5
$\begingroup$

I have been told in several answers that the term "relativistic mass" is no longer considered adequate wrt. the mass/energy increase in kinetic energy.

Yet, I read that 1 kg of gold's mass increases by 10^-14 kg if its temperature is raised by a couple of degrees, and that the very "invariant mass" of the proton is actually only 1% proper mass and the rest is energy in the form of virtual photons/ gluon soup.

I am baffled; can you please clarify when something qualifies for the attribute of "mass"? Can you explain why energy bound as heat in gold is more "mass" then energy bound as kinetic energy in motion? Former "relativistic" mass is sensitive to gravity as is "heat mass" in gold; it determines a larger trajectory at LHC, just as a proton does (an electron with 1836 "relativistic" masses has the same distance from the center of LHC as a proton, doesn't it?) So, what properties of mass has kinetic energy mass lacking in order to qualify for that title?

$\endgroup$
  • $\begingroup$ The increase in potential energy due to excited electrons of gold, more frequent collisions between molecules, and greater stretching of bonds, might add to the increase in its mass. $\endgroup$ – philip_0008 Jun 9 '16 at 18:04
3
$\begingroup$

Suppose you start with your (stationary) 1kg block of gold. If you raise its temperature you have to add energy to it, and that means it's different after you've raised its temperature. For example you could shine an infrared lamp on it, in which case you've added the energy from the IR lamp. The mass changes because if you add an energy $E$ the mass goes up by $E/c^2$.

However suppose instead of heating the gold you go shooting off in a rocket at $0.999c$. In your rest frame the gold is now moving at $-0.999c$ so in your rest frame the momentum of the gold is given by the relativistic momentum formula:

$$ p = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$

where $m$ is the invariant mass i.e. 1 kg.

The old idea of relativistic mass came from writing the momentum as:

$$ p = \frac{m}{\sqrt{1 - \frac{v^2}{c^2}}} v = m_r\,v $$

where $m_r$ is defined as the relativistic mass:

$$ m_r = \frac{m}{\sqrt{1 - \frac{v^2}{c^2}}} $$

But you need to be clear that the relativistic mass is a computational device and it does not mean the mass is changing. Nothing happens to the gold. If I am sitting by the gold while you speed off in your spaceship I will see nothing happen to it.

This is the key point. When we heat the gold all observers, no matter what their speed, will agree that the gold has changed because we all see it absorbing the light from the IR lamp.

The invariant mass is the parameter $m$ in the relativistic energy momentum equation:

$$ E^2 = p^2c^2 + m^2c^4 $$

This energy $E$ is one component of the four-momentum and the four-momentum is a coordinate independent object. So any observer, moving at any speed in any way, can start with the four-momentum transformed into their coordinates and calculate the invariant mass $m$. And all such observers will end up with the same value for $m$.

$\endgroup$
  • $\begingroup$ @ally: If you use the Newtonian expression for the centripetal acceleration required to keep the proton in the LHC then you'll find it gives the wrong result. The relativistic mass fans would insist that the Newtonian expression is correct and the mass of the proton has increased. The rest of us would point out the the Newtonian expression doesn't apply to a relativistic object (duh!) so the mass has stayed the same but we need to use the full relativistic expression for the centripetal force. $\endgroup$ – John Rennie Jun 9 '16 at 8:37
  • $\begingroup$ @ally: In circular motion the radius does not Lorentz contract because it's always normal to the direction of motion. Both observers agree on the radius of the motion. Also note that magnetic and electric fields look different to observers moving at different speeds. $\endgroup$ – John Rennie Jun 9 '16 at 9:27
  • $\begingroup$ @ally: Special relativity says that every observer who measures the speed of any ray of light anywhere in the universe will measure that speed to be $c$. You might be interested in discussing this in the chat room. $\endgroup$ – John Rennie Jun 10 '16 at 8:22
  • $\begingroup$ We should discuss this in the chat room $\endgroup$ – John Rennie Jun 10 '16 at 8:38
1
$\begingroup$

For an indivisible particle, probably the simplest operational definition of "mass" in your context is $\sqrt{E^2 - (p c)^2}/c^2$, where $E$ is its energy and $p$ its momentum. Equivalently, the mass of an indivisible particle at rest is $E/c^2$ in that frame.

A lump of gold and a proton are not indivisible particles, because their constituent parts can move with respect to each other. (In the case of the lump of gold, this movement comes from the thermal kinetic energy of the individual atoms in the center-of-mass frame. In the case of an atomic nucleus, which is highly quantum-mechanical, talking about the "motion of the quarks" is really an inappropriately classical picture, but it will do for this discussion.) We therefore need to be a little more careful about how we define their mass. There are two reasonable options:

(1) We could define the mass to be the sum of the individual masses of each indivisible constituent piece. Under this definition, the mass of the lump of gold does not depend on its temperature, and the mass of the proton is simply the sum of the rest masses of its three constituent quarks (since gluons are massless) - the "proper mass" you mention above.

(2) Alternatively, we could "lump everything together" (or "coarse-grain") the composite object and give it an effective mass defined to be $E/c^2$, where $E$ is the total energy of the composite object, including the kinetic energy of its constituents. Under this definition, the effective mass of the gold does increase when it's heated, because its constituent atoms gain kinetic energy. And the effective mass of the proton becomes much heavier than the sum of its quark rest masses, because it turns out that the quarks are so relativistic that their kinetic energy dwarfs their rest mass energy.

In practice, the second definition is pretty much always a more useful definition of the "mass" of a composite object. This is because in general relativity, the kinetic energy contribution gravitates, just as much as the rest energy contribution does. Similarly, it's more difficult to push something with a high effective mass than something with a low one, even if their masses are the same under definition (1). So the second definition is more experimentally accessible. Moreover, the first definition raises the question, "where do you stop?" We treated the gold atom as being made up of atoms, but of course those atoms are themselves made up of protons, etc. If we only counted the rest mass of those sub-constituents, the mass (under definition #1) would decrease further. So it's better to treat the kinetic energy of the constituent particles as contributing to the rest mass of the composite object, as counterintuitive as that idea may be.

As for a single particle moving at relativistic speeds, the unambiguous fact is that it gravitates more strongly than an equivalent particle at rest. You can tell two different stories to explain this: (1) only mass gravitates, but the relativistic motion of the particle actually increases its mass, causing the additional gravitation, or (2) mass and kinetic energy both gravitate, so the particles mass has not increased but its new kinetic energy causes the additional gravitation. Both stories result in equivalent equations, so which one you like to tell is largely a matter of taste. Special relativity was originally thought of along the lines of the first story, but now most people find it easier to keep things straight if they tell the second story. As you become more comfortable with special relativity, you'll eventually stop worrying too much about what counts as "mass" and what counts as "energy," and realize that at the end of the day they're really the same thing.

$\endgroup$
  • $\begingroup$ By the way, I didn't mean for that last sentence to be condescending, but instead reassuring. These issues are subtle and it's completely normal to be very confused at first. But I find that it's not like you have to crack each question one by one and then start over with the next one. Instead, the various pieces come together gradually, and in answering one question you'll find that you've also answered others you weren't expecting to be directly related. $\endgroup$ – tparker Jun 9 '16 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy