Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: geordief on 23/06/2018 14:37:44
-
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?
-
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.
-
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?
-
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
-
Energy is a property of something measurable, be it proper mass or not. A 'coin of exchange' as JP called it.
-
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"
:(
-
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.
-
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.
-
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?
-
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.
-
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.
-
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.
-
I have never understood "rest energy" (perhaps above my pay grade).
Is "rest energy" perhaps always calculated in its own FOR?
-
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.
-
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.
-
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.
-
Thanks Pete. My memory let me down.
-
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!
-
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.
-
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.
-
Yes, there is a centre of mass frame of the universe //
Wrong. No such frame exists. When physicists use that term it really means zero momentum frame.
-
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"
:(
Nah, being slow is cool. If people use it as a negative then that says more about them than about you. Take your time, it's yours.
-
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.
True, but I said naively. There is no rest frame as there can't be a observer. And that makes the other proposition with light not 'propagating' a better one to me. You see, as long as we talk about a propagation, and 'particles' you naively can imagine such a circumstance. Doesn't make it fit though, does it :)
-
Actually, talking about 'propagation'. Imagine two lasers (set together) pulsing at a same rhythm towards some measuring device. According to that device the two lasers 'photons' hit it simultaneously. Are those 'singular corpuscles/photons', or not, 'at rest' with each other under their 'propagation'?
If not, what makes it 'simultaneous' at the measuring device?
And of course we will have to assume the 'system' of lasers/measuring device as well as however you will define a observer, to be 'at rest' with each other too
=
Remember that this is a 'ideal system', so we will presume that everything is ideally as well as perfectly set up at a same distance etc etc.
-
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.
There is an issue with the Lagrangian that I brought up in the pendulum thread:
https://www.thenakedscientists.com/forum/index.php?topic=73606.0
The understanding/description of physics is in trouble here.
-
If not, what makes it 'simultaneous' at the measuring device?
My initial thought is that the photons arrive simultaneously relative to the measuring device. It says nothing about simultaneity in the RFs of the photons, since we cannot define such RFs.
-
That's one answer. Anyone else prepared to define why you can't talk about them 'being at rest with each other' in whatever frame now 'propagating photons' might be said to exist :)
=
Do notice that I'm not using 'rest frame' above. From a 'frame' of a photon time is nonexistent, and as a distance/speed is described by a ruler and a clock, that distance/speed disappear from a photon's view. Expressed otherwise you need a arrow of time to use that ruler to measure a distance.
It becomes a contradiction in terms as from the observers point of view you emit a 'photon' from the laser, or in our case one 'photon' each from our co-joined lasers, that then slightly later registers at the measuring device. That's how we get to 'c', well, we use the 'two mirror experiment' but still, you want that clock and ruler and then measure how long it took that photon to 'propagate'.
Never the less, all distances must be non existent from that photons point of view, and so emitting and absorbing should be one and a same event. And if we then look at frames in where we find 'things' being at rest with each other, as being in a same frame of reference, then all photons must share that same frame of reference. And actually, there is no 'motion' in that frame, is there :)
Only relative proper mass aka matter
===
And yes, photons are indeed 'timeless', to the best of our knowledge. We can see them from the Big Bang, as good as new :) 13. 7 ~ billion years new that is. So asking oneself whether a 'photon' can decay is a very meaningless question. Another very interesting point is then if you can attach some sort of mass to a photon, http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html
You might want to argue that as we find a defined speed then it has a 'mass' of sorts, relative us. But if we by it would mean 'rest mass' we no longer would be able to argue a 'time less ness', and yes, now photons should 'decay' as I think about it. Which then would question everything we know from studying early light.
Arguing that it has a energy isn't that expressive actually. Everything has, or at least is expected to have, even a 'perfect vacuum'. And thinking of that one might notice that 'time' has a lot to do with the definition of 'virtual particles', too transient to make a difference
-
Never the less, all distances must be non existent from that photons point of view, and so emitting and absorbing should be one and a same event.
That's where my "logic" took me before I realised that I was extrapolating SR to include a mass-less particle, travelling at c. I think that's outside the scope of SR.
-
Then your logic did you a good turn Bill. Because it fits, a 'photon' is time less, we can actually confirm that by being able to 'see' 13.7~ly. While we can't be 'at rest' with it other 'photons' definitely should be able to be, presuming everything mainstream being correct, about it being a 'particle' 'propagating' at 'c'
=
Also, presuming that 'c' is correct, there can only be one 'frame' for those photons, unless someone wants to argue that the energy content differs and so then should the 'frames'? But being 'at rest' or sharing a same 'frame of reference' doesn't discuss the 'energy content', as far as I understand.
=
Let me widen that definition slightly, Presuming that 'c' is 'observer dependent, and from that finding that depending on somethings 'uniform motion' relative some other objects proving' that 'c' differs, that as both come to 'c' locally which 'can't be correct'. That as you can prove that you and the others objects 'uniform motion' definitely differ from each other, but it will still be a fallacy. The main reason for it being so is that you then must presume there to be a 'golden standard'. A 'absolute frame' from where we can define all other motion.
There is none.
-
In fact, to get it right we need to stop looking at 'c' as a speed defined. It's a lot more than just a 'speed'
-
Energy is not observer dependent. (Except kinetic energy)
If many types of energy are observer independent and one type of energy is observer dependent, is energy then observer dependent? I would say no.
If we manage to understand why the one type of energy is observer dependent, then we understand all of energy's observer dependence. And that seems quite easy to understand.
-
oh yes, energy is definitely observer dependent Toffo :)
Presuming that a photon hits your retina its energy will differ depending on direction of your motion, relative the photons 'propagation'
-
Presuming that a photon hits your retina its energy will differ depending on direction of your motion, relative the photons 'propagation'
Well that tells us that energy of a photon is kinetic energy.
-
You need to define your terms there. What kind of 'energy' do you consider to be observer independent?
-
You need to define your terms there. What kind of 'energy' do you consider to be observer independent?
Do you know what energy is? Do you know what kinetic energy is?
Observer independent energy is energy, except kinetic energy.
Things with momentum p and speed v have kinetic energy p*v.
Things with energy E and kinetic energy 0 have observer independent energy E.
-
Hmm :)
Would that make your definition a absolute frame then
=
Maybe you're thinking of the energy one define while being 'at rest' or in 'same frame of reference' with whatever you measure. And as we can't be at rest with light it has no other definition than 'observer dependent' from our point of view. But when we come to relative motion all uniform motions becomes the same and there is no golden standard defining what is a 'null motion' in this universe.
-
Relativity has two definitions, accelerations/decelerations and uniform motion. Of those two only the first definition is (locally) provable. You can prove to your satisfaction that there exist different uniform motions relative yourself, but all uniform motions express themselves equivalently, locally defined.
=
Actually it's not that I think you wrong, I have the same sort of feeling that there should be a 'golden standard', and the way you go about defining it is similar to mine, because it's 'local'. Locally defined, as in being at rest with something, or sharing the same frame of reference. But doing it that way turns it up side down, suddenly we lost the 'whole universe' and the 'objective 'global' perspective', instead going for local definitions. Which I actually (again) think makes sense :)
Although doing it that way introduce a new way of looking, and should lead to new definitions too.
We'll see
-
Actually. if 'energy' is a coin of exchange it has to have a consistent value. Other wise that 'coin' will just give you a headache