Gravitational vs. electromagnetic waves

[soarce]

Let's suppose that on Earth I have two devices which permits me to detect the gravitation produced and electromagnetic waves (photons) emitted by the Sun. Then, the Sun suddenly disappears. Which of the two devices will notice first that the Sun has gone, i.e. which wave propagates faster, the gravitational or the electromagnetic ?

 

[pervect]

Let's suppose that on Earth I have two devices which permits me to detect the gravitation produced and electromagnetic waves (photons) emitted by the Sun. Then, the Sun suddenly disappears. Which of the two devices will notice first that the Sun has gone, i.e. which wave propagates faster, the gravitational or the electromagnetic ?

There isn't any mathematically valid solutions to the equations of General Relativity in which the "sun suddenly disappears". Therefore there isn't any way to answer mathematically "what happens if it does". What GR says is that this is impossible, and the sudden disappearance of the sun would falsify GR.

Also please note that the gravity of the sun is not due to "gravitational waves". The attractive gravity of the sun is much more similar to the attraction of unlike charges than it is an electromagnetic waves. The attraction of unlike charges is not due to photons, though it can be thought of as the exchange of "virtual photons".

 

[PeterDonis]

which wave propagates faster, the gravitational or the electromagnetic ?

Pervect is correct that the scenario as you state it is impossible according to GR. However, one could imagine a possible situation that would give rise to something fairly similar. Suppose, for example, that a pair of huge laser beams were fired at the Sun, in a direction perpendicular to the Earth-Sun line, in such a way as to change both the Sun's brightness and its total energy (and hence the mass seen by the Earth as the "source" of the Sun's gravity), without changing its momentum (beams are equal in intensity and opposite in direction). In this scenario, according to GR, observers on Earth would see the Sun's brightness change at the same time as they saw its gravity change; both changes would propagate to Earth as waves, and the waves would travel at the same speed.

Also see this page on the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

 

[pervect]

Pervect is correct that the scenario as you state it is impossible according to GR. However, one could imagine a possible situation that would give rise to something fairly similar. Suppose, for example, that a pair of huge laser beams were fired at the Sun, in a direction perpendicular to the Earth-Sun line, in such a way as to change both the Sun's brightness and its total energy (and hence the mass seen by the Earth as the "source" of the Sun's gravity), without changing its momentum (beams are equal in intensity and opposite in direction). In this scenario, according to GR, observers on Earth would see the Sun's brightness change at the same time as they saw its gravity change; both changes would propagate to Earth as waves, and the waves would travel at the same speed.

Also see this page on the Usenet Physics FAQ:

http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_speed.html

I agree with almost all of the points made, but I'm not sure that the above experiment actually shows the speed of gravity. It would if one assumed that the laser beams didn't produce gravitatational effects of their own, but they do produce such effects.

I'm pretty sure that the gravity wave associated with the pulse would be some variant of the Aichelberg-sexl ultraboost, http://en.wikipedia.org/wiki/Aichelburg–Sexl_ultraboost, an impulsive plane gravity wave.

Depending on the geometry, one of the gravity waves from either of the laser beam could hit the Earth before the two laser beams collided.

 

[PeterDonis]

Depending on the geometry, one of the gravity waves from either of the laser beam could hit the Earth before the two laser beams collided.

Yes, good point. Another illustration of why it's tough to come up with an experiment at all that measures the speed of gravity--anything you could think of using in the experiment gravitates. You can't make a "gravitational Faraday cage" to isolate just the effects you're interested in.

 

[Naty1]

Soarce: pervect and PeterDonis are experts here. Me: one of the 'dummies'.

But sometimes a 'dummy answer' [a simplified one] is a reasonable place to start thinking about things: Electromagnetic waves and insofar as I have read, gravitational waves, propagate at the speed of light. So I'd say the detections should be simultaneous.

I might even be able to take that one step further, but maybe not so far as the other two guys:

If a gravitational source moving along suddenly stops at it's current position [also not so realistic for our sun, but maybe ok for a thought experiment] the field at a
distant location will continue to point toward its “extrapolated” position—it's anticipated position...even though the source never actually reaches that extrapolated position. At the time the source stops the field at the source will abruptly switch direction to point toward the true fixed and present position of the source. This sudden change in the field propagating outward at the speed of light is true for both the gravitational and the electromagnetic radiation. A test particle [or approximately, earth, or your detectors if you wish] at a distant position will continue to accelerate toward the extrapolated position of the source until the finite time it takes for the gravitational signal of position change to propagate over the separation distance.

I've read multiple places the gravitational interaction propagates at the speed of light and that it can be theoretically proven no gravitational influence in general relativity can travel faster than the speed of light. I'm sure pervect and/or PeterDonis can better explain or refute all this better than I.

One thing I am aware of is that time, distance and speed at different locations in curved spacetime is rather ambiguous and so the above explanation does take a few liberties. But this was a perspective that helped me get a feel for what was happening. Also, I don't think anyone has yet experimentally verified that gravitational waves propagate the lightspeed.

One possibly interesting thing to think about with regard to the answer to your question is has to do 'unwinding' of the gravitational wave, that is the changing of spacetime between the sun and earth; the electromagnetic wave will follow a straight line path when it can..otherwise as close as it can get, a geodesic, which is like a straight line when spacetime is curved. So if the two waves are actually propagating together, in synch at lightspeed, which path is the electromagnetic path taking...the old or the new... and why. THAT is beyond my meager paygrade!

 

[PeterDonis]

If a gravitational source moving along suddenly stops at it's current position [also not so realistic for our sun, but maybe ok for a thought experiment]

Not quite, because you still have to account for the energy and momentum contained in whatever it is that causes the source to stop. (Similar to what pervect pointed out about the laser beams--you have to include their energy and momentum, and hence their gravitational effects, in your analysis, and I hadn't.)

However, I believe there is a somewhat similar thought experiment in Steve Carlip's classic paper on the speed of gravity:

http://arxiv.org/abs/gr-qc/9909087

I recommend reading this as he carefully goes into the subtleties we've been getting into here. The only piece of background you need to be aware of is that he wrote the paper in response to repeated attempts by Tom Van Flandern (who I won't discuss here, Google will quickly show plenty of items) to claim that the lack of aberration for gravity, as compared to light, somehow showed that the speed of gravity must be much faster than the speed of light, if not infinite. But the material he covers is general and not limited to the aberration issue.

the field at a
distant location will continue to point toward its “extrapolated” position—it's anticipated position...even though the source never actually reaches that extrapolated position. At the time the source stops the field at the source will abruptly switch direction to point toward the true fixed and present position of the source. This sudden change in the field propagating outward at the speed of light is true for both the gravitational and the electromagnetic radiation. A test particle [or approximately, earth, or your detectors if you wish] at a distant position will continue to accelerate toward the extrapolated position of the source until the finite time it takes for the gravitational signal of position change to propagate over the separation distance.

This is actually a fair summary of the conclusion Carlip reaches in his paper about the thought experiment that is similar to yours. So your intuition was good!

I've read multiple places the gravitational interaction propagates at the speed of light and that it can be theoretically proven no gravitational influence in general relativity can travel faster than the speed of light.

Carlip's paper discusses this in some detail, and gives good references for finding out more.

One thing I am aware of is that time, distance and speed at different locations in curved spacetime is rather ambiguous and so the above explanation does take a few liberties.

Not really; the main thing to be aware of is that it only really works for a limited class of spacetimes, where gravity is weak enough and the source is close enough to being static.

Also, I don't think anyone has yet experimentally verified that gravitational waves propagate the lightspeed.

No, in fact we haven't even detected GWs at all yet. But there are plenty of indirect measurements, such as the binary pulsar, that give good bounds on the "speed of gravity" within the framework of GR. Carlip goes into that in his paper.

So if the two waves are actually propagating together, in synch at lightspeed, which path is the electromagnetic path taking...the old or the new... and why. THAT is beyond my meager paygrade!

Both waves will take the same path. Remember that, whatever contribution either of the waves make to the overall spacetime, it must already be taken into account in determining the path both waves follow. They both follow null geodesics in the same overall spacetime.

 

[Eimacman]

Greetings:

In another thread called "Black hole growth paradox question" I purposed that gravity is not effected by gravitation but photons are. Because of the in falling of space-time and the gravitational effects of a black hole. When a mass falls into the singularity gravitational effects are immediately felt, should be around c. The speed of photons out of the black hole would be nil. Therefor for this scenario the speed of gravity would be much greater than the electromagnetic.


Eimacman.

 

[PeterDonis]

gravity is not effected by gravitation

What does this even mean? Gravity *is* gravitation.

If what you are trying to say that gravitational *waves* are not affected by gravitation, that's wrong. They are.

When a mass falls into the singularity gravitational effects are immediately felt

This is wrong. The effects of a mass falling into the BH are felt *before* it crosses the horizon. If you are hovering above the hole and a mass falls in, you observe the effects of the change in gravity at the same time you see light coming from the infalling mass (light emitted before it crosses the horizon, obviously); in other words, the change in gravity from the infalling mass propagates at the speed of light, just as the light from it does. By the time the mass crosses the BH's horizon, you have already observed all of the change in gravity associated with it falling in.

 

[Naty1]

This is actually a fair summary of the conclusion Carlip reaches in his paper about the thought experiment that is similar to yours. So your intuition was good!

Peter...you bestow WAY too much credit...If only my 'intuition' could take me so far!...I read the explanation somewhere...liked it and because it seemed insightful if not perfectly correct, saved it... I always post sources when I have them, but did not record this one.
Maybe it was Carlip...or these forums...

Not quite, because you still have to account for the energy and momentum contained in whatever it is that causes the source to stop.

Picky,picky,picky...now Looky here:
If Einstein could 'picture' catching up to an electromagnetic wave when he was 16, and wondering what he would find, as a senior I am entitled to some leeway too...so cut me a smidgen of slack in this approximation! [ok, maybe a significant 'smidgen'! We 'dummies' have to start somewhere.

 

[PeterDonis]

If Einstein could 'picture' catching up to an electromagnetic wave when he was 16, and wondering what he would find, as a senior I am entitled to some leeway too

Well, Einstein's conclusion was that that picture didn't make sense, since Maxwell's equations have no solution corresponding to a wave that is "standing still". I'm basically saying the same thing here: the EFE has no solution corresponding to a gravitating body "just stopping". Even Einstein only got a finite amount of slack.

 

[EskWIRED]

Well, Einstein's conclusion was that that picture didn't make sense, since Maxwell's equations have no solution corresponding to a wave that is "standing still". I'm basically saying the same thing here: the EFE has no solution corresponding to a gravitating body "just stopping". Even Einstein only got a finite amount of slack.

To be fair, I have seen similar thought experiments in popular literature to explain the difference between Newton's and Einstein's theories of gravity. In the article, it was said that if the sun suddenly disappeared, then under Newton's theory, the gravitational effect on Earth would be immediate, while under Einstein's theory the effect would not be noticed on Earth until an interval had elapsed proportional to the distance and the speed of light.

So, while the experiment is impossible, it is nevertheless instructive to imagine, as it can yield greater understanding of the two theories.

 

[PeterDonis]

So, while the experiment is impossible, it is nevertheless instructive to imagine, as it can yield greater understanding of the two theories.

Right up until someone asks the obvious next question, "What would the Earth's orbit look like if the Sun disappeared? Would it fly off in a straight line?" And then you have to explain how the Sun disappearing isn't really possible in the first place.

I understand that popular literature has to strike a balance between being strictly correct and being understandable. But one still has to allow for the fact that people don't just take what's said in popular literature in isolation. They draw inferences from it; they try to use it, not just as a source of convenient sound bites to describe the theory, but as a source to begin developing an understanding of the theory. Saying a bunch of things that have to be unlearned as soon as you take the next step in understanding doesn't strike me as a good strategy, when viewed in that light. But that's just my opinion; I understand that lots of people have different opinions on this.

 

[Eimacman]

Greetings PeterDonis:

I meant that gravity is not effected by its self, gravitons are another matter.

You make an interesting point about the gravitational effects of a black hole being felt before a mass falls into the event horizon, that would allow the the speed of electromagnetic waves and gravitational waves to be the same. Could you cite me a reference to this effect I would like to study this further. I have a problem with this point in that it implies that a black hole's gravitational field to be asymmetrical, at least slightly. An asymmetrical gravitational field would make a black hole with in falling matter 'noisy' gravitationally.

Also there must be some kind 'gravitational communication' between the horizon and the singularity where the gravitational field density →∞.

Eimacman

 

[soarce]

Thank you for your answers, they helped me to clarify my dilemma.

 

[Naty1]

Eimacman posts:

You make an interesting point about the gravitational effects of a black hole being felt before a mass falls into the event horizon... Could you cite me a reference to this effect I would like to study this further.

This refers to the absolute horizon characterized by a smoothly increasing area in contrast to the [older concept] apparent horizon with discontinuous jumps in size.

[I never understood the Wikipedia version, maybe it will make sense to you]

Kip Thorne has some 'not mathematical' discussion on this in BLACK HOLES AND TIME WARPS, Chapter 12, with easy to interpret spacetime diagrams. Apparently Penrose 'got there first' with the apparent horizon; Hawking later work with black hole entropy apparently envisioned the absolute horizon and that's the one most frequently utilized.

It seems the apparent horizon is relative, depending on the observers frame, and can jump discontinuously. Hawking 'absolute' horizon is frame independent. Kip Thorne says
the aapparent horizon is 'jerky' and the absolute horizon an unusual situation where a telelogical evolution of the absolute horizon is helpful; the latter depends on final causes,whether light signals from the black hole will eventually escape to the distant universe [infinity]. Thorne says the two view produce the same experimental predictions for the outcomes of experiments, but they strongly differ in the ease with which behaviors can be deduced from Einsteins field equations.

 

[PeterDonis]

Eimacman, I thought I had responded to your latest post before, but apparently it didn't get posted somehow.

I meant that gravity is not effected by its self, gravitons are another matter.

Depending on how you interpret "gravity affecting itself", independently of gravitational waves carrying energy, gravity may or may not affect itself. It is true that the "source" of gravity in the Einstein Field Equation, the stress-energy tensor, does not include any stress-energy due to "gravity itself". However, it is also true that the EFE is nonlinear, which shows up in a number of ways that many people interpret as "gravity affecting itself" or "gravity gravitating". One way is that in many situations of interest, the nonlinearity of the EFE can be expressed as energy being stored in the "gravitational field", and you can define a conserved energy for the spacetime that includes the energy stored in the gravitational field. Since "energy" is the "source" of gravity, this can be interpreted as gravity causing more gravity. Another way is that the nonlinearity of the EFE sets limits on what static configurations of matter, such as planets, stars, white dwarfs, and neutron stars, can be stable, limits which are not there in Newtonian gravity, which is linear. This can also be interpreted as gravity causing more gravity.

You make an interesting point about the gravitational effects of a black hole being felt before a mass falls into the event horizon, that would allow the the speed of electromagnetic waves and gravitational waves to be the same. Could you cite me a reference to this effect I would like to study this further.

Try this page on the Usenet Physics FAQ for a start:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html

It actually addresses something you say further down in your post:

Also there must be some kind 'gravitational communication' between the horizon and the singularity where the gravitational field density →∞.

But the answer to this sets the framework for what I said about when the gravitational effects of an object falling into the hole are felt.

First, about the "gravitational communication" in general: as the FAQ entry says, the observed "field", meaning the effects of gravity, at any event (a point in space at a given instant of time) are entirely due to the presence of "sources" at other events that can communicate with that event at speeds less than or equal to c. The brief way of saying this is that what happens at a given event is determined entirely by what happens within the "past light cone" of that event. So the "gravity" you feel when you are outside the black hole isn't coming from inside the hole: it's coming from the past, from the object that originally collapsed to form the hole.

This is actually true, btw, for *all* gravitating bodies, not just black holes. For example, the "gravity" that the Earth feels from the Sun at this instant is not determined by the Sun "right now"; it's determined by the Sun 500 seconds ago, the time it takes for light to travel from the Sun to the Earth. A good reference for how this works is this paper by Steve Carlip:

http://arxiv.org/abs/gr-qc/9909087

Carlip explains how the interaction between objects like the Sun and the Earth can look like an instantaneous "Newtonian" force even though it's actually time-delayed because of the finite speed of light.

The difference between the Sun and the black hole is that it only takes eight minutes for gravity to get from the Sun to the Earth, so to speak--or, looking at it from our point of view on Earth, we only have to go eight minutes into the Earth's past light cone to find the "source" of the gravity the Earth feels from the Sun at this instant. But if the Sun were a black hole, it might have collapsed a billion years ago, so we would have to go a billion years into the Earth's past light cone to find the "source" of the gravity the Earth feels at this instant--but the gravity itself, the effect, would be the same if the Sun's mass were the same, because the way the field from the collapsing object "propagates" through the empty vacuum region outside it is static--it stays the same for all time (again, this is in the idealized spherically symmetric case), so it doesn't matter how long ago the object collapsed to form the hole.

Now, in the light of the above, consider the case of an object with significant mass of its own falling into the black hole. You are orbiting the hole at some distance and the object falls into the hole some distance away from you (this is so what you see isn't distorted by the object's own gravity--we want to see only the effect of the object+hole combining). As soon as the object is closer to the hole than you are, you will start feeling the gravity in your vicinity change. Put another way, as soon as the object is in your past light cone, as well as the hole, the object's gravity will start adding to the hole's gravity. In an idealized case, where the effects are spherically symmetric, you will see *all* the effect as soon as the object is closer to the hole than you are; in a real case, you will see a more gradual change, but by the time the object is inside the hole's horizon, all the change will have occurred--because once the object is below the horizon, it has left all the "imprint" in your past light cone that it's ever going to leave (since the horizon and everything inside it can never be inside your past light cone if you are outside).

I have a problem with this point in that it implies that a black hole's gravitational field to be asymmetrical, at least slightly.

If you mean by this that a spacetime in which a massive object collapses to a black hole is asymmetric in time, you are correct: the collapse being "before" the black hole defines a direction of time. However, this does not make the field "asymmetrical"; once the object collapses, the gravitational field is stationary and spherically symmetric (at least in the idealized case we're considering).

An asymmetrical gravitational field would make a black hole with in falling matter 'noisy' gravitationally.

Not sure what you mean by this.

 

[PeterDonis]

This refers to the absolute horizon characterized by a smoothly increasing area in contrast to the [older concept] apparent horizon with discontinuous jumps in size.

Actually I wasn't talking about this; the effects on you, outside the horizon, of the object falling in will be seen *before* the horizon itself moves. See my response to Eimacman just now.

(What you say about the horizons is also important, btw; I just wanted to make clear that it's a separate thing from what I was saying.)

It seems the apparent horizon is relative, depending on the observers frame,

Actually, no, there is a frame-independent definition of the apparent horizon, using the concept of a "trapped surface". See here:

http://en.wikipedia.org/wiki/Trapped_null_surface

and can jump discontinuously.

Yes, this is correct.

Hawkings 'absolute' horizon is frame independent.

Yes. In fact, the only way to know for sure where the absolute horizon is is to know the entire future history of the spacetime, so that you know which events can send light signals that escape to infinity (you note Thorne's comment on this). Since that property of events is frame-independent, the absolute horizon is too.

 

[Eimacman]

Greetings PeterDonis:

What I meant by 'gravitationally noisy' is the accretion disk around a black hole can not be perfectly uniform, and that in falling matter in such a disk will be moving and the gravitation of such a imperfect disk will cause gravitational waves in per portion with such an imperfection. This is not taking into account any space-time effects.

It is like two neutron stars orbiting around a common center of mass, only instead of large gravitational waves, the waves that I was thinking of would be much smaller, possibly undetectable.

Oh and thanks for the references. It will take me a wile to study them. I can not make large replies for now, for I have shattered my left arm and the titanium plate makes typing difficult.

Eimacman

 

[PeterDonis]

What I meant by 'gravitationally noisy' is the accretion disk around a black hole can not be perfectly uniform, and that in falling matter in such a disk will be moving and the gravitation of such a imperfect disk will cause gravitational waves in per portion with such an imperfection. This is not taking into account any space-time effects.

Actually, gravitational waves can be thought of as "spacetime effects", since they are waves of changing spacetime curvature. You are correct that in general GWs will be produced when matter falls into a black hole, and the process won't be symmetric. However, the hole itself will still be symmetric in its end state, after some matter has fallen in and all GWs have been radiated away; part of what the GWs do is take away any asymmetry in the hole's horizon. After that happens the hole will be again spherically symmetric (in the probably rare case where there is zero angular momentum) or axially symmetric (in the probably more common case when the hole has some angular momentum, or acquires some from the infalling matter).

 

[EskWIRED]

After that happens the hole will be again spherically symmetric (in the probably rare case where there is zero angular momentum) or axially symmetric (in the probably more common case when the hole has some angular momentum, or acquires some from the infalling matter).

Have any Schwarzschild black holes ever been identified? Do they really exist, or are they an oversimplified model that is useful but nonexistent?

It seems to me that unless all infalling matter is directed precisely towards the singularity, and unless all off-center hits are exactly counteracted by other hits, the BH will rotate.

Am I wrong? Have nonrotating black holes been identified?And which of the various Forums the is the best one for discussions of black holes? Relativity? Cosmology?

 

[Eimacman]

Greetings PeterDonis:

As you said:

Actually, gravitational waves can be thought of as "spacetime effects", since they are waves of changing spacetime curvature. You are correct that in general GWs will be produced when matter falls into a black hole, and the process won't be symmetric. However, the hole itself will still be symmetric in its end state, after some matter has fallen in and all GWs have been radiated away; part of what the GWs do is take away any asymmetry in the hole's horizon. After that happens the hole will be again spherically symmetric (in the probably rare case where there is zero angular momentum) or axially symmetric (in the probably more common case when the hole has some angular momentum, or acquires some from the infalling matter).

I agree, and I do occasionally neglect space-time effects as a part of a thought experiment.

I am still studying the references you have cited, the math is difficult and I can not remember the meaning of some the terms.

Eimacman

 

[Austin0]

Originally Posted by Naty1 View Post

It seems the apparent horizon is relative, depending on the observers frame,

Actually, no, there is a frame-independent definition of the apparent horizon, using the concept of a "trapped surface". See here:

http://en.wikipedia.org/wiki/Trapped_null_surface

having looked at the linked article it appears that Naty 1 is correct.
there may be a frame independent definition of a null surface but any manifest apparent horizon is itself a frame dependent entity.


wiki Apparent horizons

Apparent horizons are not invariant properties of a spacetime. They are observer-dependent, and in particular they are distinct from absolute horizons.

 

[PeterDonis]

there may be a frame independent definition of a null surface but any manifest apparent horizon is itself a frame dependent entity.

It's not just the "definition" of a trapped null surface that's frame independent; whether or not a particular null surface (which is a frame independent entity) is a trapped null surface is also frame independent. (At least, I'm almost positive that's true--see below.)

The quote you give from the wiki page on apparent horizons is based on a paper by Wald, who is certainly an expert. The Wiki page says this right before footnoting Wald's paper:

Apparent horizons depend on the "slicing" of a spacetime. That is, the location and even existence of an apparent horizon depends on the way spacetime is divided into space and time. For example, it is possible to slice the Schwarzschild geometry in such a way that there is no apparent horizon, ever, despite the fact that there is certainly an event horizon.

I'll have to think some more about this (the actual paper is behind a paywall so I can't read it--I'll look to see if there's a copy on arxiv.org). It seems like it's saying that, while the property of a particular null surface being a trapped null surface is frame invariant (as I said above, I'm almost positive that's true), somehow putting together a whole set of such surfaces into something that meets Wald's definition of "apparent horizon" is frame dependent.

 

[Austin0]

It's not just the "definition" of a trapped null surface that's frame independent; whether or not a particular null surface (which is a frame independent entity) is a trapped null surface is also frame independent. (At least, I'm almost positive that's true--see below.)



I'll have to think some more about this (the actual paper is behind a paywall so I can't read it--I'll look to see if there's a copy on arxiv.org). It seems like it's saying that, while the property of a particular null surface being a trapped null surface is frame invariant (as I said above, I'm almost positive that's true), somehow putting together a whole set of such surfaces into something that meets Wald's definition of "apparent horizon" is frame dependent.

wiki null surface

a null surface is a 3-surface whose normal vector is everywhere null (zero length with respect to the local Lorentz metric), but the vector is not identically zero.

This one is beyond me. Could you make sense of everywhere zero but not identically zero. Please! ;-(

 

[PeterDonis]

Could you make sense of everywhere zero but not identically zero. Please! ;-(

A null vector is a vector that is not zero (i.e., it is not the zero vector), but has zero length. Since the length of a vector is the square root of its norm, i.e., its contraction with itself, this means that null vectors and null surfaces have some counterintuitive properties:

(1) A null vector is orthogonal (i.e., normal) to itself (because the contraction of two vectors being zero is the condition for two vectors being orthogonal);

(2) A null surface has both tangent vectors (vectors lying in the surface) that are null everywhere, *and* normal vectors (vectors orthogonal to the surface) that are null everywhere.

This is the sort of thing the quote you gave is referring to.