Hi guys. I'm currently working through Special Relativity by N.M.J Woodhouse I'm stuck, or rather confused. Exercise 5.7.iii on page 100 reads [Show that] every four-vector orthogonal to a timelike vector is spacelike. (1) Where orthogonality of two four-vectors \( X,Y\) has been defined by the condition \( g(X,Y) = X_{\mu} Y^{\mu} = 0 \), a timelike four-vector is defined as a four vector \(U\) s.t \(U_{\mu}U^{\mu} > 0\) and a space-like four-vector is defined as a four vector \(V \) s.t \( V_{\mu}V^{\mu} < 0 \) Is claim (1) even true? I can see how using an ICS where the time-like vector is given by \((a,0,0,0)\) forces any orthogonal four-vector, \(V\) to have \(V^0 = 0\), and seperately we have the result, If \(X\) is a space-like four-vector, then there exists an ICS s.t \(X^0 = 0 \), but this result is not an if and only if statement, so (1) does not follow. In fact surely the null vector (0,0,0,0) is orthogonal to any time-like 4 vector and is not itself space-like? I think I maybe have some definitions confused or something as I can not see how the claim is true, or what the claim might be trying to get at. Any help would be much appeciated. EDIT: Even more confusingly I just found this problem (http://web.mit.edu/8.962/www/probset/pset01.pdf) : "Show that a timelike vector and a null vector cannot be orthogonal" What about the null vector 0=(0,0,0,0), it is null right? it does force g(X,0) = 0 right? :s
Why don't you start with something easier to chew, like linear vector spaces perhaps, then work from there?
But if it helps, a nonzero spacelike vector is orthogonal to an equally nonzero timelike vector if their Minkowski dot product is zero.
I've done an undergrad in mathematics which covered the recommended topic. I'm doing this in my free time.
You say a nonzero space/timelike vector, but surely if they are space/timelike then by definition they are nonzero? So is the correct definition of orthogonality: g(X,Y)=0 AND X,Y =/= 0 ? Because that isn't how it is defined in Woodhouse, but if it were then everything makes sense.
It's been a while since I sat down and learned this stuff... so i might make a mistake, but, in the case you gave, you can write it down as: \(g(X,Y)=||\mathbf{X}||||\mathbf{Y}||cosQ, 0 \le Q \le \pi\)
It's easy to check that it works the other way round too. If \(x^{0} = 0\) then \(x_{\mu} \,=\, (0, \, x_{i})\) and \(x^{\mu} x_{\mu} \,=\, - (x_{i})^{2} \,\lt\, 0\), provided \(x\) isn't the zero vector. They meant light-like vectors, ie. such that \(x^{\mu} x_{\mu} = 0\). You can always choose an inertial coordinate system in which they take the form \(x^{\mu} \,=\, (a,\, 0,\, 0,\, a)\).
This doesn't hold generally in Minkowski space. For example, if \(x\) and \(y\) are two time-like vectors then the correct analogue of this relation is \(x \cdot y \,=\, || x || \, || y || \, \cosh(\phi) \;,\)where \(\phi\) is a rapidity.
I diagree, because in this case, we are interested in treating X and Y as orthogonal, and they would cease to be orthogonal, if X and Y were not treated on different footing.
Thank you for the help, it's very much appreciated. I think my confusion lies with the classification of the vector (0,0,0,0). Is it spacelike, timelike or null? I think the original question in Woodhouse would be better stated as: Show that every nonzero four-vector that is orthogonal to a timelike vector is spacelike.
Are you sure? I wasquite certain that if X and Y were both timelike or spacelike, then nothing would be orthogonal about them? If X is to be orthogonal to Y, one surely needs to be timelike, and the other spacelike?
Well if you just apply the definitions the zero vector is light-like, so it's just your last null-vector problem that would have to be reworded if you insist on that level of accuracy.
Sure about what? \(X \cdot Y = 0\) is what it means, by definition, for two vectors to be orthogonal. No, not necessarily. Two time-like vectors can't be orthogonal, but two space-like vectors can. For example, \(X^{\mu} \,=\, (0,\, 1,\, 0,\, 0)\) and \(Y^{\mu} \,=\, (0,\, 0,\, 1,\, 0)\) are space-like and orthogonal.
Ah yes, well that is what i meant by my post before you refuted it. I meant a timelike vector. Though you did jog my memory, you're quite right concerning the spatial dimensions. In any case, my relation holds above because in this case, we are specifically talking about treating X and Y as orthogonal.
The relation \(x \cdot y \,=\, ||x|| \, ||y|| \, \cos(\theta)\) is generally true, for all vectors, in a Euclidean space. It is not generally true in Minkowski space. It technically "holds" for orthogonal vectors in a Minkowski space only in the trivial sense that there's a value of \(\theta\) you can plug in that makes it reduce to \(x \cdot y = 0\). The made-up relations \(x \cdot y \,=\, \sqrt{\frac{\sin\bigl(||x|| \, ||y|| \, \cos(\theta) \bigr)}{e^{\sqrt{\pi}\theta}}}\)and \(x \cdot y \,=\, \theta \,-\, \frac{\pi}{2}\)also "hold" for orthogonal vectors for exactly the same reason.
I only meant it in the strongest technical sense. Please Register or Log in to view the hidden image!
Actually, it is true in a lot more than Euclidean space. It is in a sense a definition of the angle between two vectors in any inner product space -- assuming, that is, that that "\(\cdot\)" represents an inner product. The problem isn't so much that Minkowski space is not Euclidean. The problem is that that \(\cdot\) is not an inner product. There is just a smidgeon of abuse of notation in special relativity and general relativity. The Minkowski inner product is not an inner product and the Riemannian metric is not a metric. This might just be a red herring, but then again red herrings usually are neither red nor herrings.
This caused me lots of confusion first. Now the situation is a bit better. I think the explanation has to do with that in Euclidean space, the Euclidean inner product has (at least) two roles: i) They define the topology of the space, in the sense that you can say A is closer to B than C is to B. ii) They also define the symmetry of the Euclidean geometry, in the sense that transformations leaving the Euclidean inner product invariant considered to be rigid body motions. In Lorentzian geometry (e.g. of special relativity) the metric is used only in ii). The underlying topology is still the sort of topology that the Euclidean geometry uses.
Thanks for the clarification, though I thought any inner product (at least in the real and finite dimensional case) could be reduced to a Euclidean product.
