# Schwarzchild radius and escape velocity?

• qaovxtazypdl
In summary, we know that the escape velocity at the schwarzchild radius is c, which is defined as the velocity needed to escape from the gravitational field and reach a total energy of 0 at infinity. However, this does not mean that an object falling from infinity starting at rest will have a velocity of c at the schwarzchild radius of a black hole. This is because the infalling object will reach a velocity of dr/dτ = c at the horizon, which is a light-like surface and not a valid, physical reference frame. Thus, the concept of velocity becomes undefined at the horizon, and the object will continue to fall towards the singularity with a velocity less than c.
qaovxtazypdl
We know that the escape velocity at the schwarzchild radius is c.

Since the escape velocity is defined as the velocity needed to escape from the gravitational field, to reach a total energy of 0 at infinity:

Doesn't this mean that an object falling from infinity starting at rest, to the schwarzchild radius of a black hole will then have a velocity of c? This is obviously wrong, but I can't see where my mistake was.

qaovxtazypdl said:
We know that the escape velocity at the schwarzchild radius is c.

Since the escape velocity is defined as the velocity needed to escape from the gravitational field, to reach a total energy of 0 at infinity:

Doesn't this mean that an object falling from infinity starting at rest, to the schwarzchild radius of a black hole will then have a velocity of c? This is obviously wrong, but I can't see where my mistake was.
I think you're right. The infaller from infinity reaches a velocity $dr/d\tau= c$ on the horizon.

( Welcome to the forum - you're the first Klingon I can remember here).

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I thought an object with mass could not reach the speed of light in any frame of reference though? Or is that invalid in a curved spacetime?

qaovxtazypdl said:
I thought an object with mass could not reach the speed of light in any frame of reference though? Or is that invalid in a curved spacetime?

In any local inertial frame, the relative speed of two massive objects must be less than the speed of light. In the flat spacetime of special relativity, a local inertial frame can cover the entire universe... But not so in curved spacetime, as you suspected.

In GR, there are possibly global coordinate systems, but only local frames of reference. For the infaller, they will be move at less than c for compared to any nearby material body. Compared to a distant observer, the correct statement is that relative velocity is undefined because you can't compare vectors at different points in curved spacetime in any unique way.

As for the meaning of dr/dτ, you the relevant issue is that the r coordinate is changes from spacelike to timelike across the horizon. So dr/dτ does not represent a spatial velocity at the horizon. It does represent the r component of a 4 velocity in a particular coordinate basis; however, to discuss a spatial velocity, you must define some local spatial basis vector to take the dot product with, of the 4-velocity of the infaller. This will always be less than c, all the way to singularity.

A simplified version of what Mentz said:

First, let's describe a static observer. A static observer is one who maintains a constant Schwarzschild "R" coordinate. Or if you don't mind a bit of distortion, a static observer "hovers" at a constant distance above the black hole. (If you want to be precise, the static observer hover's at a constant R coordinate value, which is not really a distance). Static observers exist at all values of R ABOVE the event horizon. Static observers do not exist at the event horizon.

If you look at the infalling velocity (measured with local clocks and rulers of a static observer), it will indeed approach c as a limit.

However, static observers can't exist at the event horizon, so while the velocity will exist in the limit, and approach 'c', no physical observer will ever see such a speed.

From the point of the view of the infalling person, the event horizon is light-like. It's a surface traced out by light. For someone familiar with SR, who realizes that light can't have a point of view, this is another demonstration of why you can't have an observer at the event horizion. The event horizon is like a light beam, it's what we call a "null surface". As such, while light can "hover" at the event horizon, no physical observer can.

If you're not familiar with SR, there are some FAQ's on the topic of why light doesn't have a "point of view". Hopefully they will help, I get the feeling some people still come away confused by this. But the FAQ is the simplest resoucre I know.

It's perhaps not critical to understand how the two things (light not having a POV and the lack of static observers at the evet horizon) are related - it's sufficient to know that the later doesn't exist.

If you want a detailed mathematical analysis, https://www.physicsforums.com/showpost.php?p=602558&postcount=29 has one, though it does use geometric units where c=1.

Another poster has a similar analysis in the next post in the thread (#30) which comes to the same result.

qaovxtazypdl said:
I thought an object with mass could not reach the speed of light in any frame of reference though?
The Schwarzschild horizon is Inseln a light-like surfte, i.e. each point on the horizon defines s light-like trajectory. The horizon is the surface from which photons (emitted radially outwards) neither escape not fall into the singularity; so they stay at the horizon. So it's trivial that every (massive) object crosses the horizon whith v = c w.r.t. the horizon.

But the Lorentz transformation becomes singular for v = c which means that it doesn't allow us to introduce a reference frame moving with v = c w.r.t. a physical reference frame defined by a massive object. So the horizon is not a valid, physical reference frame.

(Think about a photon as a reference frame; a massive particle moves with v = c w.r.t. this frame; this s problematic, but not due to the particle but due to the frame)

qaovxtazypdl said:
We know that the escape velocity at the schwarzchild radius is c.

Since the escape velocity is defined as the velocity needed to escape from the gravitational field, to reach a total energy of 0 at infinity:

Doesn't this mean that an object falling from infinity starting at rest, to the schwarzchild radius of a black hole will then have a velocity of c? This is obviously wrong, but I can't see where my mistake was.

this radius is computed with the classical formula mv*v/2 equals Gm/r when v equals c.
Hower if one uses the relativistic expression of the cinetic energy this radius turns out to be zero.

## 1. What is the Schwarzchild radius?

The Schwarzchild radius, also known as the gravitational radius, is the distance from the center of a non-rotating black hole at which the escape velocity exceeds the speed of light. It is named after German physicist Karl Schwarzschild, who first derived the concept in 1916.

## 2. How is the Schwarzchild radius calculated?

The Schwarzchild radius is calculated using the formula Rs = 2GM/c^2, where G is the gravitational constant, M is the mass of the black hole, and c is the speed of light. This formula shows that the Schwarzchild radius is directly proportional to the mass of the black hole. The higher the mass, the larger the Schwarzchild radius will be.

## 3. What is the significance of the Schwarzchild radius?

The Schwarzchild radius is significant because it marks the point at which the gravitational pull of a black hole becomes so strong that even light cannot escape from it. Anything that crosses the Schwarzchild radius will be pulled into the black hole and cannot escape, making it the point of no return.

## 4. What is the escape velocity?

The escape velocity is the minimum speed required for an object to escape the gravitational pull of a celestial body, such as a planet, star, or black hole. It is calculated using the formula Ve = √(2GM/r), where G is the gravitational constant, M is the mass of the body, and r is the distance from the center of the body.

## 5. How is the escape velocity related to the Schwarzchild radius?

The escape velocity and the Schwarzchild radius are related because the Schwarzchild radius is the distance at which the escape velocity reaches the speed of light. This means that any object that crosses the Schwarzchild radius will need to travel faster than the speed of light to escape the gravitational pull of the black hole, which is impossible according to the laws of physics.

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