# Time as a dimension, can spatial length be applied to it?

1. Dec 20, 2015

### mcjosep

Could you use the formula "planck length/c=planck time" to convert time into a spatial distance? Using that to tell how far something is through time.

2. Dec 20, 2015

### Hornbein

Sure. I don't see why not.

3. Dec 20, 2015

### alw34

4. Dec 20, 2015

### mrspeedybob

The equivalence of time and distance is much simpler then that. 1 second = 0.3 gigameters (Approximately)
Using that equivalence you can say that all objects travel at the same rate, C, at least as far as special relativity is concerned, I suspect GR changes that.
If a clock is not moving through space, it is moving through time at 1 second / second, or 0.3 Gm/s. If it is moving through space at ,for example, .1 Gm/s then you can compute the time dilation as a simple right triangle problem with space as one leg, time as the perpendicular leg, and the hypotenuse as a constant. The space side in my example is .1 Gm, the Hypotenuse is 0.3 Gm, so the time side, which is perpendicular to the space side must be about .283 Gm. So for every second, or .3 Gm the stationary observer's clock measures, the traveling clock measures about .943 seconds, or .283 Gm.

I suspect GR changes and complicates this somewhat. I don't understand GR well enough to say, but I'd suspect that time and space may not always be perpendicular, moreover, the triangle I just described may have to be constructed in non-euclidean space.

Last edited: Dec 20, 2015
5. Dec 20, 2015

### alw34

only for massless objects, like the photon
they are never 'perpendicular'

6. Dec 20, 2015

### mrspeedybob

What's mass got to do with it?
I think you must have misunderstood what I wrote. If you can explain why you think mass is relevant I might be able to re-explain myself more clearly.

In the real universe, that may be true. I'm assuming flat space-time for simplicity's sake.

7. Dec 20, 2015

### Svein

For those of us who are into computers and high-speed electronics, 1ns corresponds to 0.3m in vacuum and about 0.2m along a cable. That is why layout becomes very critical at Gigahertz clock speeds.

Last edited: Dec 20, 2015
8. Dec 20, 2015

### Staff: Mentor

Yes. Or you can use any units you like. For instance you can say that a typical human is about 7E17 m long in time compared to about 2 m long in height. The conversion factor is c in whatever units you use.

9. Dec 20, 2015

### Staff: Mentor

He is talking about four-vectors, $(ct,x,y,z)$. Where $(1,0,0,0)\cdot(0,1,0,0)=0$ indicates that time is perpendicular to space, although "orthogonal" may be a better word.

10. Dec 20, 2015

### mcjosep

So could you say I am gravitationally attracted to something in a different time than right now and calculate that attraction by using this conversion formula and saying that length is the my radius to that object?

11. Dec 20, 2015

### mrspeedybob

No.

I know you want more of an answer then that, but the instant you start considering gravity in the context of relativity, things get HUGELY more complicated. GR is way beyond my math skills.

12. Dec 20, 2015

### Staff: Mentor

It is not clear from your question, but I assume that you are thinking of "calculate that attraction" being Newton's law of gravitation. Unfortunately, Newtonian gravity is not compatible with relativity, so that won't work. Instead you would need to calculate gravity using the Einstein field equations.