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The hyperbolic tangent function and physics

  1. Sep 22, 2009 #1
    Kind of an odd question, but here goes: can anyone offer any examples of physical systems that include the use of the hyperbolic tangent function in their mathematical solution?
     
  2. jcsd
  3. Sep 22, 2009 #2

    sylas

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    Velocity in special relativity with a constant proper acceleration.

    If the proper acceleration is "a" (this is the acceleration experienced by the accelerated observer), and the proper time is "τ" (this is the time on the clock of the accelerated observer), and we want to find the velocity "v" and location "x" of the accelerated particle in the inertial frame where it is at rest at time 0, then we get parametric equations
    [tex]\begin{align*}
    t &= \frac {c}{a} \sinh \frac {a\tau}{c} \\
    x &= \frac {c^2}{a} \cosh \frac {a\tau}{c} \\
    v &= \frac{dx}{dt} \\
    &= c \tanh \frac{a\tau}{c}\end{align*}[/tex]​

    See also the proof in [post=1784759]msg #13[/post] and following of thread "Questions about acceleration in SR"

    Cheers -- sylas
     
  4. Sep 22, 2009 #3
    Thank you for your reply! I haven't had the pleasure of studying relativity, but if I understand what the equations are saying it is that with respect to an inertial observer, if the acceleration times the time is small, then [tex]c \tanh \frac{a\tau}{c}[/tex] (remembering the Taylor series expansion of tanh) will just reduce to [tex] \frac{a\tau}{c}[/tex] and velocity will equal acceleration times time, just as in standard Newtonian physics. If the fraction is large, however, tanh might be something like 0.99999, which shows that the velocity of the particle will asymptotically approach c.
     
  5. Sep 22, 2009 #4

    sylas

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    Exactly right. Note that the parameter inside the tanh refers to "proper" time, which means the time on the clock being carried by the accelerating observer. This is also running slower and slower from the perspective of the outside observer; so for the outside observer it takes much longer for an accelerating spaceship to get up near to light speed. But while the speeds are low, it all reduces to the normal classical case.

    Cheers -- sylas
     
  6. Sep 22, 2009 #5
  7. Sep 22, 2009 #6

    arildno

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    Dearly Missed

    For ideal fluids (i.e, without friction) the speed c for a monochromatic, linearized wave-train (wave-number k) over a flat bottom, depth h, is given by:
    [tex]c=\sqrt{gh}\sqrt{\frac{Tanh(\kappa)}{\kappa}}, \kappa=kh[/tex]

    In the shallow water limit, [itex]\kappa\to{0}[/itex], we get the non-disbersive wave speed [tex]c=\sqrt{gh}[/tex], whereas at deep water, [itex]h\to\infty[/itex], we get [itex]c=\sqrt{\frac{g}{k}}[/itex]
     
    Last edited: Sep 22, 2009
  8. Sep 22, 2009 #7
    Hyperbolic tangent can also pop up in statistical mechanics.
     
  9. Sep 22, 2009 #8
    The "tanh" function occurs when analyzing an emitter coupled differential pair of bjt devices used in the front end of an op amp, without emitter degeneration resistors. The functional math relationship between the terminal voltages (inverting & non-inverting op amp inputs, which are the bases of the ec pair), and the collector currents (output of 1st stage, input to 2nd stage), is a tanh finction. Namely

    Ic = Ie*tanh(Vd/(2*Vt)).

    Ic = collector current, Ie = emitter current, Vd = differential voltage across op amp 2 input terminals, Vt = thermal voltage = nkT/q.

    I've used op amps for 30 yrs., as well as discrete amp stages, so this came to mind immediately. In real devices, there is always some emitter degeneration resistance. Even if none is intentionally added, the emitter region bulk silicon possesses inherent resistance. The tanh is an "S" shaped curve. The emitter resistance flattens out the S making the slope shallower. Real world ec pairs approach a tanh function, but are not exactly a tanh. When using discrete bjt parts, I always add discrete emitter degeneration resistors. This skews the tanh curves reducing the overall gain.

    Claude
     
    Last edited: Sep 22, 2009
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