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Relativity and Calculus Help

  1. Mar 2, 2004 #1
    This is an extra-credit problem. The question covers materials some students may not have had prior, hence not being mandatory.

    Question posed:

    Force is given by
    F=dp/dt
    F=d/dt (mv)
    F=m dv/dt
    F=ma
    and a=F/m is the slope of the line on a v versus t graph. This does not take into account the 'speed limit' of c.

    Where
    F=m dv/dt

    dv=a dt

    [tex]\int_{0}^{v} dv = \int_{t=o}^{t} a dt[/tex]

    [tex]v = a \int_{t=o}^{t} dt[/tex]

    v=at

    **I am not sure the significance of v=at, but it was given**

    So taking into account p=mv[tex]\gamma[/tex]

    We have
    F=dp/dt
    F=d/dt (mv[tex]\gamma[/tex])
    F=[tex]\gamma[/tex]m dv/dt
    (F/m) dt=[tex]\gamma[/tex] dv
    [tex]\gamma[/tex] dv=a dt

    **then when integrating those terms, I have a problem with...**

    [tex]\int_{0}^{v} \frac{dv}{\sqrt{1 - v^2/c^2}}[/tex]

    ...on the left side of the equation.

    I have only had two semesters of calculus, so I am unsure if I am simply forgetting something, of if there is something here I don't completely know. I was told by a classmate that doing this involves differential equations, of which I have little knowledge.

    The problem is to find the function that shows the line that asymptotically approaches c. I believe it involves a exponential, which is where I am faltering on my calculus.

    Thanks to anyone who can help/explain or point me in the right direction!

    ***apologies if LaTeX does not work on the first posting***
     
    Last edited: Mar 2, 2004
  2. jcsd
  3. Mar 2, 2004 #2
    You need to be careful with what you're considering as a constant. [itex]\gamma[/itex] is a function of v, which is a function of t, so it can't be treated as a constant with respect to time.

    But other than that, what are you trying to solve for? Velocity?

    cookiemonster
     
  4. Mar 2, 2004 #3
    The professor solved for the Newtonian v... however, I don't know why exactly.

    I am trying to solve for the function that gives the asymptotic line, which is linear and a=F/m classically.
     
  5. Mar 2, 2004 #4
    [tex] F = \frac{dp}{dt} = \frac{d}{dt}\Big(\gamma mv\Big)[/tex]
    [tex] \frac{F}{m} = \frac{d}{dt}\Big(\gamma v\Big) = \frac{d}{dt}\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]
    [tex] \frac{F}{m} dt = d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]

    Notice how I treated gamma as a function of velocity and left it in the derivative. Now integrate and then solve for v. Then let t-> infinity.

    cookiemonster
     
    Last edited: Mar 2, 2004
  6. Mar 3, 2004 #5
    The first thing that comes to mind for integrating
    [tex]\int_{0}^{v} \frac{dv}{\sqrt{1 - v^2/c^2}}[/tex]
    is inverse trig substitution.

    Post back if you need more help.
     
  7. Mar 3, 2004 #6
    Actually, you never need to evaluate that integral. The two integrals you actually have to do are remarkably easy. In fact, they're both the integral of 1!

    cookiemonster
     
  8. Mar 3, 2004 #7
    The more I try to do this, the more confused I get!

    So, you are saying to integrate

    [tex] \frac{F}{m} dt = d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]

    and that each side is an integral of one, which should give

    [tex] \int \frac{F}{m} dt = \int a dt = at [/tex] on the left hand side,

    and the right hand side, being an integral of 1, becomes

    [tex] d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big) = \Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]

    and solving for v gives

    [tex] v = \Big(\frac{c a t}{\sqrt{c^2+a^2 t^2}}\Big)[/tex]

    which compares to v=at for the classical value.

    So now, my question is: for the slope of the line on the graph of v versus t, the classical value given was a=F/m. With this new derivation for a relativistiv v, how do I acquire the slope formula for the relativistic slope value which asymptotically approaches c?
    THIS IS AN EXAMPLE OF THE GRAPH.

    The professor mentioned exponential functions would be used in deriving the equation. So am I looking to integrate something of the form dx/x = ln x?

    Lastly, one fellow student mentioned differential equations being used in the derivations... if so, where might they be? I have not yet taken a course in DE, however I have several resource books upon which to reference.

    Thank you all for your help!

    ***Gnome: that was my initial thought as well, however, the equation I would get from doing so would not really work here, at least to my knowledge. But thanks!
     
    Last edited: Mar 3, 2004
  9. Mar 4, 2004 #8
    Just to make something clear real quick:

    In general,
    [tex]F \neq ma[/tex].

    However, it is always true that
    [tex]F = \frac{dp}{dt}[/tex].

    That being said, you can't make the substitution F/m = a. But you're really working too hard anyway! You don't need to make that substitution. Just integrate it the way it is treating F as a constant.

    As for the differential equation and exponential term business, you've already used the DEQ (you used F = dp/dt) and I really don't see the need for an exp term. I worked it out already and I never used one.

    cookiemonster
     
  10. Mar 4, 2004 #9
    The professor states that there is an exponential involved because the function of the line on the v versus t graph asymptotically approaches c, compared to the linear function (given to us to be a=F/m) which goes unbounded past c.

    So that means I need a function of a=(something-with-an=exponential) for the equation of the line, right? Or maybe hyperbolic like tanh?

    So here I would take

    [tex] \frac{F}{m} dt = \frac{1}{m} \frac{dp}{dt} dt[/tex]

    To clarify, the left side should be integrated as

    [tex] F \int \frac{1}{m} dt[/tex] ....correct???

    and

    since F is constant, and F=dp/dt, and p=[tex]\gamma[/tex}mv, then it follows that the 1/m is constant, so we integrate the dt term remaining, giving the term dp/dt t/m as the remainder on the left side.

    And the right side is correct as I had it before simplifying both terms and solving for v?

    Further, I honestly don't know exactly why we are solving for v if we need a=(something) for the slope of the line... the professor didn't go into much detail about this -- he merely showed us that way to demonstrate the classical form. I feel like I am doing two different things which, for this purpose, are not related... although I am probably wrong.

    Thanks again for all your help!
     
  11. Mar 4, 2004 #10
    It seems like you're not quite understanding what you've already done to get to this ([itex] \frac{F}{m} dt = d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/itex]) point.

    Go back to my second post and see how I got here. You'll notice that this equation is the same as F = dp/dt, just in a different more useful form. Using F = dp/dt to get an equation and then substituting in F = dp/dt would be rather circular and quite pointless.

    Here, I'll do the integration for you and you come back and tell me if you understand it.

    [tex] \frac{F}{m} dt = d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]
    [tex] \int \frac{F}{m} dt = \int 1 \,d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big)[/tex]

    [itex]\frac{F}{m}[/itex] is not a function of t (an assumption, but a harmless one), so for the left hand side we get

    [tex]\int \frac{F}{m} dt = \frac{F}{m} \int 1\,dt = \frac{F}{m}t[/tex].

    The right hand side is even easier.

    [tex]\int 1 \,d\Big(\frac{v}{\sqrt{1-v^2/c^2}}\Big) = \frac{v}{\sqrt{1-v^2/c^2}}[/tex]

    Putting it all together,

    [tex]\frac{F}{m}t = \frac{v}{\sqrt{1-v^2/c^2}}[/tex]

    Notice that this is an equation that relates velocity to time in terms of only constants (assuming F is constant). So you can make a graph similar to the one you showed above. So now you just have to solve for velocity and see if it exhibits any asymptotic behavior.

    cookiemonster
     
  12. Mar 4, 2004 #11


    Ah-ha! That's where things weren't making sense in what I was doing.

    Ok, I managed to solve for v. The equation v approaches c like it is supposed to.

    However , how is the equation related to the slope of a line? I believe the professor (also) wants an equation which can be inputed into a computer/calculator that would draw the asymptotic line. I believe that is where the exponential comes in.

    Classically, we were given that

    [tex] v = a t [/tex]

    for the velocity, and

    [tex] a = \frac{F}{m} [/tex]

    for the slope of the line in the v versus t graph.

    Now that I have velocity, I am not sure how it relates to the graphical equation of the slope. Likewise, I do not even see how v=at relates to a=F/m in going from velocity to a slope equation.

    Thanks again for the assistance!
     
    Last edited: Mar 4, 2004
  13. Mar 4, 2004 #12
    That equation is related to the slope of the line via the derivative. Take the derivative of v with respect to t, and you'll get the slope (which is also the acceleration, dv/dt = a, remember?). If you want more proof of asymptotic behavior, you'll notice that the derivative goes to 0 as t goes to infinity, as well. I still haven't seen an exponential, though.

    As for the a = F/m business again, keep in mind that a = F/m is a purely classical expression! Since we've moved into the realm of special relativity now, a is not equal to F/m! It really has nothing to do with the entire problem. If you want to find a, differentiate the expression you got for v.

    I looked at the graph you posted (http://uwoshkosh.elitefitness.us/v vs t graph.htm), and it seems to be a graph of velocity versus time. You already have an expression of velocity in terms of time, so I'm really not sure what more you could ask for if you're going to draw velocity versus time.

    cookiemonster
     
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