# On what things can two observers agree or disagree?

This is not uncommon idea from beginners: if relativistic mass increases without bound, then at some speed, a blackhole should be formed.

Let's say we're talking about a uniform density prolate spheroid with major axis $a$ and minor axis $b$. Also, specify that:

$$ a = b^2 \times \frac{c^2}{2Gm} $$

If you boost that along its major axis with Lorentz factor:

$$ \gamma = \frac{c^2b}{2Gm} $$

the object is now a sphere of radius $b$ and "relativistic" mass:

$$ M = \gamma m = \frac{c^2b}{2G} $$

The Schwarzschild radius for that is:

$$ r_s = \frac{2GM}{c^2} = \frac{2Gc^2b}{2Gc^2}=b$$

*et* *voilá*, it's a blackhole.

Does that violate special relativity? Well, if the spacetime as defined by $g_{\mu\nu}$ around this moving blackhole transforms, under a boost into the mass's rest frame, back into a nominal non-blackhole spacetime, then the answer is probably: NO.

I will state, without proof, that this is not the case.

If it were the case, it would pose problems for special relativity.

It would mean that you and I, all of us, are blackholes in valid inertial frames, right now. Also: in much slower frames, we would not be blackholes, but the Earth (with its great mass) would be a blackhole. It's just a mess that contradicts not only observation, but also reason.

If it is not that case (that blackholes can be boosted away), but it remains true the relativistic mass causes blackhole formation, then: there is an absolute rest frame in special relativity.

All you need is a bunch of identical masses. You boost them into all directions and measure the velocity at which they become blackholes, as a function of direction:

$$ v(\theta, \phi) $$

If space (and I mean absolute space) makes any sense, then there is some boost $\vec u$ such that:

$$ v'(\theta, \phi) = {\rm constant}$$

and that is the absolute rest frame of space.

An important idea of special relativity is that motion is relative (we've all heard that many times). What it means is that you can't move relative to space. No matter how fast you go, your (empty) space slice of spacetime looks the same, and your time ticks the same. It is only when you have other objects and various events and different frames that observers disagree on what constitutes "space" and "time".

The things that relativistic observers agree on is much like classical physics. You agree on what is real. Things that are real are events that happened. How many objects there are. The order of causes and effects. Perhaps it can be summarized as "You don't change an object by running past it."

As in classical physics, there are many things you do not agree on. Whether one object is to the left or right of another. How fast a given object is moving and how much kinetic energy it has. How long the X component of an object's length is. These are things that depend on frame of reference. Different inertial frames of reference may move with respect to each other at constant velocity and have different orientations. You can also choose different length and time units, but we will gloss over that.

Different frames of reference sometimes make it hard to see the things that are real, such as conservation of energy. Different observers see two objects with different energies. They bounce off each other. Both observers measure the same amount of energy after the collision as before. Frames of reference may disguise something as simple as how big an object is.

In special relativity, one major difference is that time works differently than your classical intuition leads you to expect. This is causes confusing ladder paradoxes, twin paradoxes, and so on. Most of the time, people start with the fact that the speed of light is constant and arrive at all of these. This is because you only need one postulate, and it quickly leads to the correct math. If you have gone though that, it may help to look at how time is different.

In classical physics, time is absolute. The history of the universe can be divided up into a series of moments. All observers see the same moment as happening now. They can identify any slice of history by its time coordinate. Time is an intrinsic property of that slice.

By contrast, space is not like this. In my coordinate system, I can divide space up into planes that all have the same $x$ coordinate. I am sitting at the origin, where $x = 0$ at time $t = t_0$. A while later at $t_1$, I am sitting in the same place. However. Another observer who is moving with respect to me disagrees. He says that I am at $x = x_0$ at time $t_0$, but at $t_1$, I am somewhere different where $x = x_1$.

We are used to this. It causes no confusion that what I consider to be one place, he considers to be two places

The most counter-intuitive concept of special relativity is that time is really like space in this regard. We just move too slowly to see it. This violates our notions of what time is.

To get another point of view on this, consider that light moves at $3 \times 10^8$ m/s. I am comfortable at $3$ m/s. Imagine a world where the fastest motion is $3 \times 10^{-8}$ m/s. This is about a meter/year, the speed of a glacier. A glacier world physicist would have the same problems understanding our everyday classical physics, that I have understanding relativity.

A glacier world physicist thinks that an object's spatial coordinates never change. If all observers choose the same origin and orientation, all will agree on the coordinates of the object. Spatial coordinates are an intrinsic property of the of the object. Space is absolute.

Suppose Bob and Alice are moving at an everyday velocity with respect to each other. At $t = t_0$, they agree that they are both at the origin. $t_1$ Bob says he is at the origin, but Alice is at $x = x_1$. Alice says she is at the origin, but Bob is at $x = -x_1$. Neither has changed their unchangeable position, but now they are at different places. And both of those different places are the same place as the origin. They call this highly counter-intuitive result the failure of sameplacity.

If you have started from the constant speed of light and arrived at space time diagrams, you can see our confusion about measuring the length of a relativistic ladder.

All observers agree on the proper length of the ladder. That is the length in a frame where the ladder is at rest. To get the length, you measure the $x$ coordinates of the ends and subtract.

In a moving frame, an observer measures a contracted length. He can then do a calculation to figure out the proper length. Why does he need to? The ends have to be measured at the same time, this is where it gets tricky.

In a spacetime diagram, the $x$ axis is the line where $t = 0$. In a classical spacetime diagram, all observers use the black $x$ axis. There is no problem measuring the ends at the same time. But at relativistic speeds, the line where $t = 0$ is just as motion dependent as the line where $x = 0$. We have a glacier world like confusion. In the garage, you measure the position of the head and the foot of the ladder at the same time on the black $x$ axis. In the ladder's rest frame, you measure them at the same time on the blue $x$ axis. We have two different events where the head of the ladder is at the same time as the foot.

The head and foot of the ladder follow world lines. Time flows in each world line. But there is no universal way of matching up times in different world lines. It is motion dependent, just like matching up same places is motion dependent. Time is very different than we think.

To calculate the proper length, you need to figure out the blue $x$ coordinate when the head's world line crosses the blue $x$ axis. Part of the calculation is to figure out the black $x$ and $t$ coordinates. And part is to understand that if you drew the $1$ meter point on the blue $x$ axis, it would not be where you would expect from looking at the black $1$ meter point.

The point is that the garage frame does not measure the proper length because the head position is measured earlier that the blue frame foot. The head hasn't traveled far enough yet. This makes the length shorter.

So why does it matter? Isn't the black frame idea of same time just as good as the blue frame? Yes, the events where the head and foot really are simultaneous in the black frame. And the ladder really is shorter in the black frame.

This is why it is useful to make a 4D spacetime and create a metric for it. The metric calculates the universal proper length of the ladder. The calculation takes both $x$ and $t$.

It is a little like measuring the length of the ladder from $x$ coordinates while it is laying on the ground vs leaning on a wall. The shorter $x$ coordinate separation is correct, but it doesn't represent the length of the ladder.

Getting back to black holes, you can see that length contraction does not change the ladder. It does not become more dense, and more prone to forming a black hole.

(i)Why does that violate the principle of relativity? Which law is violated?

(2) On what can observers actually agree and what decides that (i.e what quantities are invariant and what decides that)?

I try to answer both of the questions at once. The observers will always agree on $ds^2$ both in special and general relativity.

They both agree that the speed of the light is locally c in vacuum.

They also agree on number of the particles.

Notice that not every reference frame is allowed and meaningful in relativity. For example, you can never go to a reference frame in which the speed of the light is zero. Going to a reference frame in which black holes disappear (or appear) is meaningless similarly to a reference frame in which the speed of the light is zero.

So the observers have to agree on number of the black holes.