## Wednesday, April 20, 2011

### A problem with Special Relativity Theory for perdurantists

There seems to be a problem for the conjunction of Special Relativity and perdurantism.  Maybe this is a standard problem that has a standard solution? Let's say that being bent is an intrinsic property. Perdurantists of the sort I am interested in think that Socrates is bent at a time in virtue of an instantaneous temporal part of him being bent (I think the argument can be made to work with thin but not instantaneous parts, but it's a little more complicated). Therefore:
1. x is bent at t only if the temporal part of x at t is bent simpliciter.
The following also seems like something perdurantists should say:
1. x is bent simpliciter only if every temporal part of x is bent simpliciter.
Now, we need to add some premises about the interaction of Special Relativity and time.
1. There is a one-to-one correspondence between times and maximal spacelike hypersurfaces such that one exists at a time if and only if one at least partly occupies the corresponding hypersurface.
Given a time t, let H(t) be the corresponding maximal spacelike hypersurface. And if h is a maximal spacelike hypersurface, then let T(h) be the corresponding time. Write P(x,t) for the temporal part of x at t. Then:
1. P(x,t) is wholly contained within H(t) and if z is a spacetime point in H(t) and within x, then z is within P(x,t)
and, plausibly:
1. If a point within x is within a maximal spacelike hypersurface h, then P(x,T(h)) exists.
Now suppose we have Special Relativity, so we're in a Minkowski spacetime. Then:
1. For any point z in spacetime, there are three maximal spacelike hypersurfaces h1, h2 and h3 whose intersection contains no points other than z.
1. No object wholly contained within a single spacetime point is bent simpliciter.
Finally, for a reductio, suppose:
1. x is an object that is bent at t.
Choose a point z within P(x,t) and choose three spacelike hypersurfaces h1h2 and h3 whose intersection contains z and only z (by 6). Now define the following sequence of objects, which exist by 4 and 5:
• x1=P(x,t)
• x2=P(x1,T(h1))
• x3=P(x2,T(h2))
• x4=P(x3,T(h3))
Observe that xis wholly contained in the intersection of the three hypersurfaces h1h2 and h3, and hence:
1. x4 is wholly at z.
2. It is not the case that x4 is bent simpliciter.
Now:
1. x1 is bent simpliciter. (By 1 and 8)
2. x2 is bent simpliciter. (By 2 and 11)
3. x3 is bent simpliciter. (By 2 and 12)
4. x4 is bent simpliciter. (By 2 and 13)
Since 14 contradicts 10, we have a problem.  It seems the perdurantist cannot have any objects that are bent at any time in a Minkowski spacetime. This is a problem for the perdurantist. If I were a perdurantist, I'd deny 2, and maintain that an object can be bent simpliciter despite having temporal parts that are bent and temporal parts that are not bent. But I would not be comfortable with maintaining this. I would take this to increase the cost of perdurantism. What is ironic here is that it is often thought that endurantism is what has trouble with Relativity.

#### 1 comment:

Alexander R Pruss said...

See the discussion here.