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DaveBeal
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Is it theoretically possible for the expansion of the universe to red-shift the energy of a photon all the way to zero? If so, what happens to the photon? Or does the photon's energy only approach zero as an asymptote?
No. The "energy" you refer to, which is redshifted by expansion, is the energy measured by a comoving observer who is co-located with the photon. This energy is the inner product of the comoving observer's 4-velocity with the photon's 4-momentum, and it is impossible for that to be zero. Mathematically, that is because it is impossible for the inner product of a timelike vector and a lightlike vector to be zero.DaveBeal said:Is it theoretically possible for the expansion of the universe to red-shift the energy of a photon all the way to zero?
Mathematically, the zero vector would like to know where it belongs.PeterDonis said:Mathematically, that is because it is impossible for the inner product of a timelike vector and a lightlike vector to be zero.
Tell it we don't put up with its kind here.Orodruin said:Mathematically, the zero vector would like to know where it belongs
It's not a valid timelike 4-velocity vector for an observer, and it's not a valid 4-momentum vector for a photon. So it doesn't belong anywhere in this discussion.Orodruin said:Mathematically, the zero vector would like to know where it belongs.
In the physics discussion, no. In a purely mathematical statement about timelike and lightlike 4-vectors though.PeterDonis said:It's not a valid timelike 4-velocity vector for an observer, and it's not a valid 4-momentum vector for a photon. So it doesn't belong anywhere in this discussion.
"Timelike" and "lightlike" are physical definitions. I don't think the zero vector meets the physical definition of a "lightlike" vector even though its norm is zero.Orodruin said:In a purely mathematical statement about timelike and lightlike 4-vectors though.
Not sure I agree with this. The nomenclature is obviously derived from physics, but that doesn’t make the definition that a timelike vector is a vector with positive square (or negative depending on convention) any less mathematical.PeterDonis said:"Timelike" and "lightlike" are physical definitions. I don't think the zero vector meets the physical definition of a "lightlike" vector even though its norm is zero.
Yes, agreed, but you still need to show that the antecedent is true. If @Orodruin is correct that we need to include the zero vector as "lightlike" on mathematical grounds, then it would be possible in principle to have a null vector that was a zero vector at some event. My point is that that's impossible: the definition of "lightlike" does not include the zero vector. Yes, ultimately that's based on the physical claim that the zero vector can't represent any physical thing, but it's still a mathematical definition of "lightlike" that excludes the zero vector.PAllen said:the statement remains that if a null vector is not a zero vector per some basis at some event, then for any choice of basis, and any sequence of parallel transports, it remains nonzero. This precludes the possibility mentioned in the OP.
But then you would have light that had no energy or momentum ever ( at any event on its world line , or any basis choice). I guess that motivates non calling it light like. That is, a zero vector is zero in any basis and retains this property on any parallel transport.PeterDonis said:Yes, agreed, but you still need to show that the antecedent is true. If @Orodruin is correct that we need to include the zero vector as "lightlike" on mathematical grounds, then it would be possible in principle to have a null vector that was a zero vector at some event. My point is that that's impossible: the definition of "lightlike" does not include the zero vector. Yes, ultimately that's based on the physical claim that the zero vector can't represent any physical thing, but it's still a mathematical definition of "lightlike" that excludes the zero vector.
Yes.PAllen said:But then you would have light that had no energy or momentum ever ( at any event on its world line , or any basis choice). I guess that motivates non calling it light like.
Yes.PAllen said:That is, a zero vector is zero in any basis and retains this property on any parallel transport.
PeterDonis said:Yes, agreed, but you still need to show that the antecedent is true. If @Orodruin is correct that we need to include the zero vector as "lightlike" on mathematical grounds, then it would be possible in principle to have a null vector that was a zero vector at some event. My point is that that's impossible: the definition of "lightlike" does not include the zero vector. Yes, ultimately that's based on the physical claim that the zero vector can't represent any physical thing, but it's still a mathematical definition of "lightlike" that excludes the zero vector.
Just to add, this is not correct. Parallel transport needs to conserve not only the norm of vectors, but also the inner product between different parallel transported vectors. This makes it impossible for a non-zero null vector to be parallel transported into a zero vector.PeterDonis said:then it would be possible in principle to have a null vector that was a zero vector at some event.
What I intended to state is correct, but I stated it ambiguously. Let me try again:Orodruin said:this is not correct
I don't think defining a lightlike vector as one with zero norm dictates anything about the actual usage of the vectors. It is a definition and as far as all definitions go, you may like or dislike it for various reasons, but it does not inhibit your ability to use the maths in any way or form. Nothing states that you are forced to use a zero vector as a 4-momentum just because it may be classified as lightlike.PeterDonis said:Physicists are supposed to use math as a tool, not let math dictate things that make no physical sense.
I disagree. His actual statement is: "It is particularly important to understand that a null vector is not a zero vector." The emphasis on "not" is in Schutz. That indicates to me that is specifically intends the zero vector to not be included in his definition of a null vector. His other remarks, I think, are with the same intent that I describe below with reference to Carroll--to emphasize to new students that the metric of spacetime is not Euclidean, but Minkowski, so that non-zero vectors can still have zero norm.Orodruin said:Schutz: Defines lightlike/null as having zero norm, then makes a comment that a null vector is not a zero vector. This may indicate that he is excluding the zero vector from the definition, but the way it is written gives the impression that the main takeaway should be that a null vector need not be the zero vector.
His actual statement is: "A vector can have zero norm without being the zero vector." That statement is ambiguous. It could mean that the zero vector is included in the definition of a null vector, but also that other vectors that aren't the zero vectors are; or it could be a way of pointing out to a new student who isn't familiar with the Minkowski metric that the definition of "null vector" just given is not vacuous: that it actually means those vectors which aren't the zero vector but still have zero norm, because the metric is not the Euclidean metric that the student is used to.Orodruin said:[Carroll] then makes a particular comment that a null vector does not need to be a zero vector
Can you give some others?Orodruin said:this is not the only possible physical use of vectors with zero norm
No, you are talking about a light pulse, but the usage of the terminology "lightlike" does not have this restriction.PeterDonis said:Can you give some others that are relevant in this context? Remember we are talking about the energy and momentum of a "photon" ("light pulse" would be a better term here).
Agreed, but as far as the definition goes, we can only read what he has actually written, which is that a lightlike vector is one with zero norm.PeterDonis said:So I don't think we can say that Caroll intended the zero vector to be a null vector based on that statement.
This is personal inference. Is MTW also to be considered as intended as an introduction to the subject? It is a rather extensive introduction in that case.PeterDonis said:is perfectly possible that he simply didn't want to go into such technical details in a book that was intended as an introduction to the subject
Why? Unless A and B are the same--which makes no sense since then you would not even talk about "the separation of the emission event and the event on the worldline of B", or even about a "light signal" at all, since there aren't two events, only one--the vector that connects the two events in question will not be the zero vector.Orodruin said:we could imagine talking about two inertial observers in Minkowski space and ask ourselves when a light signal from observer A would reach observer B. The reply would be "when the separation of the emission event and the event on the worldline of B is lightlike". If you do not include the zero vector in lightlike, then you would have to include "or the zero vector" in this statement.
Sure. One has to do that when reading any text.Orodruin said:This is personal inference.
The entire book is of course not just an introduction. But the definitions of "timelike", "spacelike", and "null or lightlike" are given in Chapter 2, which is intended as an introduction to SR.Orodruin said:Is MTW also to be considered as intended as an introduction to the subject? It is a rather extensive introduction in that case.
That's just being unreasonable. If it was that important to single out the zero vector from the set of lightlike vectors, they would have done so at some point in the book. If not, we have to conclude that they are perfectly fine with keeping the zero vector as "lightlike". As I have said, it changes absolutely nothing when discussing physics.PeterDonis said:The entire book is of course not just an introduction. But the definitions of "timelike", "spacelike", and "null or lightlike" are given in Chapter 2, which is intended as an introduction to SR.
Physics is full of approximations. In this case the approximation is that A and B can actually be colocated.PeterDonis said:Why? Unless A and B are the same--which makes no sense since then you would not even talk about "the separation of the emission event and the event on the worldline of B", or even about a "light signal" at all, since there aren't two events, only one--the vector that connects the two events in question will not be the zero vector.
If A and B are colocated (which means that there is just one worldline, not two) then you can't send a light signal between them, which was your example.Orodruin said:In this case the approximation is that A and B can actually be colocated.
Orodruin said:You can also have a series of Lorentz transformations whereby a lightlike vector has the zero vector as a limit
These seem more relevant, since, as you say, excluding the zero vector would be a complication for such cases.Orodruin said:Integration over loop momenta in Feynman diagrams includes integrating over the zero vector
This is an incorrect inference. That they are colocated at a particular time does not imply that they must be colocated at all later times.PeterDonis said:If A and B are colocated (which means that there is just one worldline, not two)
A sends continuous signals. B receives them. This includes the signal emitted as they pass each other to the approximation that A and B can be colocated.PeterDonis said:then you can't send a light signal between them, which was your example.
To add to this, consider soft photons in the out states. In order to get convergent amplitudes, you need some IR cutoff. This includes excluding photons having the zero vector as 4-momentum as well as photons having small, but non-zero, 4-momentum. In that sense, the zero vector is not different from other "small" light like vectors.PeterDonis said:These seem more relevant, since, as you say, excluding the zero vector would be a complication for such cases.
This was not at all clear from your previous post about this. I thought you were only talking about a single light signal.Orodruin said:A sends continuous signals. B receives them.