Post by: Sally Le Page on 06/09/2021 15:29:10
Since the Universe is expanding and light stretches across it as it does so becoming more red, what happens to the lost energy when the shorter wavelength, higher energy light towards the blue end of the spectrum is shifted into lower energy red wavelengths?
Do you know the answer?
Post by: Halc on 06/09/2021 15:47:15
Quote from: Mark
Since the Universe is expanding and light stretches across it as it does so becoming more red, what happens to the lost energy when the shorter wavelength, higher energy light towards the blue end of the spectrum is shifted into lower energy red wavelengths?Energy conservation is a property of (among others) an inertial reference frame, and is not conserved in a metric with expanding space, so the energy is gone. Similarly a rock moving at a nonzero peculiar speed will slow down without any external forces or reaction. The kinetic energy is lost.
Consider the same light in an inertial reference frame, say the 'redshifted' light from a galaxy 2 billion light years away, in the inertial frame of our solar system, and that light didn't lose energy at all and was redshifted in that frame from the moment it was emitted due to the recession velocity of the emitting galaxy.
Heat is another example. Uniformly distributed heat of stationary material in an inertial reference frame has nowhere to go and will not cool over time, but uniformly distributed heat of stationary material in an expanding frame will cool due to the dropping of the pressure/heat-density of the material. This thermal energy doesn't go anywhere since it is already everywhere in both cases.
Post by: hamdani yusuf on 08/09/2021 04:21:31
Similarly a rock moving at a nonzero peculiar speed will slow down without any external forces or reaction. The kinetic energy is lost.What does it mean?
Will the rock eventually stop?
Post by: Halc on 08/09/2021 04:44:30
In the absence of external forces on it (and there is always nearby mass accelerating it, so the condition is unrealistic), the rock will over time approach arbitrarily close to zero peculiar velocity but never reach it. Similarly, light will redshift to arbitrarily low energies but never reach zero.Similarly a rock moving at a nonzero peculiar speed will slow down without any external forces or reaction.What does it mean?
Will the rock eventually stop?
Interestingly, I cannot see angular kinetic energy reducing, so a spinning rock (at say 10 RPM) will seemingly spin at 10 RPM forever in the absence of external torque. The expanding universe doesn't drain that.
I'm not sure of this. Counter-arguments welcome.
Arguably, matter eventually breaks down and thus our spinning rock cannot be a rock forever, so eventually it becomes something other than a spinning rock (radiation??) whose energy is in fact consumed by the expanding universe.
Post by: Eternal Student on 08/09/2021 14:48:29
Energy conservation is a property of (among others) an inertial reference frame, and is not conserved in a metric with expanding space, so the energy is gone.That's OK and expresses the essence of the issue to the OP.
Consider the same light in an inertial reference frame, say the 'redshifted' light from a galaxy 2 billion light years away, in the inertial frame of our solar system, and that light didn't lose energy at all and was redshifted in that frame from the moment it was emitted (due to the recession velocity of the emitting galaxy).Minor note: I think you (Halc) accidentally missed a word like "emitted" or "released" in your statement. I've taken some liberties and edited your quote. Please check you're happy with it.
Anyway, if that is what you meant then I'm afraid it's still little bit suspect, Halc. In a spacetime with a metric for expanding space, we can only define inertial frames locally. We cannot extend an inertial frame centred on our own solar system out to cover a galaxy that was 2 billion light years away even if we wanted to. In particular, objects at a great distance from the origin of that frame would not behave as if they were in an inertial frame.
Hence, we are UNABLE to determine the frequency of the radiation at the point of release by a distant galaxy using one consistent frame for both the place of emission and the place of reception. For example, a redshift is not adequately explained by a Doppler effect and special relativity alone.
We must use General relativity to fully explain the redshift of distant galaxies. In GR, inertial reference frames are defined differently from SR, they are local and cannot be extended to a universal inertial frame (except for some trivial examples of a metric, which the expanding space metric is not one of).
If you were going to talk about inertial frames in answer to the OP, then you could try something like this:
We know that all inertial frames centred around the distant galaxy will be different to any inertial frame centred around our own solar system. (That difference is not quite as simple as one frame having a velocity relative to the other, they just are different). Therefore, there is no reason to assume energy would be conserved since we do not have (and cannot satisfy) the requirement you (Halc) stipulated no matter how hard we try. There isn't a consistent inertial frame in which to measure the energies at emission and later at reception.
It just so happens that we can identify local inertial frames at both events (emission and reception). Using the local co-ordinates would be the most natural way for a person to measure the energy of the light. This is what most PopSci articles are talking about when they discuss how light loses energy and red-shifts while travelling through expanding space. It's just unfortuante that there isn't a universal inertial frame and so there's no better way to measure the energy of the light (even if we wanted to). What is being challenged is not just the principle of conservation of energy but something more fundamental than that - Energy can't always be defined universally or measured independantly of the local environment and the local frame of reference for the object.
Just to be clear, on anything less than Astronomical scales none of this matters. Energy can be defined and the principle of conservation of energy can be assumed.
Best Wishes.
Post by: Halc on 08/09/2021 16:23:02
I've taken some liberties and edited your quote.Typo fixed, thanks.
Quote
We cannot extend an inertial frame centred on our own solar system out to cover a galaxy that was 2 billion light years away even if we wanted to. In particular, objects at a great distance from the origin of that frame would not behave as if they were in an inertial frame.While I agree with your entire post, my upshot was that it was close enough for the purpose of answering the OP. It's why I chose 2 billion and not 20. You need perhaps at least 3 digits of precision to measure the difference between actual measurements and those assuming some inertial frame because the scalefactor is currently quite linear due to the effects of mass density nearly cancelling those of dark energy. It helps if the observed matter is at a similar gravitational potential as here where we measure it.
Post by: Eternal Student on 08/09/2021 20:57:48
Interestingly, I cannot see angular kinetic energy reducing, so a spinning rock (at say 10 RPM) will seemingly spin at 10 RPM forever in the absence of external torque. The expanding universe doesn't drain that.This is complicated and interesting. This is what I think would happen (but I've not read anything that would tackle this question directly).
I'm not sure of this. Counter-arguments welcome.
The expansion of the universe would not change the angular speed, or RPM of the spinning object. However it could change the rotational kinetic energy it had.
We need to know if the spinning object is bound so that the distance from one part of the object to another part of the object never changes with time. The main alternative would be that the object co-expands with the universe, which I won't consider here.
If the object was bound and can be treated as a rigid body, then after each complete rotation we can take a snapshot of the object and see how it would look if we overlay a grid that shows the co-moving co-ordinates.
Now, since space is expanding, the grid showing co-moving co-ordinates must expand in our snapshots as time evolves (the distance as determined by the metric between two co-moving points increases with time).
So we can see that the spinning object has a constant size in terms of length we can measure intrinsically using the metric BUT it has a reducing size when expressed in the co-moving co-ordinates. So the rotational kinetic energy (in the co-moving co-ordinates) decreases.
? Well, that's my best guess anyway.
Post by: Just thinking on 12/09/2021 05:10:32
Since the Universe is expanding and light stretches across it as it does so becoming more red, what happens to the lost energy when the shorter wavelength, higher energy light towards the blue end of the spectrum is shifted into lower energy red wavelengths?The red shifted light is weakened but all the photons are still there blue shifted light is strengthened as it is compressed all the energy remains the same only arriving at different times.
Post by: Eternal Student on 14/09/2021 12:14:39
@Just thinking, I'm not sure what you were saying there.
- - - - - - - -
Usually Sally Le Page tells us when the question is answered in a podcast but hasn't done so this time, perhaps she's on holiday.
Anyway, it does seem to have been answered here: https://www.thenakedscientists.com/sites/default/files/media/podcasts/episodes/Naked_Scientists_QotW_21.09.13.mp3
It's about three and a half minutes and by the time the scene has been set, there's about 1 minute of actual answer. Their basic answer is that there is a lack of time translation symmetry. It ends with one sentence of waffle (vague comment) about space-time absorbing the energy. This seems to be an attempt to restore the public confidence that energy is conserved, it's just different. While the podcast was disappointingly short, that last sentence was actually harmful and you might have been better off without it. It completely eroded the importance of the only key statement you made - that the conservation of energy does not apply.
It's easy to criticise and we should spend a moment to recognise the good stuff in the podcast: It was pleasant and refreshing to have a podcast that focuses on Physics. It was quite courageous to (almost) publically challenge a long held view that energy is always conserved.
Best Wishes.
Post by: Just thinking on 14/09/2021 13:20:08
I'm not sure what you were saying there.Thanks, Eternal Student for your reply it is just that the question that has been asked is when light is stretched where does the extra energy go I don't understand what the extra energy is but I can only say that the energy that is there remains the same as stretching only delays the arrival time and takes nothing away. The opposite for blue shift this would mean that the light would be compressed and will arrive earlier than expected. No energy gained or lost either way.
Post by: Kryptid on 14/09/2021 15:34:58
The red shifted light is weakened but all the photons are still there blue shifted light is strengthened as it is compressed all the energy remains the same only arriving at different times.
That's not true. Photons with longer wavelengths have less energy than those with shorter wavelengths. The speed of light also does not vary with wavelength, so a red-shifted photon will arrive at the same time as a blue-shifted photon.
Post by: Just thinking on 14/09/2021 16:49:44
That's not true. Photons with longer wavelengths have less energy than those with shorter wavelengths. The speed of light also does not vary with wavelength, so a red-shifted photon will arrive at the same time as a blue-shifted photon.Not so true if a star is moving away at a high velocity the light emitted from it will be less extreme this is what makes the redshift. Each photon is arriving later and later due to the star moving away this is the stretching of the light.
Post by: Kryptid on 14/09/2021 16:51:38
Not so true if a star is moving away at a high velocity the light emitted from it will be less extreme this is what makes the redshift. Each photon is arriving later and later due to the star moving away this is the stretching of the light.
All right, so long as you understand that light of all wavelengths moves at the same speed.
Post by: Just thinking on 14/09/2021 17:07:27
All right, so long as you understand that light of all wavelengths moves at the same speed.Yes, that is true it is just if the source of the light is moving the light will be either compressed or stretched making the light more intense or weaker. This is the shift and nothing is lost or gained by the shift.
Post by: Eternal Student on 15/09/2021 00:00:38
@Just thinking,
It's difficult and possibly repetitive to explain where you might have gone slightly wrong.
First of all, there's nothing wrong with what you've said at all in situations where a Doppler effect applies and it is just that the source and receiver have some velocity relative to each other. If space was just flat Minkowski space then what you've said would be perfectly fine.
The main problem for the expansion of space is that there is no way to determine the velocity of the source relative to the receiver. Even in situations where both source and receiver seem to have 0 velocity through space (they are co-moving) so that you might reasonably expect them to be at rest with respect to each other, some redshift is still observed. To paraphrase this, there is no way to know how much of this "compression" that you describe can be explained just as the movement of the source relative to the receiver. All we know is that where space expands, each individual photon will be redshifted as time evolves whenever it's frequency is determined in the co-moving co-ordinate system. It could be that we are using the wrong co-ordinate system but there just isn't a better frame we can use.
This is complicated, so I'll try to phrase everything another way:
We can divide the universe into a large number of small patches of space. In each little patch, everything behaves just fine, special relativity and/or flat Minkowski space describes the situation just perfectly.
--|----------- |-----------|----------- |--
| patch 1 | | |
| | patch 2 | |
| | | patch 3 |
--|-----------|----------- |----------- |--
We can't easily extend a frame of reference from one patch over to cover another patch but it doesn't matter too much: In each patch there is a frame of reference that seems perfectly natural and the most sensible one to use. For an expanding universe, what we observe is that when a photon moves from one patch to a neighbouring patch, it is redshifted or seems to lose energy. The original frequency left the first patch, taking an amount of energy E= hf out of that patch but the photon entered the other patch with a lower frequency v < f and a corresponding amount of energy E = hv. So not all of the energy that left the first patch made it into the second patch and we don't really know where the difference in energy went, it didn't seem to go anywhere.
It could very well be that we just weren't using the right reference frame in the new patch but there is nothing that we can do about this. This is where General relativity significantly differs from Special relativity. We can't pick up the reference frame from patch 1 and use it in patch 2. For example, we might try and use a different reference frame in patch 2 that is moving toward patch 1. We can give the photon coming in from patch 1 exactly enough blueshift in this new frame to counter the redshift it would otherwise have had when it moves from one patch to another. That's great - BUT if you glance at the diagram above you'll see that we have made the problem much worse for a photon coming in from the East side (from patch 3 to patch 2). A photon moving in that direction would now suffer twice the redshift it needed to (one shift because it changed patches + another shift because our frame for patch 2 is now moving away from the photon). There is no easy way to define an inertial reference frame for patch 2 where a redshift doesn't occur for a photon coming in from at least some direction.
Best Wishes.
Post by: Kryptid on 15/09/2021 00:14:49
This is the shift and nothing is lost or gained by the shift.
Energy is.
Post by: Just thinking on 15/09/2021 08:02:24
Post by: Just thinking on 15/09/2021 08:15:19
This is complicated,Thanks Eternal Student for your time and effort trying to explain this problem. It may well be as you have shown but I find it very difficult to see this problem any other way anyway if I am wrong maybe the fog will lift and I will see things differently.
Post by: Eternal Student on 15/09/2021 12:34:57
If we take a given galaxy let's say a billion light years away and we conclude that it is redshifted and we conclude that it is moving away from us at 10% the speed of lightFirstly, what do you mean by moving away from us at 10% the speed of light?
Most people would assume we can measure the distance between us and the distant galaxy and then see how this changes with time. In school you would have been told that this formula holds:
Speed =
Is that what you mean? The rate of change of distance with time = 10% speed of light = 0.1 c
For flat space, this is a perfectly sensible idea of speed and it is easily converted into a velocity. The velocity of an object is just a vector with magnitude equal to the speed and direction given by the direction of travel.
However, in curved space things are more complicated. We can still calculate something we can call a speed of recession and it is exactly as described above - it is the rate of change of distance with time. However, that speed does not convert into a velocity in any sensible way. It is possible for the distant galaxy to move in a direction that just doesn't exist as a direction we can have in our solar system. The velocity of the distant galaxy is meaningless when regarded as a velocity that something could have here in our solar system.
Let's take a simple example of curved space. The usual example is to consider the surface of a ball. This will be a 2-Dimensional space. It's useful because we can embed this space in our usual 3-Dimensional space and then we have a way to see or visualise the curvature.
Here's a ball:
(https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcSI6SAvgRBj1Q7lyNMgBosBf8LendEk_sZ0og&usqp=CAU)
Our space is going to be the surface of that ball. We exist like little ants, forced to move in ("on" but it will be easier to say "in") the surface of the ball. We can move along the surface but we cannot move out of the surface. The 2-D space of the surface is the only thing we can experience. Now consider a point in our space which we will say is at the North pole of the ball. At that point in space we can move, or have velocities that are shown in the diagram above. This is the tangent plane to the ball at the North pole. The only velocities we can have are contained in what we can call the horizontal plane. We'll take the North pole to be the location of our own solar system in the universe.
Next consider a point along the equator of our ball. I've shown that on the diagram and marked the corresponding tangent plane at that point. At that point (which we will take to be our distant galaxy), we can only move (or have velocities) that are contained in the vertical plane. In particular, there is one direction of movement (straight up and down) that we can have at the equator but we cannot have at the North pole. If we had this vertical velocity at the North pole it would move us (the ants) straight out of the surface of the ball. It's not allowed, it's not a velocity that exists at the North pole, we are constrained to move in the surface always.
Now, this is where you should start to recognise some of the problems we run into in curved space. It is possible for a distant galaxy to be moving in direction that makes no sense at all (like straight down) when we are considering space from our location at the North pole. It can be moving in a "direction" that does not exist in our solar system. So we are unable to say that the distant galaxy is "moving away" from us. There is no direction we can point to and say it was moving in that direction. It is moving in a direction that is "invisible", "inaccessible" or just "nonexistent gibberish" in the local space that we are located in.
None the less, we can measure the distance between our solar system and the distant galaxy - we do this by stretching a piece of thread from us to the distant galaxy and keeping it within the surface of the ball all the way. So we can observe a change in that distance as time progresses - so we can calculate the quantity described above and called the recession speed. The recession speed makes sense just as the rate of change of distance with time, however, it is not the magnitude of any velocity that the distant galaxy had.
Anyway, these are the problems we run into with space that isn't flat Minkowski space. For an expanding universe, the distant galaxies aren't really moving "away" from us. At best, they are moving in a direction that doesn't exist or makes no sense in our local solar system. All we can establish is that the distance between us and the distant galaxies increases with time. This is something we can call a recession speed but we must be careful not to assume that the entirety of space in the universe is flat and therefore not to treat that recession speed as if it is a velocity through ordinary flat space.
I hope this helps or makes some sense. Best Wishes.
Post by: Just thinking on 15/09/2021 13:06:13
Energy is.How is energy lost as a result of redshift?
Post by: Just thinking on 15/09/2021 13:21:36
Firstly, what do you mean by moving away from us at 10% the speed of light?If light speed is one billion miles per hour the ten % of that one hundred million miles per hour. Thanks for your second post it's a lot to get my head around. When referring to the curvature of space I fail to understand. If we look at a galaxy that is one billion light years away we see the light as it was one billion years ago the galaxy is most likely in reality not in the location that we are seeing it in now it may be many degrees in any direction away from what we are seeing now. Is this related to the curviture of space or is it not related at all.
Post by: Eternal Student on 15/09/2021 21:54:07
Quote from: Kryptid on Today at 00:14:49
Energy is.
How is energy lost as a result of redshift?
Obviously I'm not Kryptid. I'm just assuming you won't mind an answer from anyone.
Don't consider light as a wave for a moment, put all your ideas about compressing a wave of electromagnetic radiation on hold for a moment. Instead let's consider light as something comprised of particles.
It it is possible to imagine a single photon travelling through space. Let's not allow it to interact with anything, get absorbed by anything or do anything interesting other than travel through space.
A photon carries a certain amount of energy which is given by the formula Energy = hf where h is plancks constant and f is the frequency of the photon.
The general idea is that if the photon redshifts as it travels through expanding space then this means it's frequency drops and the amount of energy it is carrying decreases. Does that make sense or seem reasonable?
We haven't allowed the photon to interact with anything, it's had no opportunity to pass on or transfer that energy to anything else. Energy has just been lost as it travelled.
So, just by considering the travel of a single, isolated photon, we are in a situation where the redshift of that photon would imply that energy has been lost. Conservation of energy seems to have been broken. Does that make sense? We don't really need to worry about other situations like a wide beam of light as you described earlier. Yes, a wide beam might scatter more, spread out the wave or "thin it down" as you described earlier. It could be exactly as you described earlier. We don't need to worry too much about that. The important thing is that there is at least ONE situation where the redshift of light from a distant galaxy would imply that energy was lost. That's all we need to worry Physicist's. The conservation of energy doesn't usually break in ANY situation. Does that make sense?
The redshift of photons as they travel through expanding space implies a situation where energy is not conserved. The earlier posts in this thread all relate to explaining how this happens; why it happens; how it can or can't be explained just by the movement of the source relative to receiver etc. etc.
- - - - - - - - - - -
If we look at a galaxy that is one billion light years away we see the light as it was one billion years ago the galaxy is most likely in reality not in the location that we are seeing it in now it may be many degrees in any direction away from what we are seeing now.This is mostly true. I haven't fact checked it, so it might be entirely true.
Is this related to the curviture of space or is it not related at all.It's not really related. I mean all movement and evolution of the universe is related somehow but I don't think this is a useful way of understanding curvature.
Explaining the curvature of spacetime is a big topic, it might be better in a new thread, although to be honest it might be faster just to look for some existing articles, books and/or videos that explain this.
Best Wishes.
Post by: Kryptid on 15/09/2021 22:05:01
Energy is.How is energy lost as a result of redshift?
The energy of a photon is directly correlated with its wavelength. Greater wavelengths have less energy than shorter ones. So if red-shift causes a photon's wavelength to double, that photon now has only half of the energy that it started with.
Post by: hamdani yusuf on 16/09/2021 04:18:55
Where does the half of the energy go?Energy is.How is energy lost as a result of redshift?
The energy of a photon is directly correlated with its wavelength. Greater wavelengths have less energy than shorter ones. So if red-shift causes a photon's wavelength to double, that photon now has only half of the energy that it started with.
Let's say a transmitter send a pulse of laser beam containing 1 Joule of energy. A faraway receiver gets the pulse, but the wavelength has been doubled. Does it mean that only half Joule of energy arrives at the receiver? Where does the other half Joule go? Is it absorbed by the space between transmitter and reciever?
Post by: Halc on 16/09/2021 05:27:36
Energy is conserved only in a static metric (one where space doesn't change over time) such as an inertial coordinate system. It is not conserved in an expanding metric. So it doesn't go anywhere. The total energy reduces over time relative to that metric.The energy of a photon is directly correlated with its wavelength. Greater wavelengths have less energy than shorter ones. So if red-shift causes a photon's wavelength to double, that photon now has only half of the energy that it started with.Where does the half of the energy go?
Let's say a transmitter send a pulse of laser beam containing 1 Joule of energy. A faraway receiver gets the pulse, but the wavelength has been doubled. Does it mean that only half Joule of energy arrives at the receiver? Where does the other half Joule go? Is it absorbed by the space between transmitter and reciever?
Quote from: Sean Carroll
When the space through which particles move is changing, the total energy of those particles is not conserved.https://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/
So if the transmitter and receiver are both stationary, but with space between them expanding by a factor of 2 while the light is in transit, the pulse loses energy and arrives at the receiver at a quarter the power, twice the duration, and half the energy.
Same situation using inertial coordinates:
In the inertial frame of the receiver, the light leaves the receding transmitter already at half power and loses no energy along the way.
In the inertial frame of the transmitter, the light has full energy all the way but still takes twice the duration to hit the receding receiver, but the full dose of energy is delivered to it.
This is one more way to illustrate that energy of light and other things (kinetic energy of a rock for instance) is very frame dependent, and not actually a property of the thing itself.
The example above is idealized and takes into account neither the mass of anything nor dark energy, both of which complicate the arithmetic (and mostly cancel each other) but not invalidate it other than the fact that since these things do exist, an inertial frame is only an approximation over such non-local distances.
Post by: Just thinking on 16/09/2021 08:29:44
Post by: hamdani yusuf on 16/09/2021 12:51:48
So if the transmitter and receiver are both stationary, but with space between them expanding by a factor of 2 while the light is in transit, the pulse loses energy and arrives at the receiver at a quarter the power, twice the duration, and half the energy.Will it make a difference if the change of wavelength is caused by expansion of space, compared to physical movement/velocity difference between transmitter and receiver?
Post by: Eternal Student on 16/09/2021 15:21:27
Will it make a difference if the change of wavelength is caused by expansion of space, compared to physical movement/velocity difference between transmitter and receiver?This thread does seem to be repeating itself often, I can only apologise if I say something again.
1. The receiver just receives radiation. If it gets a frequency of 100 Mhz for that radiation, that is all it cares about. The radiation will be 100 MhZ radiation regradless of how it came to be that frequency. We can redshift radiation in several ways. The receiver doesn't care if it was caused by space expanding or a relative velocity between the source and the receiver.
That is where half the problems or misconceptions come from about an expanding universe. We notice that what should be spectral emission/absorption lines from elements like Hydrogen seem to be redshifted when we look out to distant galaxies. We are reasonaby sure that they were Hydrogen emission lines, so we know what frequency they had at the point of creation and in the frame of reference where the atom was stationary. We know that we observe them at lower frequency when we receive them here on earth (or on a satellite out in space near earth). We want to explain that redshift.
2. It's natural to suggest that space behaves just like Minkowski space and the source was moving away from us but it just doesn't work well. Our theories remain more consistent if we assume the redshift was caused by the expansion of space rather than some velocity of the source and/or receiver through space. One of the main points of evidence for this would be the Hubble law. This suggests that the recession speed of distant galaxies can exceed the speed of light. We would prefer to believe that there is something "wrong" with the way we measure distances over astronomical scales. What is wrong with the way we measure distances is that the geometry of spacetime is not flat. General relativity offers an explanation for why the change in distance between us and distant galaxy with time can exceed the speed of light. This quantity
is called the recession speed but it is NOT a velocity that anything has, it is just the rate of change of distance with time. In flat space, that quantity would have to be the magnitude of a velocity - it is essentially the definition of what a velocity is.I'm not going to repeat any more of this, this general discussion is available in many other articles, threads, videos, books etc.
3. The key issue being discussed in this thread seems to be the conservation of energy. We already know that total energy is a somewhat arbitrary quantity. We are only interested in CHANGES in energy and that is all that the conservation of energy concerns itself with. For example, if we have a system that consisted of some atoms then the kinetic energy of the atoms depends on the inertial reference frame we wish to use. We can give the atoms more energy just by using a different frame and boosting the velocity of all those atoms. This doesn't matter, we are only concerned with changes in the energy that might happen as the atoms interact, we don't claim to know or care about the actual energy that the system really had. Hopefully, that makes sense. If we don't consider everything in our system under one consistent inertial frame then the conservation of energy doesn't apply.
Now, imagine a small number of photons travelling through space, let's keep them all close together and travelling parallel just to make it easy to visualise, they were all made at the same time from one source and they will all (after say 3 million years have passed on a satellite called "the receiver") be absorbed or received at the same place. We'll draw a box around a regon of the universe that includes the source and receiver and consider this as our closed system. Now we expect the conservation of energy to apply to our system. However we're going to be careless and apply a different reference frame after each millienium (million years). On the first millenium we use a reference frame such that the source was stationary in that frame. On the second millenium we'll use a reference frame that has velocity +0.1c relative to the day 1 frame. On the final millenium we'll use a ref. fame with velocity +0.2c rel. to day 1 frame. Obviously, the energies of the group of photons change each millenium (a different redshift applies). We might claim that energy was somewhow being lost each millenium but of course someone will point out that if we had used a single consistent reference frame then this wouldn't have happened.
One of the key ideas in General Relativity is that we CANNOT apply a single consistent inertial reference frame to cover all of the universe. There are always Local inertial reference frames that we can find and use but (for a general geometry of space) we cannot extend these inertial frames beyond the local region. So we can reply to those people telling us to use one consistent frame by just saying "For an expanding universe, we can't. There isn't one. The geometry of space is not Euclidean."
4. It was mentioned much earlier in this thread that the issue of redshift in expanding space isn't simply regarded as a challenge to the conservation of energy. It is just not possible to determine the energy of a photon at the time of release by a distant galaxy as it would be measured in our own local frame of refeence here on earth. So the wider issue is that energy cannot meaningfully be defined universally, only locally.
Best Wishes.
Post by: Eternal Student on 16/09/2021 15:31:15
If we have a radio tower and it is transmitting a signal let's say 27 MHz and now we move the tower at a very high speed away from us the 27MHz will be transformed into a lower frequency say 20MHz we will still receive the transmission but at a different wavelength. Am I with it now or still off the page?Yes, that's correct.
Post by: Halc on 16/09/2021 17:52:34
Excellently put. This was the gist of my post above.Will it make a difference if the change of wavelength is caused by expansion of space, compared to physical movement/velocity difference between transmitter and receiver?1. The receiver just receives radiation. If it gets a frequency of 100 Mhz for that radiation, that is all it cares about. The radiation will be 100 MhZ radiation regradless of how it came to be that frequency. We can redshift radiation in several ways. The receiver doesn't care if it was caused by space expanding or a relative velocity between the source and the receiver.
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2. It's natural to suggest that space behaves just like Minkowski space and the source was moving away from us but it just doesn't work well.We seem to disagree on this point. It seems to work well enough if you keep it sufficiently local (where the scalefactor stays reasonably linear). A separation of 2 BLY seems to fall within this 'local' range, with the deviations being minor secondary effects. At larger distances, I agree it falls apart. You go out 7 BLY and suddenly the secondary effects start being significant. The universe expansion was still slowing 7 billion years ago. Out twice that far you start dealing with event horizons that cannot be described with an inertial frame. The secondary effects begin to dominate the primary ones, and the inertial model completely falls apart. I'm not proposing the Minkowski spacetime as a model for the universe at large. That is indeed quite easily falsified.
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One of the main points of evidence for this would be the Hubble law. This suggests that the recession speed of distant galaxies can exceed the speed of light.Under the Milne model (which has been falsified), Hubble's law still stands, where Hubbles constant is exactly 1/time at all times. Galaxies don't have velocities greater than light because there are no galaxies further away than a fixed figure. Recession rates in the expanding metric can exceed c, but those rates are expressed as celerity, not as a velocity. Celerity has no upper limit, so in expanding space, the distance between a pair of galaxies can increase at arbitrarily high rates. In the inertial frame, none of these galaxies has a relative velocity greater than c.
I'm saying this not because I support the Milne model, but because Hubble's law is not in fact evidence for expanding space. The acceleration is. You can't have that in flat space. Deceleration either for that matter since the symmetry must be broken.
So real visible galaxies like GN-z11 have recession rates over 2c, and since it is far enough away that an inertial model simply cannot be applied, one cannot really assign an inertial velocity to that, but if you could, it would be around 0.98c
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General relativity offers an explanation for why the change in distance between us and distant galaxy with time can exceed the speed of light. This quantityWe seem to be in agreement about that much.
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In flat space, that quantity would have to be the magnitude of a velocity - it is essentially the definition of what a velocity is.Not that quantity, no. To convert a recession rate r to a flat velocity v of one object relative to the other: v=tanh(r), but I agree, it is fairly inappropriate to apply such a transformation over non-local distances since the transformation assumes spacetime is flat.
The pop articles never really explain the difference between recession rate and the speed something is moving, or more precisely, between celerity and velocity.
3. The key issue being discussed in this thread seems to be the conservation of energy. We already know that total energy is a somewhat arbitrary quantity. We are only interested in CHANGES in energy and that is all that the conservation of energy concerns itself with. For example, if we have a system that consisted of some atoms then the kinetic energy of the atoms depends on the inertial reference frame we wish to use. We can give the atoms more energy just by using a different frame and boosting the velocity of all those atoms. This doesn't matter, we are only concerned with changes in the energy that might happen as the atoms interact, we don't claim to know or care about the actual energy that the system really had. Hopefully, that makes sense. If we don't consider everything in our system under one consistent inertial frame then the conservation of energy doesn't apply.[/quote]Exactly, and since the universe is not a system that can be considered under any one consistent inertial frame, there is no necessary conservation of energy in the universe, something which is rarely admitted, which is why I like the Carroll article, who comes right out and says it rather than trying to sweep the embarrassment under the rug.
Post by: Eternal Student on 16/09/2021 22:18:55
Obviously, overall there is a lot of agreement. I didn't really consider the Milne model but that's probably because I didn't start studying cosmology until we had computers and the internet. Don't get me wrong, I'm old enough to remember the days before this but I just wasn't studying cosmology then. In recent times, the Milne model barely gets a mention in the textbooks. Anyway, I've enjoyed spending a couple of hours to read something about it.
It seems to work well enough if you keep it sufficiently local (where the scalefactor stays reasonably linear)OK. It does work well enough on small scales.
Under the Milne modelOK. Seems reasonable.
We could support the validity of General relativity by various observations and bits of evidence, like the procession of the orbit of mercury etc. etc.
As far as General relativity is concerned, the Milne model could only describe space that was empty (or at the very least it is obtained as a limit of a FLRW model when energy density and pressure → 0 ). I would assume the Milne model was largely ignored many years ago just because the universe does not seem to be empty (and no one was really disputing General relativity any longer).
In flat space, that quantity would have to be the magnitude of a velocity - it is essentially the definition of what a velocity is.It's taken a while to see where you were getting this from. The formula v=tanh(r) was coming from the use of a Minkowski metric, rapidity and continued reference to the Milne model, I think. So, yes actually you'd be right.
Not that quantity, no. To convert a recession rate r to a flat velocity v of one object relative to the other: v=tanh(r), but I agree, it is fairly inappropriate to apply such a transformation over non-local distances since the transformation assumes spacetime is flat.
The pop articles never really explain the difference between recession rate and the speed something is moving, or more precisely, between rapidity and velocity.
I think the problem is from my use of the word "flat space". I made the mistake here, I should have said "Euclidean space", or the thing everyone studied in school.
My original sentence was meant to be more disposable and light weight.
In Euclidean space (which we all studied at school), where (Δs) = distance = √( (Δx)2 + (Δy)2 + (Δz)2) then you would have been told that speed is defined to be
. Hence, if the distance between two objects was changing at a rate given by v = , then it would be inescapable that v is velocity that one of the those objects has relative to the other object.- - - - - - - - -
About the Hubble Law:
Hubble's law is not in fact evidence for expanding space.There are other models that could be consistent with the Hubble law. You've mentioned the Milne model. However, General relativity and a universe with a FLRW metric that has an increasing scale factor is the best candidate.
I mean let's be fair the Hubble law is more properly called the "Hubble-LeMaitre law" since Georges LeMaitre was predicting Hubble's law from General Relativity years before Hubble made made those observations.
The Milne model seems like a desperate attempt to show that special relativity and not general relativity might be enough to explain some things.
Best Wishes.
Post by: Halc on 17/09/2021 00:40:19
Which is why I reference it. It works on scales small enough where scalefactor is reasonably linear, and since mass and dark energy are reasonably balanced in the recent past, that scalefactor is sufficiently linear that the mathematics makes a good approximation.Quote from: HalcIt seems to work well enough if you keep it sufficiently local (where the scalefactor stays reasonably linear)OK. It does work well enough on small scales.
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We could support the validity of General relativity by various observations and bits of evidence, like the procession of the orbit of mercury etc. etc.You can trash the Milne model just by the very existence of Mercury (or us).
It's taken a while to see where you were getting this from. The formula v=tanh(r) was coming from the use of a Minkowski metric, rapidity and continued reference to the Milne model, I think. So, yes actually you'd be right.I always wondered if 'Euclidian space' is appropriate when applied to spacetime. Yes, flat space is Euclidean if spatial triangles add up to 180°, but Euclidean spacetime would seem to be more the Newtonian concept with the intervals being computed as the sum of the squares of the 4 components instead of subtracting space distance from the time (t²-d²) that Minkowskian spacetime does. In other words, does that difference make Minkowskian spacetime non-Euclidean? I don't know the rules. You're more the formal math guy and I'd take your word on it over my guess.
I think the problem is from my use of the word "flat space". I made the mistake here, I should have said "Euclidean space", or the thing everyone studied in school.
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In Euclidean space (which we all studied at school), where (Δs) = distance = √( (Δx)2 + (Δy)2 + (Δz)2) then you would have been told that speed is defined to beBut ds/dt is still frame dependent. Is an expanding metric applied over Euclidean spacetime (a transformation done by Milne) make it no longer Euclidean? It's just that distance (Δs) is not measured the same way in an inertial frame as it is in an expanding frame over the same space. There's different simultaneity between the two objects, and hence a very different proper separation between the two of them at any give time.
For instance, in comoving coordinates, stationary galaxy X might be 20 BLY away from stationary us (proper separation along the line of constant cosmological time) and that distance increasing at a rate of about 1.45c, and a clock on galaxy X reads the same time (since the BB) as here on Earth.
Relative to Earth's inertial frame, that same galaxy is currently a proper distance of only 12.35 BLY and moving at a rate of about .895c and a clock at galaxy X currently reads only 6.15 BY.
Both measure a proper distance, but to two completely different events separated by over 7 billion years in X time. Yay relativity of simultaneity. Yes, I know I've totally violated my definition of 'local enough' with these large figures, but we were talking Euclidean space and not real space for this example.
So same object, same Euclidean spacetime, yes vastly different velocities (rate of change of proper separation) due to different ways of drawing abstract lines of simultaneity in that Euclidean space.
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I mean let's be fair the Hubble law is more properly called the "Hubble-LeMaitre law" since Georges LeMaitre was predicting Hubble's law from General Relativity years before Hubble made made those observations.No better evidence for a theory than a prediction like that being made before there was evidence that necessitated it.
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The Milne model seems like a desperate attempt to show that special relativity and not general relativity might be enough to explain some things.That's pretty much what it tried to do: GR used to describe local wiggles in a Minkowskian universe on the largest scales, rather than SR being used to describe local flat space in an expanding universe at the largest scales. It rightly fell flat, and I only use it 'locally' where SR still reasonably applies.
Example: I have two objects bolted to a long stick (several light years long) way out in deep space between galaxy clusters. There are super sensitive force sensors that measure force of the objects on the stick. At some point in the universe history, there might be zero tension or compression on the stick, but if it's sensitive enough it will detect tension now, and compression a long time ago. Milne model doesn't allow that. That's the most local test I can think of.
Interesting exercise to work out exactly when the stick has zero force on it, which is a function of all kinds of things like the mass of the objects, length of stick, and the local gravitational repulsion due to tidal effects from the nearest galaxy clusters. Yes, gravity can repel objects, something that can be illustrated with the rubber sheet analogy.
I have to shut up now. I'm well into just posting thoughts that go through my head and I've digressed.
Post by: Zer0 on 17/09/2021 21:56:40
🙏
Hiya Sally!
🙋
If the Universe is a Closed ended system with no way out.
Then clearly everything just dissolves & dilutes but Remains!
Ps - Wonder from where do Virtual Particles appear.
🤔
Post by: Eternal Student on 18/09/2021 00:31:52
Formal definitions of Euclidean and/or Flat: I had written something here but I've deleted it. It's boring and you can look up the official definitions whenever you need them. It won't help much anyway, most of us mess it up slightly and someone else will always have a different idea of the terms.
But ds/dt is still frame dependent. Is an expanding metric applied over Euclidean spacetime (a transformation done by Milne) make it no longer Euclidean?This is the more interesting bit. Technically, a Euclidean metric has all positive eigenvalues (or it just adds all the components and never takes one away, if you want it in plain language). One of the nice things about it is that this "metric" is actually a bona fide, real, honest, straight-up, satisfies the formal mathematical definitions of........ a metric. In particular, the distance between two points is zero if and only if the points are the same.
Meanwhile the main alternative is called a Lorentzian metric and has signature (- + + +) or (+ - - -) if you prefer. Although the physicists call it a "metric" it isn't really. It's a pseudo-metric or semi-metric and you can see why: Two points can be different points but still have 0 metric distance between them. This can help to remind you that Minkowski space or anything with a Lorentzian metric isn't formally a Euclidean space, it isn't even a proper metric space.
Yet, some people will still say that Minkowski space is Euclidean - but that's people for you. It's because they are talking about the purely spatial dimensions in Minkowski space. Minkowski space does have all the properties of a Euclidean space provided we keep time quite separate and don't try to mess up the metric by including some time differences in it.
Let's just consider a photon travelling through Minkowski space. It travels on a null path. The school definition of it's speed is as follows:
Speed = (Distance travelled) / (Time taken).
If we use the full extent of the Lorentzian metric then that distance is actually what we normally call a spacetime interval. So a photon covers 0 metric distance in 1 unit of time, so it has a speed of 0. This is technically correct, the rate of change of that Lorentzian measure of distance travelled with time is zero. However, this is not what we have in mind for the speed of a photon, it is certainly not it's speed through 3-D space. We would normally restrict our definition of the distance travelled to a purely spatial measure of distance, if we do this then the photon covers a spatial distance of ct in time t and so it has a speed of c. Provided we keep time and space separate like this, then Minkowski space has Euclidean properties (and you can understand why people might say that it is Euclidean, I mean they are wrong - but you can understand why they are saying it).
This is essentially where the issues appear in the Milne model. The underlying space never was Euclidean anyway. It had a Lorentzian metric (so that the time component was subtracted). More than this, there was no attempt to keep time separate from the spatial dimensions. Quite the opposite, the whole thing rests entirely upon the ability to mix up some time with some space and apply Lorentz transformations. So that's it straight from the word go. Minkowski space used this way was never going to be Euclidean or show properties we would expect in a Euclidean space. On top of this they introduce a second metric which also has a non-Euclidean signature and it's just a mess. Nothing in the the model is Euclidean by the end. However, it wasn't really the introduction of the second metric that was the problem. Minkowski space isn't Euclidean to start with and they (Milne et.al.) were determined to exploit the Lorentzian nature of spacetime.
I can barely remember how the original use of ds/dt came up....
Best Wishes.
Post by: Halc on 18/09/2021 03:17:38
This is the more interesting bit. Technically, a Euclidean metric has all positive eigenvalues (or it just adds all the components and never takes one away, if you want it in plain language). One of the nice things about it is that this "metric" is actually a bona fide, real, honest, straight-up, satisfies the formal mathematical definitions of........ a metric. In particular, the distance between two points is zero if and only if the points are the same.So I suspected. Thanks for that. I've always hesitated to use the term Euclidean when I just mean flat.
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Meanwhile the main alternative is called a Lorentzian metric and has signature (- + + +) or (+ - - -) if you prefer.Always the latter I think since timelike intervals are expressed as positive/real and spacelike intervals as negative/imaginary. I can't find a site that uses the -+++ standard, but maybe I didn't look hard enough.
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This is essentially where the issues appear in the Milne model. The underlying space never was Euclidean anyway. It had a Lorentzian metric (so that the time component was subtracted). More than this, there was no attempt to keep time separate from the spatial dimensions. Quite the opposite, the whole thing rests entirely upon the ability to mix up some time with some space and apply Lorentz transformations.The transformation I did was not a Lorentz transformation, but rather one from an inertial frame to a hyperbolic frame (a frame anchored on an event instead of a time axis). Energy is not conserved in the latter sort of frame, but any arbitrary event can be used to define the frame.
Post by: Eternal Student on 18/09/2021 22:41:35
The transformation I did was not a Lorentz transformation, but rather one from an inertial frame to a hyperbolic frame......... (and stuff)....It's actually quite difficult to find a lot of good texts about the Milne model anymore but I think I've got an overview of the ideas. I'm mostly onboard with what was done in the transformation. The Milne universe seems to be using something like Rindler co-ordinates.
Your original question was something like this:
......... Is an expanding metric applied over Euclidean spacetime (a transformation done by Milne) make it no longer Euclidean?To which the main thing to note is that it wasn't an honest Euclidean space to start with. The underlying spacetime was Minkowski which isn't Euclidean and won't have Euclidean properties unless you are extremely careful to try and keep time and space separate. Applying a second metric made it worse but it just wasn't Euclidean to start with. As far as I can see, the entire methodology relies upon exploiting the non-Euclidean nature of the Minkowski spacetime right from the start.
Late addition: The (- + + +) convention is used in the texbook I have. Sean Carroll, An introduction to spacetime and Geometry, page 9.
He defines (Δτ)2 = -(Δs)2, and states "the interval is defined to be (Δs)2 not the square root of this quantity". So he avoids imaginary numbers all the time anyway.
I'm sure that many other sources use the other convention and also define the interval to be the square root.
Best Wishes.
Post by: Halc on 19/09/2021 02:07:52
Hi again Halc.Hyperbolic coordinates, not Rindler. The latter is for references with constant proper acceleration, which is not true of hyperbolic coordinates. The Milne universe is only one special case of the general hyperbolic coordinates. One can assign such coordinates to any event in spacetime, especially say one where grenade explodes in a vacuum.The transformation I did was not a Lorentz transformation, but rather one from an inertial frame to a hyperbolic frame......... (and stuff)....It's actually quite difficult to find a lot of good texts about the Milne model anymore but I think I've got an overview of the ideas. I'm mostly onboard with what was done in the transformation. The Milne universe seems to be using something like Rindler co-ordinates.
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Agree to all, but the question still seems to stand:Quote from: HalcIs an expanding metric applied over Euclidean spacetime (a transformation done by Milne) make it no longer Euclidean?To which the main thing to note is that it wasn't an honest Euclidean space to start with. The underlying spacetime was Minkowski which isn't Euclidean and won't have Euclidean properties unless you are extremely careful to try and keep time and space separate.
Is an expanding metric applied over Minkowskian spacetime (a transformation done by Milne) make it no longer Minkowskian?
That's what I meant to ask, and I'm not actually sure what I meant be the equation. The arbitrary coordinate system I choose to slice up the spacetime has no physical effect on the spacetime, but it isn't the Minkowskian way of assigning the coordinates, so I guess the answer is no if you take the question as an abstract one, and yes if you take the question as a physical one.
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Late addition: The (- + + +) convention is used in the texbook I have. Sean Carroll, An introduction to spacetime and Geometry, page 9.Yes, all the texts I have also show the interval to be the square and not just 's', but I don't usually see the -+++ part. So a 10 second duration has an interval of 100, or -100 if you're Carroll.
He defines (Δτ)2 = -(Δs)2, and states "the interval is defined to be (Δs)2 not the square root of this quantity". So he avoids imaginary numbers all the time anyway.
If I parse his line correctly, the interval is (Δs)² and it is the negation of (Δτ)², which seems only true for an interval between two events at the same location in space, but by assigning it as the negation like that, he uses the -+++ convention. Functionally it matters not so long as we know which convention is being used.
Post by: Eternal Student on 20/09/2021 11:57:27
.....but the question still seems to stand:
Is an expanding metric applied over Minkowskian spacetime (a transformation done by Milne) make it no longer Minkowskian?
That's what I meant to ask, and I'm not actually sure what I meant be the equation. The arbitrary coordinate system I choose to slice up the spacetime has no physical effect on the spacetime, but it isn't the Minkowskian way of assigning the coordinates, so I guess the answer is no if you take the question as an abstract one, and yes if you take the question as a physical one.
It's hard to find texts on the Milne model as I mentioned earlier.
Page 341, Sean Carroll, Spacetime and Geometry indicates the following:
The Riemann curvature tensor of the Milne Universe = 0 (has all components =0). Therefore, it is locally equivalent to flat space (Minkowski). In this case, it can be shown to be quivalent to a small patch of Minkowski space - the interior of a future light cone of some fixed point of Minkowski space, foliated by negatively curved hyperboloids.
This is the treatment of the Milne Universe in General Relativity. It doesn't seem that the Sean Carroll's book cares at all for the TWO metrics that Milne used in his original treatment which only required special relativity. There is one Milne metric as far as that textbook is concerned. This seems to be the one you describe as "the expanding metric applied over Minkowski spacetime".
Carroll considers this metric:
which is equivalent to the Milne metric stated in Wikipedia when the curvature, k is normalised (set k=-1 and adjust r and a(t)) and then identifying new co-ordinates 
Anyway, it seems that using that metric, the space (with the Milne metric as described above) is equivalent to a small patch of Minkowski space.
I don't doubt that Milne developed his model without GR and using two metrics, there's even this section written about it in Wikipedia:
Milne developed this model independent of general relativity but with awareness of special relativity. As he initially described it, the model has no expansion of space, so all of the redshift (except that caused by peculiar velocities) is explained by a recessional velocity associated with the hypothetical "explosion".
So, he may have had two metrics and potentially treated them as separate things. However, we already know that in the first metric the space was assumed to be Minkowski space. So I think a fair and reasonable answer to your question is as follows: Yes, both metrics describe a space with 0 curvature and are locally equivalent to flat Minkowski space. That's all you (a Physicist) could physically determine if they were in those spaces anyway, they have 0 curvature everywhere and local co-ordinates exist which would seem natural to you and make the space behave like Minkowski space.
(I can't sensibly use the two different metrics simultaneously - anything with two metrics defined on it can't be Minkowski space by definition, it's some new mathematical object that I'm not sure we have a name for).
Best Wishes.
Post by: EvaH on 20/09/2021 13:46:43
https://www.thenakedscientists.com/podcasts/question-week/what-happens-lights-lost-energy
Post by: hamdani yusuf on 17/12/2021 11:29:12
What's the power of the laser that we will receive? what's the wavelength? how long is the pulse?
What if the space ship is moving toward us at the same speed?
Post by: Eternal Student on 01/01/2022 05:12:21
I'm not sure how this thread has progressed, I've been away for a while.
Anyway, it seems that Hamdani's question hasn't been answered. When I say "answered", I mean pushed around, pulled apart, mulled over and generally considered. If you (Hamdani) are just after a straight answer, you could probably try a textbook or another forum.
The moving spaceship thing is quite interesting and I can see how various people can imagine things to be one way or another, coming up with differing explanations as to why redshifts may or may not occur and if energy is actually lost or not. I've not spent too much time considering the changes in and relationships between the frequency, amplitude, power and time duration of the pulse myself, so I might as well have a look at it now.
Let's take a particle model for EM radiation to start with. In this model the laser light is just a discrete collection of particles called photons.
It sends a 1 Watt 600 nm laser pulse, one second long in its frame of reference.In the particle model, we can regard the "intensity" of the light as if it is just the intensity of bombardment by photons. Specifically, we can say that Intensity is proportional to the number of photons received per second. What is the constant of proportionality? It's just the energy carried per photon: Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation. Without loss of generality, we're going to ignore the area of the receiver in this example: All emitted photons from the spaceship will eventually be absorbed by the receiver. Let's say that it's all happening in a narrow beam that is precisely directed from the spaceship to the receiver. Additionally we aren't going to trouble ourselves too much over the energy or power (measured in Joules or Joules per second). We are just going to count the number of photons that are received per second. We know that each photon has an energy E = hf associated with it, so if we know the number of photons per second coming in then we can determine the energy coming in per second if we wanted to just by multiplying by hf, where h is Plancks constant and f is the frequency for the photon(s) being received.
Just to clarify then, when I say "Intensity" I will generally mean the number of photons per second and not the energy per second. This is what will link to your (Hamdani) mention of the power being 1 Watt, there is 1 units worth of photons being created per second (or, for simplicity, we could just work with 1 photon per second and scale it up at the end).
OK. let's make a start.
1. The ship has a constant velocity of 0.1 c relative to the receiver. This produces a gamma factor of
1 / root(0.99) ≈ 1.005. The gamma factor is the same regardless of whether the ship is moving toward or away from the receiver.
2. The ship emitted photons for 1 second as far as the crew of the ship were concerned (in the rest frame of the ship). Standard time dilation stuff implies the receiver considers the ship to have been emitting photons for 1.005 seconds (in the rest frame of the receiver). This remains the case for both movement toward or away from the receiver. This is the first interesting result. The receiver believes the ship emitted photons for slightly longer (in time) than the crew believe they emitted photons.
If you (Hamdani) wanted an answer as to how long in time the pulse lasted for the receiver then we can already offer one answer: The ship was emitting photons for 1.005 seconds as far as the receiver is concerned. However this is NOT how long (in time) it will take the receiver to collect or absorb all those photons. If the ship was moving away from the receiver then the last emitted photon travelled from a more distant location than the first emitted photon, so it will take the receiver slightly longer to receive that last photon. Conversely, if the ship was moving toward the receiver then the last photon has got a bonus of being already brought slightly closer to the receiver at the time of it's emission. Let's consider this effect next...
3. The photons eventually come in and bombard the receiver. We are interested in the time elapsed from the first photon coming in to the last photon coming in. Without loss of generality set up a standard reference frame for the receiver. Put the receiver at the origin (at all times, we do want a rest frame for the receiver) and put the ship over on the right hand side which we will consider as having a positive x- co-ordinate. If the ship is coming towards the receiver then the velocity is negative, the x- co-ordinate will start large and decrease as time progresses. If the ship is moving away the opposite applies. All movement is precisely along the x- axis (if it wasn't then choose a different axis so that it is, that's why we say "without loss of generality").
Here's a diagram:
Ship-reciever.png (15.42 kB . 1258x648 - viewed 3048 times)Now, we know that light always travels at the speed c in any reference frame.
We also know that the ship has a velocity 0.1c relative to the receiver's rest frame.
Let the first photon be emitted by the ship at x and t co-ordinates given by the event written as (x,t) in the receivers rest frame. [lower case x and t]
Let the last photon be emitted at the event (X,T) in the same frame [that is upper case X and T].
We already know that (T - t ) = (gamma factor) . (1 second) = 1.005 seconds from part 1. above (the usual time dilation stuff).
We also know that X and x are just the positions of the ship at different times and they are related by the velocity of the ship (in the receivers rest frame) and that elapsed time of 1.005 seconds over which it was emitting laser light (in the receivers rest frame). Specifically X = x + 0.1c. (T-t) for the ship moving away from the receiver. While X = x - 0.1c . (T-t) for the ship moving towards the receiver.
Now the first emitted photon won't just have stayed at position x all the time while the spaceship was moving and completed its emissions. Instead the first photon will have travelled toward the receiver at speed c over the elapsed time (T-t). So at the final time T in the receiver's rest frame the first photon will actually be located at x - c(T-t) in both cases (ship moving toward or away from the receiver). Meanwhile the final photon emitted is located precisely at the starships position of x + 0.1c(T-t) or x - 0.1c(T-t) at time T depending on whether it is moving toward or away from the receiver. All we have to do now is find the difference in x -co-ordinates at time T.... just subtraction....
x + 0.1c(T-t) - [ x -c(T-t) ] = 1.1 c (T-t) for the ship moving away.
Similarly we get 0.9 c (T-t) for the ship moving toward the receiver.
If you want numbers... put T-t = 1.005 and we get: 1.1055 light-seconds when the ship is moving away.
0.9045 light-seconds when the ship is moving toward the receiver. (For comparison recall the ship's crew, in the ships rest frame, would believe the pulse is physically 1 light-second long since it was travelling at speed c away from their ship and they emitted for 1 second). That's how long (in distance) each e-m pulse (or the total stream of all the photons) will be at time T, however it stays at that length from then onwards since each individual photon within that stream of photons travels at the same speed and in the same direction.
Anyway, that's how long (in distance) the pulse of e-m radiation will be as it is observed in the receivers rest frame. If you want to know how long (in time) it takes the receiver to absorb the pulse we can do that easily enough: e-m radiation always travels at speed c in any reference frame. So if the pulse is physically 1.1055 light-seconds long then divide that by c and we see it takes 1.1055 seconds to absorb it. Similarly it takes the receiver 0.9045 seconds to absorb all of the transmission if the ship had been moving toward the receiver.
This is interesting because it's not what you get if you just applied standard distance contraction stuff to the length of the pulse as observed by the ship's crew. Recall the ship's crew believe the pulse is physically 1 light-second long. The standard distance contraction formula would imply this reduces to about 0.9 light second in the receivers rest frame regardless of whether the ship was moving toward or away from the receiver BUT it doesn't. Why not, or what prevented the formula from working, you may ask? Well, I think it's best explained by noting that the pulse was NOT moving with the ship, the pulse had a non-zero velocity relative to the ship. The standard distance contraction formula only applies to objects that would be moving with the ship or appearing to be stationary to the crew on the ship - but it's interesting and took me a good long moment to think about and check that what we have is actually right.
4. What about the energy or intensity of the pulse?
What's the power of the laser that we will receive? what's the wavelength?The wavelength of the e-m radiation is determined by the usual relativistic Doppler shift formula.
This post is already too long. So here's the Wiki article about it: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect
The photons are red-shifted (have their freqency reduced or the wavelength increased) when the ship is moving away from the receiver. The converse happens when the ship is moving toward the receiver.
Now there's an interesting result in special relativity that rarely get's mentioned and just taken for granted. There is a conservation of events (please note that is not a real name or principle, I made it up). Specifically the Lorentz transformation L : (x,t) → ( x', t' ) is injective or one-to-one. If (x1, t1) and (x2, t2) are two distinct events (so that at least one of x1 ≠x2 or t1 ≠ t2 is true) in reference frame F then they get mapped to different events (x1', t1') and (x2', t2') in the Frame F' . Why is this useful or important? It means that events are never lost or squashed down into one event in the other frame, instead they just appear somewhere else and/or at some other time when you change frames of reference but you can always identify an event in one frame uniquely with an event in the other frame. So we can have a set of events in Frame F and count them to get their total number, or we can consider their corresponding events in F' and count their number in that frame, we get the same total number of events.
Anyway, this "conservation of events" idea will hold when the Lorentz transformation is well defined, which just means when the velocity v of the frame F' relative to F is less than c - or to paraphrase this further - whenever the situation is actually sensible. (If you consider frames moving at light speed relative to each other then many events in one frame will be squashed down to the same indistinguishable event in the other frame).
As I said, this is often taken for granted, events don't just disappear or two events merge into one when you change frames of reference. Anyway, we're going to use this to count the number of photons in the stream of photons that were emitted by the ship.
In the rest frame of the ship, there were N total photons emitted. Each can be identified with an event (x,t) where x = x-co-ordinate of the ship at the time of the photons creation (so that will be 0 in the rest frame of the ship since the ship can be assumed to always be the origin in that frame). The time co-ordinate t is the time at which the photon appeared or was created. These events are mapped by the Lorentz transformation to other events in the rest frame of the receiver but the important thing is that there are still precisely N of these events. To say that another way there were N photons created in any frame of reference you care to use. Since we assume all the photons are directed toward the receiver and eventually absorbed the receiver will receive a total of N photons.
OK, so we know the receiver always absorbs N photons eventually, the only difference is the amount of time it takes to absorb those photons (see part 3 discussed above). We can now determine the bombardment intensity for the case where the ship moves toward or away from the receiver.
It's N / 0.9045 ( I can't recall if I used the right numbers... I got it from part 3 above) with the ship coming toward the receiver. While it's N / 1.1055 with ship moving away from the receiver. So it's more intense but lasts less time with the ship coming toward the receiver.
If you prefer to think about your intensity as a power (in Joules per second) instead of as a number of photons coming in per second then recall that each photon has an energy hf with f found from the relativistic Doppler shift (I think I discussed this earlier and I'm just repeating myself, sorry). Overall, the intensity (in terms of power) is much greater when the ship is coming toward the receiver but lasts only 0.9045 seconds. It's a much lower power but lasts longer, 1.1055 seconds when the ship is moving away.
5. So how does this relate to conservation of energy?
The total energy received by the receiver is always given by N . hf where N = number of photons received, h is plancks constant and f is the frequency received.
N is the same for the ship moving toward or away (see earlier discussion). h is obviously always the same. Meanwhile f changes as discussed above, the radiation received has a higher frequency if the ship is coming toward the receiver.
So, if the ship is coming toward the receiver then the receiver believes the ship emitted an amount of energy Nhf1 and the receiver will eventually absorb that same amount of energy.
If the ship was moving away then the receiver believes the ship emitted less energy Nhf2 which the receiver will also eventually receive.
In the ships rest frame, the crew believe that a middling amount of energy was released by them and will be eventually absorbed by the receiver.
So in all frames of reference, there is a conservation of energy across the whole system when the same reference frame in used throughout the experiment. However, there are differences in the actual size of the changes in energy that occurred at various places and times between the different frames of reference.
This is quite conventional in Physics. Total Energy and the changes in energy that occur are frame dependant, however the conservation of energy holds in all frames.
6. Caveats and general limitations:
This thread originally started with some discussion of GENERAL relativity but this post uses only special relativity. The change in frequency for photons is not given by the conventional relativistic Doppler shift formula when GR is used. However, locally it all still applies as it is.
Generalizing the results to GR and a more complicated spatial geometry we should note what was said in item 5. above. Energy is frame dependant. Changing frames of reference can adjust the changes in energy that occur in what would physically represent the same place and time within the system. For a complicated spatial geometry we cannot extend an inertial reference frame to cover all of the universe. As a result we are forced to use only local frames of reference. When light has travelled long distances through space, it has been necessary to change local frames many times. It should, perhaps, come as less of a surprise to find that while a stream of photons required a certain amount of energy E to create in the local frame where they were created, when they are finally absorbed they may only deliver an amount of energy e < E to the receiver in the local frame of the receiver.
7. Using a wave model of e-m radiation instead of considering it as a stream of individual photons, we need to re-interpret the bombardment intensity of the photons (number of photons incoming per second). This would now represent an amplitude of the e-m wave received. So the situation is as follows:
When the ship moves toward the receiver these things happen:
(i) The receiver takes only 0.9045 seconds to absorb the entire stream or pulse, exactly as before.
(ii) The physical length of the pulse is exactly as stated before, 0.9045 light-seconds.
(iii) The amplitude of the wave is higher then that measured by the crew of the ship in their own frame.
(iv) The freqency of the wave is also higher than the crew observe - exactly as stated before (use relativstic Doppler shift).
... similar stuff for the ship moving away... the amplitude of the wave is now LOWER than that observed by the crew.
It's interesting that we usually refer to the Doppler shift as a frequency shift but for light (not usually for sound) there should be an equally interesting shift in the amplitude of the wave that also occurs. There's some mention of the Doppler effect on intensity here: https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Doppler_effect_on_intensity
Anyway, that's it. It's late on New years eve and I've probably made some mistakes. Best wishes and happy new year to everyone.
Post by: Eternal Student on 01/01/2022 17:46:40
Request for assistance or verification from someone please.
There is very little reliable information about the Doppler effect causing a change in the amplitude of an e-m wave only that it causes a shift in frequency. (Well, not much reliable info that I can find from Google anyway).
I'd be grateful if someone can check this, provide the benefit of their experience or produce a link to some credible source backing this up.
I wouldn't like to have suggested something that is incorrect.
Thanks, E.S.
Post by: Halc on 01/01/2022 19:41:20
I didn't go through the entire post, but it seems wrong to mix a quantum explanation with a classic one.
The power of a light beam from a receding source can be computed by considering a one-second burst, but the answer is, in the end, not a function of the duration of the signal.
The moving spaceship thing is quite interesting and I can see how various people can imagine things to be one way or another, coming up with differing explanations as to why redshifts may or may not occur and if energy is actually lost or not.Relative to an inertial frame (which is what we're considering in a scenario with a nearby source of light receding at 0.1c), energy cannot be lost per conservation of energy, but total energy of the one-proper-second signal is frame dependent. So this is not really discussing the topic of this thread, which is energy actually not being conserved relative to a non-static frame such as the cosmological frame that describes the universe as a whole.
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In this model the laser light is just a discrete collection of particles called photons.This is wrong, since the intensity of light is also a function of the energy of each of those photons, which in turn is frame dependent. You don't seem to take that into account.
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Specifically, we can say that Intensity is proportional to the number of photons received per second.
The energy of the photon goes down due to redshift, and the power goes down more due to the pulse being spread out over more time than just a second.
So it seems the power is (reduced via redshift energy)/(increased duration).
Post by: Eternal Student on 02/01/2022 00:43:03
This is wrong, since the intensity of light is also a function of the energy of each of those photons, which in turn is frame dependent. You don't seem to take that into account.Yes, I'm on-board or in agreement with that.
I did mention it here:
.... We are just going to count the number of photons that are received per second. We know that each photon has an energy E = hf associated with it, so if we know the number of photons per second coming in then we can determine the energy coming in per second if we wanted to just by multiplying by hf, where h is Plancks constant and f is the frequency for the photon(s) being received....... and later on again, I did specifically mentioned using different frequencies for the different cases...
....(from item 4)... If you prefer to think about your intensity as a power (in Joules per second) instead of as a number of photons coming in per second then recall that each photon has an energy hf with f found from the relativistic Doppler shift (I think I discussed this earlier and I'm just repeating myself, sorry).
However, I fully appreciate that the post was long and no one can read it all. I deliberately flagged the unusual use of the word "intensity" in the post...
Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation.
About mixing quantum and classical theories:
I didn't go through the entire post, but it seems wrong to mix a quantum explanation with a classic one.Possibly. It was done deliberately just to see how the amplitude might be affected. Evaluating the amplitude of the wave recieved directly using classical theory throughout the derivation was hard while using a particle model for the radiation it just fell off the tree with ease.
But if the final result holds, then it should hold regardless of wheteher we look at the e-m radiation as a classical wave or a collection of particles. If you choose to regard the wave ultimately recieved as a classical e-m wave then you are forced to conclude the amplitude has changed.
If we look upon this a different way, for a classical (non-quantised) electro-magnetic wave, the energy intesity in the wave is proportional to the square of the amplitude (of either the electric or magnetic field, your choice, or use the product of the two if you prefer) and it turns out to be completely independent of its frequency of oscillation. I know that sounds weird but we can pull out the references if required. It seems that to reconcile the notion of energy intesity in a classical e-m wave with the energy content of an individual photon you need to imagine that if you could run a suitable set of detectors over an individual photon then you would observe these two things happening TOGETHER when the energy of the photon is increased: (i) The frequency of oscillation in the electric and magnetic field you observe will increase with increasing energy of the photon but also (ii) The amplitude or peak electiric field strength you would observe will also increase in proportion with the frequency. To say that another way, more energetic photons not only have a higher frequency they would display (on our imaginary oscilliscope) as if they also had a correspondingly higher amplitude.
[References available but many of them are books and not online references. Try the section around page 347 of Electrodynamics by Giffiths where the Poynting vector is discussed, if you've got this on your shelf. The energy flux in the wave is given by E x B so has magnitude proportional to E2 which varies with time (and position) since E and B can be assumed to be sinusoidal - taking an average intensity over time yields intensity proportional to Amplitude squared (and the frequency ω matters not the slightest). For a quick on-line guide (a little less authoritative) try physics libretexts https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)/16%3A_Electromagnetic_Waves/16.04%3A_Energy_Carried_by_Electromagnetic_Waves ].
So, if you do accept that the total energy in the pulse is different in the ships rest frame and the recievers rest frame, it's inescapable that either the classical amplitude is higher in the receivers frame or else the pulse is physically longer and therefore can be absorbed by the reciever for more time. However, it just isn't longer when the ship is coming toward the receiver, it's shorter (than the 1 light-second in the ships rest frame), so the classical amplitude is actually forced to be very much higher.
OK, that's probably drifted off topic enough, sorry for that. I've had a bit more time to think about it and I'm a bit happier to leave Hamdani with the impression that the (classical) amplitude of the e-m wave received will change. I can only apologise if it's incorrect. (Hamdani, you may want to take some care and double check this yourself).
Best Wishes.
Post by: Halc on 02/01/2022 05:33:46
I did mention it here:OK, I was in a severe rush at the time of that posting and didn't see it. I just saw a statement equating power to photons/sec. Was just leaving on last leg of a big road trip. 4 straight days driving. It's over now :)
Intensity is formally defined as the amount of energy received per second and per unit area of the receiver that is being bombarded by the radiation.Right. Since the beam was concentrated on a fixed area, that means intensity and power are essentially equivalent. I wasn't considering it being spread out, which would have been even more off-topic.
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It was done deliberately just to see how the amplitude might be affected.I've never thought of it in terms of amplitude.
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So, if you do accept that the total energy in the pulse is different in the ships rest frame and the recievers rest frame, it's inescapable that either the classical amplitude is higher in the receivers frame or else the pulse is physically longer and therefore can be absorbed by the reciever for more time. However, it just isn't longer when the ship is coming toward the receiver, it's shorter (than the 1 light-second in the ships rest frame), so the classical amplitude is actually forced to be very much higher.Agree with that, but I think the question concerned a receding emitter, not a ship approaching.
The topic was about energy actually disappearing (not being conserved), and that has nothing to do with ships emitting light behind them, but rather has to do with coordinate systems where space is not static over time.
Post by: Eternal Student on 02/01/2022 11:48:31
I hope your trip went well.
Post by: Iannguyen on 09/02/2022 11:19:17
Breaking and reforming molecular bonds from a less stable configuration (wood and oxygen) to a more stable configuration (ash and water vapor) releases energy.
If you look at the quantity of energy released and apply Einstein's famous conversion, E = mc2, you'll notice that the mass of the product and reactant molecules differs by a very little amount.
In anything like a nuclear event, such as one that occurs in the Sun, the mass difference is considerably more significant. In fact, if you calculated the Sun's mass from birth to now, you'd find that it has lost roughly the mass of Saturn in those 4.5 billion years of energy emission.
It's possible that a stronger definition of energy will emerge in a richer (i.e., quantum) theory of gravity, and we'll be able to tell whether it's conserved or not. However, in the lack of a formal definition, we must make do with what we have, which are the tools and definitions that we currently have. Yes, photons lose energy, but that energy does not vanish forever; the quantity of energy lost (or gained, for that matter) in the expanding (or shrinking) Universe adds up to precisely what it should.
Post by: hamdani yusuf on 10/02/2022 05:22:20
In fact, if you calculated the Sun's mass from birth to now, you'd find that it has lost roughly the mass of Saturn in those 4.5 billion years of energy emission.Have you taken solar wind and solar flares in to account?
Post by: Eternal Student on 10/02/2022 22:40:31
You've made an interesting post. Thank you and well done. There's a lot about your post that is perfectly good and makes solid scientfic sense.
Usually that doesn't get mentioned, instead someone will just go straight in to challenge one or more points in the post. So this is where you get a slightly harder time, sorry...
As the Universe expands, light loses energy, which is then used to expand the Universe itself in the form of labor.This point for example.
Someone could ask for the evidence to show that this is the correct situation. How were the calculations done? How do you begin to determine how much energy it takes to expand the universe? Having got the amount of energy required for that, how do you determine that the amount of energy lost by all the radiation is sufficient to match that? Is it actually just a reasonable guess that something like your suggestion is happening (lost energy from radiation is accounted for in the energy required to expand the universe)?
Finally, does your suggestion mean that the expansion should slow down as the density of radiation falls off and all of it is moved further to the red (low energy) range? (Because that is precisely the opposite of what does actually seem to be happening).
Don't worry too much, your post seemed quite sensible and discussion is very much welcomed. I'm just deliberately focusing in on one detail because it did seem to be presented as if it was a fact rather than just a possibility or a speculation.
Best Wishes.
Post by: alancalverd on 11/02/2022 00:17:34
The gravitational potential of "deep space", infinitely distant from any mass, is zero, and the gravitational potential close to a mass is negative.
A ballistic missile or a photon travelling upwards from the earth's surface gains gravitational potential energy (moving away from the negative potential) and loses an equal amount of kinetic energy. In the case of a photon, this appears as red shift.
If the source and receptor are moving relative to each other, the additional kinetic energy due to that motion is expressed in the Doppler frequency shift.
Post by: Eternal Student on 11/02/2022 02:22:19
You could also mention that there was a podcast based on the OP and it is available here: https://www.thenakedscientists.com/sites/default/files/media/podcasts/episodes/Naked_Scientists_QotW_21.09.13.mp3
The gravitational potential of "deep space", infinitely distant from any mass, is zero, and the gravitational potential close to a mass is negative.OK - but is there any place in space that is "infinitely distant from any mass"?
The usual cosmological principle implies space is quite homogeneous on large scales. There should be just as much mass and it should be just as close to the photon "out there" in deep space as it was right here. On average the environment for the photon does not change no matter where it goes or how long it travels. It's then hard to say that the photon is moving to a region with higher gravitational potential.
The Pound-Rebka experiment certainly shows that a photon leaving a planet or a star experiences a red-shift for a while - but that effect should stop when it gets into the gravitational well of some other celestial object and it might actually start to experience a blue-shift then. The red-shift that seems to happen to all light due to the expansion of space isn't just something that happens for a while, it seems to happen eternally.
Best Wishes.
Post by: alancalverd on 11/02/2022 09:41:08
OK - but is there any place in space that is "infinitely distant from any mass"?Physicists are quite used to dealing with idealised infinities. Indeed I was thinking about this very problem whilst shaving this morning. I can imagine an infinite space with a finite density of particles. On a big scale with a few fluctuations this looks like the universe, and on a small scale with regular spacing it looks like an ideal crystal. We can solve all sorts of equations for an infinite lattice and add boundary conditions for a finite one, so the concept of "deep space" gets us close enough to the observed energy of photons emitted close to the detector, and a boundary condition of emission from a massive object, or travelling through a finite-density universe, points us in the direction of the Pound-Rebka result.
P-R does indeed demonstrate blue shift if you invert the apparatus, and there are celestial objects with blue shift too. Problem is that this planet, and anything we make on it, is too small to induce a significant gravitational blue shift of a photon emitted from a distant star, which was necessarily a lot bigger.