Hot and cold

[Maxwell511]

It doesn't have more mass, but it has more energy, which is analogous to mass. It's an ongoing question as to whether a very hot object of mass M is inherently heavier or more gravitationally attractive than a very cold object of mass M - that is, does energy warp space?

How is that an ongoing question?

We know that temperature is the measure of the kinetic energy of particles of the system. We know that kinetic energy is related to velocity. We know from Einstein that as velocity of an object increases so does its mass.

The logical conclusion of all this knowledge is that a hot object has more mass than a cold one.

The only way that this could be an ongoing question is if we assume all physics (classical/modern) is wrong. If we assume it is right than the answer is obvious.

Also a compressed spring has more mass than a uncompressed spring.

 

[Maxwell511]

Relativity contradicts quantum mechanics, and physicists tend to hedge their bets on QM rather than GR.

The mass-energy equalivence has nothing to do with GR. It has to do with SR which is totally inline with quantum mechanics. See Dirac's equations, the existence of anti-matter etc.

 

[Wiccan_Child]

How is that an ongoing question?

We know that temperature is the measure of the kinetic energy of particles of the system. We know that kinetic energy is related to velocity. We know from Einstein that as velocity of an object increases so does its mass.

That's a matter of opinion. Relativity says that to accelerate to a high velocity requires more and more energy, to the point that you need infinite energy to accelerate to lightspeed (i.e., the energy requirement as a function of velocity asymptotes to v = c).
This can be seen as the object gaining mass as it accelerates, hence the term 'rest mass', but that's as subjective as, say, your choice of an inertial frame, or zero potential.

The logical conclusion of all this knowledge is that a hot object has more mass than a cold one.

The only way that this could be an ongoing question is if we assume all physics (classical/modern) is wrong. If we assume it is right than the answer is obvious.

Also a compressed spring has more mass than a uncompressed spring.

That's probably because you're pushing down on the spring when you're compressing it on the scales

The mass-energy equalivence has nothing to do with GR. It has to do with SR which is totally inline with quantum mechanics. See Dirac's equations, the existence of anti-matter etc.

SR is GR; specifically, the mathematical premises of SR are special cases of the more general (hint hint) model. Special Relativity is the special case of the more general General Relativity, in much the same way as Classical Mechanics is a special case of GR and QM.

 

[Maxwell511]

That's a matter of opinion. Relativity says that to accelerate to a high velocity requires more and more energy, to the point that you need infinite energy to accelerate to lightspeed (i.e., the energy requirement as a function of velocity asymptotes to v = c).
This can be seen as the object gaining mass as it accelerates, hence the term 'rest mass', but that's as subjective as, say, your choice of an inertial frame, or zero potential.

Mass is relative and therefore subjective that does not mean it does not exist. We are talking about the perception of observers and not an objective universe. You cannot "weigh" an electron and have everyone agree on the mass. It is not a matter of opinion it is a matter of how the universe works and how we perceive it.

That's probably because you're pushing down on the spring when you're compressing it on the scales

[insert witty retort]

SR is GR; specifically, the mathematical premises of SR are special cases of the more general (hint hint) model. Special Relativity is the special case of the more general General Relativity, in much the same way as Classical Mechanics is a special case of GR and QM.

With the noted exception that SR and QM are not only compatible but their union made important predictions which have been experimentally verified. GR may be wrong, SR is almost certainly* true.

*In the sense that the sun will almost certainly come up tomorrow

 

[Frumious Bandersnatch]

How is that an ongoing question?

We know that temperature is the measure of the kinetic energy of particles of the system. We know that kinetic energy is related to velocity. We know from Einstein that as velocity of an object increases so does its mass.

The logical conclusion of all this knowledge is that a hot object has more mass than a cold one.

The only way that this could be an ongoing question is if we assume all physics (classical/modern) is wrong. If we assume it is right than the answer is obvious.

Also a compressed spring has more mass than a uncompressed spring.

I think this must be correct. One can calculate the change in mass. Carbon steel has a heat capacity of 473 J/(kg*K) so to heat 1 kg to 1000 degrees C requires 473,000 J = 473,000 kg*m^2/s^2). C^2 = 9*10^16 m^2/s^2) so from the formula E = mass*C^2 we get mass = 473,000/C^2 so the energy required is equivalent to about 5*10^-12 kg of mass thus if the steel bar weighed 1.000000000000 kg at 0 C it would weigh 1.000000000005 kg at 1000 C. I don't think we have the technology to weigh the difference but in a Gendanken experiment it is there.

 

[Wiccan_Child]

Mass is relative and therefore subjective that does not mean it does not exist. We are talking about the perception of observers and not an objective universe. You cannot "weigh" an electron and have everyone agree on the mass. It is not a matter of opinion it is a matter of how the universe works and how we perceive it.

Objects at rest have a certain, immutable mass. When they accelerate, they behave as if their mass is increasing - but this isn't the only way to explain said behaviour.

If X is increasing, and we know X is Y times Z, then the sheer fact that X increases doesn't prove that Y is itself increasing - it could very well be that Y is an immutable constant, and it's Z that's increasing.

With the noted exception that SR and QM are not only compatible but their union made important predictions which have been experimentally verified. GR may be wrong, SR is almost certainly* true.

*In the sense that the sun will almost certainly come up tomorrow

True. But if GR were disproven, SR would still take quite a blow. A lot of the evidence for SR comes by proxy from evidence for GR.

 

[Maxwell511]

Wiccan_Child[/LEFT];56420188]Objects at rest have a certain, immutable mass. When they accelerate, they behave as if their mass is increasing - but this isn't the only way to explain said behaviour.

If X is increasing, and we know X is Y times Z, then the sheer fact that X increases doesn't prove that Y is itself increasing - it could very well be that Y is an immutable constant, and it's Z that's increasing.

I understand your point however but I think it is based on a linear algebra sort of reasoning. I would think that since we are dealing with change of more than one variable that multivariable calculus comes in, partial derivatives can give completely different answers if Y is constant.

True. But if GR were

disproven​

, SR would still take quite a blow. A lot of the evidence for SR comes by proxy from evidence for GR.

I know that you are a theoretical physicist so I am going to let this comment slide.

The evidence for SR is mostly empirical. While GR may provide some theoretical evidence for SR, the question of the correctness of SR is not based on the question of the correctness of GR.

 

[Wiccan_Child]

I understand your point however but I think it is based on a linear algebra sort of reasoning. I would think that since we are dealing with change of more than one variable that multivariable calculus comes in, partial derivatives can give completely different answers if Y is constant.

Sure, the 'X = YZ' was pulled from the top of my head. I'm just saying that the thing we know increases (velocity, energy, four-momentum, w/e) is a multi-variable function. Contrast this with an increase in a photon's energy, which does necessitate an increase in frequency, etc.

I know that you are a theoretical physicist so I am going to let this comment slide.

The evidence for SR is mostly empirical. While GR may provide some theoretical evidence for SR, the question of the correctness of SR is not based on the question of the correctness of GR.

Ah, that's not what I meant. Actual, direct empirical evidence for GR (e.g., gravitational lensing, the sharipo (sp?) effect) must necessarily constitute evidence for SR, since if GR is true then SR must also be true.
That said, though GR implies SR, SR doesn't imply GR: evidence for specifically SR doesn't really support GR, even though evidence for GR must support evidence for SR (until such time that GR, but not SR, is mathematically or empirically disproven).

So. Any evidence for GR constitutes evidence for SR. We have a heck of a lot of direct evidence for GR (its clash with QM notwithstanding). So, to disprove GR would mean that this evidence no longer supports SR - and, thus, SR would lose a heck of a lot of evidence.
It still has its own mountain of evidence, but still.

I'm surprisingly lucid for someone who's on his second bottle of Christmas wine (sparkling white, fyi). Cheers!

 

[Resha Caner]

I understand your point however but I think it is based on a linear algebra sort of reasoning. I would think that since we are dealing with change of more than one variable that multivariable calculus comes in, partial derivatives can give completely different answers if Y is constant.

Sure, the 'X = YZ' was pulled from the top of my head. I'm just saying that the thing we know increases (velocity, energy, four-momentum, w/e) is a multi-variable function. Contrast this with an increase in a photon's energy, which does necessitate an increase in frequency, etc.

It sounds like this comes down to a parallel postulate sort of thing, where the quandry over mass (or maybe gravity is the more basic quandry) allows a multiplicity of assumptions to proceed from that point. Any one of them would produce a mathematically consistent answer, but the issue is which is more phenomenological.

And that requires making a conclusion to connect datum to equation, i.e., attaching a meaning to a symbol, does it not?

 

[Wiccan_Child]

It sounds like this comes down to a parallel postulate sort of thing, where the quandry over mass (or maybe gravity is the more basic quandry) allows a multiplicity of assumptions to proceed from that point. Any one of them would produce a mathematically consistent answer, but the issue is which is more phenomenological.

And that requires making a conclusion to connect datum to equation, i.e., attaching a meaning to a symbol, does it not?

Indeed. It's very much like the problem over the wavefunction in quantum mechanics: mathematically, it works very well. But what is the wavefunction? Every particle has one, and the particle's behaviour can be ascertained by analysing its wavefunction, but the wavefunction isn't the particle itself. Most peculiar.

 

[Resha Caner]

Ah, yes. The wave function. Especially odd as it pertains to tunneling. So do you have a preferred explanation for that? Do you think the particle actually tunnels? Or does it high jump or something else?

 

[Wiccan_Child]

Ah, yes. The wave function. Especially odd as it pertains to tunneling. So do you have a preferred explanation for that? Do you think the particle actually tunnels? Or does it high jump or something else?

Whatever happens, it goes to places that, classically, it shouldn't be able to go. Particles appear outside nuclei when by all rights they should remain locked inside (which is, incidentally, how alpha-decay works).

I'm of the opinion that the particle actually 'pops' outside spontaneously and without prior cause; we know it doesn't physically traverse the distance, and there's something cathartic about sticking up two fingers to those philosophers who stubbornly refuse to let go of causality

 

[mzungu]

Maybe this should have gone in the "ask a physicist anything" thread, but I was afraid it would get lost there.

Does a hot object have more mass than a cold object? I think the answer is yes, but thought I would check.

The days of "PHLOGISTON" have passed away a long time ago. Check out physorg forums and ask there for a more detailed and scientific answer.

 

[Resha Caner]

I'm of the opinion that the particle actually 'pops' outside spontaneously and without prior cause; we know it doesn't physically traverse the distance, and there's something cathartic about sticking up two fingers to those philosophers who stubbornly refuse to let go of causality

Hmm. Cathartic though it may be for you, I still consider it a cop out. It seems you have backed yourself into a belief that can't be challenged (though I'm sure you don't see it that way).

How do you know it doesn't physically traverse the distance?

I've had this simple little toy I've been playing with for some time. Maybe you can help me see what holes it may have. Since it is rather classical in nature, I suppose one must consider it a macro example rather than quantum, but I'll give it anyway. Consider two particles of mass m. One has initial velocity v and the other has initial velocity 0. The first collides with the second, and the collision is perfectly elastic. What are the solutions for the velocities after the collision?

Again, very simple. Hopefully this will get interesting after the first question is answered ... though you may also find a way to quickly deflate the balloon.

 

[Wiccan_Child]

Hmm. Cathartic though it may be for you, I still consider it a cop out. It seems you have backed yourself into a belief that can't be challenged (though I'm sure you don't see it that way).

How do you know it doesn't physically traverse the distance?

For the same reason I can't fly to the Moon (quantum tunnelling notwithstanding).

I've had this simple little toy I've been playing with for some time. Maybe you can help me see what holes it may have. Since it is rather classical in nature, I suppose one must consider it a macro example rather than quantum, but I'll give it anyway. Consider two particles of mass m. One has initial velocity v and the other has initial velocity 0. The first collides with the second, and the collision is perfectly elastic. What are the solutions for the velocities after the collision?

Again, very simple. Hopefully this will get interesting after the first question is answered ... though you may also find a way to quickly deflate the balloon.

Now I'm furiously trying to find the balloon to pop

'Velocity' is relative, so I'm going to assume our inertial frame is taken to be where the second particle initially is (Einstein be damned).

The 'solutions' are that, post-collision, the hitherto moving particle is at rest, and the hitherto resting particle is moving at velocity v.

In other words, two equal masses that perfectly elastically collide effectively 'swap' velocities (as per Classical Mechanics).

 

[mzungu]

If I am not mistaken, something gains mass when accelerating and not when it is on a constant speed.

 

[Resha Caner]

For the same reason I can't fly to the Moon (quantum tunnelling notwithstanding).

That's not really an answer.

Now I'm furiously trying to find the balloon to pop.

Yes, and I'm cringing because I expect it to be popped, leaving me looking like a fool. (Shrug) At least I'll learn something in the process. The little niggling in my head is that I'm about to learn something new about uncertainty ... and that is what I expected to lie in the content of your answer to the previous question. But, to learn the Moon is made of cheese is also interesting.

'Velocity' is relative, so I'm going to assume our inertial frame is taken to be where the second particle initially is (Einstein be damned).

OK.

The 'solutions' are that, post-collision, the hitherto moving particle is at rest, and the hitherto resting particle is moving at velocity v.

In other words, two equal masses that perfectly elastically collide effectively 'swap' velocities (as per Classical Mechanics).

Yes ... sort of. The answer comes very quickly and very intuitively, without math. But if one writes out the equations and solves them, they produce two solutions. The final velocity of the first particle could be v or 0. So, how was it that we chose 0?

 

[Wiccan_Child]

That's not really an answer.

On the contrary, the reason I can't travel to the Moon is because I do not have the energy to escape the potential well. Neither does the particle. Moreover, if I did begin to traverse the distance, you can bet someone would notice

Yes, and I'm cringing because I expect it to be popped, leaving me looking like a fool. (Shrug) At least I'll learn something in the process. The little niggling in my head is that I'm about to learn something new about uncertainty ... and that is what I expected to lie in the content of your answer to the previous question. But, to learn the Moon is made of cheese is also interesting.

Wouldn't it be fun if it was.

OK.

Yes ... sort of. The answer comes very quickly and very intuitively, without math. But if one writes out the equations and solves them, they produce two solutions. The final velocity of the first particle could be v or 0. So, how was it that we chose 0?

They produce only one solution: given the scenario, we've arbitrarily decided that the first particle has velocity v, and so it must end at rest.

The same physical scenario can be redone from another inertial frame, giving different results (you can, of course, choose your inertial frame to give whatever final velocity you want (Relativity notwithstanding).

 

[Resha Caner]

They produce only one solution: given the scenario, we've arbitrarily decided that the first particle has velocity v, and so it must end at rest.

The math produces 2 solutions. An assumption must be added to choose which of those applies. Just for the sake of completeness:

1) p = m*v + m*0 = m*v1f + m*v2f (where v1f = the final velocity of particle 1, and v2f = the final velocity of particle 2)

Also, 2) E = 0.5*m*v^2 + 0.5*m*0^2 = 0.5*m*v1f^2 + 0.5*m*v2f^2

So:
1) v = v1f + v2f
v2f = v - v1f
v2f^2 = v^2 -2*v*v1f + v1f^2

2) v^2 = v1f^2 + v2f^2

Substituting 1) gives:
v1f^2 - v^2 + (v^2 - 2*v*v1f + v1f^2) = 0
2*v1f^2 - 2*v*v1f + 0 = 0
v1f^2 - v*v1f + 0 = 0
v1f = (v +/- sqrt(v^2 - 4*1*0)) / 2
v1f = (v +/- v) / 2 = [v,0]

So, what assumption is used to select v1f = v? I believe the assumption is that the collision requires a change in the particles. Why must this be so? Maybe particle 1 tunneled through particle 2 to appear on the other side.

 

[Wiccan_Child]

The math produces 2 solutions. An assumption must be added to choose which of those applies. Just for the sake of completeness:

1) p = m*v + m*0 = m*v1f + m*v2f (where v1f = the final velocity of particle 1, and v2f = the final velocity of particle 2)

Also, 2) E = 0.5*m*v^2 + 0.5*m*0^2 = 0.5*m*v1f^2 + 0.5*m*v2f^2

So:
1) v = v1f + v2f
v2f = v - v1f
v2f^2 = v^2 -2*v*v1f + v1f^2

2) v^2 = v1f^2 + v2f^2

Substituting 1) gives:
v1f^2 - v^2 + (v^2 - 2*v*v1f + v1f^2) = 0
2*v1f^2 - 2*v*v1f + 0 = 0
v1f^2 - v*v1f + 0 = 0
v1f = (v +/- sqrt(v^2 - 4*1*0)) / 2
v1f = (v +/- v) / 2 = [v,0]

So, what assumption is used to select v1f = v? I believe the assumption is that the collision requires a change in the particles. Why must this be so? Maybe particle 1 tunneled through particle 2 to appear on the other side.

Your logic is flawed .

(1) says v = v[sub]1,f[/sub] + v[sub]2,f[/sub].

This means v[sup]2[/sup] = v[sub]1,f[/sub][sup]2[/sup] + v[sub]2,f[/sub][sup]2[/sup] + 2v[sub]1,f[/sub]v[sub]2,f[/sub].

Which contradicts (2), which states: v[sup]2[/sup] = v[sub]1,f[/sub][sup]2[/sup] + v[sub]2,f[/sub][sup]2[/sup].

(1): v = v[sub]1,f[/sub] + v[sub]2,f[/sub]
(2): v[sup]2[/sup] = v[sub]1,f[/sub][sup]2[/sup] + v[sub]2,f[/sub][sup]2[/sup] + 2v[sub]1,f[/sub]v[sub]2,f[/sub]

(Bolding is for emphasis, not because they're vectors )