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Math Help - Solving equation for deeply embedded variable

  1. #1
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    Solving equation for deeply embedded variable

    Iím trying to derive an equation that provides a mathematical definition of time. The following equation is for the kinetic energy of a body moving at a substantial portion of the speed of light.

    e = \frac{mc^2}{\sqrt{1 - (v/c)^2}} - mc^2

    Where e is kinetic energy, m is mass, c is the speed of light, and v is velocity. In the above equation,

    v = d/t

    Where d is distance and t is time. Substituting d/t for v in the first equation, we have

    e = \frac{mc^2}{\sqrt{1 - (d/tc)^2}} - mc^2

    The problem is, solve the last equation for t. Does anyone have the mathematical chops to do this?




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  2. #2
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    Probably there are quite a few people on this forum who have the "mathematical chops" to solve for t. But that's not really relevant. The question is, do you have the chops? If you don't, we can help you get the chops. But we're not going to simply give you the answer. That's too easy and doesn't help your understanding.
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  3. #3
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    Ackbeet,

    Well, I meant it as sort of a challenge. It is a challenging problem, is it not? It's been a while since I solved complex algebraic equations, and I know there are various tricks to isolate a variable on one side of the equation, such as adding or subtracting the same term to both sides, taking the square root or squaring both sides, multiplying both sides by the same factor, etc. In this case, the machinations seem quite intricate, elaborate, and involved, so I thought someone with great facility in mathematics might show me the way.

    I see you are a physicist, and I welcome the opportunity to seek your opinion on the implications of this equation which can provide a mathematical definition of time. That is, if we can isolate t on one side of the equation.
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  4. #4
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    By the way, I did the work necessary to solve for t, using the simpler classical equation for the kinetic energy of a body in motion. Starting with that equation,

    e = \frac {mv^2}{2}

    Substituting d/t for v gives us

    e = \frac {m(d/t)^2}{2}

    Which I was able to solve for t

    t = \frac {d\sqrt{m}}{\sqrt{2e}}

    Here are the machinations I used to get there.

    2e = m(d/t)^2

    2e = \frac {md^2}{t^2}

     t^2 = \frac {md^2}{2e}

    t = \frac {d\sqrt{m}}{\sqrt{2e}}
    Last edited by StevenBrown; June 22nd 2010 at 10:24 AM.
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  5. #5
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    I agree with your equation for the relativistic kinetic energy, in the context of special relativity, that is. What would you do first?

    As for the physical interpretation, I would have to know which time you mean here. Proper time? Is this time associated with the reference frame you're observing, or the one you're in?
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  6. #6
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    sorry
    How can i create a new thread for my question?
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  7. #7
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    cdniki: First browse to the appropriate sub-forum, and then click the Post New Thread button at the top. That should do it.
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  8. #8
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    I don't know if I should address you as Ackbeet, Adrian, or Dr. Keister.

    Do you agree with my math in post #4 of this thread?

    I would enjoy getting into a discussion on the physics, but let us solve the equation first. I'm willing to take this step by step. What would I do first? Hmmm. I'll be back....
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  9. #9
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    Starting with

    e = \frac{mc^2}{\sqrt{1 - (d/tc)^2}} - mc^2

    So far, I have two steps toward a solution for t.

    e + mc^2 = \frac{mc^2}{\sqrt{1 - (d/tc)^2}}

    and then


    \sqrt{1 - (d/tc)^2} = \frac {mc^2}{e + mc^2}

    Now I am thinking square both sides of the equation. That would simplify the left side, but would seem to obfuscate the right side. What do you think, Dr. Keister?
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  10. #10
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    Looks good so far. As for squaring both sides, I really don't think you have much of a choice. That is, you have to do that. You can choose not to multiply out the RHS, if you wish.
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  11. #11
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    Plodding along... squaring both sides gives us

    {1 - (d/tc)^2 = (\frac{mc^2}{e + mc^2})^2

    Square the part within parentheses on the LHS:

    {1 - \frac{d^2}{t^2c^2} = (\frac{mc^2}{e + mc^2})^2

    Subtract 1 from both sides:

     - \frac{d^2}{t^2c^2} = (\frac{mc^2}{e + mc^2})^2 - 1

    Multiply both sides by -1:

    \frac{d^2}{t^2c^2} = 1 - (\frac{mc^2}{e + mc^2})^2

    Continued next post....
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  12. #12
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    Now I'm stabbing in the dark.... Multiply both sides by t^2:

    \frac{d^2}{c^2} = t^2 [1 - (\frac{mc^2}{e + mc^2})^2]

    This is getting hairy. Isolate t^2 on the RHS:

    \displaystyle{\frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]} = t^2}

    Swap sides (easy):

    \displaystyle{t^2 = \frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]}}

    Take square root of both sides:

    \displaystyle{t = \sqrt{\frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]}}}

    Now try to simplify this MATHEMATICAL MONSTROSITY.

    Continued next post....
    Last edited by StevenBrown; June 22nd 2010 at 05:49 PM.
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  13. #13
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    Quote Originally Posted by StevenBrown View Post
    Now I'm stabbing in the dark.... Multiply both sides by t^2:

    \frac{d^2}{c^2} = t^2 [1 - (\frac{mc^2}{e + mc^2})^2]

    This is getting hairy. Isolate t^2 on the RHS:

    \frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]} = t^2

    Swap sides (easy):

    t^2 = \frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]}

    Take square root of both sides:

    t = \sqrt{\frac{d^2}{c^2[1 - (\frac{mc^2}{e + mc^2})^2]}}

    Now try to simplify this MATHEMATICAL MONSTROSITY.

    Continued next post....
    So time has to be nonnegative then?
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  14. #14
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    Let's start by putting the radical in the denominator:

    \displaystyle{t = \frac{d}{\sqrt{c^2[1 - (\frac{mc^2}{e + mc^2})^2]}}}

    The square root of c^2 must be c:

    \displaystyle{t = \frac{d}{c\sqrt{1 - (\frac{mc^2}{e + mc^2})^2}}}

    Now it appears I can no longer avoid squaring what is within the parenthesis.

    Continued next post....
    Last edited by StevenBrown; June 22nd 2010 at 05:51 PM.
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  15. #15
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    Well since you didn't respond to my previous post yet, I'll explain:

    In general for real numbers x and y, it is false that

    x^2=y \Rightarrow x=\sqrt{y}

    Rather you must write

    x^2=y \Rightarrow x=\pm\sqrt{y}

    So I was responding to your step in which you discarded negative values for t.
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