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Delta Symbol confusion

  1. May 13, 2007 #1
    Okay, so how do you know when to put a delta symbol in front of a variable? For example, [tex] U=\frac{3}{2}nRT[/tex] however, it can also be written as [tex]\Delta U=\frac{3}{2}nR\Delta T[/tex] --Change in Internal energy of a system

    I'm pretty sure that this isn't the case for many equations such as: [tex] \Delta V=Ed[/tex]--Potential difference, where there is no delta signs on the right hand side of the equation right?

    Are these deltas just suppose to be memorized or is there a way to add or remove them from equations while still being able to achieve the correct answer?
     
  2. jcsd
  3. May 13, 2007 #2

    Office_Shredder

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    No, you can't remove or add deltas to an equation, unless you happen to get really lucky and what you were subtracting happened to be zero. For example

    [tex]\Delta U=\frac{3}{2}nR\Delta T[/tex]

    We know [tex]U=\frac{3}{2}nRT[/tex]

    So if we have intitial internal energy [tex]U_i[/tex] and final internal energy [tex]U_f[/tex] then [tex]U_f-U_i[/tex] is by definition the change in internal energy. But [tex]U_f=\frac{3}{2}nRT_f[/tex] where [tex]T_f[/tex] is the final temperature (assuming the quantity of material does not change, and of course R can never change) and [tex]U_i=\frac{3}{2}nRT_i[/tex] where [tex]T_i[/tex] is the initial temperature. So [tex]U_f-U_i=\frac{3}{2}nRT_f-\frac{3}{2}nRT_i=\frac{3}{2}nR(T_f-T_i)[/tex]
    So the fact that you added delta symbols isn't mysterious, it's just basic algebra
     
  4. May 13, 2007 #3

    Ohhh lol. What about the potential difference equation? How come there is a delta in front of the (V)?
     
  5. May 13, 2007 #4

    cepheid

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    There's no need to memorize anything.

    If the quantity on the left hand side has a sensible definition on its own (like U in terms of T), then the change in that quantity would have to be expressed in terms of the change in the variable it depends on (T in this case).

    Potential functions and potential energy functions are different, in the sense that the potential at a single point has a completely arbitrary definition (i.e., the reference point with respect to which you measure the potential could change, adding or subtracting a constant value to all potentials at all points). This doesn't matter, because the *difference* in potential between two points will remain the same regardless. This is what is given by Ed and what is physically relevant.

    A good analogy is elevation. Let's say you're on an inclined plane (to keep the math simple). Then let's call the slope, or gradient of the incline E (it plays an analogous role to the electric field in this analogy). Let's call the distance you have travelled up the plane d. Then the total gain in elevation is just the slope times the horizontal distance:

    [tex] \Delta h = Ed[/tex]

    But what are you measuring the elevations with respect to (i.e. where do you define h = 0 to be)? It could be relative to sea level (as is the case in real life), or it could be relative to where you are standing now (in which case sea level would have a negative elevation). The point is, it really doesn't matter. It doesn't affect the change in elevation [itex] \Delta h [/tex] between two points on the incline separated by distance d. That will always be given by Ed. So if all you are concerned with is your gain in elevation with horizontal distance travelled, then this will be your formula of interest.
     
    Last edited: May 13, 2007
  6. May 13, 2007 #5
    Voltage, like gravitational potential, is in reference to an arbitrary zero point. You can think of the change in voltage in a uniform electric field as the field times the change in distance. But were usually only concerned with the relative change in potential between 2 points
     
  7. May 13, 2007 #6

    cepheid

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    The answer to your question lies in the question itself. The potential difference is (not surprisingly) the difference in electric potential.
     
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