# How can an outward normal vector point inwards？

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GR191511
I'm following 《A First Course In General Relativity》.On page 72,it says"If the surface is spacelike,the outward normal vector points outwards.If the surface is timelike,however,the outward normal vector points inwards"I wonder why and how?

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《A First Course In General Relativity》
So, presumably Schutz’s book, but generally please also state the author and edition. Many books have similar names.

Also, please give more context. Many, like myself at the moment, do not have direct access to their copy of the book (if they even have one). Mine is currently on a shelf in my office and it is Sunday.

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Schutz, 2nd ed, it would seem (my copy was 2m away).

Schutz defines the outward normal one-form on p66, just before section 3.4. Outward normal one-forms point outward in the sense that their action on outward pointing vectors is positive. Consider some closed surface with a timelike outward normal one-form at some event on its future-facing surface. If you consider the coordinate system in which that one-form has components (1,0,0,0) and use the -+++ convention you will immediately see that the associated vector is past pointing - i.e. inwards. Similarly we can choose coordinates so that a spacelike outward normal one-form has components (0,1,0,0) and see that the associated vector points outwards.

Clearly this depends on your metric signature - with the +--- signature the timelike vectors point outwards and the spacelike ones inward. That's one part of why Schutz notes in the last sentence of section 3.5 that it's better to work with one-forms here.

• • Demystifier, Dale and vanhees71
GR191511
So, presumably Schutz’s book, but generally please also state the author and edition. Many books have similar names.

Also, please give more context. Many, like myself at the moment, do not have direct access to their copy of the book (if they even have one). Mine is currently on a shelf in my office and it is Sunday.
Thanks...Yes，it is Schutz’s book,2ed,2009...The more context：
A three-dimensional surface is said to be timelike,spacelike,or null according to which of these classes its normal falls into.In Exer.21,3.10,we explore the following curious properties normal vectors have on account of our metric. An outward normal vector is the vector associated with an outward normal one-form,as defind earlier.This ensures that its scalar product with any vector which points outwards is positive.If the surface is spacelike,the outward normal vector points outwards.If the surface is timelike,however,the outward normal vector points inwards.And if the surface is null,the outward vector is tangent to the surface!These peculiarities simply reinforce the view that it is more natural to regard the normal as a one-form,where the metric doesn't enter the definition.
...This section is named《normal vectors and unit normal one-forms》

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Ok, so ultimately this has to do with insisting on associating a tangent vector as the "normal". In actuality, the surface element is not a tangent vector but an ##N-1##-form, which are the natural elements to integrate over on an ##N-1##-dimensional surface.

• vanhees71
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I should probably qualify that a bit. A flux integral ##\int_S J^\mu dS_\mu## in an ##N+1##-dimensional spacetime is generally given by integrating the volumes spanned by ##J## and the tangent vectors ##\dot \gamma_i## in the surface.* The directed volumes are given by ##\eta(J, \dot\gamma_1,\ldots,\dot\gamma_N)## where ##\eta## is the volume ##N+1##-form (in Minkowski space ##\eta = dt\wedge dx\wedge dy\wedge dz## - or, in components, ##\eta_{abcd}=\varepsilon_{abcd}##). The surface element ##dS_\mu## is naturally a 1-form
$$dS_\mu = \eta_{\mu\nu_1\ldots\nu_N} \dot\gamma_1^{\nu_1}\ldots\dot\gamma_N^{\nu_1N} ds_1\ldots ds_N$$
with ##s_k## being the surface parameters.

Now, ##J^\mu dS_\mu## is positive if ##J## points out of the volume and negative if it points into the volume. This so far has nothing to do with the metric apart from the metric giving the natural volume form ##\eta##. The surface element remains a one-form rather than a tangent vector. The ”issue” you describe only apoears if you insist on associating this one-form to a tangent vector through the metric inverse. However, this is a rather artificial thing to do in itself because any time we want to use it to actually integrate we are going to lower the index again to use it as a one-form.

* If one wishes to be a bit more formal, the flux integral is an integral of the N-form ##i_J\eta## over an N-dimensional submanifold. Here, ##i_J \eta## is the N-form obtained by inserting ##J## as the first argument of the volume form ##\eta##. By Stokes’ theorem (the one for differential forms, not the curl theorem)
$$\oint_S i_J\eta = \int_V di_J\eta = \int_V (\nabla\cdot J) \eta$$
is the general equivalent of the divergence theorem, where ##S## is the boundary of ##V##.

• • vanhees71 and Dale
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The ”issue” you describe only apoears if you insist on associating this one-form to a tangent vector through the metric inverse. However, this is a rather artificial thing to do in itself because any time we want to use it to actually integrate we are going to lower the index again to use it as a one-form.
Just to add, at this point Schutz is introducing covectors and index raising and lowering. So the wider context is that he's explaining that some things we've casually called vectors (like vector area) in Euclidean spaces are actually covectors. Even though the metric provides a one-to-one correspondence between a vector and a covector, some things are naturally covectors and the distinction matters when the metric is not trivial.

• vanhees71
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$$dS_\mu = \eta_{\mu\nu_1\ldots\nu_N} \dot\gamma_1^{\nu_1}\ldots\dot\gamma_N^{\nu_1N} ds_1\ldots ds_N$$
with ##s_k## being the surface parameters.

Now, ##J^\mu dS_\mu## is positive if ##J## points out of the volume and negative if it points into the volume.
That depends on your definition of the Levi-Civita symbol. Not mentioning this clearly leads to big confusion (e.g., a famous error by Kinoshita in calculating corrections to the anomalous magnetic moment of the muon). Obviously your convention is that ##\epsilon_{0123}=+1##. I think in the literature you find as many sources where the opposite convention, i.e., ##\epsilon^{0123}=+1##, is used ;-).

• dextercioby
GR191511
Schutz, 2nd ed, it would seem (my copy was 2m away).

Schutz defines the outward normal one-form on p66, just before section 3.4. Outward normal one-forms point outward in the sense that their action on outward pointing vectors is positive. Consider some closed surface with a timelike outward normal one-form at some event on its future-facing surface. If you consider the coordinate system in which that one-form has components (1,0,0,0) and use the -+++ convention you will immediately see that the associated vector is past pointing - i.e. inwards. Similarly we can choose coordinates so that a spacelike outward normal one-form has components (0,1,0,0) and see that the associated vector points outwards.

Clearly this depends on your metric signature - with the +--- signature the timelike vectors point outwards and the spacelike ones inward. That's one part of why Schutz notes in the last sentence of section 3.5 that it's better to work with one-forms heWhat is "timelike outward normal one-form"and" future-facing surface"?

Schutz, 2nd ed, it would seem (my copy was 2m away).

Schutz defines the outward normal one-form on p66, just before section 3.4. Outward normal one-forms point outward in the sense that their action on outward pointing vectors is positive. Consider some closed surface with a timelike outward normal one-form at some event on its future-facing surface. If you consider the coordinate system in which that one-form has components (1,0,0,0) and use the -+++ convention you will immediately see that the associated vector is past pointing - i.e. inwards. Similarly we can choose coordinates so that a spacelike outward normal one-form has components (0,1,0,0) and see that the associated vector points outwards.

Clearly this depends on your metric signature - with the +--- signature the timelike vectors point outwards and the spacelike ones inward. That's one part of why Schutz notes in the last sentence of section 3.5 that it's better to work with one-forms here.
What is "timelike outward normal one-form"and" future-facing surface"?

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What is "timelike outward normal one-form"and" future-facing surface"?
Timelike normal one-form is defined in the text you quoted, given the definition of a normal one-form on page 66.

If you have a closed 3d volume part of its surface faces upwards, right? If you have a closed 4d volume part of its surface faces the future. As a trivial example, draw a square on a Minkowski diagram. The top surface of the square is the future facing surface of the square. The bottom edge faces the past.

• vanhees71 and GR191511
GR191511
Timelike normal one-form is defined in the text you quoted, given the definition of a normal one-form on page 66.

If you have a closed 3d volume part of its surface faces upwards, right? If you have a closed 4d volume part of its surface faces the future. As a trivial example, draw a square on a Minkowski diagram. The top surface of the square is the future facing surface of the square. The bottom edge faces the past.
Thanks...but I still don't understand:(
I think I can regard the surface of timelike as T axis(O frame is moving in the x direction relative to o frame ) then the outward normal vector is X axis...but X axis doesn't look inward？What did i do wrong?

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Thanks...but I still don't understand:(
I think I can regard the surface of timelike as T axis(O frame is moving in the x direction relative to o frame ) then the outward normal vector is X axis...but X axis doesn't look inward？What did i do wrong?
I think there is a confusion here because on p66 Schutz defines a one-form as an outward normal one-form if "...its value on vectors which point outwards from the surface is positive". However, on p72 he applies the inverse metric to the normal one-forms to obtain what he calls an "outward normal vector", which sometimes actually points inwards. So the term "outward" is doing double duty referring to vectors here - sometimes literally meaning outwards and sometimes meaning the dual of the outward one-form (and while the outward one-form points outwards, its dual vector may not). This is why you should stick with one-forms for this job!

So let's work through it, sticking with the (1+1)d Minkowski diagram with a square drawn on it. You can work out the tangent vectors to the surfaces of the square, right? They're just little arrows pointing in one or other direction of the lines. I've illustrated them below in red. Then you can use Schutz's definition to write down the normal one-forms - these must be orthogonal to the tangent vectors, so for the sides of the square they must point left or right and for the top and bottom they must point up or down. On p66, Schutz further defines the outward normal one-form as the one that has a positive inner product with an outward pointing vector. I've drawn the outward pointing vectors below in green. Hopefully you can see that the top green vector has (t,x) components (1,0), the right green vector has components (0,1), the left green vector has components (0,-1), and the bottom green vector has components (-1,0). If the one forms are to have a positive action on them, then, the components of the one forms must be (1,0), (0,1), (0,-1), and (-1,0) (i.e., the same) so that ##v^a\omega_a>0## and they are normal to their red vectors. Now all you need to do is raise the index on those one-forms to get the p72 "outward normal vector". Using the -+++ metric signature convention, that gives you outward normal vectors with components (-1,0), (0,1), (0,-1), and (1,0) (i.e., the signs on the t components have flipped). Those vectors are illustrated in blue below. Note that with the +--- sign convention, it is the x components that would have flipped sign. So finally we have the p66 "vectors which point outwards" in green, which have the natural sense of pointing outwards, and the p72 "outward normal vectors" in blue, which don't always have the natural sense of pointing outwards. Schutz is here, I think, illustrating that some things work naturally as one-forms not vectors, and surface normals are one of them. @Orodruin explained why that is so in some detail above.

Finally, I've been working in (1+1)d here. I think the only extension for a full (3+1)d spacetime is that there are three linearly independent red vectors lying in each "surface" of the 4d cube. There is still only a single one-form that is orthogonal to them (up to a constant multiple). Note also that the whole argument applies to arbitrary shapes - you just can't use a single normal for anything more than an infinitesimal part of the surface in general. They're also much harder to illustrate meaningfully.

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GR191511
I think there is a confusion here because on p66 Schutz defines a one-form as an outward normal one-form if "...its value on vectors which point outwards from the surface is positive". However, on p72 he applies the inverse metric to the normal one-forms to obtain what he calls an "outward normal vector", which sometimes actually points inwards. So the term "outward" is doing double duty referring to vectors here - sometimes literally meaning outwards and sometimes meaning the dual of the outward one-form (and while the outward one-form points outwards, its dual vector may not). This is why you should stick with one-forms for this job!

So let's work through it, sticking with the (1+1)d Minkowski diagram with a square drawn on it. You can work out the tangent vectors to the surfaces of the square, right? They're just little arrows pointing in one or other direction of the lines. I've illustrated them below in red.
View attachment 302443
Then you can use Schutz's definition to write down the normal one-forms - these must be orthogonal to the tangent vectors, so for the sides of the square they must point left or right and for the top and bottom they must point up or down. On p66, Schutz further defines the outward normal one-form as the one that has a positive inner product with an outward pointing vector. I've drawn the outward pointing vectors below in green. Hopefully you can see that the top green vector has (t,x) components (1,0), the right green vector has components (0,1), the left green vector has components (0,-1), and the bottom green vector has components (-1,0). If the one forms are to have a positive action on them, then, the components of the one forms must be (1,0), (0,1), (0,-1), and (-1,0) (i.e., the same) so that ##v^a\omega_a>0## and they are normal to their red vectors.
View attachment 302444
Now all you need to do is raise the index on those one-forms to get the p72 "outward normal vector". Using the -+++ metric signature convention, that gives you outward normal vectors with components (-1,0), (0,1), (0,-1), and (1,0) (i.e., the signs on the t components have flipped). Those vectors are illustrated in blue below. Note that with the +--- sign convention, it is the x components that would have flipped sign.
View attachment 302445
So finally we have the p66 "vectors which point outwards" in green, which have the natural sense of pointing outwards, and the p72 "outward normal vectors" in blue, which don't always have the natural sense of pointing outwards. Schutz is here, I think, illustrating that some things work naturally as one-forms not vectors, and surface normals are one of them. @Orodruin explained why that is so in some detail above.

Finally, I've been working in (1+1)d here. I think the only extension for a full (3+1)d spacetime is that there are three linearly independent red vectors lying in each "surface" of the 4d cube. There is still only a single one-form that is orthogonal to them (up to a constant multiple). Note also that the whole argument applies to arbitrary shapes - you just can't use a single normal for anything more than an infinitesimal part of the surface in general. They're also much harder to illustrate meaningfully.
Thank you!I get most of your points and I have another problem:What kind of closed surface guarantee all of outward normal vectors point inwards ?What does it look like？I can't image it.Just like your last figure,not all of outward normal vectors point inwards.

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Thank you!I get most of your points and I have another problem:What kind of closed surface guarantee all of outward normal vectors point inwards ?What does it look like？I can't image it.Just like your last figure,not all of outward normal vectors point inwards.
Again, you should not think of "surface normals" as tangent vectors. The natural description is in terms of one-forms. Associating them to tangent vectors is just resulting in confusions such as the ones you are experiencing.

• vanhees71
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What kind of closed surface guarantee all of outward normal vectors point inwards ?
The outward surface normal one-forms all always point outwards, whatever. To have their duals all point inwards the one-forms would all have to be timelike (or all spacelike, depending on your metric signature). You could draw a diamond, I suppose, so that all the normals are timelike (or all spacelike). You would be cheating in some senses because the normal isn't well defined at the corners, and if you make them anything other than mathematically perfect ideal intersections you'd find normals of other character there.

I do agree with Orodruin here - the point of Schutz's discussion is to explain why vectors are a bad way to characterise surfaces in non-Euclidean spaces. The lesson you should take away is that you can freely convert a one-form to its dual if the maths requires it, but the natural way to think about some things is in terms of covectors (even if we skated over the distinction and just used vectors in Euclidean spaces).

• vanhees71
GR191511
Again, you should not think of "surface normals" as tangent vectors. The natural description is in terms of one-forms. Associating them to tangent vectors is just resulting in confusions such as the ones you are experiencing.
Thank you very much

GR191511
The outward surface normal one-forms all always point outwards, whatever. To have their duals all point inwards the one-forms would all have to be timelike (or all spacelike, depending on your metric signature). You could draw a diamond, I suppose, so that all the normals are timelike (or all spacelike). You would be cheating in some senses because the normal isn't well defined at the corners, and if you make them anything other than mathematically perfect ideal intersections you'd find normals of other character there.

I do agree with Orodruin here - the point of Schutz's discussion is to explain why vectors are a bad way to characterise surfaces in non-Euclidean spaces. The lesson you should take away is that you can freely convert a one-form to its dual if the maths requires it, but the natural way to think about some things is in terms of covectors (even if we skated over the distinction and just used vectors in Euclidean spaces).
Thanks！I got it.I will keep that in mind.
And...Are the closed timelike surfaces inside of the future light cone or the past light cone？