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I Relativistic momentum

  1. Apr 5, 2017 #1
    I tried to get a relativistically correct expression of ##\int_0^t\frac{dp}{ds}dt## similar to the derivation of relativistic energy expression but I got a result which is not defined:
    $$\int_0^t\frac{dp}{ds}dt=\int_0^v\frac{d\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{v}$$
    $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v-\int_0^v\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}d\left(\frac{1}{v}\right)$$
    $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v-\int_0^v\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\frac{-1}{v^2}dv$$
    $$=\left|\frac{m}{\sqrt{1-\frac{v^2}{c^2}}}\right|_0^v+\int_0^v\frac{m}{v\sqrt{1-\frac{v^2}{c^2}}}dv$$
    Now, the second term is:
    $$I=\int_0^v\frac{m}{v\sqrt{1-\frac{v^2}{c^2}}}dv$$
    $$=mc\int_0^v\frac{dv}{v\sqrt{c^2-v^2}}$$
    Putting ##v=c\sin{x}## and ##dv=c\cos{x}dx##, we get:
    $$I=mc\int_0^v\frac{c\cos{x}dx}{c^2\sin{x}\cos{x}}$$
    $$=m\int_0^v\csc{x}dx$$
    $$=m|\log{|\csc{x}-\cot{x}|}|_0^v$$
    $$=m\left|\log{\left|\csc{sin^{-1}\left(\frac{v}{c}\right)}-\cot{sin^{-1}\left(\frac{v}{c}\right)}\right|}\right|_0^v$$
    $$=m\left|\log{\left|\frac{c-c \sqrt{1-\frac{v^2}{c^2}}}{v}\right|}\right|_0^v$$
    But, this expression is not defined at the lower limit ##v=0##. Have I done something wrong in the maths or does the relativistically correct expression of ##\int_0^t\frac{dp}{ds}dt## not exist?
     
    Last edited: Apr 5, 2017
  2. jcsd
  3. Apr 5, 2017 #2

    haushofer

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    Maybe it's just me, but I have absolutely no clue what your expressions mean. Can you give any reference?
     
  4. Apr 5, 2017 #3
    I'm just trying to get a relativistically correct expression of ##\int_0^t\frac{dp}{ds}dt##
    The non-relativistic formula is:
    $$\int_{t_1}^{t_2}\frac{dp}{ds}dt=\int_{v_1}^{v_2}m\frac{dv}{v}$$
    $$=m(\ln{v_2}-\ln{v_1})$$
    Note that this is similar to the work-energy equation:
    $$\int_{s_1}^{s_2}\frac{dp}{dt}ds=\frac{1}{2}m(v_2^2-v_1^2)$$
    except the qunatity ##\frac{dp}{ds}## is used in place of ##\frac{dp}{dt}##, and the integration is done wrt time instead of distance.
    So, I think the relativistically correct ##\int_0^t\frac{dp}{ds}dt## should be similar to the mass-energy equivalence relation. But, if my maths is correct, then the relativistically correct expression is not defined at the lower limit ##0##.
     
  5. Apr 5, 2017 #4

    vanhees71

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    When using non-common notation, you have to define your quantities. What, for instance, is ##s##?

    In the following I use natural units with ##c=1##.

    The standard derivation for energy and momentum of a classical massive particle is to introduce proper time ##\mathrm{d} \tau^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}## (along the worldline of the particle; ##(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)##). Then the energy-momentum four-vector is defined as
    $$p^{\mu} = m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},$$
    where ##m## is the invariant mass of the particle (and there is no other mass in SR necessary; for this issue, please search for "relativistic mass" in the forums).

    The four-momentum by definition fulfills the on-shell condition
    $$p_{\mu} p^{\mu}=m^2.$$
    A more convincing argument is to use Noether's theorem to derive the action of the free particle, which leads to
    $$A=-m\int \mathrm{d} t \sqrt{\eta_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$
    The generators of space-time translations is by definition momentum, which leads to the same result as the hand-waving short-cut argument given above.
     
  6. Apr 5, 2017 #5
    I'm sorry I don't understand these things. What was this about? And ##s## is displacement. I thought displacement was commonly given the symbol ##s##.
     
  7. Apr 5, 2017 #6
    But displacement is a vector. Does ##\frac{{dp}}{{ds}}## means something like ##\left( {\nabla _s \cdot p^T } \right)^T## or is your calculation limited to the one-dimensional case?
     
  8. Apr 5, 2017 #7

    Mister T

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    Note that this is also undefined when when ##v_1=0##. So your problem may be that you are setting the lower limit on the integral to zero instead of leaving it as a variable.
     
  9. Apr 5, 2017 #8

    mfb

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    You cannot directly plug in v=0 in the result, but you can evaluate it for small positive v. Expanding around v=0+ and setting c=1, we get

    $$\log{\left|\frac{1- \sqrt{1-v^2}}{v}\right|} \approx \log \left| \frac {\frac{v^2}{2}}{v}\right| = \log \left| \frac{v}{2}\right|$$
     
  10. Apr 5, 2017 #9
    I thought that since ##m\ln v## looks something analogous to energy, so I thought setting ##v=0## in the relativistically correct expression of ##m\ln v## would give me something analoguous to the mass-energy equivalence. Because, suppose name the quantitity ##\frac{dp}{ds}=force'##. And, we call the quantity, ##\int_0^tforce'dt=energy'##. So, if a variable ##force'## acts on a body for sometime starting from rest, and we want to calculate change in ##energy'##, then I think since the body starts from rest, so the initial velocity should be set to ##0##.
     
    Last edited: Apr 5, 2017
  11. Apr 5, 2017 #10
    So, can we take the limit of the second integral as lower limit tends to zero and say that:
    $$\int_0^t\frac{dp}{ds}dt=\frac{m}{\sqrt{1-v^2}}-m+m\log{\frac{v}{2}}$$
    And,why are you putting ##c=1##?
     
  12. Apr 5, 2017 #11

    robphy

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    One feature of kinetic energy is that it doesn't depend on direction. So, for example, if you reverse the direction of the velocity vector, the kinetic energy is unchanged.
    In your "non-relativistic version of your alternative-universe kinetic-energy" do you mean ##m\ln |v|##?

    Can you restore the units to "##m\ln v##"? The argument of ##\ln()## has to be dimensionless.
    The result of the definite integral could be written as ##m\ln\frac{v2}{v1}##.

    Of course, just because you construct something doesn't mean that it can fly.
    It's also peculiar to ask for a "relativistically correct expression" of something that might not be meaningful in relativity.
    You could translate your definite integral into rapidities [using hyperbolic trig functions where ##v=\tanh\theta##,
    then do the calculations, then try to interpret.
    ##\displaystyle
    \int_0^t\frac{dp}{ds}dt=\int_0^v\frac{d\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{v}
    =\int_0^v\frac{d(m\sinh\theta)}{\tanh\theta}=\int_0^\theta\frac{m\cosh\theta\ d\theta}{\tanh\theta}##
    You could feed your integral into (say) WolframAlpha
    http://www.wolframalpha.com/input/?i=m*int(+cosh(x)/tanh(x),x)
     
  13. Apr 5, 2017 #12
    I don't know what ##\left( {\nabla _s \cdot p^T } \right)^T## means. So, I think my calculation is limited to one dimension. I've a small book which used this approach to derive mass-energy equivalence. I'm just doing the same:
    $$KE=\int_0^sFds=\int_0^vvd(\gamma mv)$$
    $$=\int_0^vvd\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)$$
    $$=\frac{mv^2}{\sqrt{1-\frac{v^2}{c^2}}}-m\int_0^v\frac{vdv}{\sqrt{1-\frac{v^2}{c^2}}}$$
    $$=\frac{mc^2}{\sqrt{-\frac{v^2}{c^2}}}+\left|mc^2\sqrt{1-\frac{v^2}{c^2}}\right|_0^v$$
    $$=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2$$
    $$KE=(\gamma -1)mc^2$$
    Now, ##\gamma mc^2## is interpreted as total energy, so ##mc^2## is the rest energy. I tried to do the same thing for deriving ##\int_0^t\frac{dp}{ds}dt##.
     
    Last edited: Apr 5, 2017
  14. Apr 5, 2017 #13
  15. Apr 5, 2017 #14
  16. Apr 5, 2017 #15
    Nope.. It's exactly the same integral ## \int_0^t\frac{dp}{ds}dt=\int_0^v\frac{d\left(\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{v} ##
     
  17. Apr 6, 2017 #16

    vanhees71

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    Define it! I've never heard about it. Usually ##s=c \tau## is the invariant Lorentz-pathlength defined by
    $$s=\int_{t_0}^t \mathrm{d} t' \sqrt{\eta_{\mu \nu} \dot{x}^{\mu}(t') \dot{x}^{\nu}(t'))}.$$
     
  18. Apr 6, 2017 #17

    mfb

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    It simplifies the equations without changing the physics. There is no need to have different units for time and length in special relativity. A unit system where c=1 is the natural choice, as it simplifies all equations.
     
  19. Apr 6, 2017 #18

    Mister T

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    To me the analogous derivation starts with ##p=\int_0^tFdt##. The expression you're starting with makes no sense to me. How did you come up with it? What does it equal? It doesn't have dimensions of energy or momentum?
     
    Last edited: Apr 6, 2017
  20. Apr 6, 2017 #19
    It is the gradient of p(s) which is not a scalar but a matrix in more than one dimension.

    In one dimension ##\frac{dp}{ds}## is ##\frac{F}{v}## and puzzled fish already posted you the wolframalpha solution for the relativistic case.
     
  21. Apr 6, 2017 #20
    And why do you think the non-relativistic formula ##m(\ln{v_2}-\ln{v_1})## is defined at the lower limit ##v_1=0##?
     
  22. Apr 6, 2017 #21
    I'm just saying that ##\frac{dp}{ds}## can also be a measure of how hard you push or pull. If you push something hard, it's momentum will change by a larger amount in a small displacement. If you push something with feeble effort, then, for the same change in momentum, the body will have to move a larger distance. So, clearly ##\frac{dp}{ds}## is also a measure of force. If you use ##\frac{dp}{ds}## as force F, then ##\int_0^sFds## will be the change in momentum. And ##\int_0^tFdt## will be something similar to energy. This energy has the dimensions of mass but is not the same as mass.
     
  23. Apr 6, 2017 #22

    Mister T

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    Okay. But do you know why rectilinear acceleration is defined as ##\frac{dv}{dt}## instead of ##\frac{dv}{ds}##?

    Not in the most important way: It will not be a conserved quantity.
     
  24. Apr 7, 2017 #23

    vanhees71

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    It's hopeless! If you don't give a clear definition of ##s##, there's no way to clarify what you want to derive. If you want to derive something like the "work-energy theorem" in Newtonian physics, you have to start from a relativistic dynamical equation. That's best done in terms of the manifestly covariant formulation. For the only well-defined case of a single particle in some given external field the equations of motion takes the form
    $$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau} = F^{\mu}(x,p), \quad p^{\mu}=m \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.$$
    Here ##\tau## is the proper time of the particle along its trajectory and ##m## its invariant mass. Due to the on-shell constraint (discussed in some posting above in this thread) you have the contraint
    $$p_{\mu} \frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=p_{\mu} F^{\mu}=0,$$
    and that's already the work-energy theorem in relativistic form, as can be seen by splitting it into spatial and temporal components.
     
  25. Apr 7, 2017 #24

    Mister T

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    ##s## is, I gather from the context, the position coordinate in one dimension.

    (There is at least one author of an introductory physics textbook, Randy Knight, who does it this way. Thus when the position axis is horizontal he'll let ##s=x##, and when it's vertical ##s=y##. Note that the OP isn't going beyond freshman level physics with his questions or responses.)
     
  26. Apr 7, 2017 #25

    berkeman

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    Thread closed temporarily for Moderation...
     
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