Gravity effect on observed speed of light

[D H]

Let's get back to the original question.

An observer (in zero gravity) witnesses two parallel but widely separated beams of light that are transmited at the same time. One of the beams experiences zero gravity. The other beam travels between two massive objects with a very strong gravitational field, but arranged so that the beam direction is not changed.

I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

 

[PAllen]

Let's get back to the original question.I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

[Edit: Further thought: the image of the affected pulsar may become some complex shape due to graviational lensing, around the same time the graviational effects on the light become significant. This may cause the pulse to appear smeared in space and time. However, assuming perfect symmetry, there should remain some central point of the smeared image that has not appeared to shift in the sky. Will this part of the pulse image remain synchronized? ]

 

[atyy]

Isn't this obviously yes (?). KE and momentum are frame dependent, whether for light or bullets. In an accelerating frame, inertial objects appear to continuously change momentum and KE, and light appears to gain/lose energy. So, you can describe gravity well effects similarly.

How about the point of view of the inertial twin?

 

[PAllen]

How about the point of view of the inertial twin?

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

To make this better defined, assume the emitted pulses are directed, not spherical. Then, the turnaround twin changes momentum and KE a tiny bit after each pulse. Further, note that the angle of receiver needed to capture a complete directed pulse will change from the red shift case to the blue shift case. Assume receiver is big enough to capture whole pulse in both cases (relativistic beaming effects eliminated). Then, clearly the receiver will, as I said above, receive red pulses of lower total energy compared to the blue pulses. Small momentum and KE changes to the emitting, turnaround twin, preserve conservation of energy and momentum.

 

[atyy]

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

So from the inertial twin's point of view, the light lost energy?

 

[PAllen]

So from the inertial twin's point of view, the light lost energy?

See edited explanation above.

 

[Q-reeus]

As you point out - the position of the start and finish lines for both beams are seen, by the distant observer, to be the same ergo on the basis that the beams travel identical distances and, having been emitted simultaneously, arrive at the finish line simultaneously the observer is fully justified in concluding that, from HIS point of view, the beams took the same amount of time to make their trips...
..As above. The situation as depicted is that there are NO gravity affected local values and he bases his conclusion on observation alone (simultaneous emission; the position of the start and finish lines for both beams are seen to be the same; simultaneous arrival at the finish line)...

Again, you are not accepting that gravity alters distance and time scales. Identical start and finish lines does not mean identical path traversal in between. Two open ended rubber bands bridge across an equal width gap. One is stretched straight, the other is depressed down - 'saggy'. Two insects crawl across at equal local crawl speed, one on each rubber band. Do both arrive together because the gap is the same (your logic), or will the one on the saggy band arrive last (GTR = gravity warps spacetime logic)? Think before answering.

Alternately, as previously, he bases his determination of the identical distances using his meter stick held at arm's length and uses his clock in order to determine the time that it takes for the beams to make their trips. He cannot see the far distant clocks and rods...
..Let us assume that the OP involves a semblance of reality and there are two black holes between which a beam of light travels. There are NO clocks or sticks dangling in space so the distant observer can only determine his measurements in accordance with his own equipment however all of a sudden clocks and sticks appear along the path of the beam...
Does that affect (alter) the distant observer's measuring rod or clock in any way? Of course it does NOT!

So you genuinely believe that literal clocks and rulers somehow need to be present locally in order to effect things?! No - it is a very useful 'figure of speech' to refer to clocks and rulers as indicative that time and distance measure are effected by gravity in this case. You accept that clocks are used to measure time, and rulers to measure length, right? Get used to these terms as simply representative of the 'actual' quantities.

You wrote "...the booster is used up and he slows down to the same initial speed." My depiction is that he does not slow down to the same initial speed but that his speed is then slower than its initial speed thereby enabling the other beam to catch up.

So here in #31, you finally let it be known entry and exit speeds are in your universe unequal, without giving any hint in either #3:
"I believe that the latter beam will accelerate as it approaches those objects then slow down as it travels away from same thereby arriving at the destination simultaneously with the uninterrupted beam. ", or #16:
"You may prefer to look at things from the point of view that light decelerates while falling into a gravity well and accelerates while climbing out however on the basis that this viewpoint leads to some apparently paradoxical results I prefer to look at things from the point of view that light accelerates while falling into a gravity well and decelerates whilst departing.", or #18:
"In my interpretation the train accelerates as it travels toward the objects then slows down as it departs same thus presenting an overall time equivalence as determined by the distant observer's gravity unaffected clock and rod. Will the two trains arrive together? Yes."

My strong suspicion is you meant equal in and out as suggested and inferred in the above three entries of yours, but when confronted with the inescapable logic that means necessarily unequal arrival times, you have shifted ground in order to preserve your assertion of equal arrival times. OK, you now want unequal in and out speed? Then realize this necessarily implies some kind of 'draggy' gravity theory at odds with both Newtonian theory and GTR. Recalling we are dealing with static, non-rotating masses as per OP, gravity is an entirely conservative effect - same in = same out, no if's, but's, or maybe's. You doubt that? Well argue it out with http://en.wikipedia.org/wiki/Conservative_force, and nothing changes qualitatively in GTR.

As I have previously pointed out - the Shapiro delay is relevant to a SINGLE gravitational field! When somebody conducts a dual gravitational field experiment please let me know.

No. The reference makes it plain curvature is insignificant, leaving only time delay. In the OP's setup (we have gone over this before), by symmetry of the twin masses curvature is canceled out, which only makes it worse from your pov - there is nothing but time delay as effect. Where do you go from there?

I freely admit that I am academically unqualified in the subject of physics but please stop treating me like a naive child.

Well so am I unqualified, but our difference is in the willingness to recognizing that 2+2 has to add up to 4, and if our pet idea disagrees, be man enough to acknowledge that, and move on a lit bit the wiser. That will make you a real adult. Cheers!:grumpy:

 

[PAllen]

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

[Edit: Further thought: the image of the affected pulsar may become some complex shape due to graviational lensing, around the same time the graviational effects on the light become significant. This may cause the pulse to appear smeared in space and time. However, assuming perfect symmetry, there should remain some central point of the smeared image that has not appeared to shift in the sky. Will this part of the pulse image remain synchronized? ]

Anyway, the upshot remains (based on the Shapiro effect analysis), that the pulsar whose signal path is affected by the converging neutron stars would fall out of synch, falling behind the other pulsar.

Is this the consensus of the experts (I'm not an expert)?

 

[Dale]

Using his meter stick (held at arm's length) he determines that the beams travel identical distances - from their source to the target. He does not measure 'a longer spacetime path' for one of those beams.

Remember, we are talking about curved geometry here. In curved geometry it is possible for a figure to have 4 straight sides with 4 right angles and still have one side which is longer than the opposite side. Of course the observer may use a coordinate system such that the coordinate distance is the same, but all distances based on the metric (measuring rods or light pulses) will be longer for the path through the gravity well.

 

[cos]

I recognise nothing at all of your issues in the text that you did not bother to read.

In that case I'm glad I didn't waste my time reading it.

 

[cos]

Let's get back to the original question.

I'm going to rephrase it slightly in a way that I hope eliminates some of the ambiguity and simultaneity issues. I'll use our universe and a near carbon copy of our universe.

Suppose an observer simultaneously sees two bursts of light in different parts of the night sky. The path from each source to the observer is more or less flat (no significant gravity wells along the way). Now imagine an alternate universe that is exactly the same as ours except that now one of the signals travels between two massive objects with very strong gravitational fields arranged so that the beam direction is not changed. Does our alternate observer still see the bursts as arriving simultaneously?

Yes.

 

[cos]

I'll throw in an ammendment to elimintate the alternate universe complication. Two pulsars in different parts of the sky are magically synchronized. The path from both initially encounter no significant graviy well. However, two neutron stars are heading for a collision directly on the path from one of the pulsars, perpendicular to its line of sight from us. As they get close to collision on the path from one pulsar, do we see the synchronization break down, and in what direction?

The synchronization does not break down.

 

[harrylin]

In that case I'm glad I didn't waste my time reading it.

OK then, if you prefer to keep your issues, as you wish! I won't try anymore to help you get rid of them.

 

[cos]

Again, you are not accepting that gravity alters distance and time scales. Identical start and finish lines does not mean identical path traversal in between. Two open ended rubber bands bridge across an equal width gap. One is stretched straight, the other is depressed down - 'saggy'. Two insects crawl across at equal local crawl speed, one on each rubber band. Do both arrive together because the gap is the same (your logic), or will the one on the saggy band arrive last (GTR = gravity warps spacetime logic)? Think before answering.

Your having resorted to personal insults will be dealt with later!

Where, and in which direction, is the 'downward' (?) 'depression' of the path followed by that specific beam of light? Does that route 'sag' or bend in any direction?

Alternately, as previously, he bases his determination of the identical distances using his meter stick held at arm's length and uses his clock in order to determine the time that it takes for the beams to make their trips. He cannot see the far distant clocks and rods...
..Let us assume that the OP involves a semblance of reality and there are two black holes between which a beam of light travels. There are NO clocks or sticks dangling in space so the distant observer can only determine his measurements in accordance with his own equipment however all of a sudden clocks and sticks appear along the path of the beam...
Does that affect (alter) the distant observer's measuring rod or clock in any way? Of course it does NOT!

So you genuinely believe that literal clocks and rulers somehow need to be present locally in order to effect things?!

Typical ill-thought out response aimed specifically at obfuscation and deception!

I point out that the presence or absence of local clocks and rules can have NO effect on the distant observer's measuring devices and you snidely insinuate that I believe the opposite!

(<snip>Similar ill-thought out response aimed specifically at obfuscation and deception.)

...be man enough to acknowledge that, and move on a lit bit the wiser. That will make you a real adult.

It is inevitable that when a person realizes that they are losing an argument they start posting obfuscatory and deceptive material as well as personal insults to which I have, above, responded in kind.

Our discussion is accordingly terminated.

 

[cos]

Remember, we are talking about curved geometry here. In curved geometry it is possible for a figure to have 4 straight sides with 4 right angles and still have one side which is longer than the opposite side. Of course the observer may use a coordinate system such that the coordinate distance is the same, but all distances based on the metric (measuring rods or light pulses) will be longer for the path through the gravity well.

Trepidatiously anticipating eventual ad hominem attacks as have inevitably previously taken place - I submit that a person is looking out into the night sky; he sees, off to his left-hand side, one object that will eventually represent the start line of the beam that will pass between two (yet-to-appear) massive objects and, off to his right-hand side, another object that will eventually become the finish line.

He sees, initially, a beam of light cross the start line when his clock reads zero seconds and arrive at the finish line when his clock reads 5 seconds.

What is his estimation of the distance between the start line and the finish line? Is it, as I suspect it to be, (close to) 1·5 million Ks?

Having thereby determined the distance between the start and finish lines our observer then 'sees' two neutron stars appear on either side of the original beam's route which, as per my assumption below regarding the OP, generate no gravitational 'pull' on the start/finish lines.

(I assume that, in the OP, it is accepted that the intervening masses have no effect on the distance between the start and finish lines; that the start /finish lines are hypothetically immovable objects.)

According to responses - it will then take longer for the beam to traverse that distance as measured by his clock.

Does he conclude that the finish line and the starting line have physically moved further apart?

Prior to the appearance of the masses - another observer, located next to the start line, attaches a very long tape measure to it then travels over to the finish line feeding out the tape as he goes. No surprise - he determines a distance of 1·5 million Ks.

The gravitational masses make their appearance (again without having any physical affect on the distance between the immovable start/finish lines) and he repeats his test.

As he enters progressively stronger gravitational tidal areas the tape measure, attached at the other end to an immovable object, proportionally stretches! It becomes spaghettified! The physical distances between the mm; cm; m; km divisions increase!

As he moves away from that midpoint the tape remains stretched albeit by progressively lesser amounts ergo, overall, he measures that the distance between the start/finish lines is shorter than it was before the masses appeared not longer as insisted upon in this thread.

I have no doubt whatsoever that it can be mathematically 'proven' that the physical results of his hands-on experiment are 'erroneous' - that the distance increases 'due to the Shapiro delay' however I know which version I would place my money on.

 

[Dale]

What is his estimation of the distance between the start line and the finish line?

The problem is that this question is ambiguous in curved spacetime. There are many different ways you could determine the distance, and in GR they can lead to different results. For example, you could simply subtract the coordinate positions, but since the coordinate positions are arbitrary so is this procedure. You could do a radar ranging experiment or you could use a tape measure, but those only work in a static spacetime. You could connect them by a spacelike geodesic and determine the spacetime interval, but there may be more than one geodesic. Etc.

In any case, you do not need to answer this question to answer the OP. It is clear that the pulse through the gravitational field will arrive later. If you prefer to attribute that to a changed (radar) distance or a changed (coordinate) speed of light is up to you. My preference is the former.

 

[atyy]

Again, seems inevitable. Red light has different energy than blue, per photon. If turnaround twin is sending pulses of constant energy from his/her point of view, inertial twin will receive red pulses with less energy and blue pulses with more energy. How could it be otherwise?

To make this better defined, assume the emitted pulses are directed, not spherical. Then, the turnaround twin changes momentum and KE a tiny bit after each pulse. Further, note that the angle of receiver needed to capture a complete directed pulse will change from the red shift case to the blue shift case. Assume receiver is big enough to capture whole pulse in both cases (relativistic beaming effects eliminated). Then, clearly the receiver will, as I said above, receive red pulses of lower total energy compared to the blue pulses. Small momentum and KE changes to the emitting, turnaround twin, preserve conservation of energy and momentum.

How is the equivalence of energy and frequency justified in classical physics?

 

[harrylin]

How is the equivalence of energy and frequency justified in classical physics?

Here we are discussing special relativity, according to which:

f'/f = E'/E

"It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law."

- http://www.fourmilab.ch/etexts/einstein/specrel/www/ section 8 (Einstein, 1905).

 

[atyy]

Here we are discussing special relativity, according to which:

f'/f = E'/E

"It is remarkable that the energy and the frequency of a light complex vary with the state of motion of the observer in accordance with the same law."

- http://www.fourmilab.ch/etexts/einstein/specrel/www/ section 8 (Einstein, 1905).

But for anyone observer, there is no change in energy.

 

[PAllen]

But for anyone observer, there is no change in energy.

For any inertial observer there is no change in energy after the light is emitted. But two different observers disagree on energy of emitted light. Emitter thinks they emitted white light, total energy E, receding observer thinks they emitted red light of energy less than E.

On the other hand, for any observer whose state of motion is changing, they see a change in the energy of light they are passing through, as their state of motion changes.

I am really confused - I don't think any of this is controversial. If we just replace light with bullets, nobody disagrees; light is not fundamentally different. And for bullets you can certainly make an analogy between an accelerated observer seeing bullets lose KE, and bullets fired up gravity well lose KE. Specifically imagine a long rocket accelerating, bullets fired from its back to front. Someone at the front of the rocket will find tham less energetic than when they were fired at the bottom. You can explain this as accelerating toward the bullets or a fictitious gravity field. Now place same rocket on the surface of a planet. Bullets fired from the bottem will be less energetic at the top. You can say the lost energy going up a gravity well or that the rocket is accelerating upwards (to maintain static position in gravity well). Principle of equivalence says the situations are equivalent.

 

[atyy]

For any inertial observer there is no change in energy after the light is emitted. But two different observers disagree on energy of emitted light. Emitter thinks they emitted white light, total energy E, receding observer thinks they emitted red light of energy less than E.

On the other hand, for any observer whose state of motion is changing, they see a change in the energy of light they are passing through, as their state of motion changes.

I am really confused - I don't think any of this is controversial. If we just replace light with bullets, nobody disagrees; light is not fundamentally different. And for bullets you can certainly make an analogy between an accelerated observer seeing bullets lose KE, and bullets fired up gravity well lose KE. Specifically imagine a long rocket accelerating, bullets fired from its back to front. Someone at the front of the rocket will find tham less energetic than when they were fired at the bottom. You can explain this as accelerating toward the bullets or a fictitious gravity field. Now place same rocket on the surface of a planet. Bullets fired from the bottem will be less energetic at the top. You can say the lost energy going up a gravity well or that the rocket is accelerating upwards (to maintain static position in gravity well). Principle of equivalence says the situations are equivalent.

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! So somehow classical gravity knows about QM? (Yes, but how?)

The point of view from GR is just red shifts are due to light going along geodesics, and the local acceleration of the observer, and works even in cases where there is no conserved energy.

 

[PAllen]

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! Some somehow classical gravity knows about QM? (Yes, but how?)

The point of view from GR is just red shifts are due to light going along geodesics, and the local acceleration of the observer, and works even in cases where there is no conserved energy.

Harrylin's link earlier show's how Einstein derived that different frames (similarly changing frames) see light energy and frequency proportionally changed, purely from SR and Maxwell. To be successful, QED had to incorporated SR + Maxwell in the classical limit; so it did, as a result of which E=hf can be said to follow.

My meta point is that there were features 'hidden' in classical theories that became clearer in the carry over to quanum theories. The classical theories didn't 'know about' quantum theories, instead the quantum theory clarified the classical theory.

An exampel is radiation in SR + Maxwell can be a very subtle issue, as radiation is non-local in this framework. In QED many of these tricky cases are very straightforward because radiation in QED is locally defined.

 

[harrylin]

No, I am confused. What's amazing to me is that the energy argument is not an equivalence principle argument. To me it is an argument from conservation of energy and semiclassical "old" quantum physics argument. From QM we have E=hf. From classical gravity, we have dE=g.dh, which implies df. No relativity even (all in one Galilean frame)! Some somehow classical gravity knows about QM?
[..]

E/E0=f/f0 => E ~ f.
Thus I would say, QM knows SR!

Edit: Oops I was delayed in replying, meanwhile PAllen said it all.

 

[atyy]

But the classical theories are complete in themselves. If we want to bring in QM, then we could just as well say QED is a low energy approximation to a non-relativistic theory.

Anyway, going back to Harrylin's point that purely clasically E and f transform the same way, does that necessarily mean that E~f, purely classically?

 

[PAllen]

But the classical theories are complete in themselves. If we want to bring in QM, then we could just as well say QED is a low energy approximation to a non-relativistic theory.

Anyway, going back to Harrylin's point that purely clasically E and f transform the same way, does that necessarily mean that E~f, purely classically?

It seems to. Assume E=g(f) for some arbitrary g, and that by SR, E0=g(f0). We also have E/E0 = f/ f0 = expression of v and c = constant relative to E and f. Call it k. Then E0=kE, f0=kf, then:

kE=g(k f)

k g(f) = g (k f)

This implies g'(f) = g'(kf) for all f (g' as derivative). Then g'(f)=g'(f/k)=g'(f/k^2)... If g continuous, the g'(f) = g'(0). Thus g' constant, thus g(f) = c f. QED (not quantum electrodynamics).

 

[atyy]

It seems to. Assume E=g(f) for some arbitrary g, and that by SR, E0=g(f0). We also have E/E0 = f/ f0 = expression of v and c = constant relative to E and f. Call it k. Then E0=kE, f0=kf, then:

kE=g(k f)

k g(f) = g (k f)

This implies g'(f) = g'(kf) for all f (g' as derivative). Then g'(f)=g'(f/k)=g'(f/k^2)... If g continuous, the g'(f) = g'(0). Thus g' constant, thus g(f) = c f. QED (not quantum electrodynamics).

Hmmm, very interesting. A universal constant. By dimensional analysis, the constant cannot be formed from G and c, so SR implies a new constant of nature?

 

[PAllen]

Hmmm, very interesting. A universal constant. By dimensional analysis, the constant cannot be formed from G and c, so SR implies a new constant of nature?

Maybe this doesn't really say quite that much. I think all it says is that fixing a volume of spacetime, holding field amplitudes constant, the EM energy content of waves of contained waves is proportional to frequency. It doesn't say there is some smallest unit energy. To make the factor more fundamental, you would have to state is (energy * time)/( field amplitude), which would be a different sort of animal than plank's constant.

On further thought, I wouldn't be suprised if this result follows in a fairly direct way from Maxwell's equations, though I am not able to do this myself. An intuitive argument is simply that a wave of given amplitude has given energy per wave front area. Then, it would follow immediately that the shorter the wavelength, the more energy per unit volume, thus E~f.

 

[atyy]

The surprise to me is not that this implies a quantum of energy. The surprise is that it seems to imply a relationship between energy and frequency, purely classically. I had always thought that energy was related to amplitude in classical terms (eg. the formula for the Poynting vector says nothing about frequency), and that E~f is a purely quantum mechanical relation. So if E~f purely classically, and the constant of proportoinality is a universal constant, then by dimensional analysis, it seems to predict a new universal constant with the same units as Planck's constant - so classical physics would seem to know about quantum physics. It's because of this implication that I'm skeptical that classical physics implies E~f, though I cannot find any mistake in your reasoning.

 

[harrylin]

The surprise to me is not that this implies a quantum of energy. The surprise is that it seems to imply a relationship between energy and frequency, purely classically. I had always thought that energy was related to amplitude in classical terms (eg. the formula for the Poynting vector says nothing about frequency), and that E~f is a purely quantum mechanical relation. So if E~f purely classically, and the constant of proportoinality is a universal constant, then by dimensional analysis, it seems to predict a new universal constant with the same units as Planck's constant - so classical physics would seem to know about quantum physics. It's because of this implication that I'm skeptical that classical physics implies E~f, though I cannot find any mistake in your reasoning.

Perhaps you overlooked the number of photons? I think that the energy is of a light wave is not hf but nhf. There's also an old thread about this, here:
https://www.physicsforums.com/archive/index.php/t-63023.html

Harald

 

[PAllen]

I went back and reviewed derivations of energy density and flux of plane waves, and also read the prior sections of Einstein's 1905 paper. I believe all the mysteries can be resolved.

1) The mistake in my argument that E~f could be derived classically is the very assumption that E=g(f) for some unknown g. Classically, E is simply independent of f, and depends only on amplitude.

2) The classical derivation of the change in E for a traveling wave for change in motion of an observer involves a change in amplitude. Since the amplitude, energy, and frequency change in tandem, then when looked at from a quantum point of view, number of photons is preserved, each having different energy for different observers. (Same number of lower energy photons is represented classically as lower amplitude).

3) I believe all my prior statements about the effect of motion, acceleration, and gravity on the the energy of light, and the analogies with bullets, are essentially correct, if not always precisely worded. I was never setting out to reason strictly classically. Obviously, any statements about classical justification for E~f are incorrect.

 

[harrylin]

I went back and reviewed derivations of energy density and flux of plane waves, and also read the prior sections of Einstein's 1905 paper. I believe all the mysteries can be resolved.

1) The mistake in my argument that E~f could be derived classically is the very assumption that E=g(f) for some unknown g. Classically, E is simply independent of f, and depends only on amplitude.

2) The classical derivation of the change in E for a traveling wave for change in motion of an observer involves a change in amplitude. Since the amplitude, energy, and frequency change in tandem, then when looked at from a quantum point of view, number of photons is preserved, each having different energy for different observers. (Same number of lower energy photons is represented classically as lower amplitude).

3) I believe all my prior statements about the effect of motion, acceleration, and gravity on the the energy of light, and the analogies with bullets, are essentially correct, if not always precisely worded. I was never setting out to reason strictly classically. Obviously, any statements about classical justification for E~f are incorrect.

I'm afraid that I don't follow you here; I think that the discussion is simply a different topic, although related.

In a follow-up paper of the same year[1], Einstein elaborated further on the passage that I cited earlier. He started with stressing that the energy of an emitted wave as measured in a system that is moving relative to the source, is different from the energy as measured in the "rest" system.

He thus elaborated on the fact that E'/E = f'/f.

I thought that this was the discussion topic following your post #38 (elaborating on gravitation effects on light with PoE), but I must admit that I did not follow the discussion carefully.

[1] http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

 

[PAllen]

I'm afraid that I don't follow you here; I think that the discussion is simply a different topic, although related.

In a follow-up paper of the same year[1], Einstein elaborated further on the passage that I cited earlier. He started with stressing that the energy of an emitted wave as measured in a system that is moving relative to the source, is different from the energy as measured in the "rest" system.

He thus elaborated on the fact that E'/E = f'/f.

I thought that this was the discussion topic following your post #38 (elaborating on gravitation effects on light with PoE), but I must admit that I did not follow the discussion carefully.

[1] http://www.fourmilab.ch/etexts/einstein/E_mc2/www/

But it surprisingly does *not* follow from E'/E = f'/f over Lorentz transform, that E~f, and Einstein never makes this latter claim. My derivation of this is all fine *if* E is determined by some initially unknown g(f). However, if E is independent of f, my derivation amounts to a proof based on aerodynamics, that "If elephants could fly, then pigs certainly could too".

The key is that in section 7 of Einstein's paper he also derived the way amplitude transformed, and the way E transformed *follows* from the way amplitude transforms. As a result, the fact that f transforms the same as E is giving you no additional information. So we have the classically known fact that E is determined by amplitude and independent of f; amplitude transform under Lorentz determines E transform under Lorentz. It happens that E'/E=f'/f, but this says nothing about dependence of E on f in general.

[Edit: For emphasis: certainly one cannot say that amplitude depends on f! One can say that amplitude depends on number of photons and f, once you quantize.

The interesting consequence of the fact that E'/E=f'/f combined with the amplitude transform law, results that when light is quantized, a Lorentz transform preserves the number of photons.]

 

[PAllen]

To make the classical vs. quantum relation between energy, frequency, and amplitude concrete, consider a concrete example:

Imagine in one frame we have 3 pulses of light, each with energy E0, amplitude A0, and frequency f1, f2, and f3. Classically, there is no relation between E and f. Now the quantum situation is simply that pulse 1 has E0/hf1 photons, pulse 2 has E0/hf2 photons, etc.

Now apply a Lorentz transform. We have the following relations:

E0' is related to A0' the same as E0 is to A0.
E0'/E0 = f1'/f1 = f2'/f2 = f3'/f3 (1)

As a result of (1), E0'/hf1' = E0/hf1, so the number of photons is preserved. However, from a classical point of view, there remains no relation between E and f.

 

[atyy]

Do you think section 8 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ is correct? eg. is the R² in the same in primed and unprimed coordinates?

(Yes, maybe it's just that E is not a function of f at all, but why doesn't that "fall out" of the mathematics? I would have thought maybe E=g(f), Eo=h(g(fo), or that v(f) implicitly rather than v=constant)

 

[PAllen]

Do you think section 8 of http://www.fourmilab.ch/etexts/einstein/specrel/www/ is correct? eg. is the R² in the same in primed and unprimed coordinates?

(Yes, maybe it's just that E is not a function of f at all, but why doesn't that "fall out" of the mathematics? I would have thought maybe E=g(f), Eo=h(g(fo), or that v(f) implicitly rather than v=constant)

I didn't see any problem with section 8, which was based on the amplitude transform derived in section 7.

It seems to me that what my earlier derivation shows is the IF E=g(f), g must be linear, but that if E is independent of f, the argument shows nothing.