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My understanding of Quantum physics and String Theory is very basic and I don't yet have a grasp on the maths, but in my research I have come up with a question. Does a decrease in temperature also signify/create a decrease in mass?

I am quite willing to believe the concepts I think I am beginning to understand are totally misconceptions on my part. But, as I see it mass is a product/mark of disruption to the Higgs field as a particle moves. If these particles are ultimately composed of vibrating Strings, their mass is given by the rate at which they vibrate and how it interacts with the Higgs field.

I don't have any idea if, as the temperature (as we know it) approaches 0k, the string/quark/sparticle would change its state in any way or if the changing of the state of the String would change what type of particle it represents.

It seems I'm more full of questions than answers, but that's why I love these topics! I know what I'm learning is new if the best and brightest are still working on the 'basics' to just make their systems work :).

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  • $\begingroup$ So if the mass does decrease and F=ma with gravity being constant, so does the weight of the object (though it may be an immeasurability small, but calculable amount) decrease as well? $\endgroup$ – user34322 Nov 20 '13 at 21:48
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    $\begingroup$ The energy due to temperature is much smaller than the rest mass. $\endgroup$ – jinawee Nov 20 '13 at 22:13
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The answer is that a decrease in temperature does decrease the mass, though in most cases that change is exceedingly small. Temperature is a macroscopic phenomenon, so you can't really talk about the temperature of a single string or an atom. However consider the following analogy:

If you have an isolated string (or atom) in some excited state, then to relax into a lower energy state it must emit a photon or conversely to move to a more excited state it must absorb a photon. As discussed in the question Does the mass of a body absorbing photons increase? emitting or absorbing photons will change its mass.

For a macroscopic object the arguament is a bit more subtle. In macroscopic objects temperature is normally a measure of the kinetic energy of the vibrating atoms in your system. When you are calculating the gravitational field of the object you're probably used to using Newton's law. However general relativity tells us that the source of the field is an object called the stress-energy tensor. This does include the rest mass of the object(s) but it also includes momentum and pressure. Temperature increases the momentum of the atoms in your material and that contributes to the increase in the gravitational field.

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  • $\begingroup$ Does the expansion of the object ie stretching bonds, have any bearing? $\endgroup$ – Farcher Mar 6 '16 at 22:29
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    $\begingroup$ @Farcher: not directly, but it is related. When you expand a bond you need to do work against the interatomic forces, and that means you need to add some energy $E$. That energy increases the mass of the molecule by $m = E/c^2$. $\endgroup$ – John Rennie Mar 7 '16 at 6:28
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You seem to have overcomplicated this question quite a bit, you don't need to go further than the theory of relativity to find the answer. Heat is simply chaotic movement on the molecular and submolecular level, and as relativity dictate that any object moving relative to an observer will to that observer appear to have greater mass with greater relative speed, heat naturally increases mass.

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Thermal (statistical) quantum field theory tells us that temperature indeed does have an effect on mass. It is possible to ascribe a so-called thermal mass to a single particle, resulting from an interaction with the heat bath. Within this framework one can calculate corrections to the self energy of particles which increase with temperature.

To answer your question directly: a decrease in temperature leads to a decrease in mass.

For more detail, consider lecture notes on thermal field theory, e.g. http://arxiv.org/abs/hep-ph/0105183 or http://hep.itp.tuwien.ac.at/~aschmitt/thermal13.pdf.

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An important note to the other answers so far: the actual mass of a single particle (the rest mass as it used to be commonly referred to) does not change if the temperature rises (by temperature I mean that of the environment since temperature is a macroscopic quantity).

The (rest) mass of a particle is the $m$ in

$$E = \sqrt{|\mathbf{p}|^2c^2+m^2c^4}.$$

Even in a macroscopic object, the mass $m$ of each individual particle doesn't change when the object's temperature rises. What does change is the magnitude of the momentum $|\mathbf{p}|$ of each particle. So the mass doesn't change.

However, as John Rennie explains in his answer, general relativity teaches us that gravity doesn't couple just to mass (which is simply a form of energy, remember), it couples to all forms of energy. So the thermal energy that an object gains by heating it up - while it doesn't increase its mass - does cause it to gravitationally interact more strongly.

Now I've been scrupulously using the term mass in the sense of rest mass to be clear about what is changed by a rise/decrease in temperature. Of course, mass and energy can actually be used interchangeably because mass is really just energy. Or energy is really just mass, if you will. It's also in that sense that you will hear people state that most of your personal mass is not due to the BEH field, but due to binding energy of the atoms in your body and such.

Follow-up

Why (or in which way) is mass simply a form of energy? Well, the answer to that certainly isn't "because Einstein said so." It's because of that Higgsfield (BEH field). The BEH field has a non-zero vacuum expectation value or vev (basically: it's nowhere and never zero, even if nothing's around).

The idea is that each elementary particle is in and of itself massless. This is fine because we know due to Einstein that mass is energy and energy needs some sort of origin/cause. Now (almost) every particle couples to the BEH field and the interaction energy of this coupling is basically the particle's mass. Because the vev of the BEH field is non-zero, there's no escaping it: if you couple to the BEH field, you will be massive.

So to bring two concepts of mass together: the rest mass of a non-interacting particle is the BEH mass. When it starts interacting with e.g. an electromagnetic field, it gains energy which can be interpreted as a gain in mass because of the interchangeability of mass and energy. But remember that its rest mass hasn't changed.

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You can simply think of it in terms of special relativity. In general, $E = \sqrt{p^2c^2+m^2c^4}$, where $m$ is the invariant mass of a particle. As the name suggests, the invariant mass of a particle does not vary. It's a property associated with the particle. It is the mass of the particle measured in its rest frame. What changes, with temperature, is the momentum, $p$, and that affects $E$. Now if the mass you are talking about is not the invariant mass, but the relativistic mass, say $m_{r}$, it is Lorentz boosted by a factor of $\gamma$ $(= \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - \frac{p^2}{m^2c^2}}})$ for an inertial observer moving with a velocity $v$ with respect to the rest frame of the particle such that $E = \gamma mc^2 = m_rc^2$ So if a decrease in temperature implies a decrease in said velocity, the relativistic mass of a particle at any snapshot as seen by said observer decreases on average with the decrease in temperature.

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photon has very light mass almost negligible amount, so temperature does not effect mass of the object, but you take large object with high temperature it indicates measurable amount of mass change. example sun.

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