what if time does not exist

juju

D=(c-v)t

if v=c, then (c-v)=0

If (c-v)=0, then D=0.

I wonder if you are thinking about the space/time interval for a photon. For a photon the space interval is equal to the time interval and the space/time interval is equal to zero.

Your equation would be almost correct if it refers to the space/time interval.

Here, (ds^2)=((cdt)^2)-(dr^2)
so, (ds^2)=((cdt)^2)-((vdt)^2)
and (ds^2)=(c^2-v^2)dt^2

juju

PRyckman

first off, I'd like to point out once again, that the equation is already proven correct. Second of all if c-v=0 then D=0(t) Just looking at the equation tells you that something travelling light speed doesn't exist in time. And it's distance can only be found if you measure it. Once you measure it, you connect it to time.

Atonio, I know what it looks like on a graph. But you know you can't put it on a graph because to graph it, would show infinite. Therefor it's not the same equation when v=c and can't be graphd.

The only way to measure t for v=c is to measure the distance. Thus knowing it's exact place in the universe. Which pins it to a point in time.

One more thing, notice that if D equals zero, it makes sense because you know very well, that you can't know a photons position and momentum at the same time.

Thus, for light time=0 and distance=measured.

Antonio Lao

In order to show the duality of time, that is two time variables plotted against each other, one being the abscissa and the other as ordinate, there are two complementary graphs.

one graph is plotted by t1=t2/(1-t2), the other is plotted by t2=t1/(1+t1). The result is t1t2= t1-t2. These two time variables can be equal if and only if t1=t2=0 (both equal zero). When they are not zero, the values can be found by the two graphs depending on which one is the dependent and independent variable.

For one graph, it can show that as t1 approaches infinity, t2 approaches 1. For the other graph, as t2 varies from zero to infinity, there is a discontinuity at t2=1, where t1 approaches infinity from the left and approaches negative infinity from the right. But for all values greater than 1 and as t2 approaches infinity, t1 approaches the value of -1.

juju

Actually, if your equation is true, what it tells me is that if D=0(t), then something travelling at light speed has zero distance no matter what the value of t.

juju