Is there an answer to the question in the title of my post? If time is a dimension in the same sense as the spatial dimensions are dimensions, then I might think there should be a sensible way to convert meters into seconds and vice versa.
Well, is there?
-FrL-
I can think of measuring the amount of time an object takes to get from point A to point B in a meter. I think it most cases, movement is required to make that measurement. Otherwise, the meter (or any measure of length) is just here and now. How could you get time out of it?
As I understand it, time is not exactly the same as the spatial dimensions, so I don’t think there really is an answer to your question.
The best that I can do is tell you that the meter is currently defined as the distance light travels in 1/299,792,458 of a second.
(links to Wikipedia entry on the meter)
I was taught by friends at MIT that the answer to any question like this is “11 Bonner Units.”
I suppose that one could identify one second as equal to 299,792,458 meters, (the number of meters that light travels in 1 second in a vacuum).
c.
That is, there are 299,792,458 meters in one second.
In relativity, time and space are on the same footing, units-wise. The meters/seconds distinction is a Newtonian construct.
It takes massless particles a time of 1 unit to travel a distance of 1 unit. Since the time unit of 1 second is equal to a distance of 299,792,458 meters, light travels at 299,792,458 meters/second = 1 distance unit / 1 time unit.
Hown many inches of hieght are there in a inch of width?
I’ve got to agree with Tangent (and disagree with Pasta). In special relativity time is one dimension and space is the other three, but that doesn’t exactly mean they’re interchangeable.
Let me 'splain.
Considering just space for a moment, 1 metre “up” is the same as 100 centimetres “to the left”; specifically, if I have two objects, one of which is 1 metre “up” from the other, I can rotate the whole system rigidly in such a way that I end up with two objects, one of which is 100 centimetres “to the left” of the other.
Now consider space-time. I can conceive of two events in space-time, one of which is 1 second after the other but in the same location with respect to an observer. I can also conceive of two events which take place at the same time with respect to an observer, but one of which takes place 1 metre to the left of the other. But there’s no such thing as a “rigid” motion of the system which takes the first configuration of events to the other. You just can’t rotate time to point in the same direction as space, any more than you can accelerate an object past the speed of light.
So there’s no real need to be able to convert seconds into meters, even though they exist as two dimensions of same space-time. You can just arbitrarily declare that 1 second equals c meters, where c is the speed of light, and that simplifies certain calculations, but that just a convenience; there’s nothing physically necessary about that conversion in the same way that the rate of 1 meter to 100 centimeters is necessary.
That said, general relativity is a whole other kettle of fish, and there may be a natural answer to your question there. But I don’t claim to understand general relativity beyond the whole “bowling balls on a trampoline” thing.
Thanks for the replies so far, and thanks especially to *Orbifold.
-FrL-
Yes, space and time are not the same. That’s why I took care to say “units-wise”. But, while it’s true that there’s no spatial rotation that takes distance into time, there are Lorentz “rotations” that do. Space and time do indeed mix, you just need a more general rotation to do it. And, if you express them in the same units, they differ only by a minus sign.
Or in other words: the question in the OP has an answer only in the context of space-time, and in the context of space-time space and time are on the same footing when it comes to their units. They differ only by a minus sign in the equations (in special relativity that is.)
Special thanks extended also to Pasta.
Now I don’t know who to believe, though. 
-FrL-
As Orbifold said, just because time and locations form a four dimensional coordinate system it doesn’t mean that there is an equivalence between time and location.
For example, we agree to meet in the building at the corner of 1st Street and 1st Avenue at the top of the stairway on the 3rd floor. We have now given the location of a point in three dimensional space where we will meet. There is, however, one vital dimension missing that is required in order for us to meet. There is the little matter of when, or at what time, we are going to meet. So, in order to specify the location of a point in a universe where there is movement you really need to know the time at which an object was at a particular point in three dimensional space.
It turns out that because of the finite speed of light, when you specify the three spacial locations and the time for a moving object as seen by one observer, all four of them change when you refer then to a different observer. That is time and space are inextricably interwolven in a single coordinate system.
No, I have to disagree. Unless you’re using the term “Lorentz rotation” in a way I’m not expecting.
A rigid motion of Minkowski space…space-time as viewed by special relativity, in other words…has to preserve the “spacetime distance”. That means it has to preserve the set of light cones, because light cones are just the sets of points where the spacetime distance between any two points is zero. And light cones have a definite “direction”, in that the time axis is always on a different side of the light cone than the three spatial axes.
In short, any homeomorphism of Minkowski space which swaps the time axis with one of the space axes would not preserve the set of light cones and hence can’t possibly be an isometry with respect to the space-time metric.
More detail…
In a Newtonian world, one can rotate width (x) and depth (y) into a different arrangement like so:
x’ = xcos([symbol]q[/symbol])+ysin([symbol]q[/symbol])
y’ = -xsin([symbol]q[/symbol])+ycos([symbol]q[/symbol])
where [symbol]q[/symbol] is the angle of rotation. Note that you have two quantities (x and y) that each get multiplied by unitless numbers to create the new quantities. Everything is on the same footing: x, y, x’, and y’. Time has no place here since it’s Newtonian.
In special relativity, you could also do this rotation. But, you can also rotate space and time into one another. Taking x and t now:
ct’ = ct*[symbol]g[/symbol]+x*[symbol]gb[/symbol]
x’ = -ct*[symbol]gb[/symbol]+x*[symbol]g[/symbol]
Note the similarity between this and the purely spatial rotation. [symbol]g[/symbol] and [symbol]b[/symbol] are unitless numbers describing the extent of the rotation. The conversion constant c is only there because humans decided to use different units for space and time. In the spatial example, if I had chosen to measure x in inches and y in centimeters, and I had decided to pretend that these were fundamentally different things, I would have needed a conversion constant k=(2.54 cm/in) in the expression:
kx’ = kxcos([symbol]q[/symbol])+ysin([symbol]q[/symbol])
y’ = -kxsin([symbol]q[/symbol])+ycos([symbol]q[/symbol])
But we know that trying to add cm and inches requires an implicit conversion, so we don’t write it explicitly. The same holds for the space-time situation. If we had just expressed time in meters in the first place, the expression would look clean:
t’ = t*[symbol]g[/symbol]+x*[symbol]gb[/symbol]
x’ = -t*[symbol]gb[/symbol]+x*[symbol]g[/symbol]
Replying to items on preview:
Indeed, that is not a valid Lorentz transformation, so the light cones are busted. I’m not saying time and space are the same. But they don’t need to be interchangeable with respect to the physicical laws to require they have the same units.
Do you not agree that time and space show up in special relativity on equal unit-footing? (Not that they’re interchangeable – they’re not!) If you do not, then perhaps we’re caught on semantics, because I know we both know the physics.
Right. In space coordinates the distance between two points is invariant regardless of your coordinate system. That invariant in rectangular coordinates is x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup]. In space time it is t[sup]2[/sup] - x[sup]2[/sup] - y[sup]2[/sup] - z[sup]2[/sup]
Perfect! And that expression makes sense only if t and x have the same units! Which brings us all the way back: 1 second = 299,792,458 meters.
I don’t think it’s entirely semantics. You say that the physical laws require that they have the same units; that’s the part I disagree with. The physical laws don’t even require that we have a way to convert between the units. It’s convenient to have a conversion between the units, because then the formulas become cleaner, as you say. For example, the bilinear form which defines the metric becomes -t[sup]2[/sup]+x[sup]2[/sup]+y[sup]2[/sup]+z[sup]2[/sup].
But that’s just a convenience. It’s not necessary. There’s nothing preventing us from using the bilinear form -t[sup]2[/sup]+(x/c)[sup]2[/sup]+(y/c)[sup]2[/sup]+(z/c)[sup]2[/sup] instead, where c=299,792,458, and leaving x, y, and z in metres while t remains in seconds. This gives us exactly the same results. It’s just not as pretty, because we keep having to insert 299,792,458 into equations all over the place, but it works, and we never have to convert between meters and seconds.
We could do the same thing in just space, and try to define distance in the plane using the formula x[sup]2[/sup]+(y*2.54)[sup]2[/sup] where x is in centimetres and y is in inches. That would be a perfectally valid way to define Euclidean geometry in the plane. But in that case we have an isometry which takes the x-axis to the y-axis, and that provides a physical reason to say that 1 inch is 2.54 centimetres. There’s no isometry in Minkowski space which takes a space-like vector to a time-like one, however, so there’s no physical reason to say that 1 second is 299,792,458 metres.
It’s just really, really handy to do so for computational reasons.
There is a straightforward answer to this
I grant your point. The OP’s question can only be addressed if one is willing to place an artificial constraint on the form of the bilinear. The answer, then, is as artificial as the constraint. That is, the answer is not a physical law but a metric-dependent statement. A relevant statement, given the ubiquitousness and Occam-pleasingness of the “+/-1” metric, but yes, a non-fundamental one.
Drat, hit post in error
http://www.boatersdream.com/libarticles.cfm?ArticleID=20&CFID=1839183&CFTOKEN=18047320
So a Second of latitude is 6076 / 60 = 101.27 feet
So there are about 30 metres in a Second