studiot Posted November 29, 2018 Posted November 29, 2018 Surely the cosmological redshit is only an apparent 'loss of energy' due to the observer (us). A different observer would see a differnt 'loss' and one travelling with the photon would see none at all. Should we not be considering conservation of the energy-momentum four vector here?
swansont Posted November 29, 2018 Posted November 29, 2018 1 hour ago, Jinsuk Kim said: You said "Energy will not generally be the same if it's measured from two different frames of reference." I don' oppose. In this case, obviously I agree with you. I said "If E(t1)=E(t2), there is no problem." Please don't miss "If" But if E(t1)≠E(t2), it's also no problem, since that's perfectly consistent with physics. Really, one is comparing E(t1) with E'(t2), since they are in different frames of reference.
MigL Posted November 29, 2018 Posted November 29, 2018 OK, another example... You and I are sitting in a car that is travelling down the road at 100 mi/hr. I toss a baseball at you, and you catch it, no problem, as that ball is only moving at a couple of mi/hr. Now you are standing outside the car which is still moving at 100 mi/hr, and I, from inside the car, toss the baseball at you. That baseball will be moving at 100 mi/hr ( relative to you ), and it will hit you with enough energy to injure you severely. Did that knock any sense into you ? Do you now understand the importance of frames when dealing with energy ? 1
Silvestru Posted November 29, 2018 Posted November 29, 2018 8 minutes ago, MigL said: OK, another example... You and I are sitting in a car that is travelling down the road at 100 mi/hr. I toss a baseball at you, and you catch it, no problem, as that ball is only moving at a couple of mi/hr. Now you are standing outside the car which is still moving at 100 mi/hr, and I, from inside the car, toss the baseball at you. That baseball will be moving at 100 mi/hr ( relative to you ), and it will hit you with enough energy to injure you severely. Did that knock any sense into you ? Do you now understand the importance of frames when dealing with energy ? If the answer is no I propose we make a real life experiment of the second part.
Strange Posted November 29, 2018 Posted November 29, 2018 3 hours ago, studiot said: Surely the cosmological redshit is only an apparent 'loss of energy' due to the observer (us). A different observer would see a differnt 'loss' and one travelling with the photon would see none at all. That is another, very good, way of descring it. Quote Should we not be considering conservation of the energy-momentum four vector here? I think we need to get past the basics first! (Although I am fairly sure that is in the link I posted on the first page.)
J.C.MacSwell Posted November 29, 2018 Posted November 29, 2018 11 hours ago, MigL said: Let's see if we can put it another way... If you are standing at the injector and shoot a proton into the accelerator at CERN, you can give it enough kinetic energy so that when it collides, it can create particles that are orders of magnitude more massive/energetic than a proton's rest mass/energy ( like a Higgs boson ). Yet if you were moving along with that proton, you would measure its kinetic energy to be zero, and then you would scratch your head when these massive decay particles are produced in the collision.you would not understand where this 'extra' energy came from, and assume that conservation laws are violated. And you would be wrong ! Do you see why frames matter ? I don't believe you would. If you do all the accounting in that frame energy should be conserved. The extra energy would be in whatever your zero kinetic energy particle collided with. Energy conservation should work for any event, measured from any and every inertia frame. 5 hours ago, studiot said: Surely the cosmological redshit is only an apparent 'loss of energy' due to the observer (us). A different observer would see a differnt 'loss' and one travelling with the photon would see none at all. Should we not be considering conservation of the energy-momentum four vector here? Every observer would measure the photon to be at c. None would travel with the photon.
swansont Posted November 29, 2018 Posted November 29, 2018 3 minutes ago, J.C.MacSwell said: I don't believe you would. If you do all the accounting in that frame energy should be conserved. The extra energy would be in whatever your zero kinetic energy particle collided with. Energy conservation should work for any event, measured from any and every inertia frame. Yes. In this example, if the proton is at rest, the target is moving. You will never spontaneously see more massive particles come into existence.
studiot Posted November 29, 2018 Posted November 29, 2018 37 minutes ago, J.C.MacSwell said: 6 hours ago, studiot said: Surely the cosmological redshit is only an apparent 'loss of energy' due to the observer (us). A different observer would see a differnt 'loss' and one travelling with the photon would see none at all. Should we not be considering conservation of the energy-momentum four vector here? Every observer would measure the photon to be at c. None would travel with the photon. Have you never heard of a hypothetical thought experiment? If not then just calculate (do the transformation - you do dance?) from the photon's point of view.
J.C.MacSwell Posted November 29, 2018 Posted November 29, 2018 33 minutes ago, studiot said: Have you never heard of a hypothetical thought experiment? If not then just calculate (do the transformation - you do dance?) from the photon's point of view. OK...what would "different energy loss" mean in this hypothetical point of view?
MigL Posted November 30, 2018 Posted November 30, 2018 Yeah, not one of my best examples, JC. However, I really don't think Jinsuk would have caught on.
Q-reeus Posted November 30, 2018 Posted November 30, 2018 4 hours ago, studiot said: Have you never heard of a hypothetical thought experiment? If not then just calculate (do the transformation - you do dance?) from the photon's point of view. There can be no such thing as 'a photon's pov' i.e. 'in the rest frame of a photon'. The more you try and catch up to a photon i.e. head in it's direction of propagation, the more it redshifts in energy. In accordance with E = hf. Ignoring even that regardless of one's boost in that direction, it's still moving at c relative to you, the photon would redshift out of existence in the hypothetical sci-fi case you also 'caught up' and thus also moved at c. It's notional wavelength goes to infinity. An infinitely diffuse, zero energy phantom. In a universe of zero lateral extent. Bizarre enough? Got it?
beecee Posted November 30, 2018 Posted November 30, 2018 46 minutes ago, Q-reeus said: There can be no such thing as 'a photon's pov' i.e. 'in the rest frame of a photon'. The more you try and catch up to a photon i.e. head in it's direction of propagation, the more it redshifts in energy. In accordance with E = hf. Ignoring even that regardless of one's boost in that direction, it's still moving at c relative to you, the photon would redshift out of existence in the hypothetical sci-fi case you also 'caught up' and thus also moved at c. It's notional wavelength goes to infinity. An infinitely diffuse, zero energy phantom. In a universe of zero lateral extent. Bizarre enough? Got it? Bingo! Or to put that more simply, a photon has no rest frame so we are unable to speak of how the universe is observed from that pseudo frame.
Conjurer Posted November 30, 2018 Posted November 30, 2018 The expansion of space or dark energy hasn't been shown to be conserved with anything in physics. If energy conservation of dark energy was even going to be remotely possible, then things should lose energy when dark energy is added to the system. If things didn't lose energy from dark energy, then dark energy would be free energy. If it isn't free energy, then things would naturally lose energy from it occurring. The problem is that there is no theory that links dark energy with other energy in a system.
beecee Posted November 30, 2018 Posted November 30, 2018 7 hours ago, studiot said: Have you never heard of a hypothetical thought experiment? If not then just calculate (do the transformation - you do dance?) from the photon's point of view. I have often see the explanation from a photon's point of view, it would traverse the universe in an instant [infinite time dilation] and zero space [ infinite length contraction] I recall it was swonsont that showed me and explained why the fame of reference of a photon though is invalid as there is no rest frame for a photon. Explained nicely here.....https://www.quora.com/In-what-frame-of-reference-is-a-photon-at-rest "A reference frame is technically usually (particularly in special relativity) an inertial reference frame. That is, a frame in which an object appears at rest. Every massive object (that is not subject to an acceleration) has one, and all such frames of reference are related by Lorentz boosts. A Lorentz boost is a transformation of spacetime coordinates that takes one inertial reference frame to another, and allows us to compute what the differences would be for rulers and clocks between frames (i.e. how things would look for people moving relative to ourselves). Most importantly, there is no Lorentz boost that transforms to (or from) a frame moving at cc, the speed of light. What this means is that there is no frame in which a photon is at rest. All objects/particles without rest mass are necessarily non-inertial, and thus are moving (at cc) in every reference frame. This is part of the construction of the theory"
Q-reeus Posted November 30, 2018 Posted November 30, 2018 1 hour ago, Conjurer said: The expansion of space or dark energy hasn't been shown to be conserved with anything in physics. If energy conservation of dark energy was even going to be remotely possible, then things should lose energy when dark energy is added to the system. If things didn't lose energy from dark energy, then dark energy would be free energy. If it isn't free energy, then things would naturally lose energy from it occurring. The problem is that there is no theory that links dark energy with other energy in a system. Wrong. For instance Philip Gibbs is well known as arguing a zero-energy universe that includes all forms of matter-energy including so-called DE:https://www.quora.com/How-is-the-zero-energy-universe-hypothesis-compatible-with-the-accelerating-universe/answer/Philip-Gibbs-1?share=1
beecee Posted November 30, 2018 Posted November 30, 2018 (edited) 6 hours ago, Conjurer said: The expansion of space or dark energy hasn't been shown to be conserved with anything in physics. DE is simply the name we give to whatever it is that makes space accelerate in its expansion rate. It could loosely be termed an anti gravity force, as it is literally fighting against the gravity from the mass energy in space that was responsible for the expansion to be actually slowing down around 5 billion years ago from memory. Obviously as the universe has expanded, the density of the mass energy in a larger universe has decreased, and as a result the expansion is now accelerating. Local regions though, like our own local group of galaxies and even beyond, are decoupled from the observed accelerated expansion which is only observed over the larger scales. Edited November 30, 2018 by beecee
Strange Posted November 30, 2018 Posted November 30, 2018 6 hours ago, Conjurer said: The expansion of space or dark energy hasn't been shown to be conserved with anything in physics. I'm not sure what conservation of expansion of space means. And dark energy is not conserved. There is no reason why it should be.
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