# On what things can two observers agree or disagree?

This question was inspired from the fact whether a moving body can become a blackhole in a frame of reference. On looking it up here, I found the answer was that its not possible since it violates priciple of relativity as you can be in a frame where the black hole was never formed and was ordinary matter.

So, I concluded that since it cannot be a blackhole in one frame and a normal mass in another then its not possible. What I don't understand is this?

(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)?

For instance observers can disagree on the mass they measure (but when signaling each other they orobably agree cause I guess thats the whole point of lorentz transformtion)

Also value KE can be disagreed upon as the energy conservation is still conserved in each individual frames so signals can disagree on KE.

• Have you studied basic special relativity yet? Dec 19 '20 at 16:41
• Yes. Except Minkowski or the 4 vector treatment.
– Lost
Dec 19 '20 at 16:48
• Might I ask which part of my question motivated you to ask this? Am I asking something very silly ?
– Lost
Dec 19 '20 at 16:48
• What I was trying to ask in clear terms was...why seeing a black hole in one frame and not other violate priciple of relativity since to me energy conservation, etc etc seem intact in each of the observer's own frame
– Lost
Dec 19 '20 at 16:51
• So which physical law is being violated specifically? As far as I know, in STR , two observers can disagree on things if that doesn't violate any law.
– Lost
Dec 19 '20 at 16:52

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".

• So two basic reasons as I can deduce from your answer is: 1) That the space time around the blackhole would change and its disagreement between observers would violate relativity and 2) That an absolute frame could be defined by this. I do not get the 2nd point completely. If suppose I see a boosted blackhole I could never whether it was the blackhole moving away from me or me. So how does that define an absolute space?
– Lost
Dec 19 '20 at 19:22
• Also why can't it be true that we may be black holes from some inertial frame what stops this? It of course sounds and seems absurd ( but after studying a little STR with different times etc I leave the absurdity as a valid reason for not being able to understand sonething (: )
– Lost
Dec 19 '20 at 19:26
• (1) just doesn't work in general relativity. A singularity that can be boosted to nearly flat spacetime is not happening. (2) If you're a black hole you're moving. If you're not a black hole than you're not moving too fast. Use this fact to figure out when you're not moving at all.
– JEB
Dec 19 '20 at 21:05
• Got the first point. For second, whether I am a black hole or not is frame dependent. I can never be a black hole in my frame and other frames also don't agree whether I am a blackhole or not. Since fast-moving, slow- moving is anyways relative so does seeing a blackhole. I know I am missing something concrete here but I just want to get over this question in my mind properly. For starters, since I am not familiar with GR, I take that such a condition violates the postulates of GR... okay got it....just care to explain the second one again.
– Lost
Dec 20 '20 at 5:58

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.

• In the last paragraph you said it does not become more dense but uts mass has increased and length contracted. That ought to change the density right?
– Lost
Dec 20 '20 at 5:54
• @Lost - You are right. My bad. The length contracts and the mass m stays constant. The density increases. Dec 21 '20 at 2:33

(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.

• Agreeing on the total number of particles comes from conservation energy-mass I suppose? How does that apply 8n this case? As from an intial time to a final time everything is conserved in their own frame.
– Lost
Dec 19 '20 at 17:11
• And then 'Going to a reference frame in which black holes disappear (or appear) is meaningless' why?
– Lost
Dec 19 '20 at 17:13
• There was no postulate of relativity saying that number of black holes have to be conserved. And I can't see how any physical law is changing ...so please can you clearify nore specifically what you mean?
– Lost
Dec 19 '20 at 17:15
• Relativity has no postulation about black holes. When you do the calculation, that is what you will get. In other words, it is a consequence of the theory. Conservation of mass and energy has nothing to do with the number of the particles. Relativity does not have a mechanism to allow you to change number if the particles. Only QFT allows you change # of the particles. Everything about black holes comes mathematically from the assumption of GR which are , space is locally Minkowski and equivalent principal Dec 19 '20 at 17:19
• I am asking whether a Physical law will be changed or invariant in this scenario and if yes then how?
– Lost
Dec 19 '20 at 17:22

since it cannot be a blackhole in one frame and a normal mass in another then its not possible. [...] Why does that violate the principle of relativity?

The principle of relativity says broadly that all speeds are equivalent. In Galileo's ship, you can't do any experiment that will tell you the ship's speed. If assembling a known quantity of matter (like a known number of 1kg weights) into a planet, and observing whether it collapses into a black hole, could tell you the speed of the boat, that would violate the principle of relativity.

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

Observers don't really disagree about anything – at least not in any sense that deserves the word "disagree".

If a vehicle with a siren rushes past you, you'll hear a pitch that's first higher, then lower than the pitch heard by someone riding on the vehicle. Would you say that the two listeners disagree about the pitch? I wouldn't. They just had different experiences, as different people are wont to do. Both listeners agree that the listener on the vehicle heard an intermediate pitch, and the listener on the ground heard a higher-then-lower pitch. They agree that the latter is a Doppler effect. They agree about everything that happened. Only if you use a word to mean two different things in the same sentence, like using "pitch" to mean both "pitch heard by this listener" and "pitch heard by that other listener", can you manufacture a disagreement.

In Newtonian physics, if you use "egocentric Cartesian coordinates" with yourself at the origin, and you label the coordinates in the same way as someone else who is also using egocentric coordinates, you could claim to "disagree" with them about the coordinates of things, but it's equally silly.

In special relativity, egocentric coordinates include a time coordinate too, but it's the same story. You only "disagree" with someone in relative motion if you're both using egocentric coordinates with the same labels, and no one ever does that. You will always agree with everyone else about the UTC time of an event, and its latitude and longitude and altitude above sea level. That's a well defined system of coordinates that people actually use in real life.

• Agree on everything? Then what about length of an object etc?
– Lost
Dec 19 '20 at 17:54
• In the first part I understand that you are saying that a thing collaposing into a blackhole would become an experiment that detwrmined the speed of that frame without any reference to the external world is that correct?
– Lost
Dec 19 '20 at 17:57
• But I am not asking about something collapsing to become a blackhole. I lets say you see a blackhole which receding away from then could it be that you can see it as a blackhole because of it speed and while another observer like me is not able to? There is nothing collapsing here.
– Lost
Dec 19 '20 at 18:00
• @Lost what about length of an object - it's a difference of coordinate positions at a coordinate time. Everyone agrees on it, unless you use the same labels for different coordinate systems, then they "disagree". could it be that you can see it as a blackhole because of it speed and while another observer like me is not able to - no, because there's only one reality and everyone agrees what it is. Black holes are totally different spacetime structures from planets; they aren't just alternate perspectives on planets. Dec 19 '20 at 18:14
• How are you saying they agree on lengths? And I am not talking about change in coodinates or perspectives etc. For a rest frame it is L but for other inertial frames it is contracted. Isnt this correct? 'Reality' is defined as invariant laws aren't they? Nd two observers can see different things , I think For instance see Unruh effect.
– Lost
Dec 19 '20 at 18:19