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ravens_cry
2011-May-21, 07:44 AM
The title is pretty much the question. If the object you were falling to was massive enough, and the vacuum surrounding you was hard enough, could you fall faster than light? I don't know why I think you could and my naive and analogy based understanding of relativity says no, but for some reason my brain is wondering if this could somehow be a special case.

swampyankee
2011-May-21, 11:16 AM
No. Objects with mass cannot be accelerated to the speed of light, let alone beyond it.

Jeff Root
2011-May-21, 12:46 PM
Actually, yes, in a way. You need to fall into a black hole.
If you start free-falling into a black hole from rest at infinite
distance (or with a bit of initial speed from a close distance),
you will reach the speed of light relative to the location of
the black hole's event horizon as you pass that location.
Your speed will continue to increase after that.

However, you will not be able to measure your speed as
being the speed of light or faster relative to anything you
can observe, and no other observer will be able to measure
your speed as being the speed of light or faster relative to
them. The gravity of a black hole is so strong that
measurements over even fairly short distances are not
"local". Special relativity says that local measurements
of speed will always be less than the speed of light. But
in effect, the only measurements anyone can make are
local measurements. The best you can do is make a local
measurement of your "proper speed" relative to some
landmark, and then calculate your "coordinate speed"
relative to some other landmark that is not local to you.
When falling into a black hole, that coordinate speed may
be faster than the speed of light.

Similarly, because of the cosmic expansion, our coordinate
speed relative to very distant galaxies is greater than the
speed of light.

-- Jeff, in Minneapolis

macaw
2011-May-21, 01:28 PM
Actually, yes, in a way. You need to fall into a black hole.
If you start free-falling into a black hole from rest at infinite
distance (or with a bit of initial speed from a close distance),
you will reach the speed of light relative to the location of
the black hole's event horizon as you pass that location.
Your speed will continue to increase after that.

However, you will not be able to measure your speed as
being the speed of light or faster relative to anything you
can observe, and no other observer will be able to measure
your speed as being the speed of light or faster relative to
them. The gravity of a black hole is so strong that
measurements over even fairly short distances are not
"local". Special relativity says that local measurements
of speed will always be less than the speed of light. But
in effect, the only measurements anyone can make are
local measurements. The best you can do is make a local
measurement of your "proper speed" relative to some
landmark, and then calculate your "coordinate speed"
relative to some other landmark that is not local to you.
When falling into a black hole, that coordinate speed may
be faster than the speed of light.

Similarly, because of the cosmic expansion, our coordinate
speed relative to very distant galaxies is greater than the
speed of light.

-- Jeff, in Minneapolis

Hmm, the above seems incorrect.
1. The proper speed (what you call the "local speed) is:

dr/d\tau=c \sqrt{r_s/r}

where r_s is the Schwarzschild radius (the location of the EH). When r->r_s , dr/d\tau->c without ever reaching c.

2.The coordinate speed is:

dr/dt=c(1-r_s/r)\sqrt{r_s/r} . When r->r_s , dr/dt->0

caveman1917
2011-May-21, 01:42 PM
Hmm, the above seems incorrect.
1. The proper speed (what you call the "local speed) is:

dr/d\tau=c \sqrt{r_s/r}

where r_s is the Schwarzschild radius (the location of the EH). When r->r_s , dr/d\tau->c without ever reaching c.

2.The coordinate speed is:

dr/dt=c(1-r_s/r)\sqrt{r_s/r} . When r->r_s , dr/dt->0

It works in Gullstrand-Painlevé coordinates, there you will have no singularity at rs and you have dr/dt_r = \sqrt{r_s/r} with r < rs

ETA: though technically it's still not faster than the speed of light, since the speed of light as measured from a falling GP frame is 1 + \sqrt{r_s/r}, the difference will always be 1.

Jeff Root
2011-May-21, 02:14 PM
1. The proper speed (what you call the "local speed) is:
Actually I did call it "proper speed". :)

dr/d\tau=c \sqrt{r_s/r}
where r_s is the Schwarzschild radius (the location of the EH).
When r->r_s , dr/d\tau->c without ever reaching c.
It looks to me like it reaches c at the Schwarzschild radius.

2.The coordinate speed is:

dr/dt=c(1-r_s/r)\sqrt{r_s/r} . When r->r_s , dr/dt->0
Yes, the coordinate speed in Schwarzschild coordinates
approaches zero. I did say "When falling into a black hole, that
coordinate speed may be faster than the speed of light."
It depends on the coordinate system, naturally.

-- Jeff, in Minneapolis

macaw
2011-May-21, 02:19 PM
Actually I did call it "proper speed". :)

It looks to me like it reaches c at the Schwarzschild radius.

If the object falls from infinity. Not a practical situation.
If the object falls from r_0 then it will NOT reach c. Indeed, in this case:

dr/d\tau=c \sqrt{r_s/r-r_s/r_0}

Yes, the coordinate speed in Schwarzschild coordinates
approaches zero. I did say "When falling into a black hole, that
coordinate speed may be faster than the speed of light."

...which is clearly incorrect.

macaw
2011-May-21, 02:21 PM
The title is pretty much the question. If the object you were falling to was massive enough, and the vacuum surrounding you was hard enough, could you fall faster than light? I don't know why I think you could and my naive and analogy based understanding of relativity says no, but for some reason my brain is wondering if this could somehow be a special case.

The math says "no"

Jeff Root
2011-May-21, 03:05 PM
If the object falls from infinity. Not a practical situation.
If the object falls from r_0 then it will NOT reach c.
That is only if the objects free-falls starting from rest. Give it
a little initial shove downward and its speed will be higher than
the speed it would have free-falling from infinity all the way
down. Although in Schwarzschild coordinates, of course, its
speed will still fall to zero at the event horizon.

Indeed, in this case:

dr/d\tau=c \sqrt{r_s/r-r_s/r_0}
That's for a free-faller with zero initial speed, isn't it? See what
happens when you have an initial downward speed.

-- Jeff, in Minneapolis

macaw
2011-May-21, 03:48 PM
That is only if the objects free-falls starting from rest. Give it
a little initial shove downward and its speed will be higher than
the speed it would have free-falling from infinity all the way
down. Although in Schwarzschild coordinates, of course, its
speed will still fall to zero at the event horizon.

This is false as well. If you think this way, please provide the formula that supports the above.

That's for a free-faller with zero initial speed, isn't it? See what
happens when you have an initial downward speed.

Nothing different, the object will not reach c. If you think otherwise, please provide the formula supporting your claim. A few threads earlier I corrected grav on the same exact mistake. An easy, intuitive way to show why your claim is false is to view throwing the object from r_0 with initial speed v_0 as dropping it from r_1>r_0 with initial speed v_1=0. The result is:

dr/d\tau=c \sqrt{r_s/r-r_s/r_1}

which is clearly smaller than c. No matter how hard you try, a massive object cannot reach c, whether you are applying SR or GR.

macaw
2011-May-21, 05:18 PM
It works in Gullstrand-Painlevé coordinates, there you will have no singularity at rs and you have dr/dt_r = \sqrt{r_s/r} with r < rs

You mean r>r_s, right?

ETA: though technically it's still not faster than the speed of light, since the speed of light as measured from a falling GP frame is 1 + \sqrt{r_s/r}, the difference will always be 1.

Yes, absolutely, the proper speed of a massive object can never reach c. Interesting way of putting it.

caveman1917
2011-May-21, 05:18 PM
Strictly speaking it does reach >c inside the horizon, but so does the speed of light itself. The main point is that it will never be faster than the speed of light as seen from the same frame, which will always remain c. So even though the numbers come out as >c, it's still slower than light as it should be.

macaw
2011-May-21, 05:23 PM
Strictly speaking it does reach >c inside the horizon, but so does the speed of light itself. The main point is that it will never be faster than the speed of light as seen from the same frame, which will always remain c. So even though the numbers come out as >c, it's still slower than light as it should be.

I have not calculated the speed of light/object inside the EH. Requires switching from the external to the internal metric and redoing the calculations. Never seen it done but I think the exercise is pretty trivial. I am doubtful that the speed of light is anything but c . Otherwise all the metrics would fall apart.

caveman1917
2011-May-21, 05:27 PM
You mean r>r_s, right?

No, r < r_s.
Transform dt_r = dt - \beta \gamma^2 dr with \beta = -\sqrt{2M/r} and gamma is the standard gamma function.

Plug into the schwarzschild form of the metric and you'll have eliminated the coordinate singularity at rs. Using this mixed form you can make global expressions across the event horizon.

Proper speed will remain dr/dt_r = \sqrt{r_s/r} but you can extend this to r < rs. (c=1)
It will give you a value of >1 inside the event horizon, but that's not a problem since by the same transform light itself will have a value >1 inside the event horizon. It turns out that as taken from the same infalling frame, the speed of light is still just 1 and causality is preserved. So it doesn't go faster than light, even though its value is >c, but that's just how the numbers come out with the mixed form of the metric, there's nothing physically fishy going on.

macaw
2011-May-21, 05:30 PM
No, r < r_s.

I am calculating the drop from above the EH, when the object approaches the EH from above. This is what the OP asked.

caveman1917
2011-May-21, 05:32 PM
It's the Gullstrand-Painlevé form of the metric, see here (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates). It's an interesting "trick" since you can make global evaluations without switching metrics.

macaw
2011-May-21, 05:34 PM
No, r < r_s.
Transform dt_r = dt - \beta \gamma^2 dr with \beta = -\sqrt{2M/r} and gamma is the standard gamma function.

Plug into the schwarzschild form of the metric and you'll have eliminated the coordinate singularity at rs. Using this mixed form you can make global expressions across the event horizon.

Proper speed will remain dr/dt_r = \sqrt{r_s/r} but you can extend this to r < rs. (c=1)
It will give you a value of >1 inside the event horizon, but that's not a problem since by the same transform light itself will have a value >1 inside the event horizon. It turns out that as taken from the same infalling frame, the speed of light is still just 1 and causality is preserved. So it doesn't go faster than light, even though its value is >c, but that's just how the numbers come out with the mixed form of the metric, there's nothing physically fishy going on.

Ok, I see, you are working in "rain" coordinates (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Rain_ coordinates).

caveman1917
2011-May-21, 05:34 PM
I am calculating the drop from above the EH, when the object approaches the EH from above. This is what the OP asked.

Yes, it starts out at infinity as usual, but the mixed form allows you to continue past the event horizon, which is what i was doing since i think that was what Jeff was referring to. But it's still slower than light, even though the value is >c.

macaw
2011-May-21, 05:39 PM
Yes, it starts out at infinity as usual, but the mixed form allows you to continue past the event horizon, which is what i was doing since i think that was what Jeff was referring to. But it's still slower than light, even though the value is >c.

Yes, I understood what you are doing (this is not was Jeff was doing,his claims are about reaching the EH from outside the BH). I don't care much for GP coordinates, this is just a mathematical sleigh of hand. From what I understand, even Taylor and Wheeler dropped the "rain" coordinates from their latest edition.
I don't care much for this "the speed of light becomes larger than c inside the EH" (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light). It is pretty clear why I don't care for it.
The GP formalism might be interesting if someone decided to re-enact Pound-Rebka in a tower that has a few levels below the ground while the top is above the ground.

caveman1917
2011-May-21, 10:29 PM
Yes, I understood what you are doing (this is not was Jeff was doing,his claims are about reaching the EH from outside the BH). I don't care much for GP coordinates, this is just a mathematical sleigh of hand. From what I understand, even Taylor and Wheeler dropped the "rain" coordinates from their latest edition.
I don't care much for this "the speed of light becomes larger than c inside the EH" (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light). It is pretty clear why I don't care for it.
The GP formalism might be interesting if someone decided to re-enact Pound-Rebka in a tower that has a few levels below the ground while the top is above the ground.

Yes it's indeed a mathematical sleigh of hand, but this aspect of Jeff's explanation

If you start free-falling into a black hole from rest at infinite
distance (or with a bit of initial speed from a close distance),
you will reach the speed of light relative to the location of
the black hole's event horizon as you pass that location.
Your speed will continue to increase after that.

However, you will not be able to measure your speed as
being the speed of light or faster relative to anything you
can observe

looked to me as a description from the GP form.

I'd say GP coordinates are quite easy to use for global considerations (you don't need to switch between external and internal metrics), but one should of course remember being in a mixed form and having 'funny' stuff such as lightspeed at >c because of that.

macaw
2011-May-21, 10:54 PM
but one should of course remember being in a mixed form and having 'funny' stuff such as lightspeed at >c because of that.

actually, it is much worse than that. The speed in cause is not proper speed, it is the so-called "rain" speed. It JUMPS from:

1. dr/dt_r=1-\sqrt{r_s/r} for r>r_s

to:

2. dr/dt_r=-(1+\sqrt{r_s/r}) for r<r_s

This is totally unphysical.

caveman1917
2011-May-21, 11:00 PM
actually, it is much worse than that. The speed in cause is not proper speed, it is the so-called "rain" speed.

Yes, i should have specified that it's not the proper speed. On the other hand it's obvious by only doing "half a transform".

It JUMPS from:

1. dr/dt_r=1-\sqrt{r_s/r} for r>r_s

to:

2. dr/dt_r=-(1+\sqrt{r_s/r}) for r<r_s

This is totally unphysical.

That doesn't seem right. Are you doing the speed of the GP infalling frame or the speed of light in GP coordinates?

ETA: the reason you're getting that is because you're using light speed going outwards in the first instance and light speed going inwards in the second one.

macaw
2011-May-21, 11:02 PM
Yes, i should have specified that it's not the proper speed. On the other hand it's obvious by only doing "half a transform".

That doesn't seem right. Are you doing the speed of the GP infalling frame or the speed of light in GP coordinates?

Speed of light. It is trivial, see here (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light)

caveman1917
2011-May-21, 11:06 PM
Speed of light. It is trivial, see here (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light)

See my ETA in the previous post.

macaw
2011-May-21, 11:08 PM
See my ETA in the previous post.

Doesn't matter, the jump is still there, from 1-\sqrt{r_s/r} to 1+\sqrt{r_s/r}. To make matters worse, at the EH the jump is from 0 to 2.

caveman1917
2011-May-21, 11:11 PM
Doesn't matter, the jump is still there, from 1-\sqrt{r_s/r} to 1+\sqrt{r_s/r}. To make matters worse, at the EH the jump is from 0 to 2.

Where are you getting that from?
Light speed going outwards 1 - \sqrt{r_s/r}.
Light speed going inwards -1 - \sqrt{r_s/r}.
Both functions are continuous everywhere (except r=0).

ETA: you're switching between outwards and inwards going light at r_s, therefore getting the jump.

macaw
2011-May-21, 11:13 PM
Where are you getting that from?
Light speed going outwards 1 - \sqrt{r_s/r}.
Light speed going inwards -1 - \sqrt{r_s/r}.
Both functions are continuous everywhere (except r=0).

Come on, this is trivial, at EH r=r_s, so, one expression evaluates to 0, the other one to -2. There is a discontinuity at the EH.

ETA: you're switching between outwards and inwards going light at r_s, therefore getting the jump.

Nope. If you prefer, the jump is from 0 to +2. Same difference.

caveman1917
2011-May-21, 11:17 PM
Come on, this is trivial, at EH r=r_s, so, one expression evaluates to 0, the other one to -2. There is a discontinuity at the EH.

Outwards going light has speed 0 at the EH, inwards going light has -2. There is no discontinuity, you're comparing light going in different directions.
If i'm in minkowski and emit a photon to -x, it's velocity will be -1, a photon to +x will be +1. That doesn't mean there's a discontinuity, it just means i'm comparing velocities in different directions.

The sum of the absolute values of the two is 2, you're showing the consistency of the coordinates, not the inconsistency.

ETA: correction, it's not the sum of the absolute values that should be 2, but the difference.

macaw
2011-May-21, 11:19 PM
Outwards going light has speed 0 at the EH, inwards going light has -2. There is no discontinuity, you're comparing light going in different directions.

I give up, look at the absolute values of the term dr/dt_r. Compare 0 with +2, if it makes it easier for you. Your comparison with the case +1 , -1 does not hold.

caveman1917
2011-May-21, 11:27 PM
Let's do outwards and inwards seperately.
Inwards going light -1 - \sqrt{r_s/r}.
For r = 1.1 * r_s: -1.95
For r = r_s: -2
For r = 0.9 * r_s: -2.05

Similar for outwards going light.
There is no discontinuity.

macaw
2011-May-21, 11:32 PM
Let's do outwards and inwards seperately.
Inwards going light -1 - \sqrt{r_s/r}.
For r = 1.1 * r_s: -1.95
For r = r_s: -2
For r = 0.9 * r_s: -2.05

Similar for outwards going light.
There is no discontinuity.

But this is NOT what the wiki page (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light) says. It says clearly that

dr/dt_r=1-\sqrt{r_s/r} for r>r_s

dr/dt_r=0 for r=r_s

dr/dt_r=-(1+\sqrt{r_s/r}) for r<r_s

This is NOT what you are calculating above. You are using the SAME function on BOTH sides of the EH. Of course you get continuity this way.

Now, even if the wiki page is wrong and , in reality , the "rain" speed has nothing to do with the EH, so we interpret the two speeds as the anisotropic light speed, at the EH the anisotropy is very strong, 1c! Still unphysical, no matter how you look at it.

caveman1917
2011-May-21, 11:41 PM
But this is NOT what the wiki page (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light) says. It says clearly that

dr/dt_r=1-\sqrt{r_s/r} for r>r_s

dr/dt_r=0 for r=r_s

dr/dt_r=-(1+\sqrt{r_s/r}) for r<r_s

Where exactly does it say that?
They're just giving examples of different speeds, and confusingly using outwards light in one case and inwards in the other.
The first formula is for outwards light, the last is for inwards. Use r \rightarrow \infty and you'll find +1 or -1, which is the minkowski example i gave. The two formulas are the distinction between inwards and outwards, not inside or outside the horizon.

This is NOT what you are calculating above. You are using the SAME function on BOTH sides of the EH. Of course you get continuity this way.

That's the correct way. The wiki article is just confusing with its examples.

caveman1917
2011-May-22, 12:08 AM
Now, even if the wiki page is wrong

I wouldn't say wrong since they do state "light moving outwards" and "light that moves towards the center", but unclear certainly.

and , in reality , the "rain" speed has nothing to do with the EH, so we interpret the two speeds as the anisotropic light speed, at the EH the anisotropy is very strong, 1c! Still unphysical, no matter how you look at it.

The velocity of the GP frame itself is -\sqrt{r_s/r}.
Ergo the velocity of light moving inwards as seen from the GP frame is -1, for light moving outwards it's +1. The same as minkowski, nothing wrong there.

The coordinates are perfectly consistent, it just doesn't work like standard coordinate systems since it's a mixed form.

grav
2011-May-22, 01:16 AM
That is only if the objects free-falls starting from rest. Give it
a little initial shove downward and its speed will be higher than
the speed it would have free-falling from infinity all the way
down. Although in Schwarzschild coordinates, of course, its
speed will still fall to zero at the event horizon.

That's for a free-faller with zero initial speed, isn't it? See what
happens when you have an initial downward speed.A few threads ago, I posted this (http://www.bautforum.com/showthread.php/114621-Time-Dilation-Approaching-the-Event-Horizon?p=1879271#post1879271).

...the local speed measured at r for an object dropped from r0 [is]

v_local = c sqrt[(R/r - R/r0) / (1 - R/r0)]

For a distant observer, that becomes

v_distant = c (1 - R/r) sqrt[(R/r - R/r0) / (1 - R/r0)]

Okay, so let's say an object is dropped at r0. The speed the distant observer will measure when the object reaches r1 is

v1 = c (1 - R/r1) sqrt[(R/r1 - R/r0) / (1 - R/r0)]

The object will continue to fall until it reaches r2 with a speed of

v2 = c (1 - R/r2) sqrt[(R/r2 - R/r0) / (1 - R/r0)]

Now, as the object passes r1 with a speed of v1, if another object were thrown down at r1 with a speed of v1 at the same instant that the first object passes, then both objects will reach r2 with a speed of v2 simultaneously, so dropping an object from r0 is equivalent to throwing the object down with an initial speed of v1 at r1. So taking the equation for v1 and solving for R/r0, we have

v1 = c (1 - R/r1) sqrt[(R/r1 - R/r0) / (1 - R/r0)]

(v1/c)^2 (1 - R/r0) = (1 - R/r1)^2 (R/r1 - R/r0)

(R/r0) [(1 - R/r1)^2 - (v1/c)^2)] = (R/r1) (1 - R/r1)^2 - (v1/c)^2

R/r0 = [(R/r1) (1 - R/r1)^2 - (v1/c)^2] / [(1 - R/r1)^2 - (v1/c)^2]

and substituting that into the equation for v2, we have

v2 = c sqrt(R/r2 - [(R/r1) (1 - R/r1)^2 - (v1/c)^2] / [(1 - R/r1)^2 - (v1/c)^2])
/ sqrt(1 - [(R/r1) (1 - R/r1)^2 - (v1/c)^2] / [(1 - R/r1)^2 - (v1/c)^2])

which now gives us the final speed v2 measured at r2 by a distant observer for an object thrown down at r1 with an initial speed of v1. I notice I forgot to bring down the (1 - R/r2) part from the original equation for v2 for what the distant observer measures, so the resulting v2 is the locally measured speed, while the distant observer would measure

v2 = c (1 - R/r2) sqrt(R/r2 - [(R/r1) (1 - R/r1)^2 - (v1/c)^2] / [(1 - R/r1)^2 - (v1/c)^2])
/ sqrt(1 - [(R/r1) (1 - R/r1)^2 - (v1/c)^2] / [(1 - R/r1)^2 - (v1/c)^2])

This resulting equation is far from elegant, too many parentheses, so rather than working through that when plugging in the numbers, I would recommend solving first for R/r0 in the second to last equation, then plugging the result for R/r0 into the original equation for v2 in the center, giving the speed as measured by a distant observer, or using just v2 = c sqrt[(R/r2 - R/r0) / (1 - R/r0)] for the locally measured speed.

macaw
2011-May-22, 01:23 AM
Where exactly does it say that?

Here (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light).

They're just giving examples of different speeds, and confusingly using outwards light in one case and inwards in the other.
The first formula is for outwards light, the last is for inwards. Use r \rightarrow \infty and you'll find +1 or -1,

We aren't talking about r \rightarrow \infty, we are talking about the vicinity of the EH, i.e. r=r_s. This is where the discontinuity occurs. I think this is the 4-th time I am pointing it out.

That's the correct way. The wiki article is just confusing with its examples.

I never pay much attention to wiki anyways. The problem still stands. See above.

macaw
2011-May-22, 01:30 AM
The velocity of the GP frame itself is -\sqrt{r_s/r}.
Ergo the velocity of light moving inwards as seen from the GP frame is -1, for light moving outwards it's +1. The same as minkowski, nothing wrong there.

This looks more plausible but still unsatisfactory. The speed of light is dr/dt_r=1+\sqrt{r_s/r}. Nothing entitles you to subtract the speed of the rain frame \sqrt{r_s/r}.

If you look at the Schwarzschild (reduced) metric:

ds^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)

and you try the same exercise, i.e. determining the light speed, then things are very straightforward:

0=(1-r_s/r)dt^2-dr^2/(1-r_s/r)

therefore:

dr/dt=+(1-r_s/r)
dr/dt=-(1-r_s/r)

cjl
2011-May-22, 01:41 AM
Nothing different, the object will not reach c. If you think otherwise, please provide the formula supporting your claim. A few threads earlier I corrected grav on the same exact mistake. An easy, intuitive way to show why your claim is false is to view throwing the object from r_0 with initial speed v_0 as dropping it from r_1>r_0 with initial speed v_1=0. The result is:

dr/d\tau=c \sqrt{r_s/r-r_s/r_1}

which is clearly smaller than c. No matter how hard you try, a massive object cannot reach c, whether you are applying SR or GR.

But this is not always true. Throwing the object from r0 with an initial speed v0 can only be replicated by dropping it from r1 > r0 if v0 < vescape at r0. If an object is thrown from r0 with v0 > vescape @ r0, then that situation could not be replicated by a stationary drop from any height.

macaw
2011-May-22, 01:50 AM
But this is not always true. Throwing the object from r0 with an initial speed v0 can only be replicated by dropping it from r1 > r0 if v0 < vescape at r0. If an object is thrown from r0 with v0 > vescape @ r0, then that situation could not be replicated by a stationary drop from any height.

Interesting point. There exists r_1=\frac{r_s}{r_s/r_0-(v_0/c)^2} such that dropping the object from r_1 is equivalent to throwing it from r_0<r_1 with velocity v_0. You can easily show that r_1>r_0.
The condition for r_1 to be greater than 0 is that r_s/r_0>(v_0/c)^2, i.e v_0<c \sqrt{r_s/r_0} . Is this what you are referring to?

I hope that you agree that, if v_0>c \sqrt{r_s/r_0}, it doesn't follow that the object reaches a proper speed higher than c, anywhere during its fall. A different approach needs to be used (different from replacing the throw from r_0 with the drop from r_1>r_0).

cjl
2011-May-22, 04:08 AM
Yes, that's the case I was referring to. I also agree that it doesn't necessarily imply that the object reaches a proper speed >c anywhere, but it does require a different analysis than the one already performed.

macaw
2011-May-22, 04:18 AM
Yes, that's the case I was referring to. I also agree that it doesn't necessarily imply that the object reaches a proper speed >c anywhere, but it does require a different analysis than the one already performed.

Do you know a simple treatment for the case v_0>c \sqrt{r_s/r_0} ?

cjl
2011-May-22, 04:44 AM
No, I don't unfortunately. I would be interested to see the results though.

macaw
2011-May-22, 04:46 AM
No, I don't unfortunately. I would be interested to see the results though.

I have never seen it done, actually everything on this subject I've done it myself, the books treat only the case v_0=0.

grav
2011-May-22, 05:27 AM
But this is not always true. Throwing the object from r0 with an initial speed v0 can only be replicated by dropping it from r1 > r0 if v0 < vescape at r0. If an object is thrown from r0 with v0 > vescape @ r0, then that situation could not be replicated by a stationary drop from any height.Strictly speaking, that is true, and that is a good observation, but I see no reason it couldn't be applied to the more general case using the same formula, the only real limit for the local speed for an object thrown down being c according to a stationary observer there. We would need a more rigid proof to show it definitively, though, perhaps something dealing with the energy of the object, maybe having something to do with the energy being lost while escaping with a speed v_loc > c sqrt(r_s / r), whereas of course the same energy would also be gained while falling over the same distance.

caveman1917
2011-May-22, 10:13 AM
Here (http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates#Speed s_of_light).

It doesn't say that, it's using two examples with different directions. (my bold)

At the event horizon, r=2M\,\! , the speed of light shining outward away from the center of black hole is

Inside the event horizon, r < 2M, the rain observer measures that the light moves toward the center with speed greater than 2

It's just using confusing examples, there is no discontinuity.

We aren't talking about r \rightarrow \infty, we are talking about the vicinity of the EH, i.e. r=r_s.

No, that is me showing you a way to see your mistake.

ETA: let's go into this. reductio ad absurdum.

You're saying that dr/dt_r = 1 - \sqrt{r_s/r} is only valid outside the EH.

Therefor if we take the limit r \rightarrow \infty we get only +1.
So according to you, in the asymptotic minkowski limit we can only have light moving in one direction? Light moving in the other direction has an undefined velocity?
Your initial premise is incorrect, it has nothing to do with outside or inside the EH, but about the direction. It's the velocity of light, not the speed. Therefor there's the +1 or -1. The speed is just |v| = 1.

This is where the discontinuity occurs. I think this is the 4-th time I am pointing it out.

And for the 4th time, there is no discontinuity, only in a faulty reading of the wiki page.

I never pay much attention to wiki anyways. The problem still stands. See above.

If you disagree, please show another reference where the problem occurs other than a badly phrased wiki page.

caveman1917
2011-May-22, 10:50 AM
This looks more plausible but still unsatisfactory. The speed of light is dr/dt_r=1+\sqrt{r_s/r}.

For an accelerating observer, of course there's an anisotropy there in that case. If you want to point to a non-physical anisotropy you need to consider things from an inertial (ie freefalling) frame, which is what...

Nothing entitles you to subtract the speed of the rain frame \sqrt{r_s/r}.

...entitles me to substract.
As seen from the inertial frame the speed of light is +1 or -1. There is nothing unphysical.

If you look at the Schwarzschild (reduced) metric:

ds^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)

and you try the same exercise, i.e. determining the light speed, then things are very straightforward:

0=(1-r_s/r)dt^2-dr^2/(1-r_s/r)

therefore:

dr/dt=+(1-r_s/r)
dr/dt=-(1-r_s/r)

Why don't we do it in the relevant GP form? For radial light

0=(dr + (1 + \sqrt{r_s/r})dt_r)(dr - (1 - \sqrt{r_s/r})dt_r)

Giving
dr/dt_r = +1 - \sqrt{r_s/r}
dr/dt_r = -1 - \sqrt{r_s/r}

for an accelerating shell observer.

To get the speed of light for an inertial free fall observer, substract
dr/dt_r = - \sqrt{r_s/r} and get +1 or -1.

If you don't like working in mixed forms (where you can just substract), let's use the full transformation
dt_r = dt - \beta \gamma^2 dr
dr_r = \gamma^2 dr - \beta dt

Obtain the reduced metric
d \tau^2 = dt_r^2 - dr_r^2

For light
0 = dt_r^2 - dr_r^2

Giving
dr_r/dt_r = +1
dr_r/dt_r = -1

There is nothing wrong here.

macaw
2011-May-22, 01:46 PM
It doesn't say that, it's using two examples with different directions. (my bold)

It's just using confusing examples, there is no discontinuity.

No, that is me showing you a way to see your mistake.

ETA: let's go into this. reductio ad absurdum.

You're saying that dr/dt_r = 1 - \sqrt{r_s/r} is only valid outside the EH.

No, I am not saying that, the wiki page is saying that.

Therefor if we take the limit r \rightarrow \infty we get only +1.
So according to you, in the asymptotic minkowski limit we can only have light moving in one direction? Light moving in the other direction has an undefined velocity?
Your initial premise is incorrect, it has nothing to do with outside or inside the EH, but about the direction. It's the velocity of light, not the speed. Therefor there's the +1 or -1. The speed is just |v| = 1.

I have already accounted for that by saying that "If the wiki page is wrong then the light speed will have an anisotropy (my bold) of \sqrt{r_s/r}". This anisotropy is huge at the EH, because it equals 1c! Compare against the example of calculation in Schwarzschild coordinates.

And for the 4th time, there is no discontinuity, only in a faulty reading of the wiki page.

You need to pay attention to what I posted. See above for a repeat of that.

If you disagree, please show another reference where the problem occurs other than a badly phrased wiki page.

I don't disagree, you need to read my post in its entirety.

macaw
2011-May-22, 01:57 PM
For an accelerating observer, of course there's an anisotropy there in that case. If you want to point to a non-physical anisotropy you need to consider things from an inertial (ie freefalling) frame, which is what...

...entitles me to substract.

No, nothing entitles you to subtract. Subtracting would be tantamount to saying that the speed of light adds algebraically with the speed of the frame, we both know that this isn't true.

Why don't we do it in the relevant GP form? For radial light

0=(dr + (1 + \sqrt{r_s/r})dt_r)(dr - (1 - \sqrt{r_s/r})dt_r)

Giving
dr/dt_r = +1 - \sqrt{r_s/r}
dr/dt_r = -1 - \sqrt{r_s/r}

for an accelerating shell observer.

...which, according to my original post (http://www.bautforum.com/showthread.php/115822-Could-You-Fall-Faster-Than-the-Speed-of-Light?p=1891919#post1891919) may be interpreted as light speed anisotropy in the accelerated frame. We both know that this happens (in accelerated frames). The problems with this approach are:

1. There is no such anisotropy in Schwarzschild coordinates.
2. The anisotropy in GP coordinates is huge.

The two issues above are sufficient for preferring Schwarzschild over GP. The fact that the GP metric works across the boundary of the object (there is no "internal" metric, different from the external metric, as in the Schwarzschild case) is not reason enough to use it over the Schwarzschild metric.

To get the speed of light for an inertial free fall observer, substract
dr/dt_r = - \sqrt{r_s/r} and get +1 or -1.

Nothing entitles you to do that, you don't know that speeds add algebraically. Even worse, you don't know that the speed of light adds with the speed of the frame.

If you don't like working in mixed forms (where you can just substract), let's use the full transformation
dt_r = dt - \beta \gamma^2 dr
dr_r = \gamma^2 dr - \beta dt

Obtain the reduced metric
d \tau^2 = dt_r^2 - dr_r^2

For light
0 = dt_r^2 - dr_r^2

This is a much cleaner approach, I will buy this. Now, you are getting isotropic light speed in the the coordinates (r_r, t_r). Doesn't change the fact that light is (hugely) anisotropic in the coordinates (r,t_r). So, the unphysical effect persists.

caveman1917
2011-May-22, 08:45 PM
OK, just the anisotropy then :)

No, nothing entitles you to subtract. Subtracting would be tantamount to saying that the speed of light adds algebraically with the speed of the frame, we both know that this isn't true.

Remember that we're in mixed coordinates here, we already did half a lorentz transform to get to our GP form in the first place. It does add algebraically in this case, proof is in the latter part of my previous post (the part you say is a much cleaner approach).

1. There is no such anisotropy in Schwarzschild coordinates.

But those are not mixed coordinates.

2. The anisotropy in GP coordinates is huge.

But physically meaningless since it doesn't actually represent any observer, so it's not problematic.

The two issues above are sufficient for preferring Schwarzschild over GP. The fact that the GP metric works across the boundary of the object (there is no "internal" metric, different from the external metric, as in the Schwarzschild case) is not reason enough to use it over the Schwarzschild metric.

I'm not saying that one is "better" than the other.
My argument is
1. There is nothing unphysical or inconsistent about GP coordinates.
2. I only brought them up because that's what seemed the basis for Jeff's description (it's used a lot in layman descriptions, because of the ease of use). Look back at posts 3, 4 and 5.

Even worse, you don't know that the speed of light adds with the speed of the frame.

It does in this case.

This is a much cleaner approach, I will buy this. Now, you are getting isotropic light speed in the the coordinates (r_r, t_r). Doesn't change the fact that light is (hugely) anisotropic in the coordinates (r,t_r). So, the unphysical effect persists.

The coordinates (r,t_r) are pretty meaningless in the first place, they're mixed coordinates. If i use proper time of one observer and proper distance of another, of course i'll get funny-looking stuff. Doesn't mean anything unphysical.

If you want to talk about things being physical or not, take the infalling "rain frame" for which the full transform is used, ie (r_r,t_r), or take the shell frame.
Neither has anisotropy.

macaw
2011-May-22, 09:32 PM
Do the same thing in schwarzschild coordinates, take the frame of a stationary observer hovering close to the event horizon. Limit r \rightarrow r_s, you'll find the exact same anisotropy.

There is no reason to do that, I have derived \frac{dr}{dt} and it came out isotropic, as it should be.

The coordinates (r,t_r) are pretty meaningless in the first place, they're mixed coordinates. If i use proper time of one observer and proper distance of another, of course i'll get funny-looking stuff. Doesn't mean anything unphysical.

(r,t_r) is how the metric is expressed. In this set of coordinates all kind of unphysical things happen, leading whoever wrote the wiki article to claim that the speed of light is 2c inside the EH.

If you want to talk about things being physical or not, take the infalling "rain frame" for which the full transform is used, ie (r_r,t_r).

I have no quibble with this set of coordinates, they seem perfectly ok.

caveman1917
2011-May-22, 09:38 PM
...which, according to my original post (http://www.bautforum.com/showthread.php/115822-Could-You-Fall-Faster-Than-the-Speed-of-Light?p=1891919#post1891919) may be interpreted as light speed anisotropy in the accelerated frame. We both know that this happens (in accelerated frames).

I was incorrect about that, the coordinates (r,t_r) don't represent an accelerated frame. In fact they represent no physical frame at all. The accelerated shell frame is given by d \tau^2 = dt_s^2 - dr_s^2^ which is again isotropic as should be.

caveman1917
2011-May-22, 09:39 PM
There is no reason to do that, I have derived \frac{dr}{dt} and it came out isotropic, as it should be.

Yes i was incorrect about that, just caught the error and edited my post as you were replying.

caveman1917
2011-May-22, 09:46 PM
(r,t_r) is how the metric is expressed. In this set of coordinates all kind of unphysical things happen, leading whoever wrote the wiki article to claim that the speed of light is 2c inside the EH.

The easiness they represent is that you can consider the physics of a black hole as playing out in simple minkowski with a background "flow" of - \sqrt{r_s/r}.
I guess they're ok as long as one keeps in mind what one is actually doing, and remembers that the physical description is in the falling frames (r_r,t_r) not in the mixed form (r,t_r).

John Mendenhall
2011-May-23, 02:16 PM
Lengthy article on this in this month's Discovery. I suspect you are all correct within your relative frame of reference (pun intended). Nice thread, gentle folk, thanks, very enjoyable.

macaw
2011-May-23, 02:27 PM
The easiness they represent is that you can consider the physics of a black hole as playing out in simple minkowski with a background "flow" of - \sqrt{r_s/r}.
I guess they're ok as long as one keeps in mind what one is actually doing, and remembers that the physical description is in the falling frames (r_r,t_r) not in the mixed form (r,t_r).

Well, the author(s) of the wiki page still believes that the speed of an object freefalling inside the EH can reach 2c....what can I say? The dangers of coordinate-dependent solutions lurk everywhere.

tommac
2011-May-23, 02:49 PM
The title is pretty much the question. If the object you were falling to was massive enough, and the vacuum surrounding you was hard enough, could you fall faster than light? I don't know why I think you could and my naive and analogy based understanding of relativity says no, but for some reason my brain is wondering if this could somehow be a special case. Relative to what?

NEOWatcher
2011-May-23, 02:56 PM
Relative to what?
I think that was implied in the question... For it to be possible, there would have to be a reference frame of some sort defining it and would have to show up in the answer.
Since the overall answer is "no", then "Relative to what" has no meaning.

tommac
2011-May-23, 02:58 PM
Well, the author(s) of the wiki page still believes that the speed of an object freefalling inside the EH can reach 2c....what can I say? The dangers of coordinate-dependent solutions lurk everywhere.
2c relative to what?

macaw
2011-May-23, 03:41 PM
2c relative to what?

luckyfrank
2011-May-23, 03:53 PM
In theory is there anything possible in the universe with a stronger gravitational pull than a black hole or anything as dense

tommac
2011-May-23, 03:53 PM
I think that was implied in the question... For it to be possible, there would have to be a reference frame of some sort defining it and would have to show up in the answer.
Since the overall answer is "no", then "Relative to what" has no meaning.

I somewhat disagree that it has no meaning. The "no" would be local but could it fall at faster speeds relative to a point at infinity OR relative to the singularity.

tommac
2011-May-23, 03:55 PM
More mass closer up means more gravity ... an EH the defnition of the most packed information can be.

In theory is there anything possible in the universe with a stronger gravitational pull than a black hole or anything as dense

tommac
2011-May-23, 03:55 PM
No ... just did now.

NEOWatcher
2011-May-23, 05:24 PM
I somewhat disagree that it has no meaning. The "no" would be local but could it fall at faster speeds relative to a point at infinity OR relative to the singularity.
Isn't that the whole point of the concept of "local"?

caveman1917
2011-May-24, 05:36 AM
In theory is there anything possible in the universe with a stronger gravitational pull than a black hole or anything as dense

If you mean by "black hole" only the singularity, the answer is "no" (the entire theory breaks down because the density is so high).
If you mean by "black hole" the part of space inside the event horizon, the answer is that a black hole can be as low density as you like.