Since time is variable, how do you measure its velocity and acceleration?
seconds per ?
What do you mean by saying "time is variable"?
all depends on gravity as i see it
I get the impression, Comp, that you're thinking of time as a physical object of some kind, and are confused about relativistic time dilation due to gravity or velocity. Time is just Nature's way of keeping everything from happening at once. It has no intrinsic physical properties such as velocity or acceleration.
if it is a dimension in the universe than vectors exist in it and those vectors have slope.
Er... vectors have slope?!!!
I should have said the graphs of vectors have slope which as we all know correspond to rates of change. but please, you cannot really believe that is not what I was referring to.
Do you mean an angle?
I know I am just being confusing and non relativistic.
if you plot a velocity vector as a rate of change, that rate of change will have a slope.
since time is a dimension, vectors exist in it, since vectors exist in it, you must be able to plot them as a rate of change, even if it is constant.
so, there is some ratio for the description of the rate of change. I was curious as to what it was.
is it simply a ratio of seconds/second?
so if all measurements were taken from our standard earthly perspective, time near a massive body would slow to say 1 second/2 seconds, etc.
OK, but you do know that I can easily plot scalars, instead of vectors, and still get slopes from that too, don't you? Look at the gradient of a potential field, which is a field of scalars.
Your original question is confusing because you're asking about a "velocity", which is the time rate of change of TIME, of a quantity that is used to DEFINE the word "time". You don't do that when you ask for a velocity of an object, because the object itself is not the definition of the quantity being measured. The ball doesn't affect the definition of "time" and "displacement or space". The ball may had a change of displacement over a time, but it doesn't affect the definition of those two quantities. In other words, the ball is inside this "frame" called space-time. Your question essentially asked the "time rate of change of time".
Unless you wish to clearly define such a thing, then there's a potential of tripping over oneself in cases like this.
Yes, I realize that. I should have quoted "velocity" so that you knew to take it metaphorically rather than literally.
No, because velocity is a rate of change, it is the 'slope' (i.e. gradient function) of the thing that it is the rate of change of. In short, at time t, v(t) is the gradient of the tangent of s(t) (displacement) at that time.
A vector comprises of two perpendicular components. Time is 1-dimensional. So how can you describe something in 1 dimension as a vector? Furthermore, just because velocity is a rate of change and velocity is a vector, is does not mean all vectors are the rate of change of something. Position, for instance, is not the rate of change of anything, but can be described as a vector.
s/s = 1, so the rate of change of time is unitless. Moreover, since the rate of change of time is t/t, the value is always 1 so it is not a variable.
If you mean the rate of change of a clock from someone else's reference frame (t'/t), then this is essentially a scale and so is, again, unitless.
Then you're using the word to mean something else. It is then imperative that you DEFINE what is a metaphorical velocity.
Keep in mind that in physics, EVERY principle and ideas have clear underlying mathematical description. You can't use the word "velocity" in any form or shape that you wish, and neither can the rest of us. If not, you can already imagine the major confusion being created.
yes yes, I understand all that... I was just having a difficult time describing the relationship and I should make sure I am more clear when I am in analogy mode. :-)
Are you perhaps expressing coordinate time t as a function of proper time tau? I.e. if you specify the path of a particle, you might specify x(tau),y(tau),z(tau) and t(tau), tau being the "proper time".
Basically, if you figure out what your question is, the answer will probably be pretty clear :-)
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