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I Relativistic mass and gravitational potential

  1. Feb 13, 2019 at 9:31 AM #1
    Hello everyone,

    Any object has a gravitational potential energy as a function of the distance from the earth (R). Does this energy depend only on the rest mass of the object; or one must take into account it's relativistic mass?

    In other words, if we imagine two identical bullets on the top of a building with different initial velocities ( towards the earth); will they have the same potential energy?
  2. jcsd
  3. Feb 13, 2019 at 9:38 AM #2


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    You cannot apply concepts of Newtonian gravity on special relativity.
  4. Feb 13, 2019 at 1:28 PM #3


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    Gravitational potential energy is a Newtonian concept. General relativity has various (more than one) concept of system energy, but there's no way to split the system energy up into kinetic and potential energy.

    As Orodruin mentions, special relativity is not compatible with Newtonian gravity, this was one of the primary reasons that prompted Einstein to develop General Relativity. So this means the question as phrased cannot be answered, because one needs General Relativity to attempt an answer it, but General Relativity does not have the concept of "potential energy" mentioned in the question.

    In the case where the "bullets" are assumed to be test masses, one can compute a meaningful notion the total energy of each of the bullets. If the bullets are free-falling (following geodesic paths), there is a suitable notion of "total energy" that remains constant along the entire path of the bullet, a notion called "The Energy at Infinity" by MTW in their textbook "Gravitation". From this, one can compute the impact velocity of the bullets relative to the ground as measured by an observer on the ground. This requires a working knowledge of General Relativity to follow the details of the calculation, however.

    In the more general case where the bullets are not test masses, the above approach would not work. Getting into further details about this case would be overly complex for the point I'm trying to make.
  5. Feb 13, 2019 at 3:15 PM #4


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    Let me add an online link to the formulas that I mentioned in my last post: https://www.fourmilab.ch/gravitation/orbits/. I happen to know that these formulas are essentially identical to the ones in MTW's text, "Gravitation", but the interested reader who has access to the text might want to study the original. The web-page is more convenient and accessible, though.

    The case given in the webpage includes cases where the velocity is not central, it applies to orbiting test masses. To get the equations for when the motion is purely radial, set the angular momentum L to zero.

    Then we wind up with the expression in geometric units assuming L=0 as:

    $$\left( \frac{dr}{d\tau} \right) ^2 + 1-\frac{2M}{r} = E^2$$

    where ##E^2## is a constant of motion related to energy, r is the Schwarzschild radial coordinate, ##\tau## is proper time, and M is the mass of the massive body in geometric units. From memory, M for the sun is about 1500 meters.

    Note that ##dr / d\tau## is related to velocity, but further work would be needed to convert it into the velocity as measured by a co-located inertial observer. dr is not the same as distance measured by a co-located inertial observer due to the metric coefficients of the Schwarzschild metric, and ##d\tau## is the change in proper time, not the change in coordinate time.

    Basically, we can use "conservation of energy" arguments in the case of a test mass falling into a massive body to come up with meaningful equations, but the details of how we do this are different in General relativity than they are in Newtonian mechanics. The simple approach outlined above won't work if the bodies are not test masses, one way of describing the issue is to say that gravitational radiation is emitted when the infalling body is not a test mass, and the problem becomes significantly more complex.
  6. Feb 13, 2019 at 3:41 PM #5
    Thank you for your response.

    I understand. Lets forget the notion of the potential energy and return to the example of the 2 bullets.

    They have the same rest mass and the same distance from earth. We know the initial velocities of the bullets. We can calculate their relativistic masses: m = m0/sqrt(1 - v2), and their energies: E=mc2

    We are also able to measure the final velocities of the bullets at the bottom of the tower; and calculate corresponding relativistic masses and energies.

    Here is the question: if we calculate the energy differences between the top and the bottom of the tower, will we obtain the same value for the 2 bullets (with different initial velocities)?
  7. Feb 13, 2019 at 4:45 PM #6


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    The energy measured by a observer at constant altitude is ##g_{\mu\nu}U^\mu P^\nu=m_0g_{\mu\nu}U^\mu V^\nu##, where ##U^\mu## is the four-velocity of an observer at constant altitude, and ##P^\nu## is the four-velocity of the bullet, which is equal to the its rest mass, ##m_0##, times its four velocity ##V^\nu##. Since, in Schwarzschild coordinates, ##U^t=1/\sqrt{g_{tt}}## is the only non-zero component of ##U## and the metric is diagonal, the measured energy is ##m_0\sqrt{g_{tt}}V^t##.

    Of course, ##g_{tt}=1-R_s/r##. And for a geodesic ##V^t=dt/d\tau=E_\infty/(1-R_s/r)##, where ##E_\infty## is the energy per unit mass at infinity (see 7.43 in Sean Carroll's GR lecture notes, for example - he uses just ##E## for what I've called ##E_\infty##, but you've used that for something else). Thus the measured energy is ##m_0E_\infty/\sqrt{1-R_s/r}## and the energy difference you are asking about is $$\Delta E=m_0E_\infty\left(\frac 1{\sqrt{1-R_s/r_1}}-\frac 1{\sqrt{1-R_s/r_2}}\right)$$where the bullet has fallen from ##r_2## to ##r_1##.

    The initial velocity of the bullet is encoded in ##E_\infty##. The ##\gamma## factor measured by the hovering observer at ##r_2## is actually given by ##\gamma=g_{\mu\nu}U^\mu V^{\nu}=E_\infty/\sqrt{1-R_s/r_2}##, so we can write the energy difference as$$\Delta E=m_0\gamma\left(\frac{\sqrt{1-R_s/r_2}}{\sqrt{1-R_s/r_1}}-1\right)$$I would strongly resist using the term "relativistic mass" but, assuming I haven't mucked that up, it would appear that the "energy gain" is proportional to ##\gamma m_0##, yes.
    Last edited: Feb 13, 2019 at 5:40 PM
  8. Feb 14, 2019 at 12:28 AM #7
    Thank you so much for your answer and your time. It is so complete that I dont have any further questions.
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