Vis_viva ? (Dark Energy)

[Capiert]

Vis_viva=2*KE

 is twice the kinetic energy

 KE=m*v*va

 of a mass m

 multiplied by (both) its ((linearly) accelerated)

 speed_difference v=vf-vi,

 &

 average_speed va=(vi+vf)/2,

 for the initial & final speeds vi, vf.

 

--

(That (double KE) was)

Something I call(ed) speed_energy

SE=2*KE, (pronounced: see to key)

to get rid

of the 2 divider

(way back when .., in my notes).

I.e. Depetto's (1903) E=m*(c^2)

 looked like twice a KE formula.

("Wasn't that (rest mass energy) missing the half?")

--

 

 Gravesande (had) dropped various brass ball weights in clay mud,

 & found:

 

The product of height h

 & weight Wt=g*m,

 produced identical impressions;

 

 & so he concluded (what I call, (& nobody else))

 Leibnitz's "v's: v; va" was (=were) correct. -Pun.

 

What a joke. (=Sidetrack (away) from Newton's correct momentum physics.)

 

Now, 100's of years later,

 we've got a big problem

 with unknown (=dark) energy.

 

I.e.

Although we can show

 a linear height relation

 with speed_squared

 (apparently) supporting energy,

 physicists did not pin point

 their missing double mass 2*m (factor).

E.g. dark (=unknown) mass. 

 

Summary: 

mom^2=m*E*(v/va),

E=(mom^2)*(vf+vi)/((vf-vi)*2*m). 

 

2017_07_29_1124_Gravesande’s_vis_viva_2017 07 29 1124 PS Wi.pdf

Edited July 29, 2017 by Capiert

[Strange]

How does kinetic energy or vis viva relate to the accelerating expansion of the universe?

[Mordred]

If Kinetic energy is of a two times value and you apply the scslar modelling formula in cosmology.

Your universe would blow up long before the first stars could possibly form.

https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology)

So you don't need to take my word for this,

https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1309.4188&ved=0ahUKEwijquKl7q7VAhXJhFQKHTKRCCMQFggiMAE&usg=AFQjCNGmyIRwg87z0Dg-lQUBqV_uhS7exQ

apply your kinetic energy as 2 times to equations 1 thru 9 of the last link, which you will notice equation 9 reflects the kinetic energy term.

Not to mention conservation of energy momentum violations.

Your pdf paper needs some serious work...I will leave it at that....

Edited July 29, 2017 by Mordred

[Capiert]

On 29 July 2017 at 3:31 PM, Strange said:

How does kinetic energy or vis viva relate to the accelerating expansion of the universe?

That's a mighty fascinating question Strange.

Gravitational push (e.g. expanding matter theory=hypothesis)

 supports an accelerating expansion,

 & a steady state (=NO big bang*),

 & that black holes are an optical illusion

 (=the red shifts are observed as different speeds further from the universe's center).

I see KE as a(n incomplete) distortion of Newton's (linear acceleration) mechanics.

(E.g. How can you support both COE & COM

 (conservations of: energy & momentum),

 when the math doesn't work

 to unite them?)

 

*I just can't maintain the religious belief

 that all this universe

 was magically created

 in an instant,

 (like someone had done it, abracadabra!)

It's so rediculous (=absurd) to maintain

 the superstitious tendancies

 of our (midevil) ancestors

 (if you can pierce thru the(ir) psychology).

Tradition is popular,

 so it'( wa)s kept (to sell the idea, e.g. books).

 

 

Edited July 31, 2017 by Capiert

[Strange]

1 hour ago, Capiert said:

Gravitational push (e.g. expanding matter theory=hypothesis)

 supports an accelerating expansion,

 & a steady state (=NO big bang*),

 & that black holes are an optical illusion

There is no such theory. 

 

1 hour ago, Capiert said:

*I just can't maintain the religious belief

 that all this universe

 was magically created

 in an instant,

I don't care about your religious beliefs. 

 

1 hour ago, Capiert said:

(E.g. How can you support both COE & COM

 (conservations of: energy & momentum),

 when the math doesn't work

 to unite them?)

Except the math does work. 

[Mordred]

I see so because you don't understand how Coe and Com works with the potential and kinetic energy relations to maintain the conservation laws. 

You assume they are wrong due to not knowing how to do the math correctly.  

Got it.

[latex]ke=m×v*v*a[/latex] this is a garbage formula that doesn't even match your descriptive you give your formula.

Edited July 31, 2017 by Mordred

[Capiert]

15 minutes ago, Mordred said:

I see so because you don't understand how Coe and Com works with the potential and kinetic energy relations to maintain the conservation laws. 

You assume they are wrong due to not knowing how to do the math correctly.  

Got it

Not yet.

Please (simply) show me that math you mean

(which unites both conservation laws).

[Mordred]

please show how the equation I quoted from your paper passes a simple dimensional analysis.

 

Your units on the LHS do not match the units on the right hand side of the equal sign. Your first equation fails dimensional analysis. This makes the equation invalid.

After work If you want I will show how COM and COE is applied correctly with KE and PE. Its a basic high school lesson

Edited July 31, 2017 by Mordred

[Capiert]

1 hour ago, Mordred said:

I see so because you don't understand how Coe and Com works with the potential and kinetic energy relations to maintain the conservation laws. 

You assume they are wrong due to not knowing how to do the math correctly.  

Got it.

ke=m×v∗v∗a

this is a garbage formula that doesn't even match your descriptive you give your formula.

I agree that formula you typed is garbage,

 but don't you think you made a typo (error)?

I typed

KE=m*v*va.

va (=average_speed), NOT 

v*a (=speed_difference multiplied by acceleration).

 

But thanks for showing me my formula that you meant.

Edited July 31, 2017 by Capiert

[imatfaal]

OK - So how are you calculating average speed; because at first glance apart from constant acceleration scenarios that formula does not work.  I ride my bike 1km with at a cruising speed of 10m/s; I accelerate to that 10m/s at 1m/s^2 from 0m/s.  After 11, 12, 13 ....100 seconds I would have my KE as identical and unchanging.  You would have it slowly increasing as my average speed for the trip increased.  That is nonsense - how can an instantaneous measurement of KE depend on the length of the journey

[swansont]

2 hours ago, Capiert said:

Gravitational push (e.g. expanding matter theory=hypothesis)

 supports an accelerating expansion,

!

Moderator Note

You don't get to introduce speculation without supporting it with a model of some sort. Also, any speculation introduced must be explained with mainstream physics, not another speculation.

You've been around long enough that you can't plead ignorance of the rules

 

 

 

[Mordred]

4 hours ago, Capiert said:

I agree that formula you typed is garbage,

 but don't you think you made a typo (error)?

I typed

KE=m*v*va.

va (=average_speed), NOT 

v*a (=speed_difference multiplied by acceleration).

 

But thanks for showing me my formula that you meant.

Lets try a basic equation

[latex] 1 metre *1 metre=(unit?)[/latex]

now keep this basic example in mind does

[latex] kg*m/s*m/s*m/s^2=ke?[/latex]

joule is[latex] kg*m^2*s{-2}[/latex] 

I don't know about you but I would find such a mistake extremely embarassing

Edited July 31, 2017 by Mordred

[Capiert]

36 minutes ago, Mordred said:

Lets try a basic equation

1metre∗1metre=(unit?)

now keep this basic example in mind does

kg∗m/s∗m/s∗m/s2=ke?

joule iskg∗m2∗s−2  

I don't know about you but I would find such a mistake extremely embarassing

(I agree, but it looks like I did NOT make it (=that mistake),

 & you (accidently) did. Y/N?)

Units: meter * meter = meter_squared.

Your ke=m*v*v*a is wrong,

 & my KE=m*v*va is correct.

I do NOT know why

 you (want to) perpetuate an error.

Do you (know why)?

Do you NOT recognize the difference

 between both formulas (KE & ke)?

Surely that error must be obvious (=trivial). ?

Edited July 31, 2017 by Capiert

[Capiert]

3 hours ago, imatfaal said:

OK - So how are you calculating average speed; because at first glance apart from constant acceleration scenarios that formula does not work.  I ride my bike 1km with at a cruising speed of 10m/s; I accelerate to that 10m/s at 1m/s^2 from 0m/s.  After 11, 12, 13 ....100 seconds I would have my KE as identical and unchanging.  You would have it slowly increasing as my average speed for the trip increased.  That is nonsense - how can an instantaneous measurement of KE depend on the length of the journey

 va=(vi+vf)/2=(0 m/s + 10 m/s)/2=5 m/s

I guess (accelerated) average_speed (va) is the wrong name

 (for what I'm trying to say).

What would you propose

(for the accelerated average speed)?

 

(average accelerated speed, =average_(accelerated)_speed?)

 

 

Speed_difference

v=vf-vi=10 m/s - 0 m/s=10 m/s.

Edited July 31, 2017 by Capiert

[Mordred]

oh my the blooming SI unit for velocity is m/s the blooming SI unit for KE is kg*m^2*s^-2.

Learn how to do a simple dimensional analysis on your equations.

You cannot take a formula that passes dimensional anslysis 

[latex]ke=1/2mv^2[/latex] then add any random variable without changing the left hand side of the equation to keep it balanced

 

I posted the fricken units for the formula you provided.

you had ke=mass times velocity ×velocity×acceleration. All I did was convert that to units.

Even if you have

[latex] ke=m×v×va[/latex] as written this is mass ×velocity ×velocity×accelerationthe proper form to write average velocity is v_a or

[latex]v_a[/latex]

 

Though I have absolutely no idea why you think average velocity is needed

Edited July 31, 2017 by Mordred

[Capiert]

1 hour ago, Mordred said:

Oh my, the blooming SI unit for velocity is m/s, the blooming SI unit for KE is kg*m^2*s^-2.

Learn how to do a simple dimensional analysis on your equations.

You cannot take a formula that passes dimensional anslysis 

ke=1/2mv2 then add any random variable without changing the left hand side of the equation to keep it balanced.

 

I posted the fricken units for the formula you provided.

You had ke=mass times velocity ×velocity×acceleration.

No I did NOT.

I (really) provided

KE=mass x velocity x average_(accelerated)_velocity

(as Imatfaal pointed_out =indicated, a bit, was needed).

Quote

All I did was convert that to units.

Which is a good (standard) method to trouble_shoot.

Quote

Even if you have

ke=m×v×va

which I did (have)

Quote

as written this is mass ×velocity ×velocity×acceleration

No it's NOT because "you" are using 2 different syntaxes

 at the same time for multiplication,

 instead of being consistent

 with only 1.

ke does NOT mean k x e in your syntax, does it?

Why then should va mean v x a?

Quote

the proper form to write average velocity is v_a or

va

Well at least now we know

 where that 1 went wrong.

It's an honest error (typo).

Quote

Though I have absolutely no idea why you think average velocity is needed.

It's just convient (to simplify the formula, compacter; & make some sense of it (=interpret it, for me)), that's all.

A helpful (equality) tool?

The math tells me what it is.

It looks easier (to me).

Edited July 31, 2017 by Capiert
Format

[Mordred]

If you were trying to derive the kinetic energy formula starting with f=ma then average velocity is part of the solution. However that would lead to ke=1/2mv^2 ie from rest to final velocity.

Edited July 31, 2017 by Mordred

[Capiert]

57 minutes ago, Mordred said:

How can average velocity possibly simplify the problem? We don't care about the amount of energy to get from a to b, we care how much  energy is behind tbe impact, example how much kinetic energy is involved when a bullet hits a wall. For that we only need the mass of object and velocity at the time of impact.

If the bullet recoils (=bounces, backwards)

 then that (reversed) speed

 also plays a role

 (for the transfer).

-Newton's 3rd law.

 

(I suspect)

 the average (accelerated) velocity

 (only) simplifies the syntax (formula),

 (as an alternative).

E.g. an equivalent, or substitution.

 

(That impact)

 Energy has the ability

 to accelerate mass.

 

I guess it only helps me:

 with constant linear acceleration g

 versus fallen height h.

e.g.

va=h/t.

Edited July 31, 2017 by Capiert

[Mordred]

What you need to do is understand the work energy theorem to understand why 1/2mv^2. Here is a simple derivitave for you

https://physics.stackexchange.com/questions/35987/why-there-is-a-1-2-in-kinetic-energy-formula 

I changed my previous post as I forgot your examining vis viva which has its own treatments (also its own formula)

https://en.m.wikipedia.org/wiki/Vis-viva_equation

 

Though you will see vis viva aka living force is more commonly shown as mv^2. I had forgotten about the vis visa and kinetic energy debate

Edited July 31, 2017 by Mordred

[Capiert]

I think we're a bit off topic there.

Gravesande dropped balls into clay.

I.e. (dropped) brass balls

 were (passively, linearly) accelerated (with constant acceleration)

 from rest (velocity) vi=0 m/s

 to a final (fall) velocity vf

 (obtainable) without any calculus.

 

Your bullet was accelerated

 in the gun

 before travelling

 to strike with impact.

I don't know how it's acceleration was

 during the explosion,

 & it's too complicated

 for me to calculate (that)

 as such.

 

(Anyway: )

If I have a 1/2 in that (constant, linear acceleration) equation

 it is often simply due to an average

 using the triangle's area A=b*h/2, b=base, h=height, rule,

 when plotted graphically.

That's simple enough for me to understand.

I also know something will fall g=-9.8 m/(s^2),

 meaning (starting at rest vi=0 m/s)

 after 1 second,

 it will have fallen (the height) h=-4.9 m

 (e.g. faster).

From that,

 & using standard mechanics, displacement (distance)

 (s=vi*t + a*t*t/2)

 I can derive the rest of the free fall equation

 (at least confirm it as)

 

 h=vi*t+g*t*t/2,

 

 where g=-9.8 m/(s^2), g=v/t, so v=g*t, v=vf-vi,

 & h=hf-hi, (postscripts) f=final, & i=initial

 for (duration) time t.

 

Their (complicated?) math

 does NOT show me the connection

 between the 2 conservation laws (com & coe)

 which my math shows me

 does NOT exist.

 

Maybe you know a simpler strategy?

 

 

Edited August 1, 2017 by Capiert
typo

[Mordred]

Ok well I am not going to give a dozen examples to determine which system you want to describe kinetic energy and the conservation laws under you want a center of mass frame your automatically dealing with a multi-particle system. This alone complicates the math so you need to define the conservation of momentum and conservation of energy under a center of mass frame and account for all particles.

Your math isn't looking at the conservation laws directly kinetic energy isn't always a conserved quantity. It is in perfect elastic collisions but not in inelastic collisions. Well here is a pdf to save us some time

https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-07-dynamics-fall-2009/lecture-notes/MIT16_07F09_Lec11.pdf

 

you can see from this the subject gets complex depending on the system state..(also the effective degrees of freedom )

You don't seem to have any problem with the simple examples of kinetic energy so the pdf will help you get into multiparticle systems.

The thing is conservation of momentum and conservation of energy has two distinct definitions and as such has two distinct values.

conservation of energy being a scalar quantity

conservation of momentum being a vector quantity.

kinetic energy is a scalar quantity so you need to add the directional components in your examination to get the conservation of momentum.

This is where your vis visa arguments come into play...

though quite frankly energy regardless of type is simply the ability to perform work,  a particle with momentum naturally has the ability to perform work. What you need to do is examine how the conservation laws applies to the amount of work performed within your system. That is not strictly energy but a combination of mass and energy. It is also not strictly based on kinetic energy as you need the potential energy as well.

This is where your problem is in seeing the connection, from what I can tell from your posts and mathematics.

 

lol side note it will take a huge understanding to show that the stress tensor in GR is wrong when it comes to the dark energy as per your opening post and article.  You are nowhere even close to having the required details...in order to even begin to understand the stress energy/momentum tensor you will need to fully understand multiparticle system modelling. None of your math even comes close to describing a multiparticle system under conservation laws for an adiabatic expansion.  You need to apply the ideal gas laws and how the three not two conservation laws apply.

lets show conservation of energy under the stress tensor used by GR just to demonstrate.

[latex]T_{\mu\nu}=(\rho+p)U_\mu U_\nu+Pg_{\mu\nu}[/latex]

where

[latex]U_\mu=(1,0,0,0)[/latex] is the 4 velocity in a comoving frame

conservation of energy using the covariant derivative is therefore

[latex]\Delta T_0^{\mu}=0[/latex] gives

[latex]\frac{\dot{\rho}}{\rho}=-3(1+w)\frac{\dot{a}}{a}=-3(1+wH)[/latex]

 

see what I mean by tons of study needed to describe conservation of energy I just described? That is assuming conservation of energy even applies in Cosmology ( there is an ongoing debate on that).

In order to understand what I just posted you need to learn Gauss divergence theorem, the 4 momentum and 4 velocity under GR, the field equations under Noethers theorem as well as multi particle/field modelling as well as the ideal gas laws which these all use

Trust me your just getting started.....for example the 1/r potential conservation equation for  the FLRW metric from the Einstein field equation above is

[latex](\frac{\dot{a}}{a})^2=H^2=\frac{8\pi G}{3}\rho-\frac{k}{a^2}[/latex]

f-ma for a 1/r^2 force field is

[latex]\frac{\ddot{a}}{a}=-\frac{4\pi G}{3}(\rho+3P)[/latex]

so if your goal is to solve dark energy by looking for a fault in kinetic energy and conservation laws as to how they are applied. Well then you really have your work cut out for you......

edit I should explain the 1/r potential this is a scalar field. 1/r^2 is a point charge ie center of mass (ie gravity)

 

Edited August 1, 2017 by Mordred

[Mordred]

On ‎2017‎-‎07‎-‎29 at 11:23 AM, Mordred said:

If Kinetic energy is of a two times value and you apply the scslar modelling formula in cosmology.

Your universe would blow up long before the first stars could possibly form.

https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology)

So you don't need to take my word for this,

https://www.google.ca/url?sa=t&source=web&rct=j&url=https://arxiv.org/pdf/1309.4188&ved=0ahUKEwijquKl7q7VAhXJhFQKHTKRCCMQFggiMAE&usg=AFQjCNGmyIRwg87z0Dg-lQUBqV_uhS7exQ

apply your kinetic energy as 2 times to equations 1 thru 9 of the last link, which you will notice equation 9 reflects the kinetic energy term.

Not to mention conservation of energy momentum violations.

Your pdf paper needs some serious work...I will leave it at that....

did you even understand this as per the scalar modelling formula under the first link?

If you double the kinetic energy your universe would expand far too fast...

You want a way to connect the conservation laws under GR ??

then study the energy/momentum four vectors.

http://oyc.yale.edu/sites/default/files/relativtity_notes_2006_8.pdf

this paper discusses the conservation laws under the four vector... it will help get you started...

here is the four momentum conservation

https://www.physics.mcmaster.ca/~yavin/courses/PHYS4E03/02Mar2015.pdf

this link is of better quality

http://www.phys.ufl.edu/~acosta/phy2061/lectures/Relativity4.pdf

it will tie in your e=mc^2 relations with KE under relativity. One line of interest is this statement

"As we have learned, mass is a form of potential energy.  It can be converted into energy, or energy can be converted into mass.  Because of this, mass does not have to be conserved in reactions.  If you throw two balls at each other and they stick together (an inelastic collision), the resulting mass is not necessarily the sum of the individual masses of the two balls"

The article discusses this but as I mentioned KE isn't conserved in inelastic collisions...

Another key note..mass changes with velocity via equation

[latex]m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}[/latex]

a consequence of SR is that mass changes with velocity so momentum also changes... see why you need conservation of four momentum?

 

Edited August 1, 2017 by Mordred

[Capiert]

Thanks. You've done an awful lot of preparation.

(I've got (way too) much to tackle. ..as you said.)

"did you even understand this as per the scalar modelling formula under the first link? "

No, I don't think it registered (well enough).

 

But my major interest

 is in Gravesande's experiment

 & the energy discrepancies

for a(n energy versus momentum) calibration,

 (preferably exclusively with algebra,

 because I don't trust calculus).

I don't see why we need, tensors, calculus, or relativity,

 when (serious (algebraic)) math problems are already obvious.

(& the others are not my specialty, at all, anyway.)

 

You know a vast amount of details,

 to point me in the right direction.

 

Put simply, momentum & KE

 deal with only 2 concepts

 (at sub_light speeds):

 mass & velocity.

That means if we know those 2

 then we (always) know momentum

 & KE & know they don't agree.

Just dump the info in an excel table (spreadsheet)

& it won't lie that there is a (big) problem.

& it's simply algebra.

 

Edited August 1, 2017 by Capiert

[swansont]

31 minutes ago, Capiert said:

 But my major interest

 is in Gravesande's experiment

 & the energy discrepancies

for a(n energy versus momentum) calibration,

You have not clearly explained what you think the problem is.

[swansont]

18 hours ago, Capiert said:

Their (complicated?) math

 does NOT show me the connection

 between the 2 conservation laws (com & coe)

 which my math shows me

 does NOT exist.

 

Maybe you know a simpler strategy?

Why should there be a direct connection? Momentum is conserved when the net external force on an object or system is zero. It stems from the symmetry under coordinate transformation. Energy is always conserved within a system. It stems from  symmetry under time translation.

They are not interchangeable, and there are ways of defining systems such that either or both are not conserved, or that both are.