Alternative to relativity (split from A problem to the theory of relativity ?)

[Bjarne-7]

52 minutes ago, Bufofrog said:

WHAT?!?  That ain't close to how science works.  You can't say you have a 'new' force that has the different units that every other force.  If the 'thing' doesn't have the units of force - it ain't a force

The force is still a force, - but the the prerequisites may be different

Is this not what Newton did ?  I mean "associated / converted " mass to a force. None of his fellow students grasped a piece. He was considered odd.

Now its about space, off course the parameters must be different. Space is not matter.

In the end of the day also transformation of  serveral other equations could be necessary, - I mean if the ruler is a variable which consequences would that have ?

The only place you mathematically can "grab" the claimed elastic nature of space , - is in a process of transformation, and right here there must be a force hidden. - Like it or not.

I think a new law is necessary, no wonder if new math would be "a stranger"

Edited November 27, 2023 by Bjarne-7

[Bufofrog]

34 minutes ago, Bjarne-7 said:

The force is still a force, - but the the prerequisites may be different

I don't know what you mean by that.  Bottom line if your equation for a force does not have the units of force then your equation is wrong,

37 minutes ago, Bjarne-7 said:

Is this not what Newton did ?

No.

 

[Bjarne-7]

10 minutes ago, Bufofrog said:

Bottom line if your equation for a force does not have the units of force then your equation is wrong,

I gave the equitation the resulting unit F, so did Newton. If it is a "new" / unknown force, -then its new and unknown. And it required a completely new equation. 

I am the one that can decide the outcome of that equation, if the outcome is correct or false is a different story.

Edited November 27, 2023 by Bjarne-7

[Genady]

F.

(Academic grading in the United States - Wikipedia)

[Bufofrog]

13 minutes ago, Bjarne-7 said:

I gave the equitation the unit F, so did Newton

This is getting really out of the realm of science and math into uh, uh... I don't even know.

If you call some random equation a force it does not magically become a force.  If I say F = m * s, it doesn't mean that I have discovered a new force it just means I'm wrong.

13 minutes ago, Bjarne-7 said:

I am the one that can decide the outcome of that equation, if the outcome is correct or false is a different story.

Ok, that equation is trivially false.

Edited November 27, 2023 by Bufofrog

[studiot]

6 hours ago, Bjarne-7 said:

I think 2 new laws are necessary

     or  F/m =        

Motion Resistance force / deceleration when moving towards any absolute motion direction

------------------------------------------------------------------------------------------------

    or  F/m =     

Reduced absolutte motion resistance force, -  when moving oppesite any absolute motion direction

F = RR-Force

v = the speed by the travelling observer

m = mass of the traveling observer

c = the speed og light

This is progress.  +1

😀

The plus 1 is because this is the first time you have laid things out in proper fashion.

You see that immediately, even whilst I was writing this, folks have been able to see what you are doing and offer sensible and useful comment.

 

However listing your symbols and stating what they stand for is really good practice and what I mean by progress.

So many waste so much time and effort just writing algebra down.

(did you know that although our word, algebra was,  was named after early arabic 'al-jabr', which meant completion ?)

 

I am sorry but yes your algebra is flawed.

So let us look at your algebra. (I will only work the first equation)

You claim this expression for your RR Force.

[math]F = 1 - \sqrt {\frac{1}{{1 + \frac{{m{v^2}}}{{{c^2}}}}}} [/math]

 

To show that this does not lead to the expression for the force per unit mass as you then stae, proceed as follows.

[math]F - 1 = \sqrt {\frac{1}{{1 + \frac{{m{v^2}}}{{{c^2}}}}}} [/math]

[math]{\left( {1 - F} \right)^2} = \frac{1}{{1 + \frac{{m{v^2}}}{{{c^2}}}}}[/math]

[math]\frac{1}{{{{\left( {1 - F} \right)}^2}}} = 1 + \frac{{m{v^2}}}{{{c^2}}}[/math]

[math]\frac{1}{{{{\left( {1 - F} \right)}^2}}} - 1 = \frac{{m{v^2}}}{{{c^2}}}[/math]

[math]\frac{{1 - {{(1 - F)}^2}}}{{{{(1 - F)}^2}}} = \frac{{m{v^2}}}{{{c^2}}}[/math]

[math]\frac{{1 - {F^2} + 2F - 1}}{{{F^2} - 2F + 1}} = \frac{{m{v^2}}}{{{c^2}}}[/math]

[math]\frac{{{c^2}F\left( {F + 2} \right)}}{{m{v^2}{{(F - 1)}^2}}} = 1[/math]

We have now reached the stage where we can isolate F/m, your force per unit mass.

[math]\frac{F}{m} = \frac{{{v^2}{{(F - 1)}^2}}}{{{c^2}\left( {F + 2} \right)}}[/math]

 

And we some that not only is the expression much more complicated than yours, but it still contains F.

This is because the original expression is what is known as implicit.
 

That is it is not possible to separate the variables F and m to obtain an explicit expression between them.

An explicit equation would contain only an expression in F on one side and an expression in m on the other, which is what you are trying to do.

[Bjarne-7]

11 minutes ago, Bufofrog said:

  If I say F = m * s, it doesn't mean that I have discovered a new force it just means I'm wrong.

Why s, - This equation can transform a lot, and properly more as we think, the result is not s, but a factor, - we are allowed to use such as we want. - If that is false or correct is a different story

7 minutes ago, studiot said:

This is progress.  +1

😀

The plus 1 is because this is the first time you have laid things out in proper fashion.

You see that immediately, even whilst I was writing this, folks have been able to see what you are doing and offer sensible and useful comment.

 

However listing your symbols and stating what they stand for is really good practice and what I mean by progress.

So many waste so much time and effort just writing algebra down.

(did you know that although our word, algebra was,  was named after early arabic 'al-jabr', which meant completion ?)

 

I am sorry but yes your algebra is flawed.

So let us look at your algebra. (I will only work the first equation)

You claim this expression for your RR Force.

F=1−11+mv2c2−−−−−√

 

To show that this does not lead to the expression for the force per unit mass as you then stae, proceed as follows.

F−1=11+mv2c2−−−−−√

(F−1)2=11+mv2c2

1(F−1)2=1+mv2c2

1(F−1)2−1=mv2c2

1−(F−1)2(F−1)2=mv2c2

1−F2+2F−1F2−2F+1=mv2c2

c2F(F+2)mv2(F−1)2=1

We have now reached the stage where we can isolate F/m, your force per unit mass.

Fm=v2(F−1)2c2(F+2)

 

And we some that not only is the expression much more complicated than yours, but it still contains F.

This is because the original expression is what is known as implicit.
 

That is it is not possible to separate the variables F and m to obtain an explicit expression between them.

An explicit equation would contain only an expression in F on one side and an expression in m on the other, which is what you are trying to do.

 

 

 

 

 

 

 

 

Please wait until I have moved m as I wrote some post earlier - a change will come

Edited November 27, 2023 by Bjarne-7

[Bufofrog]

3 minutes ago, Bjarne-7 said:

Why s

No particular reason at all.

4 minutes ago, Bjarne-7 said:

This equation can transform a lot

What is that supposed to mean?

[joigus]

1 hour ago, Bjarne-7 said:

Remember none of the big names you mentioned have solved any of the huge challenges we face in komos. The modification I point out solves them ALL in one fell swoop, - IS it a coincidence?

No. It seems to be a fantasy.

1 hour ago, Bjarne-7 said:

I believe that when you introduce a new law, you decide for yourself which unit result it should give.

No. It should have a clear empirical content, or at least be possible to relate to such empirical content by the theory. Your formulae do neither.

You might want to take a look at this:

https://plato.stanford.edu/entries/operationalism/

So it's not just about writing nice equations that satisfy my sense of what this or that should be.

[swansont]

2 hours ago, Bjarne-7 said:

I believe that when you introduce a new law, you decide for yourself which unit result it should give.

Units need to be consistent with existing physics

2 hours ago, Bjarne-7 said:

For example what does the speed of light have to do with energy (E=mc^2 )
What do mass and radius have to do with Force (MG/r^2)

mc^2 has units of energy and Einstein derived the equation. He didn’t just cough it up on a whim.

Mass and distance affect the force of gravity; this is observed in nature with orbital behavior.

 

2 hours ago, Bjarne-7 said:

etc.
Also remember that the result of Lorentz transformation is unitless and is used to transform time, mass, energy

Transformations are unitless because they have to be - you don’t change the units of the quantity you are transforming 

2 hours ago, Bjarne-7 said:

Not always we are able to understand why equations are like they are. Isacc Newton was critized because he could not explian WHAT gravity was, so much that he delyed to publish his finding for years.  Einsten even did have any idear why matter and space somehow must be connected. - etc. 

etc.

You’re not being asked for the “why” in this fashion. But consistent reasoning, physically reasonable behavior of equations and proper units are required. 

[Bjarne-7]

2 hours ago, swansont said:

Units need to be consistent with existing physics

Usually Yes, but not necessarily when a new law is introduced.

2 hours ago, swansont said:

mc^2 has units of energy and Einstein derived the equation. He didn’t just cough it up on a whim.

Still why exactly c ?

2 hours ago, swansont said:

Mass and distance affect the force of gravity; this is observed in nature with orbital behavior.

Yes, we know that today, but in Newton's time, people had no idea what the cause of gravity was.
No one had thought that it was a force, - only Newton.
For years he kept his records entirely to himself because he was constantly mopped that he didn't know enough to make such claims, especially what Earth had to do with F.
I have read several books about Newton's life, so I know.

Units included in the equation GM/r^2 are: m (mass) and r (distance) - but the result is a completely different unit:  F.
Here the units clearly do not agree with each other. 
You can allow yourself to do that when you introduce a new law.

2 hours ago, swansont said:

Transformations are unitless because they have to be - you don’t change the units of the quantity you are transforming 

The result F in the equation GM/r^2 is based, as I wrote above, on M and r, - but purely mathematically these units are meaningless, they could just as well have been without units. - And precisely for this reason, in a new law / equation that also results in F, you can very well use values without unity and still get the result F. Something similar has actually happened before in history.

2 hours ago, swansont said:

You’re not being asked for the “why” in this fashion. But consistent reasoning, physically reasonable behavior of equations and proper units are required. 

I dont agree. - F=m * a  also this equation is a "transformation equation", again you see no connection between the units, - but what you see is kinetic energy converted to M, 
- it is the "same" (similar) process  as you see in the Lorentz Transformation. - It is therefore not mysterious that F can be involved in both equations. Whether one believes this claim is another matter entirely.

I've been sitting too much in front of my PC for the last week, and I've had a lot of pain in my shoulders, so I have to take a break.

Edited November 27, 2023 by Bjarne-7

[studiot]

17 minutes ago, Bjarne-7 said:

Usually Yes, but not necessarily when a new law is introduced.

Usually new work of any description extends existing knowledge, but remains compatible with where existing knowledge is known to work.

The have been a very few cases where the new work has actually contradicted existing thinking, caloric would be an example, but such examples are very few indeed.

And there has never been a case where the existing rules of mathematics has been breached in the way that you are trying to do here.

[Bufofrog]

2 hours ago, Bjarne-7 said:

Still why exactly c ?

Google "derive e=mc2".

2 hours ago, Bjarne-7 said:

The result F in the equation GM/r^2 is based, as I wrote above, on M and r, - but purely mathematically these units are meaningless, they could just as well have been without units.

Wrong if you leave off the units the answer is meaningless.  Is r in km or furlongs?

2 hours ago, Bjarne-7 said:

dont agree. - F=m * a  also this equation is a "transformation equation", again you see no connection between the units, - but what you see is kinetic energy converted to M,

F = ma says nothing about energy.  The equation says, for instance, kg times m/s^2 results in Newtons which is units of force.

 

[swansont]

4 hours ago, Bjarne-7 said:

Here the units clearly do not agree with each other. 

The agreement of units is that each side of an equation must resolve to the same units. Individual terms do not have to be the same.

Quote

You can allow yourself to do that when you introduce a new law.

Newton could, because he had a clean slate and was inventing physics. But we’re not dealing with a new branch of physics.

[Bjarne-7]

12 hours ago, studiot said:

Usually new work of any description extends existing knowledge, but remains compatible with where existing knowledge is known to work.

The have been a very few cases where the new work has actually contradicted existing thinking, caloric would be an example, but such examples are very few indeed.

And there has never been a case where the existing rules of mathematics has been breached in the way that you are trying to do here.

Let's look again at F=GM/r^2=a
Of course, all these units are important to us.
But mathematically, the units M and r are unimportant, only the numerical values are important.

The units make us understand a relationship, but basically we don't know what F is.
Units do not contribute to understanding the nature of F
In Newton's time, this question was a very hot potato, just as it will be if the law I now assert may/may not be true.

A new F - law can very well be built on a factor without unit.
Mathematically, the unit in this regard are irrelevant.
Experience will determine whether a new F law is right or not.
But that not means guarantee that we will  reach a deeper understand of the the nature of F.

One may wish that F, which is linked to mass and enerti, should contain the same units as F linked to space. But this wishful thinking is mathematically unjustified.

7 hours ago, swansont said:

The agreement of units is that each side of an equation must resolve to the same units. Individual terms do not have to be the same.

I can only agree with you, it sounds very logical, but it is not certain that nature follows our logic

7 hours ago, swansont said:

Newton could, because he had a clean slate and was inventing physics. But we’re not dealing with a new branch of physics.

Perhaps the exploration of the nature of space is a completely new slate?
Again and again we have to realize that we do not have the tools to solve plenty of big challenges that continue to pile up.

It requires enormous energy to reach high speeds, - it requires infinite energy to reach c, - therefore there must also be a corresponding resistance. - Something is preventing. - To claim that we know the cause of that is too stupid. - We should be open to possible solutions. Somewhere we have to start to dig deeper into the rabbit hole.

The nature of the new Force  (Relativist Resistance Agains Motion )

It is usually not the mathematics itself that is decisive for whether the introduction of a new law is accepted or not.

But to a much greater extent if it can solve the challenges we are facing.
It is therefore interesting to see and understand a little more about how significant an influence we are exposed to when it comes to Dark Flow.

Let's say that Dark flow is 1000 km/s - RR = 5.5e-6m/s²  
DFA is an equivalent value and therefore also = 5.5e-6m/s²
If a galaxy's inclination is 100% linear with the Dark Flow axis, this will periodically reduce RR by 300km/s = 5e-7m/s²  .  

This is then the acceleration imbalance between DFA and RR, - and this explains that it is interactions between RR and DFA that are responsible for a galaxy's energy and not the galaxy's / galaxy's own mass.
The large orbits are therefore totally controlled by RR and DFA.

When an object moves under an atronomic object, this can cancel out DFA. 
Therefore release of Dark Flow Related Tension (DFRT) happens = 5.5e-6m/s²
In a cosmic context, this is a rather significant factor. 
Therefore, we will see many more objects (than Oumuamua and Borisov)  that move in a way revealing to completely under the control of DFA,  - just as galaxies are. 
And therefore we cannot explain this energy either. 
This means that we should expect to discover far more objects that move more or less North / South, - and which we wrongly think "must" come from other solar systems.

RR is a result of converting kinetic energy to mass.
RR requires a force for a cash speed to be maintained.
Otherwise, we will see deceleration which occurs opposite to an absolute direction of movement. 
RR is not a classic force, and therefore will not be able to be combined with other forces. 
This is maybe also the reason that the mathematic not always is similar to classis Forces. 

A better understanding of the new F, - can certainly contribute to our having to recognize again and again that such a force is lurking and that we now have a tool to uncover it. It takes a long time to recognize that RR is now exposed and therefore finds it harder to hide. 

[Bjarne-7]

10 hours ago, swansont said:

The agreement of units is that each side of an equation must resolve to the same units. Individual terms do not have to be the same.

Its not a classic acceleration, because each second "v" is no longer the same, and "v" is the basic of the equation, so the acceleration its not linear,. - This is properly the reason why we dont see a "clean math", - and that the the expression "acceleration" only is a approximation.- All we have is really a snapshot of the first second. - Then we need to repeat the equation.  As I wrote this is rocket science when it is worst. 

3000 USD + guaranteed world fame to the one that does the non-linear deceleration equation. It can be done, but NOT simple.. - F is certainly correct, but acc requires math talents

Edited November 28, 2023 by Bjarne-7

[swansont]

10 hours ago, Bjarne-7 said:

Perhaps the exploration of the nature of space is a completely new slate?
Again and again we have to realize that we do not have the tools to solve plenty of big challenges that continue to pile up.

It requires enormous energy to reach high speeds, - it requires infinite energy to reach c, - therefore there must also be a corresponding resistance. - Something is preventing. - To claim that we know the cause of that is too stupid. - We should be open to possible solutions. Somewhere we have to start to dig deeper into the rabbit hole.

This does not follow.

You’re arguing that Newtonian physics (or Galilean relativity) must be correct. What evidence do you have that this is so?

Even in Newtonian physics, the energy relationship is nonlinear. The amount of energy it takes to double one’s speed is not twice as large. Relativity just has a different nonlinear relationship, and that relationship is derived from the notion that the speed of light is invariant. That invariance carries with it certain ramifications - that time and distance are relative values, that mass is a form of energy, that no massive object can attain the speed of light, and massless entities must travel at c.

Knowing the cause, or the “nature” of space, are issues of philosophy. Physics investigates how nature behaves, and physics theories have to agree with observation. As such, we quantify the behaviors under investigation. That’s why we need the math - it allows for prediction and comparison with experiment.

 

10 hours ago, Bjarne-7 said:

The nature of the new Force  (Relativist Resistance Agains Motion )

It is usually not the mathematics itself that is decisive for whether the introduction of a new law is accepted or not.

Sure it is. Theory has to match experiment, and we need to quantify the behavior.

10 hours ago, Bjarne-7 said:

But to a much greater extent if it can solve the challenges we are facing.
It is therefore interesting to see and understand a little more about how significant an influence we are exposed to when it comes to Dark Flow.

Let's say that Dark flow is 1000 km/s - RR = 5.5e-6m/s²  
DFA is an equivalent value and therefore also = 5.5e-6m/s²
If a galaxy's inclination is 100% linear with the Dark Flow axis, this will periodically reduce RR by 300km/s = 5e-7m/s²  .  

This is then the acceleration imbalance between DFA and RR, - and this explains that it is interactions between RR and DFA that are responsible for a galaxy's energy and not the galaxy's / galaxy's own mass.
The large orbits are therefore totally controlled by RR and DFA.

When an object moves under an atronomic object, this can cancel out DFA. 
Therefore release of Dark Flow Related Tension (DFRT) happens = 5.5e-6m/s²
In a cosmic context, this is a rather significant factor. 
Therefore, we will see many more objects (than Oumuamua and Borisov)  that move in a way revealing to completely under the control of DFA,  - just as galaxies are. 
And therefore we cannot explain this energy either. 
This means that we should expect to discover far more objects that move more or less North / South, - and which we wrongly think "must" come from other solar systems.

RR is a result of converting kinetic energy to mass.
RR requires a force for a cash speed to be maintained.
Otherwise, we will see deceleration which occurs opposite to an absolute direction of movement. 
RR is not a classic force, and therefore will not be able to be combined with other forces. 
This is maybe also the reason that the mathematic not always is similar to classis Forces. 

A better understanding of the new F, - can certainly contribute to our having to recognize again and again that such a force is lurking and that we now have a tool to uncover it. It takes a long time to recognize that RR is now exposed and therefore finds it harder to hide. 

The thing is, we can investigate more mundane behaviors to test this idea. But you haven’t been able to articulate the idea in a way that allows for others to investigate anything. The equations you have given don’t work in very fundamental ways. You can’t give a direction to the acceleration, and if you call something a force or acceleration, then they have to be those things, in accordance with existing physics. If they aren’t - e.g. if you have something that doesn’t have units of force, then it’s not a force. You have to fix it, or call it something else.

By making vague predictions and explanations for exotic observations, you have jumped ahead of the process of testing your idea. 

9 hours ago, Bjarne-7 said:

Its not a classic acceleration, because each second "v" is no longer the same

Acceleration is acceleration. a = dv/dt

Even with constant acceleration, velocity continually changes. That’s what an acceleration is - a change in velocity.

The issue hare is that you’re convinced that your idea is right, despite not having that math that would allow for testing, but math and scientific evidence are what’s needed to convince physicists. And that’s the bulk of the work in developing a physics theory. Asking someone to do the math for you isn’t likely to gain traction, because for someone who understand what’s going on, it’s pretty obvious that the math isn’t going to work.

[studiot]

14 hours ago, Bjarne-7 said:

To claim that we know the cause of that is too stupid.

To try something and find it doesn't work......................................That is adventurous.

And if you then learn something like why it didn't work then it may have been worth it.

 

To try the same thing again with the same result........................ is foolhardy.

 

To try the same thing a third time.............................................. Now that is just plain stupid.

[Bjarne-7]

On 11/27/2023 at 3:20 PM, studiot said:

This is progress.  +1

😀

The plus 1 is because this is the first time you have laid things out in proper fashion.

You see that immediately, even whilst I was writing this, folks have been able to see what you are doing and offer sensible and useful comment.

 

However listing your symbols and stating what they stand for is really good practice and what I mean by progress.

So many waste so much time and effort just writing algebra down.

(did you know that although our word, algebra was,  was named after early arabic 'al-jabr', which meant completion.

 

I am sorry but yes your algebra is flawed.

So let us look at your algebra. (I will only work the first equation)

You claim this expression for your RR Force.

F=1−11+mv2c2−−−−−√

 

To show that this does not lead to the expression for the force per unit mass as you then stae, proceed as follows.

F−1=11+mv2c2−−−−−√

(F−1)2=11+mv2c2

1(F−1)2=1+mv2c2

1(F−1)2−1=mv2c2

1−(F−1)2(F−1)2=mv2c2

1−F2+2F−1F2−2F+1=mv2c2

c2F(F+2)mv2(F−1)2=1

We have now reached the stage where we can isolate F/m, your force per unit mass.

Fm=v2(F−1)2c2(F+2)

 

And we some that not only is the expression much more complicated than yours, but it still contains F.

This is because the original expression is what is known as implicit.
 

That is it is not possible to separate the variables F and m to obtain an explicit expression between them.

An explicit equation would contain only an expression in F on one side and an expression in m on the other, which is what you are trying to do.

 

 

 

 

 

 

 

 

 

On 11/27/2023 at 3:15 PM, Bufofrog said:

This is getting really out of the realm of science and math into uh, uh... I don't even know.

If you call some random equation a force it does not magically become a force.  If I say F = m * s, it doesn't mean that I have discovered a new force it just means I'm wrong.

Ok, that equation is trivially false.

 

On 11/27/2023 at 3:59 PM, swansont said:

Units need to be consistent with existing physics

mc^2 has units of energy and Einstein derived the equation. He didn’t just cough it up on a whim.

Mass and distance affect the force of gravity; this is observed in nature with orbital behavior.

 

Transformations are unitless because they have to be - you don’t change the units of the quantity you are transforming 

You’re not being asked for the “why” in this fashion. But consistent reasoning, physically reasonable behavior of equations and proper units are required. 

I will read tomorrow,  - before reading you last post, I considered you previous post, and realize you was right, I will try to fix my self, if I don't succeed I will ask for help somewhere.   Its too late now.

[Bufofrog]

59 minutes ago, Bjarne-7 said:

I will read tomorrow,  - before reading you last post, I considered you previous post, and realize you was right, I will try to fix my self, if I don't succeed I will ask for help somewhere.   Its too late now.

Fair enough.

[Bjarne-7]

A NEW RELATEVISTIC LAW

Notes
This new law is ”similar to:  F = MG/r²   

The new law don’t need  r²   - because this law is about the motion resistance in space. 
Left  from the “similar” old classic law is therefore F =MG . 
G is the unit-converter factor (constant). 

In the same way the new law have MK   (K as constant instead of G) , 
Because this is a relativistic law “ f ” - is certainly needed. 
It looks like the new law only need “ f ” as “double functional” -  transformation / unit-converter. - Why ? - we can speculate, - if true properly the answer is "hidden in the math". 

Exact measurement in the future will properly cause K to drop out og the equation, if not this is a “adjustment parameter”. 

Left to do is only the test the law by measurements and mathematical calculations. 

Edited November 30, 2023 by Bjarne-7

[Bufofrog]

What are the values and units of f and K?

[Bjarne-7]

26 minutes ago, Bufofrog said:

What are the values and units of f and K?

K = 1 

f = the transformation factor

What the equation shows is that " f  " consists of a variable.
"f" represent a common proportional extension of both s and m (meter / ruler) 
The equation therefore shows that when " f " changes (increases its value) , then the RF has to follow. 

Conversely, the equation also shows that when " f " deteriorates , - and F is constant there will be an excess of F.
By other words:  " f " represents positive meter/s (or negative) depending on whether "f" increases or decreases its value.

Because the equation already shows an object in motion, " f " will repeat its value for every second that as M moves, - and therefore the movement causes M to either accelerate or decelerate.
K therefore becomes redundant, as this is already represented by " f "

Edited November 30, 2023 by Bjarne-7

[swansont]

Force is a vector. What is the vector term on the right-hand side of the force equation? What is the direction of the force?

[Bjarne-7]

1 hour ago, swansont said:

Force is a vector. What is the vector term on the right-hand side of the force equation? What is the direction of the force?

Lets say you will send a space rocket out  in space (towards an 100% true absolute motion direction)  , - now the vector is known.

As soon the rocket burns out of fuel RR- declaration will happens 180 degree opposite the absolute motion direction.

 

Edited November 30, 2023 by Bjarne-7