second component here is not correct as the factor $$\frac{1}{(1-2M/r)}$$ has been written inversely. If this was corrected, then its magnitude would be

$$||\textbf{V} '|| =||\textbf{V}||.$$​

If my reasoning above is correct, then the answer is yes there is something anaogous in Schewarzschild coordinates, if we understand gamma to mean 1/(1-2M/r)*1/(1-v^2) and not just 1/(1-v^2) as in the Special Relativity context. I have changed v to lower case in the above quote, in line with our agreed notation for 3 velocity.)

I don't know what you mean by this paragraph but the above assumption you made with an observer at infinity instead of an instantaneously at rest observer only works when the 4-velocity $$V'$$ is defined to be of the same general form as in SR. The important observation here is that the 4-velocity of the observer at infinity must be $$U'.$$ And this is nothing new but transferring the stationary observer to an infinitely large distance from where it is!

AB

 

[dx]

What exactly is the problem here? Are you trying to express the 4-velocity of an infalling observer in a basis derived from the Schwarzschild coordinates?

An object falling radially inward in a Schwarzschild spacetime with a speed v (as measured by a local inertial frame) will have a 4-velocity

$$u = \gamma (1 + m/r) \partial_t - v\gamma(1 - m/r)\partial_r$$

(ignoring quadratic terms in m/r and setting G = c = 1)

 

[yuiop]

What exactly is the problem here? Are you trying to express the 4-velocity of an infalling observer in a basis derived from the Schwarzschild coordinates?

An object falling radially inward in a Schwarzschild spacetime with a speed v (as measured by a local inertial frame) will have a 4-velocity

$$u = \gamma (1 + m/r) \partial_t - v\gamma(1 - m/r)\partial_r$$

(ignoring quadratic terms in m/r and setting G = c = 1)

The trouble with this equation is that it does not seem to work, but maybe I am just reading it wrong. To clear things up, what does your equation predict for u, when v = 0.8 and r = 2 Rs where Rs = 2GM/c^2 ?

I calculate u = dr/dtau = 0.94281 using a different method. What do you get?

(I am assuming (1 +m/r) is an approximation of 1/(1-2m/r). Is that right?)

 

[yuiop]

I have been putting off replying to this thread, because I started reading up on matrix and tensor notation, which got me hung up on a couple of things and probably got me even more confused. Anyway, Dr Greg and Altabeh have put me straight on some things and I would like to thank them for that.

To be consistent, you need to prime all the symbols:
$$d\tau^2 = (1-2m/r') {dt '}^2 - \frac{1}{(1-2M/r')}{dr '}^2 - r'^2{d\theta '}^2 - r'^2 sin^2(\theta') {d\phi '}^2 }$$

Hmmm, I have always assumed that m = m', r = r' and $\theta = \theta '$. Is that not the case? Anyway, the local observer can always make measurements in terms of locally Minkowsky coordinates and not concern himself with the values of m, r, $\theta$, m', r', or $\theta '$, no?

Second, the "length" of a 4-vector is calculated via the metric

$$||\textbf{U}|| = \sqrt{|g_{\alpha\beta}U^{\alpha}U^{\beta}|}$$​

OK, I think I understand that now. $g_{\alpha\beta}$ is the metric of the coordinate system in matrix form (usually called the metric tensor) and is required if you want to find the norm of a four vector.

Now let us say the 3 velocity of a vertically falling test particle is w, then the 4 velocity according to a local stationary observer at r is

$$\textbf{V} = (\gamma, \gamma w, 0, 0)$$​

where $\gamma = (1-w^2)^{-1/2}$, then the 4 velocity of the same test particle according to the observer at infinity is:

$$\textbf{V} ' = \left(\frac{\gamma}{(1-2M/r)}, \frac{\gamma w}{(1-2M/r)}, 0, 0\right)$$​

Taking onboard what Dr Greg and Altabeh have told me, the above statements need to be corrected and I should have said something like:

Let the coordinate 3 velocity of a vertically falling test particle in locally Minkoskian coordinate terms be w, according to a local stationary observer at at distance r from the gravitational source. The 4 velocity according to the same observer is

$$\textbf{V} = (\gamma, \gamma w, 0, 0) = \gamma(1,w,0,0)$$​

where $\gamma = (1-w^2)^{-1/2}$.
The 4 velocity of the same test particle in Schwarzschild coordinates (i.e according to the observer at infinity) is then:

$$\textbf{V} ' = \left(\frac{\gamma}{\sqrt{1-2M/r}}, \gamma w \sqrt{1-2M/r} , 0, 0\right) = \gamma\left((1-2M/r)^{-1/2}, (1-2M/r)^{1/2} w , 0, 0\right)$$​

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

$$\textbf{v} ' = \gamma\left((1-2M/r)^{-1/2}, w (1-2M/r)^{1/2}, 0, 0\right) (1-2M/r)^{1/2} = \gamma\left(1, w(1-2M/r), 0, 0\right)$$​

Hopefully, that is all a bit more correct.

 

[Altabeh]

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

$$\textbf{v} ' = \gamma\left((1-2M/r)^{-1/2}, w (1-2M/r)^{1/2}, 0, 0\right) (1-2M/r)^{1/2} = \gamma\left(1, w(1-2M/r), 0, 0\right)$$​

Hopefully, that is all a bit more correct.

More correct but not correct at all! In Schwartzschild metric, finding the coordinate velocity is a little bit cumbersome and needs some nerves actually! This can be done for a radially free falling particle in the following way:

The metric is assumed to be in the form

$$ds^2=(1+f(r))dt^2-(1+f(r))^{-1}dr^2-r^2(d\theta^2+\sin^2{\theta}d\phi^2),$$

where I use the convention $$c=1$$, for the sake of simplicity and $$f$$ being the relativistic correction to the Minkowski metric in a weak gravitational field.

First off, I have to write the time and radial geodesic equations:

$$\ddot{t}+\frac{f'}{1+f}{\dot{t}\dot{r}}=0,$$
$$\ddot{r}+\frac{f'}{2}(1+f)\dot{t}^2-\frac{f'}{2(1+f)}-r(1+f)\dot{\theta}^2-r(1+f)\sin^2\theta}\dot{\phi}^2=0,$$

where overdots are derivatives wrt the affine parameter $$s$$ and the primes stand for differentiation wrt the coordinate $$r.$$ By integrating the first equation one gets

$$\dot{t}=\frac{k}{1+f},$$

for some integration constant $$k.$$ Introducing this into the second equation and noticing that the motion is radial, then

$$\ddot{r}=-\frac{f'}{2(1+f)}(k^2-\dot{r}^2)$$

which if integrated, it would yield the equation for the coordinate radial velocity in Schwarzschild coordinates:

$$v_r=\frac{dr}{dt}=(1+f)}(1-\frac{a}{k^2}(1+f))^{1/2}.$$ (#)

Here $$a$$ is some other integration constant. The constant $$\frac{a}{k^2}$$ can be written in terms of the initial velocity $$(v_r)_{(initial)}$$ where

$$(v_r)_{(initial)}=(1+f_i)}(1-\frac{a}{k^2}(1+f_i))^{1/2}.$$

It is apparent from (#) that if $$f=-1$$, then we have $$v_r=0$$ which means that the free falling particle has a zero coordinate velocity when it hits the singularity.

AB

 

[yuiop]

The metric is assumed to be in the form

$$ds^2=(1+f(r))dt^2-(1+f(r))^{-1}dr^2-r^2(d\theta^2+\sin^2{\theta}d\phi^2),$$

where I use the convention $$c=1$$, for the sake of simplicity and $$f$$ being the relativistic correction to the Minkowski metric in a weak gravitational field.

OK, we are considering radial motion only here, so we can cut out the last term and write the above equation in this form:

$$1 = (1+f)\dot{t}^2}-(1+f)^{-1}\dot{r}^2$$

where I have written f as abbreviation for f(r) as you appear to have done in the remainder of your post. Is that correct?
GR is complicated enough without introducing unnecessary complications!

First off, I have to write the time and radial geodesic equations:

$$\ddot{t}+\frac{f'}{1+f}{\dot{t}\dot{r}}=0,$$

OK, slow down a bit here. I get that when you differentiate 1 wrt some variable you get zero, but my calculus was never very good, so can you break down how you get the geodesic equations?

$$\ddot{r}+\frac{f'}{2}(1+f)\dot{t}^2-\frac{f'}{2(1+f)}-r(1+f)\dot{\theta}^2-r(1+f)\sin^2\theta}\dot{\phi}^2=0,$$

where overdots are derivatives wrt the affine parameter $$s$$ and the primes stand for differentiation wrt the coordinate $$r.$$ By integrating the first equation one gets

$$\dot{t}=\frac{k}{1+f},$$

for some integration constant $$k.$$ Introducing this into the second equation and noticing that the motion is radial, then

$$\ddot{r}=-\frac{f'}{2(1+f)}(k^2-\dot{r}^2)$$

My algebra is better and something is not right about the above. After the substitution the end result should be:

$$\ddot{r}=-\frac{f'}{2(1+f)}(k^2-1)$$

or the initial equation (ignoring non radial spatial terms) should have been:

$$\ddot{r}+\frac{f'}{2}(1+f)\dot{t}^2-\frac{f'}{2(1+f)}\dot{r}^2 =0$$

 

[Altabeh]

OK, we are considering radial motion only here, so we can cut out the last term and write the above equation in this form:

$$1 = (1+f)\dot{t}^2}-(1+f)^{-1}\dot{r}^2$$

where I have written f as abbreviation for f(r) as you appear to have done in the remainder of your post. Is that correct?

Yes!

OK, slow down a bit here. I get that when you differentiate 1 wrt some variable you get zero, but my calculus was never very good, so can you break down how you get the geodesic equations?

I thought you're familiar with the derivation. You should put the right-hand side of the above equation in the Euler-Lagrange equations for the time coordinate, i.e.

$$\frac{d}{ds}(\frac{\partial F(x^\alpha,\dot{x}^\alpha,s)}{\partial \dot{t}})=\frac{\partial F(x^\alpha,\dot{x}^\alpha,s)}{\partial {t}}$$.

Then you'll get that equation of mine.

or the initial equation (ignoring non radial spatial terms) should have been:

Yeah, just forgot to write $$\dot{r}^2$$ in the second term.

AB

 

[yuiop]

Here $$a$$ is some other integration constant. The constant $$\frac{a}{k^2}$$ can be written in terms of the initial velocity $$(v_r)_{(initial)}$$ where

$$(v_r)_{(initial)}=(1+f_i)}(1-\frac{a}{k^2}(1+f_i))^{1/2}.$$

It is apparent from (#) that if $$f=-1$$, then we have $$v_r=0$$ which means that the free falling particle has a zero coordinate velocity when it hits the singularity.

AB

I am not sure what you mean by "the singularity". Normally this means the singularity at the center of a Schwarzschild black hole, but you seem to be talking about the coordinate singularity at the event horizon, because f = -2GM/(rc^2) and f=-1 at the event horizon. Are you are claiming that a particle can remain stationary at the event horizon?

I am also not sure why you are attaching any significance to initial velocity, because this can be arbitrarily set to any value you like between 0 and c. The useful quantity is the difference between final and initial velocities.

Also, the coordinate velocity components of the same particle in Schwarzschild coordinates is:

$$\textbf{v} ' = \gamma\left((1-2M/r)^{-1/2}, w (1-2M/r)^{1/2}, 0, 0\right) (1-2M/r)^{1/2} = \gamma\left(1, w(1-2M/r), 0, 0\right)$$​

Hopefully, that is all a bit more correct.

More correct but not correct at all!

I do not see why you think my equation is not correct. Please elaborate.

I think we are talking at cross purposes here and that is why our equations differ. I am talking about arbitrary 3 velocities being transformed from locally Minkowskian coordinates to Schwarzschild coordinates while you seem to be talking about terminal velocities of free falling particles.

 

[Altabeh]

I am not sure what you mean by "the singularity". Normally this means the singularity at the center of a Schwarzschild black hole, but you seem to be talking about the coordinate singularity at the event horizon, because f = -2GM/(rc^2) and f=-1 at the event horizon. Are you are claiming that a particle can remain stationary at the event horizon?

Here by singularity I mean the coordinate singularity or the hypersurface $$f=-1.$$ This never means that the particle remains stationary at $$f=-1$$ but that its coordinate acceleration goes to infinity while the coordinate velocity gets zero at the time that the particle crosses the surface $$f=-1$$.

I am also not sure why you are attaching any significance to initial velocity, because this can be arbitrarily set to any value you like between 0 and c. The useful quantity is the difference between final and initial velocities.

I don't even know what this is supposed to imply! Did I insist on a certain value of initial velocity and said that for example $$v_i=v_0$$ knocks spots off $$v_i=v_1$$!?

I do not see why you think my equation is not correct. Please elaborate.

I think we are talking at cross purposes here and that is why our equations differ. I am talking about arbitrary 3 velocities being transformed from locally Minkowskian coordinates to Schwarzschild coordinates while you seem to be talking about terminal velocities of free falling particles.

See, for example, excercise 9.13 at page 225 of "General relativity" by Michael Paul Hobson, George Efstathiou, Anthony N. Lasenby to find what your equation lacks in the radial coordinate.

AB