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Questions about relativistic mass

  1. Feb 18, 2016 #1
    A : If a relativistic particle hit something, does its kinetic energy derived from base mass, or relativistic mass?
    B : Do we have any data, how it affects a gravity based interaction between two relativistic objects?
    C : Lets suppose a relativistic isotope emits two particles, one in the direction of acceleration (where we accelerated it) one in the opposite direction. Will be speed difference between them? (Compared to emitter isotope.)
     
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  3. Feb 18, 2016 #2

    Ibix

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    Don't use relativistic mass. It just confuses things.

    The kinetic energy of a particle of mass (or rest mass, if you must) m is ##(\gamma -1)mc^2##.

    The rest mass is part of the stress energy tensor which is the source of gravity in general relativity. Relativistic mass does not appear. Other terms related to the energy of motion do appear, although not in the naive "makes it more massive" sense. Regarding data you may wish to look up Gravity Probe B and the LIGO experiment for interesting gravitational effects of mass in motion.

    Your third question is rather imprecisely stated. I'm assuming you mean that the isotope emits two identical particles in opposite directions at the same speed viewed from the rest frame of the particle. In that case, unless the particles are massless and move at the invariant speed of light, yes, they will have different velocities as viewed from a frame where the isotope is moving.
     
  4. Feb 18, 2016 #3
    Third one, no that triviality wasnt my question. My question was, if two identical smaller particle (two electrons for example) emitted in opposite directions, will there be speed difference between them compared to emitter. I ask this because emitter has 0.9c compared to us, add another 0.9c to electron, that were superluminal compared to us, while this isnt the case in the opposite direction
     
  5. Feb 18, 2016 #4

    HallsofIvy

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    No, you are mistaken about a fundamental principle of relativity. Speeds don't add that way. If particle A moves at speed u relative to us and particle B moves at speed v, in the same direction, relative to particle A, the particle B has speed, relative to us, of [itex]\frac{u+ v}{1+ \frac{uv}{c^2}}[/itex] which is always less than c.

    In the particular case that u= v= 0.9c, that gives [itex]\frac{0.9c+ 0.9c}{1+ \frac{(0.9c)(0.9c)}{c^2}}= \frac{1.8c}{1.81}= 0.99c[/itex].
     
  6. Feb 18, 2016 #5

    Ibix

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    In that case, in the rest frame of the isotope the two particles have the same speed u, but with opposite signs. In a frame where the isoyope is moving at speed v (and all the velocities are colinear) the particles have speed $$\frac {\pm u-v}{1\mp uv/c^2}$$This is the relativistic velocity addition formula.
     
  7. Feb 18, 2016 #6
    There is a recession speed difference between the particles, difference is as large as the time dilation difference. (So that both particles would measure the same recession speed)

    (and relativistic mass difference is as large as time dilation difference)

    (recession speed means how fast the distance between the particle and the emitter increases)
     
  8. Feb 19, 2016 #7
    Ok, thanks.
     
  9. Feb 19, 2016 #8

    vanhees71

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    A: There's no need for relativistic mass. It's an old-fasioned idea, which is obsolete since 1908, when Minkowski found the fully developed math behind special relativity. The energy and momentum of a classical particle in terms of its invariant mass ##m## and the usual three-velocity ##\vec{v}=c \vec{\beta}##(which is what you seem to call "base mass", an expression I've never heard before) are
    $$E=\frac{m c^2}{\sqrt{1-\beta^2}}, \quad \vec{p}=\frac{m c}{\sqrt{1-\beta^2}}.$$
    You can also rewrite the energy in terms of the momentum
    $$E=c \sqrt{m^2 c^2+\vec{p}^2},$$
    as you can easily check from the above conditions. This leads to the definition of the invariant mass in terms of four-vectors, showing that it is a scalar, i.e., independent of the frame of reference
    $$(p^{\mu}):=(E/c,\vec{p}) \; \Rightarrow \; p_{\mu} p^{\mu} = (E/c)^2-\vec{p}^2=m^2 c^2.$$
    This is also called the "on-shell condition".

    B: Gravity is understood in terms of General Relativity. There idea of a "relativistic mass" here becomes totally unusable! Just last week we had a breakthrough with the first ever direct detection of a gravitational wave by the LIGO and VIRGO collaborations. The accurate analysis of this signal lead to the conclusion that its source are two supermassive black holes orbiting around each other and then merging into one even larger black hole. The comparison between the measured signal and the expectation for the (numerical) solution of the corresponding general-relativistic Einstein field equations lead to a very accurate confirmation of General Relativity even under stuch extreme conditions.

    Even more accurate tests of General Relativity are possible using "pulsar timing", i.e., one measures the radio signals of pulsars very accurately. Sometimes you also have a pulsar orbiting around another object, and sometimes you are even as lucky to have such a double-star system where both stars are pulsars whose signals are observable on Earth. With these very accurate measurments of the electromagnetic signals from such systems you get also very accurate (in fact the most accurate) confirmations of General Relativity.

    C: The most simple way to analyse such reactions is to use the covariant four-vector formalism and analyze the energy-momentum conservation together with the on-shell conditions. Then you can use the most simple reference frame for the process and write it down in a way that it applies in any frame of reference. For your decay example, the most simple reference frame is where the decaying particle is at rest. Let's call the four-momenta of the mother particle ##p_1## and that of the daughter particles ##p_2## and ##p_3##. Further let's use "natural units", setting the speed of light ##c=1##. Then you have
    ##p_1=(m_1,0,0,0)## and ##p_2=(E_2,\vec{p})##, ##p_3=(E_3,-\vec{p})##. The latter follows from the conservation of energy and momentum
    $$p_1=p_2+p_3 \; \Rightarrow \vec{p}_2=\vec{p}=-\vec{p}_3.$$
    The energies are determined by the on-shell conditions:
    $$E_2=\sqrt{m_2^2+\vec{p}^2}, \quad E_3=\sqrt{m_3^2+\vec{p}^2}.$$
    Now you can determine the momentum using the invariant
    $$p_1^2=m_1^2=(p_2+p_3)^2=p_2^2+p_3^2+2p_2 \cdot p_3=m_2^2+m_3^2 + 2 (E_2 E_3+\vec{p}^2).$$
    Solving for the magnitude of the momentum gives
    $$|\vec{p}|=\frac{\sqrt{[m_1^2-(m_2+m_3)^2][m_1^2-(m_2-m_3)^2]}}{2m_1}.$$
    Since the momenta in this reference frame (the center-momentum frame of the decay) are collinear, you can calculate the relative velocity in the naive way,
    $$|\vec{v}_{\text{rel}}|=\left |\frac{\vec{p}_2}{E_2}-\frac{\vec{p}_3}{E_3} \right|.$$
    One can show (try it!) that this can be expressed in a manifest covariant way by
    $$|\vec{v}_{\text{rel}}|=\frac{\sqrt{(p_2 \cdot p_3)^2-m_2^2m_3^2}}{E_2 E_3}.$$
     
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