4 velocity in Schwarzchild metric

[Big Guy]

How do we calculate the 4 velocity of a particle that is projected radially downwards at velocity u at a radius r_a?

The condition on 4 velocity is that g_μνv^μv^ν = 1 which implies that at radius r_a we have

g^a₀₀(v⁰)² + g^a₁₁(v¹)² = 1 (eq 1)
​

So if we start from x^μ = (t,r) we get v^μ = (1/√g₀₀ , 1/√g₁₁) but this is only for a particle that fell from rest at infinity correct? If we want to give it a velocity u which suggests we Lorentz boost v^μ to

v^μ = γ(1/√g₀₀ , β/√g₁₁) (eq 2)
​

Where β is the velocity we boost it at. This does solve eq 1 but I don't think it's correct to Lorentz boost because we are not traveling between two frames in SR.

Do I have to solve the geodeisic equation for v^μ?

Thank you!

 

[Big Guy]

Ok so I made some progress with the timelike geodeisic equation \dot{t} = \frac{k}{1-\frac{2\mu}{r}} equating k as equal to gamma at infinity.

This approach basically assumes you release a particle at infinity at some velocity that gives you gamma hence k and you then measure it again when it passes the relevant observation point you desire. However if you want to look at velocities u that are below escape speed (i.e dropped form closer than infinity) you have to figure out how much further away the particle was dropped at rest from and work from there.

However it isn't very satisfactory having to employ two different methods depending on whether the particle has a speed that could have been the result of dropping from infinity or not. Is there a quick way of determining 4 velocities of particles in the Schwarzschild metric knowing their starting point and initial 3 velocity?

 

[BiGyElLoWhAt]

So, I'm not an expert on the actual maths behind GR, but I don't like this equation:
$v^{\mu}=(\frac{1}{\sqrt{g_{00}}},\frac{1}{\sqrt{g_{11}}})$
as it doesn't have norm 1, or c, depending on your unit choice. Did you mean to include a gamma in front as you did in your boost velocity equation?
Also, with one of it's spatial coordinate velocities equal to 1, in natural units that corresponds to a velocity of c, which appears to be what you are using. Assuming that in your final equation, $\gamma = (1-\beta^2)^{-1/2}$ (again, with c=1), then I don't see anything wrong with that result. The constraint then would be beta<1

Edit..
It seems to be consistent with the metric description on Wikipedia. I'm honestly not sure, if that's the case.

 

[PWiz]

It's usually always a good idea to state the coordinates you're using, and to refrain from using numbers for tensor indices. In this case, I'm guessing you're using standard Schwarzschild coordinates. I'm also assuming that you're interested in the four-velocity of the particle in the frame of reference of an observer located "far away" from the mass (i.e. in terms of $(t,r)$), because otherwise the four-velocity would be of the trivial form $U^{\mu} \rightarrow (τ,0,0,0)$.

Note that since the motion of the particle is purely radial, $g_{\phi \phi}$ and $g_{\theta \theta}$ equal zero, which you have noted:

The condition on 4 velocity is that gμνvμvν = 1 which implies that at radius ra we have

ga00(v0)2 + ga11(v1)2 = 1 (eq 1)
​

However,

So if we start from xμ = (t,r) we get vμ = (1/√g00 , 1/√g11)

I don't think this is correct. How did you get this result? I think the best way to proceed forward from here is to use the geodesic equation:
$$ \frac{d^2 x^{\lambda}}{dτ^2}+Γ^{\lambda} \ _{\mu ν} \frac{dx^μ}{dτ} \frac{dx^ν}{dτ} = 0 $$
(You'll have to compute the Christoffel symbols though, and they can sometimes get messy, but luckily they aren't over here. Go ahead and see!)

 

[Big Guy]

I figured it out! Suppose we have an observer at infinity M, an observer A at co-ordinates (t,r) in M's frame and A pitches an object O at a 3-velocity u radially downwards as shown (file at bottom inconveniently).

In the frame of A the velocity of u is the distance radially traveled divided by the time taken so \overrightarrow{u} = \frac{ds_{A}}{dt_{A}} = \frac{\sqrt{g_{11}}dr_{M}}{dt_{A}} relating to the radial distance traveled in frame M.

Now since x_{M}^{\mu} = (t,r) we get the 4 velocity of O as u_{M}^{\mu}=\frac{dx_{M}^{\mu}}{d\tau} = \frac{dx_{M}^{\mu}}{dt_{A}} \frac{dt_{A}}{d\tau}= \frac{dx_{M}^{\mu}}{dt_{A}}\gamma_{u}

EDIT: d\tau = dt_{O}, proper time is considered in the moving frame O we are interested in.

As A and O are moving at u relative to each other locally. Now we tackle the other fraction:

\frac{dx_{M}^{\mu}}{dt_{A}}\gamma_{u} = \gamma_{u}(\frac{dt_{M}}{dt_{A}}, \frac{dr_{M}}{dt_{A}})

Now we just replace the components with what we already have from the metric and u:

u^{\mu} = \gamma_{u}(\frac{1}{\sqrt{g_{00}}}, \frac{u}{\sqrt{g_{11}}})

Giving us our answer after a lot of Latex experience. As with everything in GR I needed to be more careful with which frame was differentiating which quantity etc. You can easily check the above expression is normalised to unity.

 

[PWiz]

I think your answer is wrong. $\frac{dt_A}{dτ} ≠ \gamma _u$ because O is going deeper into the gravitational potential well. This is not a simple case of applying the principle of equivalence and saying that "we can use SR because A is moving away from O with constant proper acceleration and O is not undergoing any proper acceleration", because O is moving radially, so gravitational time dilation will come into play. I really think you should use the geodesic equation.

 

[PeterDonis]

How do we calculate the 4 velocity of a particle that is projected radially downwards at velocity u at a radius r_a?

Velocity $u$ relative to whom?

If it is velocity $u$ relative to an observer who is "hovering" at $r_a$, then your expression in post #5 is correct, at least for the instant the particle is at radius $r_a$. But it doesn't say anything about what happens after that; see below.

if we start from x^μ = (t,r) we get v^μ = (1/√g00 , 1/√g11)

Knowing $x^\mu$ doesn't tell you anything about $v^\mu$. A particle at coordinates $x^\mu$ could be moving in any direction, with any speed less than that of light. There's no way to "take the derivative" of $x^\mu$ if you only know $x^\mu$ at one point; you would have to know a function that describes $x^\mu$ along an entire worldline.

Also, the $v^\mu$ you wrote here isn't even unit vector.

this is only for a particle that fell from rest at infinity correct?

Here, again, you're asking, not just about a particle's 4-velocity at a given instant, but about a particle's 4-velocity along its entire worldline; you're asking, if a particle fell radially inward from rest at infinity, what would its 4-velocity be at radius $r_a$? That requires solving the geodesic equation; the analysis you've done says nothing about it.

If we want to give it a velocity u which suggests we Lorentz boost v^μ to

v^μ = γ(1/√g00 , β/√g11) (eq 2)
​

Where β is the velocity we boost it at.

That is not what you get if you take the 4-velocity of a particle that fell from rest at infinity to radius $r_a$ and boost it. It's what you get when you take the 4-velocity of a particle that is "hovering" at rest at radius $r_a$, and boost it by velocity $\beta$ radially inward. That's why your expression in post #5 gives the correct 4-velocity for a particle that is projected inward from radius $r_a$ at velocity $u$ relative to a "hovering" observer at that radius. But, as I said above, that in itself tells you nothing about how that 4-velocity will change as the particle falls. For that, you need to solve the geodesic equation.

 

[Big Guy]

I think your answer is wrong. $\frac{dt_A}{dτ} ≠ \gamma _u$ because O is going deeper into the gravitational potential well. This is not a simple case of applying the principle of equivalence and saying that "we can use SR because A is moving away from O with constant proper acceleration and O is not undergoing any proper acceleration", because O is moving radially, so gravitational time dilation will come into play. I really think you should use the geodesic equation.

Gravitational time dilation is already taken account of due to the 1/√g₀₀ in the 4-velocity.

Velocity uu relative to whom?

If it is velocity uu relative to an observer who is "hovering" at rar_a, then your expression in post #5 is correct, at least for the instant the particle is at radius rar_a. But it doesn't say anything about what happens after that; see below.

Yes, all I wanted was the 4 velocity of a particle relative to a distance observer that is thrown by a stationary observer at A with velocity u. I know it's a pretty useless expression in terms of particle dynamics afterwards but I was just curious :)

 

[PWiz]

Gravitational time dilation is already taken account of due to the 1/√g00 in the 4-velocity.

No it isn't. You need to understand that the metric tensor has different components in different frames. In your OP, you're only talking about O's frame. In #5, you introduced a new intermediate frame A, and just carried over the metric tensor components. It doesn't work that way.

To make it more explicit, I'll write out the whole equation. Let's take the OP first. Let $x_o$ denote O's coordinates. Then $dt_o^2=(1−\frac{2GM}{r^2} )dt^2−(1−\frac{2GM}{r^2})^{−1} dr^2$. There is no way you can find an expression for $\frac{dt}{dt_0}$ in terms of $G$, $M$ and $r$ just using this information (you must use the geodesic equation).

Let's talk about #5 now. You haven't clearly defined what A's frame really is, but I'm guessing it "hovers" at constant altitude and it was the point from which O was released with some 3 velocity $u$ in the radial direction (towards $M$ ). Let $x_A$ denote A′s coordinates. So we have $dt_A^2 = (1- \frac{2GM}{r_A^2}) dt^2$, where $r_A$ is the constant altitude at which A hovers. But we want to convert to O′s coordinates, and in A′s frame of reference, O is not stationary, so we get $dt_o^2=(1−\frac{2GM}{r^2} ) (1- \frac{2GM}{r_A^2})^{-1} dt_A^2−(1−\frac{2GM}{r^2})^{−1} dr^2$. This is as unsolvable as the first scenario (unless you meant something else in your posts).

There is no real need to complicate the problem. Just stick with the OP. I'll help you get started with the geodesic equation:
Take the "far away" coordinates. The metric tensor then has the form $g_{tt} = (1−\frac{2GM}{r^2} )$, $g_{rr} = - (1−\frac{2GM}{r^2})^{−1}$, $g_{\theta \theta} = 0$, $g_{\phi \phi} = 0$ and $g_{\mu ν} = 0$ for all $\mu ≠ ν$. The inverse of the metric, i.e. $g^{\mu ν}$, will have the components $g^{tt} = (1−\frac{2GM}{r^2} )^{-1}$, $g^{rr} = - (1−\frac{2GM}{r^2} )$, and all other components would be zero, as is the case with the 'covariant' version of the metric tensor. (Notice how easy computing the inverse metric components was because the metric tensor was diagonalized in our coordinates.)

Now we can use the fact that $Γ^{α} \ _{μν} = \frac{1}{2} g^{ασ} (∂_μ g_{ν σ} + ∂_ν g_{σμ} - ∂_σ g_{μν})$, that the metric tensor components are a function of $r$ only, and that the metric tensor is diagonalized to know which Christoffel symbols are non-zero. Can you take it from here?

 

[PeterDonis]

No it isn't

As I said in a previous post, the 4-velocity given in post #5 is correct (except for one thing, which I'll mention below), for the limited purpose that the OP has said he wants: he just wants the 4-velocity of an object moving radially inward at velocity $u$, relative to an observer "hovering" at the same radius, expressed in the global Schwarzschild coordinate chart. Expressing it in the global chart means the $1 / \sqrt{g_{00}}$ and $1 / sqrt_{g_{11}}$ factors need to appear, because the norm gets computed using the Schwarzschild line element. The norm is then

$$
| u^\mu |^2 = g_{00} u^0 u^0 + g_{11} u^1 u^1 = \gamma_u^2 \left( g_{00} \frac{1}{g_{00}} - u^2 g_{11} \frac{1}{g_{11}} \right) = \gamma_u^2 \left( 1 - u^2 \right) = 1
$$

which is what is required for a valid 4-velocity.

There is no way you can find an expression for $\frac{dt}{dt_0}$ in terms of $G$, $M$ and $r$ just using this information (you must use the geodesic equation).

Not for the whole worldline of the infalling object, no. But the OP has said (post #8) that he wasn't trying to do that; he was trying to do something much more limited (see above). For that limited purpose, he only needs $dt / dt_0$ at one point, not over the entire worldline.

You haven't clearly defined what A's frame really is, but I'm guessing it "hovers" at constant altitude and it was the point from which O was released with some 3 velocity $u$ in the radially direction (towards $M$ ).

That was my understanding as well, and the OP pretty much confirms it in post #8.

So we have $dt_A^2 = (1- \frac{2GM}{r_A^2}) dt^2$, where $r_A$ is the constant altitude at which A hovers.

Yes, or, alternatively, we can say that A's 4-velocity, expressed in global Schwarzschild coordinates, is $u^\mu = (1 / \sqrt{g_{00}}, 0)$.

we want to convert to O′s coordinates

I agree that the method that the OP used in post #5, where he tries to express things in O's coordinates and then transform back, is needlessly roundabout. But, since he is only interested in the 4-velocity at one event, the event where O and A are co-located, his method gives the right answer. And since he is only looking for the 4-velocity at one event, there is no need to use the geodesic equation either.

The quick way of getting the right answer is simply to observe that O's 4-velocity is just A's 4-velocity boosted radially inward with ordinary velocity $u$. So you just take A's 4-velocity, $u^\mu = (1 / \sqrt{g_{00}}, 0)$, and boost it radially inward with ordinary velocity $u$. That gives O's 4-velocity as $u^\mu = \gamma_u (1 / \sqrt{g_{00}}, - u / \sqrt{g_{11}})$; the minus sign in the second component is the only thing missing from post #5. The $\sqrt{g_{11}}$ factor in the second component is needed because we are working in global Schwarzschild coordinates.

If that procedure seems too heuristic, it's easy enough to add a step to make it more rigorous: first, observe that in A's local inertial frame, his 4-velocity is $(1, 0)$, so O's 4-velocity at the event where they are co-located must be $\gamma_u (1, - u)$ by a standard Lorentz transformation. Then just transform back to global coordinates, which adds the square roots of the metric coefficients in the denominators.

 

[PeterDonis]

all I wanted was the 4 velocity of a particle relative to a distance observer that is thrown by a stationary observer at A with velocity u

It's worth noting that "relative to a distant observer" is not really the right way to express this. The answer you got (but with a minus sign in the second component, as I said in my response to PWiz just now) is the 4-velocity of O at the instant it is co-located with A, expressed in global Schwarzschild coordinates. But that 4-velocity is not, strictly speaking, "relative to a distant observer", because in GR there is, strictly speaking, no such thing as "relative velocity" between spatially separated objects. The only time a relative velocity has meaning is for two objects that are spatially co-located, i.e., at the event at which their worldlines intersect. So the only physically meaningful relative velocity here is $u$, the velocity of O relative to A at the instant they are spatially co-located.

 

[PWiz]

Not for the whole worldline of the infalling object, no. But the OP has said (post #8) that he wasn't trying to do that; he was trying to do something much more limited (see above). For that limited purpose, he only needs dt/dt0dt / dt_0 at one point, not over the entire worldline.

Ah, I completely glossed over that. All that Latex for nothing

Yes, or, alternatively, we can say that A's 4-velocity, expressed in global Schwarzschild coordinates, is uμ=(1/g00−−−√,0)u^\mu = (1 / \sqrt{g_{00}}, 0).

I wanted to avoid that expression because it was so similar to what the OP obtained in #1, and I thought that putting that in would simply confuse the OP further (it's also why I made all the equations explicit).

first, observe that in A's local inertial frame, his 4-velocity is (1,0)(1, 0), so O's 4-velocity at the event where they are co-located must be γu(1,−u)\gamma_u (1, - u) by a standard Lorentz transformation. Then just transform back to global coordinates, which adds the square roots of the metric coefficients in the denominators.

Yes, then the OP is correct indeed. It's just that I could've never imagined that someone would want to know the four velocity at one event.