Math symbol regarding time duration?

Discussion in 'Physics & Math' started by Quantum Quack, Feb 11, 2013.

  1. Quantum Quack Life's a tease... Valued Senior Member

    Messages:
    23,328
    I believe commonly that the symbol t is often used to denote a specific point in time. My quest though is to find the notation commonly used to denote the "duration of time" when including it in a mathematical formulation.

    example: the time it takes to go from a to b is written using what time symbol?
    any help appreciated...
    a link would be great as Google failed to yield any useful results
    If I wished to write:
    (+2)+(-2) + (time duration = >0) = 0
    how would I write it?
     
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  3. rpenner Fully Wired Valued Senior Member

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    4,833
    \(\Delta t_{\tiny AB}\) if we are talking about the elapsed coordinate time or elapsed Newtonian time.

    \(\Delta \tau_{\tiny AB}\) if we are talking about the elapsed proper time in Special or General relativity.
     
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  5. youreyes amorphous ocean Valued Senior Member

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    2,830
    usually two dots on top of a unit symbolizes derivative w respect to time.

    delta t is elapsed time in interval.
     
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  7. Quantum Quack Life's a tease... Valued Senior Member

    Messages:
    23,328
    thanks guys!
    so if i wished to write the following would the notation be correct if referring to Newtonian time

    0 = (+)1 + (-)1 + (delta t = > 0)
    make sense?
    where
    zero = positive 1 plus negative 1 plus time duration is greater than zero.
    I am not so much interested in the validity of the outcome but wishing to know how to write the argument properly.

    The argument that I am trying to write correctly for Newtonian space is:

    0=/= (+)1 + (-)1

    0 = (+)1 + (-)1 + (delta t = > 0)

    where 1 could be any value
     
  8. origin Heading towards oblivion Valued Senior Member

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    11,888
    What are you trying to do, that equation looks really strange. I assume all of the terms time based?
     
  9. James R Just this guy, you know? Staff Member

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    39,397
    All additive terms in any physical equation must have the same units. Thus, this is potentially correct:

    0 (seconds) = (+1 seconds) + (-1 seconds) + (Delta t)

    where Delta t is understood to be in seconds (and would have to be zero in this case to make the equation true).

    This, on the other hand, is meaningless:

    0 = +1 + (-1) + (Delta t)

    because 0, +1 and -1 are unitless numbers, while it is understood that Delta t has units of time (e.g. seconds).
     
  10. Quantum Quack Life's a tease... Valued Senior Member

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    23,328
    Thanks James!
     
  11. Asexperia Valued Senior Member

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    1,724
    That is correct, but both

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    and "t" are expressed equally, in a time unit. It is essentially the same.

    The point in time (B) requires an origin (A).
     
  12. Mathers2013 Banned Banned

    Messages:
    190
    The x-axis is used to denote time:x is one, two, three... etc. depending on the number of time(s) one has performed the action. Each time x is used it increases by one, even if you are wrong, so it becomes more difficult to correctly guess the time(s). Y is the value up the y-axis.

    So, an equation should read 'y=x...' and such an equation should read correct for each 'x.'
     
  13. rpenner Fully Wired Valued Senior Member

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    4,833
    No. // edit: this "No" was in response to a spam post now deleted.

    Usually, one dot on top symbolizes the derivative with respect to the single independent variable.

    If \(f\) is short for \(f(\lambda)\) then \(\dot{f}\) is short for \(\frac{df}{d\lambda} { \Large ( \lambda ) }\).

    Two dots would signify a double derivative, as in the Newtonian equation of motion: \(\vec{F} = m \ddot{\vec{x}}\) as being short for \(\vec{F}(t) = m \frac{d^2 \vec{x}}{d t^2} { \Large ( t) }\).

    If there is more than one independent variable, then the concept of of partial derivatives is needed, and different notation is needed.

    For example, if \(f\) is short for \(f(x,y)\) then some partial derivatives might be abbreviated:
    \( f_{\tiny x} = \frac{\partial f}{\partial x} { \Large ( x, y ) } \\ f_{\tiny xy} = f_{\tiny yx} = \frac{\partial^2 f}{\partial x \partial y} { \Large ( x, y ) } \\ f_{\tiny yy} = \frac{\partial^2 f}{\partial y^2} { \Large ( x, y ) }\)
    and so forth.
     
    Last edited: Feb 18, 2013

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