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Mass Dilation and Velocity Question

  1. Jun 9, 2006 #1
    Hi, we just finished talking about relativity in my physics class, and using its equations we were supposed to find a fuel efficient speed limit for our highways if the speed of light were 30m/s. This course assumes no calculus, but since i know some differential calculus I get the feeling calculus can help.

    I'm considering a graph of mass versus x where

    x=v/c where v=object's velocity and c= speed of light

    using the function m(x)=(1-x^2)^(-1/2)

    I get the graph shown, and I am wondering if the point where the derivative is 1 would be what I'm looking for. I say this because before that the mass is increasing pretty slowly,and then after that the mass increases drastically while the velocity does not increase much. Secondly if the graph were rotated as shown, the tangent line I'm considering would be horizontal, ie a critical point.

    The course isn't calculus based, and I think we were just supposed to estimate, or something obvious is slipping by me, but I'd just like to know what anyone thinks. My math teachers told me to try a forum since they don't really know...my physics teacher, although quite good, is not well versed in calculus and can just agree that generally what I'm saying makes sense. I was hoping someone here might be able to give me a hand in this, and if I'm being vague in the problem let me know I'll try and explain myself better. For the record, the extent of my calculus knowledge is basic differential calculus thanks to our awesome ministry of education.

    So far I've differentiated the equation and solved for f'(x)=1,and I got a value somewhere in the neighbourhood of x=0.5 (I don't remember exactly and don't have my work with me right now), which seems like a reasonable value.

    Thanks in advance for any insights.
     

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  3. Jun 9, 2006 #2

    Andrew Mason

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    You will have to decide what the criteria is for fuel efficiency. The most fuel efficient speed would be the slowest speed that is reasonable. What is reasonable?

    Why f'(x)=1? What is the significance of that?

    AM
     
  4. Jun 10, 2006 #3
    Part of the problem is that the question does not define what is most fuel efficient, so part of my problem is whether or not my definition makes sense.

    The reason I chose f'(x)=1 is that the angle between the tangent line and the x-axis is 45 degrees, so before that it seems to me the velocity would be increasing more rapidly than the mass, and after that it seems that the mass would be increasing more rapidly than the velocity. So at f'(x)=1 it seems to me the two variables are increasing at the same rate, neither outweighing the other.

    edit: now that I think of it, I realize that i said my function was m(x), so the derivative I'm talking about is actually m'(x). Since I didn't define a function f, f'(x) would be totally meaningless :smile:

    Thanks!
     
    Last edited: Jun 10, 2006
  5. Jun 10, 2006 #4

    Hootenanny

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    Make sense to me. However, if you have not been given a defined efficency (or a formulae for calculating it) you should state what you are taking to be the most efficent speed.
    Solving the differential I get a value of 0.564 (3sf) and the other two solutions are complex.
     
  6. Jun 10, 2006 #5

    Andrew Mason

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    But energy is proportional to mv^2.

    Also, since light from oncoming vehicles would travel to you at the speed of 30 m/sec, if the vehicle is traveling toward you at 15 m/sec, you will be having a lot of traffic issues.

    AM
     
  7. Jun 10, 2006 #6
    When I was asked to define efficiency I see part of my problem. Efficiency is definitly a ratio of energy out to energy in, and I guess I assumed that it meant efficient in a colloquial sense since it would be impossible to predict that from a classical view. Now its occurring to me to use the relativistic equations for the kinetic energy....

    relativistic total energy= t = (mc^2)/sqrt(1-v^2/c^2)


    Then, the limit of t as v approaches c is infinity. Since the relativistic kinetic energy is t minus mc^2, the kinetic energy will also approach infinity. If efficiency is energy out over energy in then as v approaches c the efficiency should be infinite regardless of the energy in, which I should mention I can't think of what it would be.

    So it looks to me like using the classical equation for efficiency might be out.

    I'm totally new to this relativity stuff, and the only time I've applied calculus to the physics we do in class is on my own so I feel like I'm in doubly over my head here.

    I'll consider traffic issues once I deal with the math, because somehow that seems like it might end up being totally subjective since the beings in this imaginary universe aren't humans so maybe their reaction time is different and blah blah blah :tongue2:

    Anyway my head hurts now... I'll work on this some more and post if I come up with anything that might be helpful, but in the meantime feel free to add some insights!

    By the way I've already handed the homework in, all this stuff I've been doing is after the fact just because its interesting. Since the course isn't calculus based I just estimated a value that looked good with a graphing calculator. But I'd like to try and solve this mathematically just for the heck of it.
     
  8. Jun 10, 2006 #7

    Hootenanny

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    I would recommend using this form of Einstein's equation and setting the rest mass to 1kg;

    [tex]E_{k} = m_{0}c^2 \left( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1 \right)[/tex]
     
  9. Jun 11, 2006 #8
    Thanks, I'll try out some calculations with it.
     
  10. Jun 13, 2006 #9
    So I took that equation for kinetic energy, and then the equation for total energy, and i divided the total by the kinetic by the total as energy ouput divided by energy input (ie efficiency). I found that the graph ended up being constantly decreasing with no critical points for 0<v<c. So apparently the most efficient speed is 0m/s, but its not a practical speed if you plan on going anywhere.

    I don't think they meant efficient in the exact sense, in a matter of fact my teacher mentioned that the point of the thought experiment was to think about relativity and not necessarily come up with any real math. So thanks for your help, but somehow I get the feeling I might be beating a dead horse. At least it was fun :smile:
     
    Last edited: Jun 13, 2006
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