Speed of light = oo mass = black hole?

[MassiveMass]

So I'm new to this, but it's one hobby that fascinates me. I figured I'd come here to listen to your thoughts. I was thinking the other day about an alternative energy and naturally "too bad we can't tap into a quasar" came up. For now I think the thermosphere seems to be the biggest source of raw clean energy here.

Here it is. What if you harnessed "some of Super Conductor of the future". Then with this Mega Massive super conductor you found "some way of utilizing power from a quasar". This is all a little out there so bear with me but this is the kind of energy you'd want to actually test this theory in theory. We'd then use that energy to power a large spacecraft that could accelerate stably to near or if say we could in theory travel fast enough to catch up with light." Now present day we believe this to be almost impossible. Now let's say in theory they are wrong. Well that ship when it reached the speed of light. Here's the question. When we reach this speed of light and our density in theory reaches the critical point would matter implode on itself? Maybe creating a small black hole or singularity like when an incredibly dense star collapses on itself after the white dwarf stage and creates a gravitational vacuum. Well I don't think it defies physics anyway, but some do. I just think there's some point like the speed of light which i think is the "speed of time or how we measure it, but also a max density as well or some sort of critical mass or the point where matter implodes because it's so dense. I'm just curious about these things.

Then there's time. Does the object appear to be moving at a standstill to an outside observer or more likely vice versa based on relativity. Wouldn't that kind of gravity at the speed of light warp time itself so what would that look like to an observer coupled with the standstill. Please help me educate myself in this.

Thanks.

 

[mathman]

When something gets close to the speed of light, relative to us, the mass appears very large to us. However in the reference frame of the object itself the mass is still the rest mass.

 

[jtbell]

FAQ: If you go too fast, do you become a black hole?

 

[MassiveMass]

Hey thanks for the help that makes way more sense. Just to be clear. A small object with low density moving very fast doesn't have more gravity then objects of higher density not moving at all. If I'm off just let me know. Thanks.

 

[Elroch]

Hey thanks for the help that makes way more sense. Just to be clear. A small object with low density moving very fast doesn't have more gravity then objects of higher density not moving at all. If I'm off just let me know. Thanks.

An object with little mass moving fast does have a greater gravitational influence, but it is because it has momentum: Einstein's equation says roughly that curvature of space-time is proportional to a thing called the stress-energy tensor. One of the components of this tensor comes from mass, all the others derive from momentum. An extreme example is light, which has zero rest mass, but does have momentum and does have a gravitational influence.

 

[MassiveMass]

That's what I'd previously understood. So the stress-energy tensor is how we actually measure gravitational pull. I did not know that. So then at what factor does the stress-energy tensor become large enough for a black hole? I.E. (put warp drive on a planet for example and accelerate with the energy of a quasar etc. etc.). What is the point at which we reach singularity? Are there other ways to reach singularity?

 

[Naty1]

Post 5 and 6 above are incorrect statements and conclusions.

Oh, I just noticed post #3...read the link posted!

A long discussion is here:

https://www.physicsforums.com/showthread.php?p=3661242&posted=1#post3661242

For a quick perspective, see my post #5 and PeterDonis post #10.

Going faster does NOT offer the opportunity for anything to turn into a singularity, that is, a black hole. Gravity is the only force strong enough to create a black hole and does NOT work that way.

 

[Elroch]

Naty1, you claimed my post was incorrect: please specify precisely to which statement you were referring unless you posted in error.

I stand by my statements.

Specifically, the momentum of a body affects its gravitational influence. The clearest example of this is light, which has zero rest mass, but which obviously causes curvature of space time according to Einstein's field equation.

Note that whether something forms a black hole is obviously independent of the frame of the observer, so is not affected by changes of relative velocity, for example.

 

[Matterwave]

FAQ: If you go too fast, do you become a black hole?

I've always found this FAQ inadequate. It sort of just says "no, but the reasons are complicated". It also implies (I think, through it's rather vague language) a 1 to 1 equivalence of black holes and singularities. Now, I'm not an expert on the singularity theorems of Hawking and Penrose, but I do know that the so called "cosmic censorship conjecture" is simply a conjecture, and therefore naked singularities at least have not been proven to not exist. There is therefore not a 1 to 1 equivalence between a singularity and a black hole. Do the singularity theorems say that for every black hole there is a singularity?

Since the curvature is expressed as a tensor, the components of which are obviously dependent on the coordinate system used, and therefore the frame of reference, it seems to me that the only thing we can say for sure is that a singularity does not form. Obviously since a singularity implies an incomplete geodesic, it cannot exist in one FoR and not exist in another. What about a black hole?

@Elroch: Because the curvature is expressed as a 256 component tensor (the Reimann), it's not so easy to just make a statement like "gravity increases". Which component are you talking about? The Einstein Field equations only specify the trace of the Reimann (the Ricci tensor basically), but do not specify the traceless part of the Reimann (the Weyl tensor). The traceless part is determined in some way by (some sort of global) boundary conditions.

Outside a distribution of mass (in the vacuum), it is only the Weyl curvature which survives. It doesn't seem legitimate to me, then, to say that the "stress energy increases therefore gravity increases" since in the vacuum region, the stress energy tensor is 0 for both a moving or non-moving mass. If you are in the "interior region" where the stress energy does not vanish, then you are obviously moving along with the particle, and not in a FoR where the particle is moving at great speeds.

 

[Elroch]

I've always found this FAQ inadequate. It sort of just says "no, but the reasons are complicated". It also implies (I think, through it's rather vague language) a 1 to 1 equivalence of black holes and singularities. Now, I'm not an expert on the singularity theorems of Hawking and Penrose, but I do know that the so called "cosmic censorship conjecture" is simply a conjecture, and therefore naked singularities at least have not been proven to not exist. There is therefore not a 1 to 1 equivalence between a singularity and a black hole. Do the singularity theorems say that for every black hole there is a singularity?

Since the curvature is expressed as a tensor, the components of which are obviously dependent on the coordinate system used, and therefore the frame of reference, it seems to me that the only thing we can say for sure is that a singularity does not form. Obviously since a singularity implies an incomplete geodesic, it cannot exist in one FoR and not exist in another. What about a black hole?

@Elroch: Because the curvature is expressed as a 256 component tensor (the Reimann), it's not so easy to just make a statement like "gravity increases". Which component are you talking about? The Einstein Field equations only specify the trace of the Reimann (the Ricci tensor basically), but do not specify the traceless part of the Reimann (the Weyl tensor). The traceless part is determined in some way by (some sort of global) boundary conditions.

Outside a distribution of mass (in the vacuum), it is only the Weyl curvature which survives. It doesn't seem legitimate to me, then, to say that the "stress energy increases therefore gravity increases" since in the vacuum region, the stress energy tensor is 0 for both a moving or non-moving mass. If you are in the "interior region" where the stress energy does not vanish, then you are obviously moving along with the particle, and not in a FoR where the particle is moving at great speeds.

Thanks Matterwave, a helpful and informed post which makes me realize how far from complete my understanding is! My wording was (partially intentionally) very loose, indicating merely that momentum contributes to the gravitational field. My understanding is that the curvature resulting from the stress-energy is the source of the large scale gravitational characteristics (metric structure?) in the region outside the matter. For example, orbit period. Is this correct?

It seems trivially obvious that changing your velocity won't turn you into a black hole. How can your velocity have any effect, since you and your physical behaviour are always the same in any frame at rest w.r.t to you?

 

[jtbell]

It seems trivially obvious that changing your velocity won't turn you into a black hole. How can your velocity have any effect, since you and your physical behaviour are always the same in any frame at rest w.r.t to you?

Or look at it this way: we can increase my velocity relative to you either by accelerating me, or by accelerating you. Suppose we do the latter. How can your acceleration turn me into a black hole?

 

[Naty1]

Elroch and Matterwave: Curvature and gravity is waaay more subtle than I thought when I started in these forums...
I've made notes from posts over several years and pieced together what may help you as it did me:

An object with little mass moving fast does have a greater gravitational influence,

is incorrect:

You should read Peter Donis explanation which I referenced above...
"Does mass really increase with speed"

except the posts are # 10, #16...

PeterDonis:

The "amount of gravity produced" by the object is not a function of its energy alone, it's a function of its stress-energy tensor [ SET] of which energy is only one component. In a frame in which the object is moving, there will be other non-zero components of the SET as well as the energy, and their effects will offset the apparent "effect" of the increase in energy, so the final result will be the same as it is for a frame in which the object is at rest.

This is the correct idea:


[QUOTE... changing your velocity won't turn you into a black hole. ...you and your physical behaviour are always the same in any frame at rest w.r.t to you?...Note that whether something forms a black hole is obviously independent of the frame of the observer, so is not affected by changes of relative velocity, for example.[/QUOTE]

also:

How can your acceleration turn me into a black hole?

likely you mean VELOCITY.


I found a different post from Peter which is close to what was referenced above:

... a very high energy electron will cause a lot of spacetime curvature

No, it won't. The spacetime curvature caused by any object is a frame-independent, invariant quantity. Since the object's velocity is not a frame-independent, invariant quantity, the curvature the object causes can't depend on the velocity. (Same argument for the kinetic energy.)

https://www.physicsforums.com/showthread.php?t=548148&highlight=speed+of+an+electron



And one of my favorite explanations relates two types of CURVATURE:


The key is that gravitational curvature IS observer independent (as already noted) and is reflected as curvature of the spacetime manifold ("graph paper" as described below). Frame dependent curvature (observer dependency) is a variable overlay on top of this fixed background curvature,


From Dr Greg:

What we call the "curvature of spacetime" has a technical meaning; the equations that describe it are very similar to the equations that describe, say, the curvature of the Earth's surface in terms of latitude and longitude coordinates, or any other pair of coordinates you might choose. This "curvature" need not manifest itself as a physical curve "in space".

For the rest of this post let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. …
In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

If we switch to a non-inertial frame ([an accelerated observer] but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

So, to summarize, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer.

 

[PeterDonis]

Specifically, the momentum of a body affects its gravitational influence. The clearest example of this is light, which has zero rest mass, but which obviously causes curvature of space time according to Einstein's field equation.

Light does cause spacetime curvature, but this is due to its energy and pressure, not its momentum.

 

[Naty1]

(put warp drive on a planet for example and accelerate with the energy of a quasar etc. etc.). What is the point at which we reach singularity? Are there other ways to reach singularity?

By now, it's likely clear that acceleration and velocity do not cause an increase gravity. They do cause an apparent ...frame dependent...curvature separate from gravity.

Gravitational collapse occurs when an object's internal pressure is insufficient to resist the object's own gravity.

http://en.wikipedia.org/wiki/Black_hole#Gravitational_collapse

Gravitational collapse results in a singularity accompanied by a black hole ...and event horizons of various sorts.

 

[Matterwave]

Which parts of my post are you critiquing specifically? I tried to look for quotes from me but I couldn't find any.

Elroch and Matterwave: Curvature and gravity is waaay more subtle than I thought when I started in these forums...
I've made notes from posts over several years and pieced together what may help you as it did me:
is incorrect:

You should read Peter Donis explanation which I referenced above...
"Does mass really increase with speed"

except the posts are # 10, #16...

PeterDonis:This is the correct idea:

... changing your velocity won't turn you into a black hole. ...you and your physical behaviour are always the same in any frame at rest w.r.t to you?...Note that whether something forms a black hole is obviously independent of the frame of the observer, so is not affected by changes of relative velocity, for example.

also:

likely you mean VELOCITY. I found a different post from Peter which is close to what was referenced above: https://www.physicsforums.com/showthread.php?t=548148&highlight=speed+of+an+electron
And one of my favorite explanations relates two types of CURVATURE:


The key is that gravitational curvature IS observer independent (as already noted) and is reflected as curvature of the spacetime manifold ("graph paper" as described below). Frame dependent curvature (observer dependency) is a variable overlay on top of this fixed background curvature,From Dr Greg:

 

[harrylin]

Or look at it this way: we can increase my velocity relative to you either by accelerating me, or by accelerating you. Suppose we do the latter. How can your acceleration turn me into a black hole?

Yes, I think that that is the best (simple, straightforward and relativistic) explanation.
Can anyone add it to the FAQ please?

PS it is mentioned in there, but in a weak form; it's not stressed that this must be so according to relativity theory.

 

[Dale]

Post 5 and 6 above are incorrect statements and conclusions.

Post 5 was good except for one small nitpick:

One of the components of this tensor comes from mass, all the others derive from momentum.

Here "mass" should be "energy". Other than that minor detail it is fine. I would also probably say "different" gravitational influence rather than "greater", but I am sure that you could justify "greater" in some coordinate systems.

 

[Dale]

So the stress-energy tensor is how we actually measure gravitational pull. I did not know that.

No, the Riemann curvature tensor represents gravity. The stress-energy tensor is the source of gravity. I.e. in Newtonian mechanics the source of gravity is mass, but in GR the source of gravity is a tensor comprised of energy, momentum, pressure, and stress.

So then at what factor does the stress-energy tensor become large enough for a black hole?

It doesn't. As you go faster your energy increases, but so does your momentum. Those two work together to prevent a horizon from forming.

 

[Dale]

Since the curvature is expressed as a tensor, the components of which are obviously dependent on the coordinate system used, and therefore the frame of reference, it seems to me that the only thing we can say for sure is that a singularity does not form. Obviously since a singularity implies an incomplete geodesic, it cannot exist in one FoR and not exist in another. What about a black hole?

This is a good and well-reasoned question. Consider null geodesics in the rest frame of some non-black hole object. Those geodesics all escape to an asymptotically flat region. Now, apply a boost such that the object is moving at relativistic speeds. The null geodesics in the original frame remain null geodesics in the new frame and escape to an asymptotically flat region. Therefore the object is not a black hole.

 

[Dale]

Light does cause spacetime curvature, but this is due to its energy and pressure, not its momentum.

Since light has momentum then it has some corresponding non-zero components in the stress energy tensor, and therefore by the EFE it does cause spacetime curvature due to its momentum also.

 

[Naty1]

Matterwave:

Which parts of my post are you critiquing specifically...

none. Reading your questions I saw you seemed to have some of the same questions I have asked and I thought the posts I have collected might help answer them.

...1 to 1 relationship between black holes and singularities...

yes there is: Roger Penrose seems sure there is a singularity with every black hole:

In the ROAD TO REALITY Roger Penrose says the following on page 712, 23.9, addressing Black Holes:

There are very comprehensive theorems which tell us that singularities cannot be avoided in any gravitational collapse that passes a 'point of no reutrn'.

In a footnote to that passge he adds:

...The occurrence of what is known as a 'trapped surface" is one useful characteristic of such a 'point of no return'. ..We expect to find trapped surfaces inside the horizon 'H' of a black hole...Once a trapped surface is formed, then singularities are inevitable [assuming very weak and reasonable conditions concerning causality andf energy positivity in the Einstein theory... Similar results apply to the Big Bang singularity

 

[nitsuj]

Curvature and gravity is waaay more subtle than I thought when I started in these forums...

Yea, I noticed the same thing the same way. I stopped thinking about GR once it was apparent it was not nearly as intuitive as SR.

With that being said I have a question, an SR one.

How is it that kinetc energy from gluons accounts for most of the mass of objects?

Said differently I am having trouble discerning between the topic discussed in this thread and the idea of kinetic energy from gluons being most of the mass of objects. Like a ship near c that becomes increasingly difficult to accelerate, why don't these gluons give at "rest" objects the same effect.

And asked differently again;
On one hand e=mc2 doesn't mean the faster you go the more massive you get (just equivalent), but on the other hand it seems that it does, such as the case of kinetic energy and gluons.



I'm finding mass tough to define I guess, right now I'm leaning more towards thinking of it as kinetic energy and not at all related to "weight" in anysense. Is the definition of mass and kinetic energy simular?

Heres excerpts from wiki that maybe out of context, i can't tell:

Mass: "... can be defined as a quantitative measure of an object's resistance to the change of its speed."

Kinetic energy: "It is defined as the work needed to accelerate a body..."

Now come to think of it, I don't understand kinetic energy either. lol

Maybe my HS level AP physics study guide will help me understand these terms, if it can't be explained simply here.

 

[jtbell]

On one hand e=mc2 doesn't mean the faster you go the more massive you get (just equivalent), but on the other hand it seems that it does, such as the case of kinetic energy and gluons.

Consider a proton at rest. Loosely speaking, you have a collection of quarks and gluons "buzzing around." That motion contributes to the mass of the proton. (When I say "mass" here, I mean the "invariant mass" a.k.a. "rest mass.")

Now put the proton in motion with some velocity v. That motion does not contribute to the mass of the proton, but the "buzzing around" motion still does, by the same amount.

 

[Naty1]

from this discussion: (3/2011)
https://www.physicsforums.com/showthread.php?p=3209674#post3209674

Are the following statements true ?
1) Increasing the kinetic energy in the form of heat of a body increases its gravitational attraction to another body
2) Increasing the kinetic energy in the form of rotational energy of a body increases its gravitational attraction to another body
3) Increasing the kinetic energy in the form of relative velocity of a body increases its gravitational attraction to another body

Dalespam answered in his post #38:
1)yes,
2)yes,
and via post#41:
3)no...

if two equal-mass objects inertially moving parallel to each other at the same speed pass a stationary massless observer at t=0 then they will collide at some t=T according to the massless observer. This time T will be greater the faster the masses are moving, meaning that the gravitational attraction has decreased.

Remember, the sign of the metric is opposite for the timelike and spacelike terms, so you generally expect the timelike and spacelike components to have somewhat opposite effects. Of course, even that is a big oversimplification and I am sure there are counterexamples.

I'd like to ask Dalespam: you sure about #3??
I thought that any question about an observer and a high velocity passing object(s) could be transformed to an equivalent slow speed object(s) and high speed observer...So if I sit myself on one of the objects, it doesn't seem I get your increase in time...??

 

[Naty1]

Dalespam, good catch, post#20...

Light does cause spacetime curvature, but this is due to its energy and pressure, not its momentum.

Since light has momentum then it has some corresponding non-zero components in the stress energy tensor, and therefore by the EFE it does cause spacetime curvature due to its momentum also.

I think light is unique in that all it's energy derives from momentum. [?]

For massive objectives, there is a distinction between a single particle and between a groups of particles moving at relatively different velocities regarding momentum: (and maybe pressure?)

A system of fast moving particles will have more gravitational attraction than a system of similar slow moving particles. For a system of particles moving rapidly in different directions, all frames will show a system of rapidly moving particles; which is moving which way will change, but you can't transform away the fact that total KE of the system is greater than for a slow moving system of similar particles.

And the old example of a compressed spring in a jack in the box also results in greater gravitational attraction: you can't transform away the compressed spring energy either.

 

[PAllen]

from this discussion: (3/2011)
https://www.physicsforums.com/showthread.php?p=3209674#post3209674



Dalespam answered in his post #38:
1)yes,
2)yes,
and via post#41:
3)no...

I'd like to ask Dalespam: you sure about #3??
I thought that any question about an observer and a high velocity passing object(s) could be transformed to an equivalent slow speed object(s) and high speed observer...So if I sit myself on one of the objects, it doesn't seem I get your increase in time...??

I have not tried to work out exactly what Dalespam described, but there are well known examples of the difference between parallel and anti-parallel motion in GR. For example, two parallel laser beams (per GR) will not attract each other at all, while to anti-parallel beams will. This is related to the fact that collective motion of 'particles' has quite different impact than random motion of particles.

 

[nitsuj]

Consider a proton at rest. Loosely speaking, you have a collection of quarks and gluons "buzzing around." That motion contributes to the mass of the proton. (When I say "mass" here, I mean the "invariant mass" a.k.a. "rest mass.")

Now put the proton in motion with some velocity v. That motion does not contribute to the mass of the proton, but the "buzzing around" motion still does, by the same amount.

Yea that's what I read. But when the proton is put into motion why doesn't the velocity contribute to the kinetic energy of the collection of quarks and gluons buzzing around, inturn making the proton increasingly difficult to move even at small velocities.

The only thing I'm considering right now is what I read in the last few posts of this thread; about how all this "buzzing around" that is kinetic energy is in many different directions.

Does a gyro scope demonstrate kinetic energy? In that it is more difficult to move in some directions compared to other directions.

Anyhoo, next on the learn it list Kinetic energy.

 

[Naty1]

Dalespam:
I just noticed in post #17 you said Post #5 was ok...I did not think this part of post #5 was ok:

An object with little mass moving fast does have a greater gravitational influence, but it is because it has momentum..

I did not think that coordinate dependent velocity affected gravitational attraction.
Can you clarify how you interpretated the above from post #5.

 

[Matterwave]

Matterwave:


none. Reading your questions I saw you seemed to have some of the same questions I have asked and I thought the posts I have collected might help answer them.



yes there is: Roger Penrose seems sure there is a singularity with every black hole:

In the ROAD TO REALITY Roger Penrose says the following on page 712, 23.9, addressing Black Holes:



In a footnote to that passge he adds:

Thanks for this quote =D

 

[jtbell]

But when the proton is put into motion why doesn't the velocity contribute to the kinetic energy of the collection of quarks and gluons buzzing around, inturn making the proton increasingly difficult to move even at small velocities.

For a single particle, the relationship between energy, mass and momentum is

[tex]E^2 = (mc^2)^2 + (pc)^2[/tex]

[tex]mc^2 = \sqrt{E^2 - (pc)^2}[/tex]

where m is the invariant mass ("rest mass"). For a collection of particles, you use the total momentum and energy to find the system mass:

[tex]m_{sys} c^2 = \sqrt{E_{tot}^2 - (p_{tot}c)^2}[/tex]

If the system (e.g. proton) is at rest, then [itex]p_{tot} = 0[/itex] and [itex]m_{sys}c^2 = E_{tot}[/itex].

If the system has an overall motion, [itex]E_{tot}[/itex] increases because of the increased kinetic energy of the component particles, but now [itex]p_{tot}[/itex] is not zero, and it counters the increase in [itex]E_{tot}[/itex], as far as calculating [itex]m_{sys}[/itex] is concerned.

Put it another way: if you increase the kinetic energies of the component particles by increasing the speed of the "buzzing around," it increases the total energy, but not the total momentum, because the increases in the individual particles' momenta are in different directions so they cancel out in the sum. Therefore the system mass increases.

On the other hand, if you increase the kinetic energies of the component particles by making the whole system move in one direction, it increases both the total energy and the total momentum, and the two effects cancel out as far as the system mass is concerned.

 

[Dale]

Dalespam answered in his post #38:
1)yes,
2)yes,
and via post#41:
3)no...

I said yes, it gets complicated, and it is complicated.

 

[Dale]

Dalespam:
I just noticed in post #17 you said Post #5 was ok...I did not think this part of post #5 was ok:

I did not think that coordinate dependent velocity affected gravitational attraction.
Can you clarify how you interpretated the above from post #5.

I said that I would have said "different" rather than "greater". It is complicated because it is the spatial components rather than the time-time component. Things change, but not always in an obvious way like simply getting greater.

 

[nitsuj]

For a single particle, the relationship between energy, mass and momentum isPut it another way: if you increase the kinetic energies of the component particles by increasing the speed of the "buzzing around," it increases the total energy, but not the total momentum, because the increases in the individual particles' momenta are in different directions so they cancel out in the sum. Therefore the system mass increases.

On the other hand, if you increase the kinetic energies of the component particles by making the whole system move in one direction, it increases both the total energy and the total momentum, and the two effects cancel out as far as the system mass is concerned.

WOw so it is like I questioned in the above post with kinetic energy in all different directions.

Not that it matters much to me but is kinetic energy a vector?

This direction thing is so slick, it has to be why a simple shove to an object and it goes forever. the kinetic mass "balance" has been "shifted*". is that generally right? I hope so cause it really starts piecing together from there. Ah and that delta from balance is momentum. (thinking of it more kinetic energy may not be the vector but momentum is, idk)

darn, i re-read your post, so as the velocity increases, that energy is added to the particles which are already causing the kinetic mass; in all directions. The added energy is measured as momentum(?).

Still a very cool concept.

even still yet, if i considered this "shifted (the net of kinetic energy's direction, momentum)" as relative (which makes sense cause it's basically speed I am describing) how/when is it wrong?

 

[Naty1]

jtell: Love your posts 23 and 30..never considered that previously...good insights, thanks!

This thread has helped clarify several things for me.

 

[nitsuj]

jtell: Love your posts 23 and 30..never considered that previously...good insights, thanks!

This thread has helped clarify several things for me.

I couldn't agree more.

Thanks to jtbell's easily understood descriptions I think I'm beginning to understand what momentum, mass and kinetic energy are.

very important if I'm going to try and learn SR lol