Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy of a moving mass

  1. Feb 23, 2007 #1
    My first question
    I want to know the amount of energy in a moving mass. Using the equation E=MC^2/sqrt(1-(V^2/C^2)) (Which is the full form of E=mc^2 i think) i should be able to find out the amount of energy there is of a mass plus its movement altogether. So say a person sees a 1kg object moving an .9c in a straight line in space. Now doing the calculation out -
    1*C^2/sqrt(1-((.9*C)^2/C^2) which = 2.064741605e17 joules. Is this the correct answer??

    Once this is answered i will ask my second question.
  2. jcsd
  3. Feb 23, 2007 #2
    I'm always telling people this, but if your initial data only has one significant figure (a la "1kg"), it's misleading to pretend to express 10 significant figures in your result. But sure, the formula looks reasonable, what's the second question?
  4. Feb 23, 2007 #3
    err.... according to me the answer should be:9.37526713 * 10^16 Joules.
    Also this magnitude is equivalent to the total energy of the body,ie if the body is converted to energy while it is moving,then the energy would be equivalent to the above figures.

    in the above case the person will see the person's kinetic energy to be


    I hope this helped
    Last edited: Feb 23, 2007
  5. Feb 24, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

  6. Feb 24, 2007 #5
    oh sry calculation mistake in squaring 0.9c.Sry sry
  7. Feb 24, 2007 #6
    Ok then. My second question-

    From the person's reference frame watching the object moving at .9c, could he say that the mass moving at .9c has a greater gravitational field because it has more energy?? Does this make any sense?
  8. Feb 24, 2007 #7


    User Avatar
    Gold Member

    GR states that gravity is produced by energy and momentum, not just mass.
  9. Feb 24, 2007 #8
    So it is safe to say that the person watching the object moving at .9c could say it has a greater gravitational field due to an increase of momentum.... interesting. Say the mass moving at .9c is in deep dark space has no frame of reference it can depict from to get its true speed. To that mass it could be considered not moving at all. From the mass' reference frame could you say that it has the same gravitational as before when it wasn't moving? I hope this makes sense.
  10. Feb 24, 2007 #9


    User Avatar
    Science Advisor

    Fundamental error here: there is no such thing as "true speed"! All speed is relative.
    That makes perfect sense because from the mass' reference frame is still isn't moving!
    Last edited by a moderator: Feb 24, 2007
  11. Feb 24, 2007 #10


    User Avatar
    Staff Emeritus
    Science Advisor

    A lot depends on what you mean by the term "gravitational field".

    Your question has been discussed previously - see for instance


    You might also want to check out, for background, the simpler question of what the electric field of a moving charge looks like.

    See for instance http://www.phys.ufl.edu/~rfield/PHY2061/images/relativity_15.pdf

    and if you want more detail on this related topic, try looking at different page "numbers" than 15.

    The short answer for a moving charge is that the parallel component of the electric field is not affected by motion, while the transverse components are boosted. One also has additional terms which are known as the "magnetic field".

    In Newtonian physics, the term "gravitational field" is usually implied to mean a force. Unfortunately, there is no way to directly measure this "force" unless one has a gravitationally neutral test particle as a reference. And there isn't any such thing as a gravitationally neutral reference.

    Just as GR says that it is energy, not mass, which causes gravity, GR also says that gravity is not a force, but a curvature of space-time. GR also offers a recipe to convert this curvature back into forces for slowly moving objects. Unfortunately, this recipe fails for quickly moving objects - its only an approximation.

    Under most circumstances, this curvature, mathematically described by the Riemann tensor, can by physically interpreted as a "tidal force". (This identification is actually still not quite perfect, but it's reasonably close. The identification works perfectly only when the observer measuring the tidal force is free-falling).

    So, while there turns out not to be any unambiguous way to determine the "force" of gravity of a moving object (I have a rather long-running argument with another poster on this point, BTW), it is quite possible (and in the context of GR, even natural) to talk about the "gravitational field" in terms of the Riemann tensor, or in terms of tidal forces.

    A tidal force is just the force/unit length caused by the change in gravity. For instance, if you stand in the Earth's gravity field, your head is further away from the center of the planet than your feet. This causes a slight, but measurable difference in an accelerometer mounted at your feet and and a different one mounted on your head. This difference, per unit length, is just a tidal stretching force. While it is very small in magnitude, it can be measured with a sensitive enough instrument (such as a Forward mass detector).

    Now that we've discussed the background, we can talk about the actual result of the gravitational tidal force generated by a moving mass. The result is not intuitive - it says that if you directly approach or move away from a massive object, there is no effect on the tidal force(s) at all.

    If you move tangentially to the object, the tidal force(s) increases.

    Thus, for instance, an observer falling directly towards a black hole (following a geodesic) at nearly light-speed will not see any change in the tidal force from that black hole. (This is in the textbooks, MTW's "Gravitation" discusses this).

    However, an observer moving in such a manner as to orbit a black hole at nearly light speed (still following a geodesic) will see an increased tidal force from the black hole. (This example is not in the textbooks, but can be worked out with enough knowledge and effort).
    Last edited by a moderator: Apr 22, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook