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What if time does not exist 
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#199
Sep1904, 03:01 PM

P: 294

An observer riding on a photon would see the rest of the universe moving at the local speed of light and all other photons at rest. juju 


#200
Sep2004, 01:11 AM

P: 137

D=(cv)t Obviously, if v=c then D=t You put the infinite's and zero's in there for yourself, because there is no way to solve the equation. The only way to solve the equation is to measure it, to give it a distance. Only then does it interact with time. And as for the physics 101 comment, D=(cv)t is correct, it's physics 701 The c from E=mc2 Is calculated by cross multiplying that equation, or rather, my distance is cross multiplied from equation for c One more thing juju. Don't forget that what velocity was equal to right before it was equal to c, has no bearing on D=t it is not the same equation. The only way to solve this equation is to get out there with a ruler. Hence, the uncertainty principle, the two slit experiment and "the spooky action at a distance" As Einstein said, sorry I forget it's real name, It has to do with the polarization or spins of the particles, affecting eachother across distance immediatly. 


#201
Sep2004, 09:01 AM

P: 1,443

There is an intimate relation between two quantities such that for all positive values between zero and infinity of the independent variable, two distinct graphs can be shown that the product of ab=ab.
[tex] a=\frac{b}{1b}[/tex] [tex] b=\frac{a}{1+a}[/tex] [tex] 1b=\frac{b}{a}[/tex] [tex] 1+a=\frac{a}{b}[/tex] [tex](1b)(1+a)=1[/tex] [tex] ab=ab[/tex] The graph of abscissa b versus ordinate a is a discontinuous curve at 1, with 0 to infinity and from negative infinity approaching 1. The graph of abscissa a versus ordinate b is a curve from 0 and approaching the value 1 at infinity. 


#202
Sep2004, 09:48 AM

P: 1,443

When a and b are certain quantities for time, distance, velocity, mass, charge, force, or energy, the negative nature of these physical quantities can be explained.
This gives a basis for a renormalization principle and that is to find a and b such that [tex] 1=\frac{ab}{ab}[/tex] 


#203
Sep2004, 01:09 PM

P: 4

Here are my thoughts on the subject. It occurs to me (as it probably has to many other people in the past) that the only way we know about the passage of time is through movement. You may say that this is not true; that even when everything is still around you, you have an impression of the passage of time.
Things are not still however. While bulk movements may be absent, there are still tiny movements occuring everywhere. Thermal vibrations, quantum fluctuations and the like, but the movement of neurons in your brain (well, the movement of something  however the brain works basically  i'm not a neurologist). Suppose now that we could stop these movements, disregarding QM for just a second (this argument is more about philosophy than physical observables anyway). If there were no movement  none at all, including the neurons in your brain  there would be no perception of the passage of time. Indeed, this is because your brain would not function, since it depends on the movement of neurons or whatever. I guess in short I'm saying without movement there would be no time, and further, time is just a product of the way in which humans perceive movement (I suppose an alternative way of saying this is that without time, there would be no speed). Anyways, as I say, these are just some random thoughts on the subject. (PS. I havent read all the posts for this thread, so sorry if i've copied stuff other people have said) 


#204
Sep2004, 02:49 PM

P: 1,443




#205
Sep2004, 04:18 PM

P: 294

if v=c, then (cv)=0 If (cv)=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^2v^2)dt^2 juju 


#206
Sep2004, 06:30 PM

P: 137

first off, I'd like to point out once again, that the equation is already proven correct. Second of all if cv=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. 


#207
Sep2104, 04:46 AM

P: 1,443

one graph is plotted by t1=t2/(1t2), the other is plotted by t2=t1/(1+t1). The result is t1t2= t1t2. 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. 


#208
Sep2104, 03:16 PM

P: 294

juju 


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