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Time dilation and length contraction for rotating black holes

by relativityfan
Tags: black, contraction, dilation, holes, length, rotating, time
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relativityfan
#1
Jul15-10, 12:40 AM
P: 76
hello,

how is it possible to calculate the only GRAVITATIONNAL time dilation and length contraction factor in the Kerr metric and/or the Kerr-Newman metric, for an object falling but otherwise at rest, since there is frame dragging, and the metric contains dt^2 components but also dt.dphi components?

should the time dilation factor always equal the length contraction factor?

the event horizon is for g_rr -> infinite and not g_tt=0
does that mean that g_rr should be the source of the time dilation/length contractor factor, or g_theta_theta and g_phi_phi are also involved?

please reply, this is extremely important
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relativityfan
#2
Jul15-10, 12:56 AM
P: 76
the given time dilation formula seems to work only for some approximations like the Schwarzchild metric.
I would like to know what the formula is for the Kerr metric that is requisite for most relativistic objects such as neutron stars and black holes
Jonathan Scott
#3
Jul15-10, 02:34 AM
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P: 1,159
Quote Quote by relativityfan View Post
the given time dilation formula seems to work only for some approximations like the Schwarzchild metric.
I would like to know what the formula is for the Kerr metric that is requisite for most relativistic objects such as neutron stars and black holes
If you know the metric the time dilation is [itex]\sqrt{g_{tt}}/c[/itex] as mentioned above. That is, it is simply the rate at which proper time changes with coordinate time.

Dr Chaos
#4
Jul15-10, 02:45 AM
P: 33
Time dilation and length contraction for rotating black holes

Quote Quote by relativityfan View Post
the given time dilation formula seems to work only for some approximations like the Schwarzchild metric.
I would like to know what the formula is for the Kerr metric that is requisite for most relativistic objects such as neutron stars and black holes
Yes - I would also like to have that formula.

Does nobody have a brick? I am yet to learn calculus so I suppose I shall have to ask a maths teacher how to do this next week...
relativityfan
#5
Jul15-10, 02:56 AM
P: 76
Quote Quote by Jonathan Scott View Post
If you know the metric the time dilation is [itex]\sqrt{g_{tt}}/c[/itex] as mentioned above. That is, it is simply the rate at which proper time changes with coordinate time.
well this works for metrics like the Reissner Nordstrom metric, but only for static metrics, from what I understand:
in the case of the Kerr metric, there is frame dragging, and there is not only dt^2 in the metric but dt.dphi too

furthermore, at the event horizon of a Kerr black hole, g_rr -> infinite often without g_tt =0
this means that space is infinitely contracted but time is not infinitely dilated, if only g_tt provides the time dilation.
in such case the factor of the time dilation would not be the factor of the length contraction.

in other words, the event horizon is not always the ergosphere
George Jones
#6
Jul15-10, 07:33 AM
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Relativityfan, what do you mean by "time dilation factor" and "length contraction factor"?

Outside the event horizon, the Kerr solution is stationary, but not static. In this case, coordinates exist such that all the components of metric are independent of [itex]x^0[/itex],

[tex]\frac{\partial g_{\mu \nu}}{\partial x^0} = 0.[/tex]

Suppose such coordinates are used, and suppose that light is emitted at event [itex]p[/itex] and received at event [itex]q[/itex]. Then, the relationship between the emitted wavelength as measured at [itex]p[/itex] by an observer with unchanging spatial coordinates and received wavelength as measured at [itex]q[/itex] by an observer with unchanging spatial coordinates is

[tex]\frac{\lambda_q}{\lambda_p} = \sqrt{\frac{g_{00} \left( q \right)}{g_{00} \left( p \right)}} .[/tex]

This exact result is true for [itex]p[/itex] and [itex]q[/itex] outside the ergoshere, since it is only outside the ergosphere that observers can have fixed spatial coordinates.

A completely general that is true inside ergospheres and even in non-stationary spacetimes:

[tex]\frac{\lambda_q}{\lambda_p} = \frac{g \left( \mathbf{p}_p , \mathbf{u}_p \right)}{g \left( \mathbf{p}_q , \mathbf{u}_q \right)} .[/tex]

Here, [itex]\mathbf{p}_p[/itex] is the 4-momentum of the emitted light, [itex]\mathbf{p}_q[/itex] is the 4-momentum of the received light, [itex]\mathbf{u}_p[/itex] is the 4-velocity of the observer who emits the light, and [itex]\mathbf{u}_q[/itex] is the 4-velocity of the observer who receives the light.
relativityfan
#7
Jul15-10, 08:08 AM
P: 76
thank you, I mean the classical Lorentz factor due only to the gravitationnal field and not to the momentum of the object falling inside of the field, relative to an inertial observer at rest(r-> infinite)


I do not understand the meaning of the formula and how to use it
relativityfan
#8
Jul15-10, 12:01 PM
P: 76
I would be grateful if anyone could provide this formula for rotating objects because it is essential for the genuine relativistic objects. Every neutron star or black hole necessarily rotates relativistically, so the formula providing the Lorentz factor is crucial.
starthaus
#9
Jul15-10, 12:51 PM
P: 1,568
Quote Quote by relativityfan View Post
I would be grateful if anyone could provide this formula for rotating objects because it is essential for the genuine relativistic objects. Every neutron star or black hole necessarily rotates relativistically, so the formula providing the Lorentz factor is crucial.
Start with the Kerr metric:

[tex]
c^{2} d\tau^{2} =
\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2} dt^{2}
- \frac{\rho^{2}}{\Delta} dr^{2}
- \rho^{2} d\theta^{2} -
\left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \ d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \, dt \, d\phi
[/tex]

General case:

divide both sides by [tex]dt^2[/tex]
relativityfan
#10
Jul15-10, 03:42 PM
P: 76
thank you, but the above formula seems to work only for the coordinates theta=0 or theta=pi.

for theta=pi/2, this cannot work because the ergosphere is not the same as the event horizon: once again, at the event horizon, the previously given formula for the the time dilation would not match the fact that space is infinitely contracted, for theta=pi/2

I believe that g_rr should be used as the length contraction to determine the Lorentz factor and therefore the time dilation, am i right?
starthaus
#11
Jul15-10, 03:47 PM
P: 1,568
Quote Quote by relativityfan View Post
thank you, but the above formula seems to work only for the coordinates theta=0 or theta=pi.
Err, "Make [tex]d\theta=d\phi=dr=0[/tex] (stationary observer)"
starthaus
#12
Jul15-10, 03:50 PM
P: 1,568
Quote Quote by starthaus View Post
Start with the Kerr metric:

[tex] ds^{2} =
\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2} dt^{2}
- \frac{\rho^{2}}{\Delta} dr^{2}
- \rho^{2} d\theta^{2} -
\left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \ d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \, dt \, d\phi
[/tex]

Make [tex]d\theta=d\phi=dr=0[/tex] (stationary observer)

You get the time dilation immediately:

[tex]\frac{d\tau}{dt}=\sqrt{1 - \frac{r_{s} r}{\rho^{2}}}[/tex]

If you want the time dilation between an observer and an emitter, you can use the above to recover the formula given above by George Jones:

[tex]\frac{d\tau_a}{d\tau_b}=\sqrt{1 - \frac{r_{s} r_a}{\rho_a^{2}}}/ \sqrt{1 - \frac{r_{s} r_b}{\rho_b^{2}}} [/tex]
Now, to get length contraction along the radial direction, make [tex]d\theta=d\phi=0[/tex]:

[tex]
ds^{2} =
\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2} dt^{2}
- \frac{\rho^{2}}{\Delta} dr^{2} [/tex]
starthaus
#13
Jul15-10, 04:04 PM
P: 1,568
Quote Quote by relativityfan View Post

I believe that g_rr should be used as the length contraction
Yes, see post above.

and therefore the time dilation, am i right?
No, time dilation is a function of [tex]g_{tt}[/tex] (see also above)
relativityfan
#14
Jul15-10, 05:09 PM
P: 76
well,

dtheta=dphi=dr=0 is only possible when the object falling is not at rest , i.e, if it tries to travel in the opposite sense of the frame dragging.
with only the gravitationnal field and no other momentum, then dtheta=dphi=dr=0 is not possible near the black hole.
how to get the time dilation for theta different than 0 or pi, and dtheta=dphi=dr different than 0?
this would be the general case and not only a special case
I need to know the general case of time dilation for a rotating black hole
starthaus
#15
Jul15-10, 05:23 PM
P: 1,568
Quote Quote by relativityfan View Post
well,

dtheta=dphi=dr=0 is only possible when the object falling is not at rest , i.e, if it tries to travel in the opposite sense of the frame dragging.
with only the gravitationnal field and no other momentum, then dtheta=dphi=dr=0 is not possible near the black hole.
how to get the time dilation for theta different than 0 or pi, and dtheta=dphi=dr different than 0?
this would be the general case and not only a special case
I need to know the general case of time dilation for a rotating black hole
if it is the general case you want, then look at the edited post 9
relativityfan
#16
Jul15-10, 06:29 PM
P: 76
thank you, the purpose of this post is about only a gravitationnal field without any other momentum.
this is therefore not really the general case: we should deduce from this hypothesis constraints from theta, phi, and then deduce the Lorentz factor.
I thought that for any metric:
1) if time is dilated by k then space has to be contracted by k
2) the length contraction occurs only for one dimension: the direction of movement

do you agree with this?
could anyone determine what the Lorentz factor is from these hypotheses?
George Jones
#17
Jul16-10, 07:36 AM
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P: 6,246
Quote Quote by relativityfan View Post
I need to know the general case of time dilation for a rotating black hole
Is there any particular reason that you "need to know"?
relativityfan
#18
Jul16-10, 09:59 AM
P: 76
well it is for the understanding of the structure of a black hole. for example, I still dont know if time stops at the event horizon of a rotating black hole for a distant observer...


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