Radial infall velocity in Schwarzschild metric
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Radial infall velocity in Schwarzschild metric
I'm having trouble reconciling something...
It seems that the velocity of a radially infalling massive particle (that started at 0 velocity from an infinite distance?) is given by [imath]v = c \sqrt{1  \tau}[/imath], where [imath]\tau = t \sqrt{1  \frac{2GM}{rc^2}}[/imath]. ex: Where [imath]r[/imath] is set to Mercury's semimajor axis, velocity is the expected value of [imath]47866[/imath]. Likewise where [imath]r[/imath] is very close to the Sun's Schwarzschild radius, velocity is the expected value of [imath]\sim c[/imath].
It just doesn't seem right to me though because the dimension of [imath]v[/imath] (as calculated above) doesn't seem to be in metres per second. Am I incorrect in assuming that the dimension of [imath]\tau[/imath] is "seconds", or is it actually dimensionless?
Am I making a silly mistake somewhere here?
Thank you for any help!
It seems that the velocity of a radially infalling massive particle (that started at 0 velocity from an infinite distance?) is given by [imath]v = c \sqrt{1  \tau}[/imath], where [imath]\tau = t \sqrt{1  \frac{2GM}{rc^2}}[/imath]. ex: Where [imath]r[/imath] is set to Mercury's semimajor axis, velocity is the expected value of [imath]47866[/imath]. Likewise where [imath]r[/imath] is very close to the Sun's Schwarzschild radius, velocity is the expected value of [imath]\sim c[/imath].
It just doesn't seem right to me though because the dimension of [imath]v[/imath] (as calculated above) doesn't seem to be in metres per second. Am I incorrect in assuming that the dimension of [imath]\tau[/imath] is "seconds", or is it actually dimensionless?
Am I making a silly mistake somewhere here?
Thank you for any help!
 gmalivuk
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Re: Radial infall velocity in Schwarzschild metric
[imath]\frac{GM}{r}[/imath] gives units of energy, which is velocity squared. To get c^{2} on the bottom, you're dividing by velocity squared, making what's under the radical dimensionless.
So where does the t come from at all, in your equation for [imath]\tau[/imath]? In other words, what value do you input for t when calculating your infall velocities at various radii?
So where does the t come from at all, in your equation for [imath]\tau[/imath]? In other words, what value do you input for t when calculating your infall velocities at various radii?

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Re: Radial infall velocity in Schwarzschild metric
Energy units have a mass component too:
1J = 1kg*m^2/s^2
It would seem the quantity below the radical should be in units of mass? o.O
EDIT: nvm, there's only one M in there. So the term under the radical is indeed unitless.
1J = 1kg*m^2/s^2
It would seem the quantity below the radical should be in units of mass? o.O
EDIT: nvm, there's only one M in there. So the term under the radical is indeed unitless.
Re: Radial infall velocity in Schwarzschild metric
From a purely qualitative viewpoint, I think it would make more sense if the equations were
[math]v = c \sqrt{1  \frac{d\tau}{dt}}[/math]
[math]d\tau =dt \sqrt{1  \frac{2GM}{rc^2}}[/math]
The units make more sense that way too. Dunno if the first is the right formula for a particle infalling from infinity; I've crunched numbers before but haven't tried to solve it analytically.
[math]v = c \sqrt{1  \frac{d\tau}{dt}}[/math]
[math]d\tau =dt \sqrt{1  \frac{2GM}{rc^2}}[/math]
The units make more sense that way too. Dunno if the first is the right formula for a particle infalling from infinity; I've crunched numbers before but haven't tried to solve it analytically.
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 gmalivuk
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Re: Radial infall velocity in Schwarzschild metric
Oh, right, tau as in proper time, no? Then yeah, the equations Goemon has make the most sense, because the metric is for those derivatives, and changes as you approach the mass. (Treating the two as proportional to each other assumes constant velocity and constant gravity.)
Re: Radial infall velocity in Schwarzschild metric
Goemon wrote:From a purely qualitative viewpoint, I think it would make more sense if the equations were
[math]v = c \sqrt{1  \frac{d\tau}{dt}}[/math]
[math]d\tau =dt \sqrt{1  \frac{2GM}{rc^2}}[/math]
The units make more sense that way too. Dunno if the first is the right formula for a particle infalling from infinity; I've crunched numbers before but haven't tried to solve it analytically.
Ah yes, I thought about this way... removing the "seconds" dimension from velocity by first dividing proper time by coordinate time. I didn't know if this was a "hack" or not, since it seemed too good to be true (a value of 1, a dimension of s, how perfect!). I'll run with it. Thank you!
Re: Radial infall velocity in Schwarzschild metric
gmalivuk wrote:[imath]\frac{GM}{r}[/imath] gives units of energy, which is velocity squared. To get c^{2} on the bottom, you're dividing by velocity squared, making what's under the radical dimensionless.
So where does the t come from at all, in your equation for [imath]\tau[/imath]? In other words, what value do you input for t when calculating your infall velocities at various radii?
Another way that I look at the dimensionless quality of this calculation is that the Schwarzschild radius is also equivalent (numerically and dimensionally) to [imath]R_S = \frac{2E}{F_{Planck}} = \frac{2GM}{c^2}[/imath], which has dimension of metres (or Joules/Newtons). Dividing that by [imath]r[/imath] then makes things dimensionless just like you mention.
In the equation t stands for the unit of coordinate time, which I assumed was necessary to give [imath]\tau[/imath] the dimension of seconds. The value of t is just 1, always.
Re: Radial infall velocity in Schwarzschild metric
Sorry, looks like my earlier comment was wrong. I don't think [imath]\tau[/imath] has anything to do with proper time; a most unfortunate use of the variable.
I get
[math]\tau = \frac{1}{1\ln(1\frac{2GM}{rc^2})}[/math]
which is dimensionless. It doesn't seem to match the equation you gave,though, unless the value of t is taken as the ratio between the two. Where did you get these equations?
Well, if it's always 1, what's the point of having it there?
I get
[math]\tau = \frac{1}{1\ln(1\frac{2GM}{rc^2})}[/math]
which is dimensionless. It doesn't seem to match the equation you gave,though, unless the value of t is taken as the ratio between the two. Where did you get these equations?
taby wrote:The value of t is just 1, always.
Well, if it's always 1, what's the point of having it there?
Life is mostly plan "B"
Re: Radial infall velocity in Schwarzschild metric
Hi Goemon,
I got the equation from the Schwarzschild metric: [imath]c^2{d\tau}^2 = c^2{dt}^2 (1  \frac{2GM}{rc^2})  ...[/imath]
For radial infall, gravitational time dilation and kinematic time dilation are the same thing, so I guess an equivalent formula to what you posted originally would be [imath]d\tau = dt \sqrt{1  \frac{v^2}{c^2}}[/imath], though [imath]dt[/imath] doesn't necessarily have to be [imath]1[/imath], I suppose. I'm sure you're probably aware of this. I just wanted to mention it for anyone else following along. Now that I look at it this way, simple algebra would allow me to calculate [imath]v[/imath]. Thank you.
The only reason to include [imath]dt = 1[/imath] would be to make the dimensions work out as they should.
I'm just messing around with calculations. I appreciate everyone's input!
I got the equation from the Schwarzschild metric: [imath]c^2{d\tau}^2 = c^2{dt}^2 (1  \frac{2GM}{rc^2})  ...[/imath]
For radial infall, gravitational time dilation and kinematic time dilation are the same thing, so I guess an equivalent formula to what you posted originally would be [imath]d\tau = dt \sqrt{1  \frac{v^2}{c^2}}[/imath], though [imath]dt[/imath] doesn't necessarily have to be [imath]1[/imath], I suppose. I'm sure you're probably aware of this. I just wanted to mention it for anyone else following along. Now that I look at it this way, simple algebra would allow me to calculate [imath]v[/imath]. Thank you.
The only reason to include [imath]dt = 1[/imath] would be to make the dimensions work out as they should.
I'm just messing around with calculations. I appreciate everyone's input!
Re: Radial infall velocity in Schwarzschild metric
Well in that case, you can check my math
I started with the relativistic version of the acceleration due to gravity,
[math]a = \frac{GM}{r^2}\frac{1}{1\frac{2GM}{rc^2}}[/math]
(which I didn't know offhand but derived from a somewhat complicated use of the metric and may have screwed up; if you want more details I can provide them). You'll notice it looks pretty much like the Newtonian version; just has the second term tacked on to prove Einstein knew something Newton didn't.
If you multiply this by m to get the force and then integrate from some starting radius R out to infinty, you get the gravitational potential energy for radius R. For an infalling particle, the kinetic energy as it crosses any radius R equals this potential energy (started at zero velocity out at infinity, all potental energy has been converted to kinetic energy). The kinetic energy for a relativistic particle with velocity v is
[math]E_k = mc^2 \left( \frac{1} {\sqrt{1\frac{v^2}{c^2}}}1 \right)[/math]
So all(!) you have to do is set the two equal and solve for v Lots of algebra which I did very quickly and sloppily, I wouldn't be surprised if I left out some minus signs or constants somewhere...
I started with the relativistic version of the acceleration due to gravity,
[math]a = \frac{GM}{r^2}\frac{1}{1\frac{2GM}{rc^2}}[/math]
(which I didn't know offhand but derived from a somewhat complicated use of the metric and may have screwed up; if you want more details I can provide them). You'll notice it looks pretty much like the Newtonian version; just has the second term tacked on to prove Einstein knew something Newton didn't.
If you multiply this by m to get the force and then integrate from some starting radius R out to infinty, you get the gravitational potential energy for radius R. For an infalling particle, the kinetic energy as it crosses any radius R equals this potential energy (started at zero velocity out at infinity, all potental energy has been converted to kinetic energy). The kinetic energy for a relativistic particle with velocity v is
[math]E_k = mc^2 \left( \frac{1} {\sqrt{1\frac{v^2}{c^2}}}1 \right)[/math]
So all(!) you have to do is set the two equal and solve for v Lots of algebra which I did very quickly and sloppily, I wouldn't be surprised if I left out some minus signs or constants somewhere...
Life is mostly plan "B"
Re: Radial infall velocity in Schwarzschild metric
This is fun. Thanks!
 gmalivuk
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Re: Radial infall velocity in Schwarzschild metric
taby wrote:The only reason to include [imath]dt = 1[/imath] would be to make the dimensions work out as they should.
But you don't need to do that, since it's dtau/dt in the other equation, rather than just tau.
Re: Radial infall velocity in Schwarzschild metric
gmalivuk wrote:taby wrote:The only reason to include [imath]dt = 1[/imath] would be to make the dimensions work out as they should.
But you don't need to do that, since it's dtau/dt in the other equation, rather than just tau.
I was referring to [imath]\tau = t \sqrt{1  \frac{2GM}{rc^2}}[/imath], which without [imath]t[/imath] would be dimensionless (an unacceptable situation). I should have been more explicit.
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Re: Radial infall velocity in Schwarzschild metric
I know, but where'd that equation come from? And why is it a problem if tau is dimensionless, anyway?
Re: Radial infall velocity in Schwarzschild metric
gmalivuk wrote:I know, but where'd that equation come from? And why is it a problem if tau is dimensionless, anyway?
I got the equation from the Schwarzschild metric: [imath]c^2{d\tau}^2 = c^2{dt}^2 (1  \frac{2GM}{rc^2})  ...[/imath]. It is my understanding that the rate of time [imath]t[/imath] reduces to [imath]\tau[/imath] due to gravitational time dilation [imath]\sqrt{1  \frac{2GM}{rc^2}}[/imath].
I was under the impression that proper time [imath]\tau[/imath] is a measurement of seconds, which is why I didn't want it to be dimensionless. This leads back to one of my earlier questions: is [imath]\tau[/imath] supposed to be dimensionless? More specifically, when [imath]c[/imath] is not set to [imath]1[/imath], should [imath]\tau[/imath] be dimensionless? Or am I missing the mark here by not using [imath]d\tau[/imath]?
This entire thread is a product of my bastardized (complete lack of) education. My apologies if it's frustrating the hell out of everyone.
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Re: Radial infall velocity in Schwarzschild metric
taby wrote:Or am I missing the mark here by not using [imath]d\tau[/imath]?
That's what I was thinking, yeah. Are you sure about [imath]v = c \sqrt{1  \tau}[/imath], or might it be as Goemon says with dtau/dt instead?
Re: Radial infall velocity in Schwarzschild metric
The tau in [imath]c^2{d\tau}^2 = c^2{dt}^2 (1  \frac{2GM}{rc^2})  ...[/imath] most definitely has units of seconds (it represents proper time), but the one in [imath]v=c\sqrt{1\tau^2}[/imath] has to be dimensionless. I don't think they could be the same variable; not sure how you'd get from one equation to the other.
[edit] You can write the velocity function for an infalling particle in the form [imath]v=c\sqrt{1f^2}[/imath] (see above), but the f in this equation is not proper time...
[edit] You can write the velocity function for an infalling particle in the form [imath]v=c\sqrt{1f^2}[/imath] (see above), but the f in this equation is not proper time...
Life is mostly plan "B"
Re: Radial infall velocity in Schwarzschild metric
Goemon wrote:The tau in [imath]c^2{d\tau}^2 = c^2{dt}^2 (1  \frac{2GM}{rc^2})  ...[/imath] most definitely has units of seconds (it represents proper time), but the one in [imath]v=c\sqrt{1\tau^2}[/imath] has to be dimensionless. I don't think they could be the same variable; not sure how you'd get from one equation to the other.
[edit] You can write the velocity function for an infalling particle in the form [imath]v=c\sqrt{1f^2}[/imath] (see above), but the f in this equation is not proper time...
What then is [imath]f[/imath], if not proper time [imath]\tau[/imath]?
I was just thinking [imath]v = c{\;} \sqrt{1  \LARGE \frac{d{\tau}}{dt}}[/imath] then?
If [imath]{d\tau} = 0[/imath], then [imath]r = R_S[/imath], [imath]v = c[/imath].
In the weakfield limit ([imath]d\tau \approx 1[/imath]), the radial infall velocity given for the semimajor axis of the planet Mercury's orbit roughly coincides with the planet's average orbit velocity.
It seems that gravitational time dilation and kinematic time dilation are one and the same for a radially infalling mass.
Oppositely for an orbiting mass, relativistic precession in the strongfield ([imath]r \approx R_S[/imath], [imath]d\tau \approx 0[/imath]) makes it impossible to sustain anything remotely close to a stable circular orbit at [imath]v \approx c[/imath]. If the mass is orbiting transversely, gravitational time dilation and kinematic time dilation are not equivalent, with gravitational time dilation accounting for 2/3 of relativistic precession, and kinematic time dilation accounting for the remaining 1/3.
Edit #928: Made this post too late at night.
Last edited by taby on Sun Oct 12, 2008 7:03 pm UTC, edited 1 time in total.
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Re: Radial infall velocity in Schwarzschild metric
No, dt and dtau don't work like that. You can't just declare that a differential has a set value. dtau/dt, on the other hand, does make sense, as it's simply the difference between the two time variables. (And it's not dtau that's close to 1 in a weak field, but dtau/dt.)
Re: Radial infall velocity in Schwarzschild metric
gmalivuk wrote:No, dt and dtau don't work like that. You can't just declare that a differential has a set value. dtau/dt, on the other hand, does make sense, as it's simply the difference between the two time variables. (And it's not dtau that's close to 1 in a weak field, but dtau/dt.)
That makes sense.
Re: Radial infall velocity in Schwarzschild metric
Ok, now I see what you're getting at, but I think you might be mixing formulas from flat and curved spacetime.
[math](1) \quad v = c \sqrt{1  \left( \frac{d\tau}{dt} \right)^2} \quad \mbox {gives the relationship between velocity and time dilation in Lorentzian spacetime;}[/math]
[math](2) \quad \frac {d\tau}{dt} = \sqrt{1  \frac{2GM}{rc^2}} \quad \mbox{ gives the rate of time dilation for an object at rest in a Schwarzchild frame (as seen by a distant observer). }[/math]
The version of the first equation that would be used in a Schwarzchild coordinate system (by a distant observer) is
[math](3)\quad v = c \left( 1  \frac{2GM}{rc^2} \right) \sqrt{1\left(\frac{d\tau}{dt}\right)^2 \frac{1}{1  \frac{2GM}{rc^2}}}[/math]
You'll notice that if you substitute (2) into (3), then (3) simply reduces to v = 0, since (2) is used to calculate the time dilation of objects which are at rest with respect to the gravitating mass. Also, for r >> M, (3) reduces to (1), and as r approaches [imath]R_s[/imath], a distant observer sees the velocity of all objects approaching zero  they're slowing to a stop because time itself is running so slowly there.
[math](1) \quad v = c \sqrt{1  \left( \frac{d\tau}{dt} \right)^2} \quad \mbox {gives the relationship between velocity and time dilation in Lorentzian spacetime;}[/math]
[math](2) \quad \frac {d\tau}{dt} = \sqrt{1  \frac{2GM}{rc^2}} \quad \mbox{ gives the rate of time dilation for an object at rest in a Schwarzchild frame (as seen by a distant observer). }[/math]
The version of the first equation that would be used in a Schwarzchild coordinate system (by a distant observer) is
[math](3)\quad v = c \left( 1  \frac{2GM}{rc^2} \right) \sqrt{1\left(\frac{d\tau}{dt}\right)^2 \frac{1}{1  \frac{2GM}{rc^2}}}[/math]
You'll notice that if you substitute (2) into (3), then (3) simply reduces to v = 0, since (2) is used to calculate the time dilation of objects which are at rest with respect to the gravitating mass. Also, for r >> M, (3) reduces to (1), and as r approaches [imath]R_s[/imath], a distant observer sees the velocity of all objects approaching zero  they're slowing to a stop because time itself is running so slowly there.
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Re: Radial infall velocity in Schwarzschild metric
Goemon wrote:Ok, now I see what you're getting at, but I think you might be mixing formulas from flat and curved spacetime.
[math](1) \quad v = c \sqrt{1  \left( \frac{d\tau}{dt} \right)^2} \quad \mbox {gives the relationship between velocity and time dilation in Lorentzian spacetime;}[/math]
[math](2) \quad \frac {d\tau}{dt} = \sqrt{1  \frac{2GM}{rc^2}} \quad \mbox{ gives the rate of time dilation for an object at rest in a Schwarzchild frame (as seen by a distant observer). }[/math]
The version of the first equation that would be used in a Schwarzchild coordinate system (by a distant observer) is
[math](3)\quad v = c \left( 1  \frac{2GM}{rc^2} \right) \sqrt{1\left(\frac{d\tau}{dt}\right)^2 \frac{1}{1  \frac{2GM}{rc^2}}}[/math]
You'll notice that if you substitute (2) into (3), then (3) simply reduces to v = 0, since (2) is used to calculate the time dilation of objects which are at rest with respect to the gravitating mass. Also, for r >> M, (3) reduces to (1), and as r approaches [imath]R_s[/imath], a distant observer sees the velocity of all objects approaching zero  they're slowing to a stop because time itself is running so slowly there.
It's not so much confusion as an identification of two identical things. Gravitational and kinematic time dilation only seem to be different when the body is moving perpendicularly to the field's gradient (orbit). If I'm falling toward the Earth from the roof of an infinitely tall building, is my time dilated to due to being within the Earth's gravitational field, or is it because I'm not at rest with respect to the centre of the Earth (due to the gradient of the Earth's gravitational field)? I'd say both, because they are the same thing in that very limited case of radial infall. It seems that only in perpendicular motion does gravitational and kinematic time dilation become separate things, which is perhaps why relativistic precession in the weakfield is proportional to 3GM, not 2GM, and that the radial infall velocity is related to 2GM, not 3GM.
Edit:..
I can see how this infall velocity is incorrect now, since it's only taking into account gravitational time dilation, in a special relativistic type of way... just like you said.
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