B When I jump up and down what is the Einsteinian way to describe it?

[ESponge2000]

TL;DR Summary

Is it …. In a “curved by planet earth” spacetime for me, where my geodesic aligns with earth accelerating in the straight up direction,

I obtain the ability when in full touch with earth’s force enacted on me, to apply the energy in calories from the food I eat to further exacerbate my acceleration upwards in a pulse of energy ,
a capability I lose once not in contact with the earth’s force acting upon me.

However, this expended energy from the calories I ate, is rewarded back to me by a freedom to rest in my geodesic weightlessly for roughly a fraction of a second

Ok I’m missing something in the physics here because how do we explain that when i regain contact with the ground after a jump
Up, the impact will be harder on my feet vs if I did not make a jump up?

 

[ESponge2000]

Wait before I jumped up I wasn’t at rest … I was weighted against the earth which means the impact of my food calories dose of energy only explains a portion of the force that allowed me to lift off the ground, right ? The portion coming from having been assisted by the force of the earth before expending additional force from my food calories and brain steering them to be used in a jump maneuver.

Hence why a portion of that force up is rewarded by the fraction of a second of weightlessness, and the other portion is paid for with the harder clash with the ground at the end of the jump, correct ?

No this still can’t be right …. The harder/higher i jump, the more i get both weightless time on the coming down but also the harder I hit the floor too… and the difference from a smaller jump is entirely from me, nothing to do with the earth before the jump, so help me out

 

[Ibix]

You are overcomplicating it. And you appear to be mixing force and energy as if they were interchangeable things, which they're not.

In relativistic physics, hitting the ground is the same as hitting a wall. You are moving on a free trajectory with no force acting on you. Then you slam into a solid object that exerts a force on you through the patch of you that's in contact with it. That unbalanced force slows you to a stop relative to the ground/wall.

 

[Dale]

In an inertial frame the ground (blue line) is accelerating upwards (which I have pointing to the right on this spacetime diagram). When you (orange line) jump your worldline departs from the ground's worldline. You are not accelerating while in the air, so your worldline is straight. Eventually it re-intersects with the ground's worldline and you land. From that point on you are accelerating with the ground.

If you are not familiar with spacetime diagrams, they are just position vs time diagrams. Traditionally time is put on the vertical axis and space is put on the horizontal axis. This may be a little confusing at first, but like any convention you can get used to it.

 

[A.T.]

Ok I’m missing something ...

I rather think you are already mixing too much stuff together and confusing it all. You don't seem to have a good grasp on the concepts of classical Newtonian physics, which you should learn before General Relativity.

 

[ESponge2000]

In an inertial frame the ground (blue line) is accelerating upwards (which I have pointing to the right on this spacetime diagram). When you (orange line) jump your worldline departs from the ground's worldline. You are not accelerating while in the air, so your worldline is straight. Eventually it re-intersects with the ground's worldline and you land. From that point on you are accelerating with the ground.
View attachment 360367
If you are not familiar with spacetime diagrams, they are just position vs time diagrams. Traditionally time is put on the vertical axis and space is put on the horizontal axis. This may be a little confusing at first, but like any convention you can get used to it.

This graph is the most helpful for me understanding and I see we are assuming a grid for a window of the acceleration of the ground and the vertical axis representing time , hence why the blue line is sloped parabolic as the 9.8 meters per second squared that we perceive as free fall but in Einstein GR is the ground motion up …. We are in our geodesic for the entire time we aren’t touching the ground, and removing air resistance from the equation, our velocity from the energy we use to move “up” is our new inertial frame that starts immediately after the new energy release is done and we are off the floor.

One analogy I want to question if it’s similar that would be similar is …:

In a frictionless world with no air and no gravity …
I am in a convertible with open top… the car is accelerating at a constant rate in a straight line , at which point i decide to jump out of the convertible car in the forward direction ..:

assume The floor is very very soft and harmless or if we assume no gravity there is no vertical free fall … the jump is horizontal, and assume the only object to look out for is the car itself from behind … Also assume there’s very little length between the hatch of the car and the front bumper on this imaginary car … no air ….if I jump forward ….. my start momentum will be the terminal velocity of the car at the moment i Jump forward plus the energy of my jump , correct ?

But if the car continues to accelerate at the same rate the car will catch up with my forward velocity and then impact I will be riding the bumper … that impact with the bumper could be very similar to jumping up and down if the rate of the acceleration of the convertible is 9.8m/s^2 correct ?

Also in the car analogy in order to jump forward in an accelerating car , as in this example the car is not at a constant velocity but actually in acceleration …. Here if one jumps inside the car straight up … they will fall towards the back of the car if they don’t hold on tight … to some level even they could jump forward and still end up falling backwards in the car and a certain jump energy would equilibrium keep them in the same position of the car …. A stronger jump would be neeeeded to actually position forward within the car right ? And with jumping up off the ground is that dynamic similar ?

 

[ESponge2000]

In an inertial frame the ground (blue line) is accelerating upwards (which I have pointing to the right on this spacetime diagram). When you (orange line) jump your worldline departs from the ground's worldline. You are not accelerating while in the air, so your worldline is straight. Eventually it re-intersects with the ground's worldline and you land. From that point on you are accelerating with the ground.
View attachment 360367
If you are not familiar with spacetime diagrams, they are just position vs time diagrams. Traditionally time is put on the vertical axis and space is put on the horizontal axis. This may be a little confusing at first, but like any convention you can get used to it.

One other question I have here… if we add to this diagram another line for a spaceship, we would need to redraw the blue parabola to fit the spacetime curvature the rocket experiences , right ? Which is a reduction in the acceleration curve as the spaceship reaches a certain altitude and so the curve would get less steep and gradually transform into a straight line correct? The orange line would slope to the right and in the event that its slope is flatter than the terminal blue straightened out curve , that is now we would draw that the rocket obtained “escape velocity”?

 

[jbriggs444]

In a frictionless world with no air and no gravity …
I am in a convertible with open top… the car is accelerating at a constant rate in a straight line , at which point i decide to jump out of the convertible car in the forward direction ..:

assume The floor is very very soft and harmless or if we assume no gravity there is no vertical free fall … the jump is horizontal, and assume the only object to look out for is the car itself from behind … Also assume there’s very little length between the hatch of the car and the front bumper on this imaginary car … no air ….if I jump forward ….. my start momentum will be the terminal velocity of the car at the moment i Jump forward plus the energy of my jump , correct ?

The term "terminal velocity" here is not appropriate. This may be due to a difficulty with language.

In English-speaking physics, "terminal velocity" refers to the velocity that is approached by a falling object. The typical example is a parachutists. The downward pull of [Newtonian] gravity on the such as a parachutist is constant. The upward force due to air resistance increases as the parachutist falls faster and faster. Eventually an equilibrium is approached at "terminal velocity".

My best guess is that you mean "final velocity" here. The velocity of the convertible at the moment of the forward jump.

Your starting momentum will be equal to your momentum due to the motion of the car plus the additional impulse from your jump.

If the convertible has final velocity $v_\text{f}$ and your mass is $m_\text{you}$, you will have momentum $p=m_\text{you}v_\text{f}$ just prior to your jump.

If you jump with relative velocity $v_\text{jump}$ and we ignore any recoil of the car then the impulse provided by the jump will be $\Delta p = m_\text{you}v_\text{jump}$.

As stated above, the result is $p_\text{total} = m_\text{you}(v_\text{f}+v_\text{jump})$

For this computation, it is not the energy of the jump that matters, it is the impulse.

One could certainly perform an energy computation instead. Can you do it?

But if the car continues to accelerate at the same rate the car will catch up with my forward velocity and then impact I will be riding the bumper … that impact with the bumper could be very similar to jumping up and down if the rate of the acceleration of the convertible is 9.8m/s^2 correct ?

Yes. This is the classic elevator scenario used in stating the equivalence principle:

No local experiment can detect the difference between a jumper in a 1 g accelerating elevator cab in the middle of deep space and the same jumper in a stationary elevator cab sitting on the surface of the Earth.

 

[Dale]

This graph is the most helpful for me understanding and I see we are assuming a grid for a window of the acceleration of the ground and the vertical axis representing time , hence why the blue line is sloped parabolic as the 9.8 meters per second squared that we perceive as free fall but in Einstein GR is the ground motion up

Yes, the horizontally right direction in the graph corresponds to the vertically up direction physically. That is just due to the convention of using the vertically up direction in the graph to correspond to the future time direction physically.

We are in our geodesic for the entire time we aren’t touching the ground

Yes, that is the straight segment of the orange graph.

my start momentum will be the terminal velocity of the car at the moment i Jump forward plus the energy of my jump , correct ?

You cannot add velocity plus energy to get momentum. They all have different units.

But if the car continues to accelerate at the same rate the car will catch up with my forward velocity and then impact I will be riding the bumper … that impact with the bumper could be very similar to jumping up and down if the rate of the acceleration of the convertible is 9.8m/s^2 correct ?

Yes.

 

[ESponge2000]

now let’s add in air resistance ….
This is another subject but correct me if I’m wrong, depending on an objects mass/volume and shape,

In Newtonian interpretation it’s ,

There is a velocity “relative to earth” such that above that velocity, the friction from air is powerful enough to reduce your freefall to a certain constant velocity ….

Below that velocity of fall you will accelerate at a decreasing acceleration rate till your final velocity converges on a rate where the air resistance offsets the gravity

The equilibrium velocity caused by air resistance depends on one’s density and shape where the ideal shape for a slow equilibrium velocity is an open parachute


But in einsteinian perspective this is a little harder to wrap my head around ….

If relative to earth, due to air resistance, the relationship between the ground and a space of free fall above the ground …. Is a constant velocity …. Such as when a parachute achieved slowing a skydiver down to where the rest is a constant slow earth moving up at gentle constant velocity ….
Then this must meant the force of friction from the air has matched the same acceleration as the earth surface , but for whatever reason it doesn’t close the gap on some velocity that is attributed to earth’s acceleration against one’s geodesic ? This is a little harder for me to grasp . Can this be drawn to a diagram?

And one other question i believe has to be true is that for any object such as a parachute , such as a bird or a plane (would have higher final velocity if air resistance offsets the earth’s acceleration )…. Some level of speed obtained in the horizontal direction relative to the air medium both reduces that threhold velocity and also “removes that threshold velocity) Such that a plane can not just avoid the freefall acceleration projectory but also float entirely relative to the ground

So question 1: Why does air resistance play out such that the relationship of it to the acceleration of earth yields removal of the acceleration but not the velocity ?

Question 2: How a plane escapes the velocity and acceleration without needing to apply combustion in any of the up direction at all?

 

[jbriggs444]

now let’s add in air resistance …

There is enough misunderstanding already without adding more complexity.

Then this must meant the force of friction from the air has matched the same acceleration as the earth surface

We could start with units. Do you understand that force and acceleration have different units?

 

[ESponge2000]

For the plane the lift is actually the tilt of the wings the shape of the air resistance, so it’s an engineering maneuver that allows for having a plane jet go level horizontal and not have any effects of the geodesic that aren’t offset

 

[ESponge2000]

There is enough misunderstanding already without adding more complexity.

We could start with units. Do you understand that force and acceleration have different units?

I confuse these . Isn’t air resistance a manifestation of a force (of the ground) manipulating the chain of air molecules that start at the ground and chain up above it ? Which basically allows the ground to start applying its force in increments layered above the ground because air is a medium supported by the acceleration at ground surface ?

This acts like a warning system the air saying “we are in contact with the ground , it is accelerating up… let us help you become aware of what that means for you and help get you situated before it’s too late “

 

[DaveC426913]

I confuse these . Isn’t air resistance a manifestation of a force (of the ground) manipulating the chain of air molecules that start at the ground and chain up above it ? Which basically allows the ground to start applying its force in increments layered above the ground because air is a medium supported by the acceleration at ground surface ?

This acts like a warning system the air saying “we are in contact with the ground , it is accelerating up… let us help you become aware of what that means for you and help get you situated before it’s too late “

You can simplify this. You don't need to concern yourself with air-in-communication-with-ground.

The atmosphere is part of the planet. You can consider it the same thing as ground, just ... fluffier.

 

[Dale]

But in einsteinian perspective this is a little harder to wrap my head around

So then use the easier perspective. The whole point of the principle of relativity is that you can use whatever coordinate system is most convenient. Not that you have to use free falling coordinates if they make things harder.

Can this be drawn to a diagram?

Sure. Again, the ground is blue. The orange is a skydiver that is at a constant velocity relative to the ground. They are both accelerating "upward" at $g$

So question 1: Why does air resistance play out such that the relationship of it to the acceleration of earth yields removal of the acceleration but not the velocity ?

Question 2: How a plane escapes the velocity and acceleration without needing to apply combustion in any of the up direction at all?

These questions have nothing to do with relativity. You should ask those in the classical physics section.

 

[jbriggs444]

I confuse these . Isn’t air resistance a manifestation of a force (of the ground) manipulating the chain of air molecules that start at the ground and chain up above it ?

Do you want to discuss the equivalence principle? (The equivalence between an accelerating elevator in empty space and a stationary elevator on Earth)?

Do you want to discuss Pascal's principle of hydrostatics? (How equilibrium results from a balance between gravity and pressure gradients)?

Do you want to discuss Aerodynamics? (How lift and drag result from the motion of an object through a stationary fluid irrespective of the mechanism by which the boundaries of the fluid volume are maintained)?

You need to pick one thing to understand. Your admitted confusion may be due to trying to consider too much at once.

Edit: It is perhaps worth noting that trying to understand aerodynamics in terms of molecules bouncing around is counter-productive and can even lead to false conclusions. It is more illuminating to treat the fluid as a continuum. Molecules should usually be considered only as a justification for the correctness of the continuum model.

This acts like a warning system the air saying “we are in contact with the ground , it is accelerating up… let us help you become aware of what that means for you and help get you situated before it’s too late “

That could be yet another unrelated phenomenon: ground effect.

 

[DaveC426913]

That could be yet another unrelated phenomenon: ground effect.

Indeed. That was my point about "air is just fluffy ground".

A lifting wing technically works on any non-solid medium - soil, water or air - just to slightly differing effect.

 

[PeterDonis]

For the plane the lift is actually the tilt of the wings the shape of the air resistance, so it’s an engineering maneuver that allows for having a plane jet go level horizontal and not have any effects of the geodesic that aren’t offset

A plane traveling in level flight over the Earth's surface is not traveling on a geodesic.

 

[ESponge2000]

Ok I figured that the plane is using force to evade a geodesic and I understand it at a basic level.

I have a message for all of you whenever you feel defeated in life, pat yourself on the back for having kept up with the force of the ground your whole life long.
Riding that force to defy your own geodesic your whole life long and never once falling behind is power to you is it not?

Even When I jump up according to they earlier space-time diagram I am in a linear motion path that I would be nowhere even close to if wasn’t riding this force from the ground all since I was born ! If you’re surviving and you’re reading this, you’ve been honored to keep up with the ground force and its accelerating scheme of 9.8 m/s^2 since the day you were born , isn’t that doing a great job? Imagine if you stayed to your geodesic at birth how way off the mark you’d be now

Note: I am aware the force up from earth all my life as far as mapping position over time has absolutely no practical application , is regional , and there’s no reconstruction of the universe we would ever make use of where we redraw a hypothetical imaginary geodesic line as if the force field was constant everywhere, In other words there’s no practical way to show over a longterm, a change in position in the universe due to constant force (and I believe this locks in the point made repeatedly by others that a force is not an acceleration nor a change in a velocity and hence not a change in positioning. Still a fun thought if we try to simulate the elevator equivalence principle version of gravity, since the elevator in space version would indeed equate positioning to force in the way gravity does not.

Also note the upward elevator in space simulating gravity would not entail a “constant force”, I believe it would here entail a constant “acceleration” …. The difference being that whenever an object of mass leeped off the elevator , the change in the mass of the platform would have to apply a “change in force” necessary to hold the acceleration the same . A constant force would not achieve this
*and if we want to get extremely technical the elevator in space itself would have a very very very very very small but non-zero gravitational field of its own that we ignore for purposes here as it’s very negligible

 

[ESponge2000]

In a black hole past an event horizon is it accurate to say the _____ from the geodesic exceeds c velocity?

Or is that not the correct terminology ? We know velocity above c isn’t possible . but when spacetime itself is distorted then using our spatial dimensions from our perspective we have faster than light (ex the universe is expanding out faster than c) as the space itself is expanding not any objects

 

[PeterDonis]

the proper acceleration from the geodesic

Is word salad. An object whose worldline is a geodesic has zero proper acceleration.

the proper acceleration from the geodesic exceeds c?

c is a velocity, not an acceleration.

You seem to be very confused.

 

[ESponge2000]

Is word salad. An object whose worldline is a geodesic has zero proper acceleration.


c is a velocity, not an acceleration.

You seem to be very confused.

I am confused , and what’s the proper words to explain that in a black hole, a hypothetical acceleration to the velocity of c (infinite energy) would not outweigh the relationship of the motion between the center of the black hole and the object past the event horizon ?

Doesn’t something need to be labeld
Exceeding c in order to say that going c wouldn’t get you out if it were possible, obviously not objects but the impact of the spacetime distortion on a linear perspective , or something like that ?

 

[jbriggs444]

I am confused , and what’s the proper words...

In order to phrase a coherent thought coherently, it helps to start with a coherent thought.

 

[ESponge2000]

In the case of being past the event horizon of a black hole , what is the lowest meters per second squared that leads to having an escape velocity above c?

That is the question ? And assume roughly the radius of our planet for the black hole and the lowest density needed to make it a black hole

For Infinite force to not get you out or that light can’t escape , there has to be an apparent relationship between 2 things that are converging at faster than c, correct ?

 

[PeterDonis]

in a black hole, a hypothetical acceleration to the velocity of c

There is no such thing. We can't give any meaningful response to a meaningless question.

the relationship of the motion between the center of the black hole and the object past the event horizon ?

A black hole has no center in the sense you mean.

You need to learn how black holes actually work.

Exceeding c in order to say that going c wouldn’t get you out if it were possible, obviously not objects but the impact of the spacetime distortion on a linear perspective , or something like that ?

This is all word salad.

In the case of being past the event horizon of a black hole , what is the lowest meters per second squared that leads to having an escape velocity above c?

I have no idea what you mean by this.

assume roughly the radius of our planet for the black hole and the lowest density needed to make it a black hole

First, a black hole does not have a meaningful density because it does not have a meaningful volume. The only meaningful quantity is the surface area of its horizon, which depends on its mass.

Second, if you have specified the hole's Schwarzschild radius, assuming the hole is non-rotating, you have specified everything about the hole. There is no freedom left to independently specify anything else.

Third, you do not appear to have enough understanding of black holes to even formulate a meaningful question about them. So I would strongly suggest that you stop trying to. Black holes are off topic for this thread anyway; your original question had nothing to do with them.

 

[ESponge2000]

I don’t understand how we can say

1) light can’t escape and light travels at c

2) there’s no apparent perspective of anything connected to anything whatsoever that is or appears to be going above c

What am I missing ? My current thinking is what is perceived to be above c is a bending of spacetime itself upon which nothing actually travels through

 

[Ibix]

What am I missing ?

The mathematical foundation to follow the answers you get, I suspect.

There are no timelike or null paths leading out of a black hole. It has nothing to do with acceleration. All possible paths anything can follow terminate on the singularity.

Similarly, there can be no "perspective" of something travelling faster than light because you'd require a timelike axis to be spacelike. It's a contradiction in terms.

 

[Nugatory]

I don’t understand how we can say
1) light can’t escape and light travels at c
2) there’s no apparent perspective of anything connected to anything whatsoever that is or appears to be going above c
What am I missing ?

When you're inside the horizon, no matter which direction you shine your light it will end up at the singularity.

 

[ESponge2000]

The mathematical foundation to follow the answers you get, I suspect.

There are no timelike or null paths leading out of a black hole. It has nothing to do with acceleration. All possible paths anything can follow terminate on the singularity.

Similarly, there can be no "perspective" of something travelling faster than light because you'd require a timelike axis to be spacelike. It's a contradiction in terms.

What we mean is space itself , take a very very very small particle of space , its place with respect to the center of the black hole, is not static . If staying in the particle you are in a 0 acceleration

It could also be said that a particle of space in the black hole center is also at a 0.

What’s going on? The structure of how space is connected to space is In distortion due to the mass of the black hole , is this correct ?

All things including light travel through space in accordance with the structure of spacetime and can’t do anything about its curvature which is only something mass naturally does

So that’s my current thinking and maybe I have it wrong. (An illlustration: On a sheet of paper write 1 2 3 4 5 6 7…… left to right
Imagine a current that always must follow the the direction of 1 2 3 4 5 6 7 no matter what
Then take scissors and chop the paper into loose threads but don’t cut it into 2, but start wiggling the shredded paper all around and now apply the same rule to the current must obey passing through 1 2 3 4 5 6 7 in order and what we get is a distortion of what a straight path means that can help us understand GR, am I close or way off?

 

[ESponge2000]

When you're inside the horizon, no matter which direction you shine your light it will end up at the singularity.

That is to say the rate at which spacetime is being distorted results in there not being a path for anything to connect in the reverse direction

 

[ESponge2000]

The mathematical foundation to follow the answers you get, I suspect.

There are no timelike or null paths leading out of a black hole. It has nothing to do with acceleration. All possible paths anything can follow terminate on the singularity.

Similarly, there can be no "perspective" of something travelling faster than light because you'd require a timelike axis to be spacelike. It's a contradiction in terms.

Will this appear to a traveler past the event horizon much like if the singularity contains an elephant, an elephant will display in each and every direction for them and the gray elephant perimeter converging in on a shimmering point ?

 

[Ibix]

Will this appear to a traveler past the event horizon much like if the singularity contains an elephant, an elephant will display in each and every direction for them and the gray elephant perimeter converging in on a shimmering point ?

No. The singularity is a point in time. Asking what it looks like is like asking what Monday morning looks like frim Sunday.

 

[jbriggs444]

That is to say the rate at which spacetime is being distorted results in there not being a path for anything to connect in the reverse direction

You cannot distort spacetime over time because time is already part of spacetime. There is no separate "hyper time" over which a spacetime can evolve.

We view spacetime as a four dimensional whole. Not as a continuous evolution of three dimensional spatial slices. The curvature is a property of the spacetime, not [merely] of the slices.

Will this appear to a traveler past the event horizon much like if the singularity contains an elephant, an elephant will display in each and every direction for them?

This appears to ask about the visual image that a free falling traveler would see when looking "toward the singularity".

I am no expert, but the best answer I can come up with would be "what do you see when you look toward tomorrow?".

 

[ESponge2000]

I still can’t wrap my head around it. I can’t see 4–dimensionally very well. I can imagine bending of space but bending of spacetime for me and maybe for physicists I Don’t know, is applying formulas and studying terminology that is tested , understanding it visually is very challenging

 

[jbriggs444]

I still can’t wrap my head around it. I can’t see 4–dimensionally very well. I can imagine bending of space but bending of spacetime for me and maybe for physicists I Don’t know, is applying formulas and studying terminology that is tested , understanding it visually is very challenging

You should not expect to understand it intuitively. Homo Sapiens has evolved in a low energy, low gravity environment where Newtonian mechanics and Euclidean geometry serve as a very accurate model. Our intuitions are trained in this environment.

Trying to carry those pre-relativistic intuitions over into a high energy, high gravity environment is not reasonable. Many of our common sense intuitions must simply be discarded. Or retrained using textbooks and instructors. There is, for most of us, simply no magic trick or gimmick to gaining an immediate and perfect understanding of the stuff. That's why we have schools and classes.

 

[PeterDonis]

My current thinking is what is perceived to be above c is a bending of spacetime itself upon which nothing actually travels through

Wrong.

What we mean is space itself , take a very very very small particle of space

Word salad.

its place with respect to the center of the black hole

The black hole has no center. I've already said this.

If staying in the particle you are in a 0 acceleration

I have no idea what you mean by this.

a particle of space

What does this even mean?

The structure of how space is connected to space is In distortion due to the mass of the black hole , is this correct ?

It's not even wrong.

All things including light travel through space in accordance with the structure of spacetime

No, they travel through spacetime in accordance with the structure (geometry) of spacetime.

and can’t do anything about its curvature which is only something mass naturally does

Light has energy, so it can cause spacetime curvature. "Mass" is not the only thing that does that.

that’s my current thinking and maybe I have it wrong.

Your current thinking is not even wrong.

am I close or way off?

I can't even make sense of your "illustration".

 

[PeterDonis]

Thread closed for moderation.

 

[Dale]

In other words there’s no practical way to show over a longterm, a change in position in the universe due to constant force

If you accelerate at a constant proper acceleration of 1 g for more than about a month, then you will start to need to use relativity.

Here is a page on that topic

https://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html

what is the lowest meters per second squared that leads to having an escape velocity above c?

There is none. It doesn’t matter how hard you accelerate, nor for how long you accelerate, you can never reach the speed of light because you have mass.

In spacetime a small object is described by a worldline. For example, the orange lines on the drawings I posted. Proper acceleration is how tightly the worldline bends. For example, in the first drawing, at the beginning the worldline is gradually curved and at the end also. These are the times when the accelerometer reads a nice steady 1 g upwards. While they are in the air, their accelerometer reads 0 proper acceleration, so the worldline is not curved at all. And there are sharp curves corresponding to the sudden high proper accelerations at jumping and landing.

But the geometry in spacetime is not Euclidean like the paper. No matter how sharply your worldline bends, it can never turn past the angle that $c$ makes on the diagram

 

[Dale]

I still can’t wrap my head around it. I can’t see 4–dimensionally very well.

I would recommend getting some experience reading and drawing spacetime diagrams for flat spacetime first. Once you are comfortable with flat spacetime then you can start thinking about black holes and curved spacetime

 

[PeterDonis]

After moderator review, the thread will remain closed. Thanks to all who participated.