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I know that in relativistic condition the increase in kinetic energy of an object increases its relativistic mass as $$m=\frac{m_0}{(1-v^2/c^2)^{1/2}},$$ and mass is another form of energy.

So my question is if that object have same amount of potential energy instead of kinetic energy then can we say that its relativistic mass is increased?

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The short-short answer: "Potential energy always belongs to system rather than to a single object, and the system's mass is increased when you add potential energy to the system but the component parts do not change their masses."

I know, we often say that if you raise a book off the lab bench the book gains potential energy (attributing the energy to the book). But that's a short hand, because the motion of the book alone is not sufficient for the energy to change: it must make that motion in the presence of the planet. If you take the book and the bench into deep space and perform the same action ("lift" the book 1 meter "above" the bench) nothing special happens.


As an aside: the term "relativistic mass" isn't necessary here as the invariant mass of the system is increased. You'll find that despite its popularity in pop-sci sources and secondary-school text-books most physicists in sub-fields that use relativity all the time organize the discipline with a different set of definitions that don't include "relativistic mass" and instead recognize only one mass (called the "invariant mass" when you need to be painfully clear). Personally, I feel very strongly that calling $\gamma m$ the "relativistic mass" only encourages sloppy thinking about relativity and would encourage you to find a source that doesn't use it.

In that context it is worth noting that most of the mass of ordinary matter is binding energy due to the strong force, so most of the "ordinary mass" around you is of exactly the kind you are asking about.

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    $\begingroup$ Nicely crafted aside. I'll probably copy it someday. Maybe I'll remember to give you credit. ;) $\endgroup$ – garyp Aug 16 '16 at 17:37
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    $\begingroup$ i understood that the increase in potential energy increases the mass of the whole system ,,but doesn't that mean the increase in mass of all components???how is it possible to be increased the mass of whole system without increase in mass of its component parts?? $\endgroup$ – Sagar Timalsina Sep 19 '16 at 5:45
  • $\begingroup$ @SagarTimalsina Something often overlooked in listing the ways that Einstein's world differs from that of Galileo and Newton is that the mass of a system is not necessarily the sum of the masses of the parts. The term "mass deficit" is mostly used in the context of nuclear decay, fission or fusion but it applies all the time to everything. Mathematically you get the mass of a system by adding the four-moment of the parts (including fields) and then finding the norm of the resulting sum. $\endgroup$ – dmckee --- ex-moderator kitten Sep 19 '16 at 17:14
  • $\begingroup$ so,,if an object is at surface of earth (or other planet)or it is lifted at some height ,the mass of that object will not be changed .but the amount of mass equivalent to the energy that we gave to lift the object will be added to the whole system. am i right? $\endgroup$ – Sagar Timalsina Sep 19 '16 at 18:16
  • $\begingroup$ @SagarTimalsina I think the answer to your question depends on whether you consider the actor (you) to be part of the system. If you are not part of the system then yes, you increase the mass of the system, because you are spending energy on it. On the other hand, if you are part of the system, then the answer is no--you only participated in moving energy around that was already there. Caveat: I am not a physicist. $\endgroup$ – Mark McKenna Dec 27 '16 at 5:50
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To add to Statics' answer. You can think of potential energy due to other than the gravitational interaction (e.g. electromagnetic) as the energy of the field itself. Not only this additional mass exists but you can also know where in the system it is located.

Edit: A word of caution. If you calculate the energy of the field of a point charge, you will get infinity. That's because EM fields can act on the charge they were produced by. It makes sense then: for example the electrostatic potential is $~ \frac{1}{r} = \frac{1}{0} = \infty$. This is not very relevant to reality, because to model electrons we need quantum mechanics, but you can get limited success with modelling them classically as balls of finite radius instead of point charges. I believe, the Feynman lectures have a discussion about that.

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    $\begingroup$ This approach is, of course, viable, but it will result in a rather strange answer in the case the OP is envisioning. You'd subtract the initial energy density from the final one and find the change in energy distributed largerly in the region of the motion but to a lesser extent throughout space. $\endgroup$ – dmckee --- ex-moderator kitten Aug 17 '16 at 3:18
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Disregarding your choice of word in your question i.e. "relativistic", I can go ahead and answer. Energy and mass are the same thing, proportioned by the constant speed of light. To visualize this you could rewrite the formulae $$E = mc^2$$ to

$$m = E / c^2 = (m_0c^2 + V) / c^2 = m_0 + V/c^2$$

Where V is the potential energy. Of course you can incorporate your "relativistic" mass as a function of speed ( as you have written in your question) and it will still be true. I just skipped it here to simplify.

The potential energy is not limited to type of potential energy... It could be gravitational energy, or a compressed spring. In all cases the reference of frame need to be considered and not violated.

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  • $\begingroup$ MathJax is your friend. Here’s a MathJax tutorial $\endgroup$ – garyp Aug 16 '16 at 17:35
  • $\begingroup$ Haha I know... I'm at a conference and figured I reply quickly on my phone aaaand its quite simple stuff.. I need to read up on mathjax... Thanks $\endgroup$ – niCk cAMel Aug 16 '16 at 17:42
  • $\begingroup$ There! Its now mathjaxed ;) $\endgroup$ – niCk cAMel Aug 16 '16 at 19:10
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I think that the apparent mass increase is a property of the frame of reference. If there are two inertial frames and their relative velocity is $v$ and the $\gamma$ parameter is $\frac{1}{\sqrt{1-(v/c)^2}}$ then the mass $m_0$ in any of the rest frame is perceived as $\gamma m_0$ in the in other frame.

It should be noted that this type of argument is very crude and not considered accurate in modern terminology but for the sake of argument it will work.

Rise/fall of potential energy is also relative to the reference level, Simply by changing the reference level the potential energy of an object can go up or down. Hence in my opinion the change in potential energy can not increase the so called "relativistic mass".

Also from another point of view since $v=0 ,\ \gamma=1$.

hope this will help

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In special relativity the concept of potential is not defined. In general relativity potential is related to gravitational fields. To describe a potential of a massive star one could use the Schwarzschild metric.

Assuming that the object in a potential field is not moving the total energy $E$, which should stay conserved, is its rest mass $m_0$ plus the potential plus the kinetic energy. By defining the relativistic mass as the total energy of an object at rest one could see the dependence on the potential.

Derivation

The Schwarzschild metric can be writtten as $$\rm{d}s^2=-(1-\frac{2M}{r})dt^2 + (1-\frac{2M}{r})^{-1}dr^2 - r^2(d\theta^2+\sin^2\theta d\phi^2)$$ with $M$ the mass of the star, $r$ the distance from the star and angles $\theta$ and $\phi$. The metric is independent of time, hence there is a Killing vector $K^\mu=(1,0,0,0)$ related to the momentum dual vector $p_\mu$ such that $K^\mu \cdot p_\mu=const=p_0 =p_t=E$ - the total conserved energy of the object as seen at infinity.

The momentum mass relation reads $p^\mu \cdot p_\mu=-m^2_0$. Or in components: $$p^\mu \cdot p_\mu = p_\nu \cdot p_\mu\, g^{\mu\nu}=p_t p_t g^{tt} + p_r p_r g^{rr} + p_\theta p_\theta g^{\theta\theta} +p_\phi p_\phi g^{\phi\phi}=-m^2_0.$$Let the object be at rest: $p_r=p_\theta=p_\phi=0$. Hence one obtains $$p_t p_t g^{tt}=-E^2(1-\frac{2M}{r})^{-1}=-m^2_0.$$

$E$ is the total energy of the object at rest in a grav potential and can aswell be defined as the relativistic mass which would be dependent on the distance from the star or on the potential $$m=m_0\sqrt{(1-\frac{2M}{r})}.$$

The relativistic mass of an object at rest at infinity would just be its rest mass.

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  • $\begingroup$ oho rest mass is not constant,,i thought it is constant always ,i have heard that it is the mass observed in rest frame or something like that? $\endgroup$ – Sagar Timalsina Aug 17 '16 at 0:50
  • $\begingroup$ Right. is corrected $\endgroup$ – Statics Aug 17 '16 at 11:14

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