[geordief (OP)]

I have come across this statement a few times. I do not doubt it but I do not understand it.

Am I perhaps a bit slow?Maybe it is not too hard to understand.

Is it the same as saying  that energy is not an absolute quantity? We cannot attribute a clear value for the energy of any given system?

[Colin2B]

If you are looking at potential energy it depends where you measure from. A weight on top of a tall building will have greater potential energy relative to you if you are on the ground compared to your being on a floor midway.
If your car is moving at 10mph and a car rams you from behind at 30mph the kinetic energy transferred is greater than if you were moving at 20mph.
So energy is relative, but if you define the system properly and are clear about the reference frames you can account for the energy exchanged and understand the conservation rules.

[geordief (OP)]

If you are looking at potential energy it depends where you measure from. A weight on top of a tall building will have greater potential energy relative to you if you are on the ground compared to your being on a floor midway.
If your car is moving at 10mph and a car rams you from behind at 30mph the kinetic energy transferred is greater than if you were moving at 20mph.
So energy is relative, but if you define the system properly and are clear about the reference frames you can account for the energy exchanged and understand the conservation rules.

Perhaps I was falling into the (engrained) assumption that energy is a thing rather than a property?

If it is a property ,what is it a property of ? A system ? An object?

[yor_on]

You could think of it this way, and no, it's not 'slow' to ask the right questions, or if it is?
Heck, let's be 'slow' for once, heh.

I'm always slow, that's the curse I live under :)
And it's ok being slow, I think?

'At rest' there is a defined magnitude/energy whatever, without it you will have to look to at what speed/mass etc you define something else, relative yourself. Photons/light are defined to never be 'at rest', but the way I naively look at it myself I allow for the possibility of two 'photons' propagating aside each other, and then define them as being 'at rest' with each other
======================

That is naturally from a view of propagation. which is the simpler one presumably. Without a propagation every 'photon' is equivalent too, but the theoretical proof becomes circumstantial. It's different and I would need to think about that one some more.
=

Eh, every photon of a same energy, disregarding the observer, that should mean :)
It's tricky though, as it might imply that there is a 'background' of sorts, with 'definite values'

So maybe I need to think more about that one

[yor_on]

Energy is a property of something measurable, be it proper mass or not. A 'coin  of exchange' as JP called it.

[geordief (OP)]

And it's ok being slow, I think?

For 50 years I took that description as a qualified compliment but now I am starting to see it as "just slow"
 

[Bill S]

Photons/light are defined to never be 'at rest', but the way I naively look at it myself I allow for the possibility of two 'photons' propagating aside each other, and then define them as being 'at rest' with each other

Wouldn’t this be defining the photon frame as an inertial frame?
My understanding is that it is not, so velocities with respect to it cannot be measured.

[wolfekeeper]

There is actually a minimum energy for a system, which is the total energy measured in the centre of mass frame.

OK, you probably don't know what that means. For simplicity, let's just take kinetic energy. Kinetic energy is a square law, so a vehicle going twice as fast, has four times the energy, not just twice.

i.e. Kn = 0.5 * Mn * Vn^2

Where Kn is the kinetic energy of a body, Mn is the mass of that body and Vn is the speed of that body.

So if you add up all the Kn's you can get the total kinetic energy.

So let's say for the sake of argument you want to calculate the total kinetic energy of all the cars on Earth.

The problem is that the velocities depend on what you're measuring the speed relative to - is the centre of the Earth? The centre of the sun? The centre of the galaxy? They're all moving differently and all of these give you WILDLY different answers.

But if you pick the centre of mass of all the cars for the reference point- it turns out that that's the minimum kinetic energy. It also turns out that you can define the potential energy independent of the frame of reference that you're calculating it in, so only the kinetic energy matters much.

[geordief (OP)]

There is actually a minimum energy for a system, which is the total energy measured in the centre of mass frame.

OK, you probably don't know what that means. For simplicity, let's just take kinetic energy. Kinetic energy is a square law, so a vehicle going twice as fast, has four times the energy, not just twice.

i.e. Kn = 0.5 * Mn * Vn^2

Where Kn is the kinetic energy of a body, Mn is the mass of that body and Vn is the speed of that body.

So if you add up all the Kn's you can get the total kinetic energy.

So let's say for the sake of argument you want to calculate the total kinetic energy of all the cars on Earth.

The problem is that the velocities depend on what you're measuring the speed relative to - is the centre of the Earth? The centre of the sun? The centre of the galaxy? They're all moving differently and all of these give you WILDLY different answers.

But if you pick the centre of mass of all the cars for the reference point- it turns out that that's the minimum kinetic energy. It also turns out that you can define the potential energy independent of the frame of reference that you're calculating it in, so only the kinetic energy matters much.

So ,when you measure  and sum all the velocities in the FOR of the centre of mass do you count velocities approaching that centre as negative wrt velocities  in the opposite direction?

If you do  what would it imply if the sum total (the integral?) was zero?

[wolfekeeper]

The centre of mass reference frame has the property that the sum of the momentums is zero. So if the vector sum of the momentums is zero, relative to any other frame (such as that of the Earth) then you can use that other frame instead (which is normally more convenient). So if the masses are all the same and the velocities are all equal and opposite relative to the Earth, then you can just use the Earth frame instead.

The other thing that's worth knowing is that it turns out that you can convert from the Centre of Mass frame (i.e. the zero momentum frame) to any other frame really easily by adding on half the square of the relative frame speed times the total mass.

So for example if you calculate some total energy, say 1MJ, for four cars each weighing a tonne- for simplicity, let's say it's symmetrical relative to the Earth frame, and you want to calculate it instead relative to the Sun's frame, the Earth moves at ~30km/s, so you add on 0.5 * (4 * 1000) * 30000^2 = 1.8TJ  and then you just add on the original 1 MJ = ~1.8TJ (which is quite a big number- but the Earth moves very, very fast!)

FYI it's not actually terribly hard to prove this, you mostly just substitute the Vn' = Vn+Vf in the sum of kinetic energies equation above and expand it and sum it and rearrange it a bit.

[geordief (OP)]

The centre of mass reference frame has the property that the sum of the momentums is zero. So if the vector sum of the momentums is zero, relative to any other frame (such as that of the Earth) then you can use that other frame instead (which is normally more convenient). So if the masses are all the same and the velocities are all equal and opposite relative to the Earth, then you can just use the Earth frame instead.

The other thing that's worth knowing is that it turns out that you can convert from the Centre of Mass frame (i.e. the zero momentum frame) to any other frame really easily by adding on half the square of the relative frame speed times the total mass.

So for example if you calculate some total energy, say 1MJ, for four cars each weighing a tonne- for simplicity, let's say it's symmetrical relative to the Earth frame, and you want to calculate it instead relative to the Sun's frame, the Earth moves at ~30km/s, so you add on 0.5 * (4 * 1000) * 30000^2 = 1.8TJ  and then you just add on the original 1 MJ = ~1.8TJ (which is quite a big number- but the Earth moves very, very fast!)

FYI it's not actually terribly hard to prove this, you mostly just substitute the Vn' = Vn+Vf in the sum of kinetic energies equation above and expand it and sum it and rearrange it a bit.

So what happens if we apply that to  the universe as a whole ? Is it possible that every point in the Universe  would measure all the vectors of all the  momentums ** as summing to zero?(since the centre is everywhere it seems)

** just a thought experiment I guess.

[jeffreyH]

Rest energy is defined as E = mc² and kinetic energy is defined as K = (1/2)mv². A value for total energy should then be T = m(c²+(1/2)v²). The other component is then gravitational potential energy. While rest energy does not have a component that depends on motion, the other two do have that dependence. If we compose a formula that contains all three we get E = m(c²+(1/2)v²+u²). Here u is the gravitational potential. Looked at this way rest mass is affected by gravitational potential which then affects inertial resistance to motion. This cannot be true. Since weight increases in a stronger gravitational field. If the above were true the opposite would happen. Therefore we have two distinct type of energy which cannot be mixed together. Those energies defined at rest and those energies defined in motion. This means that either rest energy or kinetic energy must contribute to the total energy of the universe but not both. Since negative mass is not as observed property of any object in the universe this means it can only be kinetic and potential energy that are considered. Therefore it is the entropy of the universe that attention is mainly focussed upon.

[geordief (OP)]

I have never understood "rest energy" (perhaps above my pay grade).

Is "rest energy" perhaps always calculated in its own FOR?

[jeffreyH]

I have never understood "rest energy" (perhaps above my pay grade).

Is "rest energy" perhaps always calculated in its own FOR?

My definition for kinetic energy above was wrong and I have just corrected it. If we use rest mass to derive an energy value then we get rest energy. A mass in motion has an implicit relationship with the Lorentz factor due to special relativity. This is also known as relativistic gamma. It implies an increase in mass. This resulted in the concept of relativistic mass. Since including rest, kinetic and gravitational potential energies in one formula affected rest mass, does applying the Lorentz factor have this effect? In this case resulting in an increasing mass term. If this was wrong in the earlier case, how can it be right now? Except that we are not including rest energy in this formula. We have what is known as a Lagrangian. This states that L = T - U. Where T is the kinetic energy of the system and U is its potential energy. Neither term involves the speed of light. Simply an actual or potential velocity.

[PmbPhy]

I have never understood "rest energy" (perhaps above my pay grade).

Is "rest energy" perhaps always calculated in its own FOR?

I created this webpage for this kind of thread: http://www.newenglandphysics.org/physics_world/cm/what_is_energy.htm

All rest energy means is the amount of energy associated with a body at rest due to its mass-energy.

[PmbPhy]

This states that L = V + U.

Correction. L = T - V, where T = Kinetic energy and V = Potential energy. If the potential is a velocity dependent one then the letter U is used. This is particularly important in electrodynamics.

[jeffreyH]

Thanks Pete. My memory let me down.

[wolfekeeper]

So what happens if we apply that to  the universe as a whole ? Is it possible that every point in the Universe  would measure all the vectors of all the  momentums ** as summing to zero?(since the centre is everywhere it seems)

** just a thought experiment I guess.

Yes, there is a centre of mass frame of the universe (it's not physically special in terms of Relativity though), which minimises the kinetic energy, and would have been the motion of the Big Bang.

It's also thought that the total energy of the universe is probably zero in that reference frame-  gravitational potential energy is negative, and kinetic and mass energy is positive. They appear to cancel out to within measurable accuracy (which isn't very highly accurate, but still, it could make sense that it would be zero.) If so, it costs no energy to make a universe!

[geordief (OP)]

Yes, there is a centre of mass frame of the universe (it's not physically special in terms of Relativity though), which minimises the kinetic energy, and would have been the motion of the Big Bang.

Are you confident about that? I have just posted the question elsewhere on the internet as I see the question has come up on other forums and ,at first glance seems to have been answered in the negative.

[jeffreyH]

If everything were heading towards a single point then there would be a global increase in kinetic energy over time. Since the gravitational potential would be decreasing over time. This would still conserve the overall energy.