Finding Function that Vanishes Only When x^mu = y^mu

[welcomeblack]

Hi all,

Just doing some hobby physics while I put off working on my research. In one dimension, the function
\begin{equation}
f(a,b)=[1-\exp(-(a-b)^2)]
\end{equation}
vanishes when a=b. In Minkowski spacetime though, such a function is not so easy to find (if you require Lorentz invariance). If we attempt the same thing with the 4-vectors x and y,
\begin{equation}
f(x,y)=[1-\exp(-(x-y)^2)],
\end{equation}
the function vanishes everywhere on the lightcone (x-y)^2=0. But I'm looking for a function that vanishes only when x is equal to y. From my perspective, this hypothetical function should approach zero as x gets *near* to y. However, *near* is hard to define for what I'm looking for since the usual Minkowski metric only tells you how close you are to the lightcone. I'd like to be able to write something like
\begin{equation}
f(x,y)=[1-\exp(-(t_x-t_y)^2-(\vec{r}_x-\vec{r}_y)^2)],
\end{equation}
but that's obviously not Lorentz invariant. So I'm stuck, and would like your help. Do any of you clever people have some ideas?

 

[bcrowell]

I think you've essentially proved that this can't be done.

To simplify the discussion a little, you can just talk about a single vector rather than the difference of two position vectors. (This is actually more rigorously well founded, since in GR the set of spacetime points doesn't even form a vector space.)

So now you want a function f(x) that takes a vector as input and gives a scalar as output, and that is Lorentz-invariant with respect to its input. The only such function is the metric, or functions of the metric. This is actually one of the fundamental assumptions of relativity. If it were violated, then spacetime would have some other piece of geometrical apparatus built into it other than the metric.

Note that any two points on each other's light cone can be made arbitrarily close to one another ("close" in terms of their Minkowski coordinates) by doing a Lorentz transformation.

 

[robphy]

(Edit: Use something related to the) Dirac delta function [distribution]?

 

[Demystifier]

(Edit: Use something related to the) Dirac delta function [distribution]?

It can be done even with true functions, if one does not require that the function needs to be analytic. For instance,
$$f(x,y)=7 \;\;\; {\rm when} \;\;\; x\neq y, \;\;\; f(x,y)=0 \;\;\; {\rm when} \;\;\; x=y$$

Unfortunately, both the delta-function and my "stupid" function do not satisfy the requirement that the function should be small when x is "near" y.

 

[HomogenousCow]

So now you want a function f(x) that takes a vector as input and gives a scalar as output, and that is Lorentz-invariant with respect to its input. The only such function is the metric, or functions of the metric. This is actually one of the fundamental assumptions of relativity. If it were violated, then spacetime would have some other piece of geometrical apparatus built into it other than the metric..

Because that basically defines the metric right? Once we put some restrictions on it, like bilinearity there should only be one unique metric. Correct me if I'm wrong.

 

[bcrowell]

Because that basically defines the metric right? Once we put some restrictions on it, like bilinearity there should only be one unique metric. Correct me if I'm wrong.

That sounds right, although the OP doesn't actually care if it's bilinear.

If this were GR, then I guess we could use scalar polynomial curvatures. For example, let R be the Kretschmann invariant. Then we could construct the function $(R(x)-R(y))^2$, which will vanish when x=y and will be nonzero at most other points. To make it a more reliable test, we could pick some other scalar polynomial curvature Q and construct $(R(x)-R(y))^2+(Q(x)-Q(y))^2$. If you keep adding in more and more such terms, then I think you could get to a point where the function is only going to vanish for $x\ne y$ if the spacetime has some symmetry.

 

[Demystifier]

The OP says s/he wants the function to be Lorentz invariant. If my #2 is a correct interpretation of what s/he means by this, then I don't think your example satisfies that condition.

How about the function
$$f(x,y)=7 \;\;\; {\rm everywhere}$$
Is that function Lorentz invariant in your interpretation?

 

[bcrowell]

@Demystifier: Sorry, I was confused. I'll delete the incorrect part of my #6.

 

[HomogenousCow]

o now you want a function f(x) that takes a vector as input and gives a scalar as output, and that is Lorentz-invariant with respect to its input. The only such function is the metric, or functions of the metric. This is actually one of the fundamental assumptions of relativity.

Actually now that I think about it, isn't this kind of circular? At least in SR, the Lorentz transforms are defined in relation to the Minkowski metric.

 

[bcrowell]

Suppose you could construct a function $f(x)$ that was smooth and Lorentz-invariant, and vanished only for x=0.

Then the gradient of this function at x=0 would have to be zero as well, because otherwise it would pick out a preferred direction in spacetime, violating Lorentz invariance.

Now suppose we pick a certain frame K, with Minkowski coordinates (t,...). For derivatives taken along lightlike directions, we would have $\partial^2 f/\partial t^2=k$, for some constant $k$, and by Lorentz invariance $k$ would have to be independent of which lightlike direction we chose. Now if we switch to a different frame K', we can define a $k'$ by forming the analogous derivative in that frame's Minkowski coordinates. But for the lightlike direction with maximal positive projection along the boost, $k'>k$, unless $k=k'=0$. This would contradict Lorentz invariance, since Lorentz invariance requires $k'=k$. So the second derivative along the light cone has to be zero as well.

It appears to me that you could probably make similar arguments for derivatives of all order, and therefore all the derivatives have to vanish on the light cone. That means that the function can't be analytic.

I think essentially what's going on is that $\nabla_i \nabla_j \ldots \nabla_k f$ is a tensor, and you can't pick out a nonzero preferred tensor without violating Lorentz invariance. I would have to think about this a little more, though, because in GR we do have the Riemann tensor, and GR does have local Lorentz invariance.