# Mass at a distance

1. Nov 13, 2008

### Denton

A nucleus is always lighter than its composite particles combined due to binding energy, does this hold true two object would be lighter when together than apart due to gravitational effects?

If this is correct than can I infer that the universe has gained mass as it expanded and will continue to gain mass?

2. Nov 13, 2008

### MeJennifer

Yes and yes.

3. Nov 13, 2008

### Denton

So then the energy of the universe has increased over time? I find this highly disturbing.

4. Nov 13, 2008

### MeJennifer

On balance it is zero as the gravitational field, which is spacetime curvature, is a form of negative energy.

By the way energy cannot be described in pure tensoral form, so really in GR there is no such thing as energy, same goes for momentum.

Last edited: Nov 13, 2008
5. Nov 13, 2008

### Phrak

In what inertial frame would you measure the mass of the universe? I heard a rumor that claimed "What is the mass of the universe?" is an ill posed question.

6. Nov 13, 2008

### Naty1

No. Making a simple extrapolaton from nuclear forces to gravitational effects over cosmic distances is unrealistic.

Is the universe is expanding, and infinite?, therefore the mass is infinite. OR

the opening paragraph points out in a homogeneous universe all space is equivalent...there is no center gravity. Seems like any implied increase in mass due to separation is illusory.

7. Nov 13, 2008

### Jonathan Scott

I would have thought you could get a reasonable theoretical definition for an effective total energy of the observable universe by simply adding up the energy of every object and non-gravitational field as seen from the observer's location, taking into account any shift due to velocity and/or the expansion of the universe. To put it another way, imagine that everything visible has been emitting spherical waves at the frequency corresponding in some frame to its energy, calculate how those waves appear to propagate in the observer's frame (taking into account velocity, gravitational and cosmological red-shift) and count the total rate at which the waves arrive at the observer.

However, this scheme doesn't appear to account correctly for gravitational energy (static or waves). The gravitational red-shift of these hypothetical energy waves effectively adjusts rest energy for the potential energy (due to the object itself or any other objects), but that adjustment is twice the correct potential energy because it means each interaction is counted both ways round, as can be seen by comparison with Newtonian potential theory.

I'd like to point out that mass at a distance can't be defined in the same way as local mass within a general coordinate system, because mass is locally equal to energy divided by the local speed of light squared, but if you try to take distant mass as being coordinate energy divided by coordinate speed of light you have problems because the coordinate speed of light can be different in different directions. You can instead take the coordinate energy divided by the standard local speed of light, that is the coordinate energy in mass units, but it should be noted that this isn't the exact coordinate equivalent of local mass, so care has to be taken when defining quantities such as the coordinate momentum.

8. Nov 13, 2008

### Phrak

I get one quarter million Google hits searching on the exact phrase "Energy of the Universe".

9. Nov 14, 2008

### Denton

On the scale of the universe or at local scale or both?

Good on ya.

10. Nov 14, 2008

### Phrak

It seems to be the customary practice in cosmology to calculate the energy or mass of the universe by adding up pieces over the atlas of nominally flat inertial frames comoving with the cosmic background radiation--although it's probably close enough use the Earth frame. If this thread where posted in the Cosmology Folder, perhaps someone could tell us. The same sort of thing has to be done when calculating the age of the universe.

Last edited: Nov 14, 2008
11. Nov 14, 2008

### Phrak

I hear echos of this statement over and over every few years. How would the binding energy of an electron to a nucleus, for instance, make the atom lighter than its constituent parts??

The atom is lighter because the 'binding energy' has been radiated away. The tighter it binds, the more energy is removed.

12. Nov 14, 2008

### Denton

Im not sure what the problem is. And from the looks of it I don't think you wouldn't know the processes that are involved in nuclear fission / fusion.

However It also seems like you have no idea. There is no binding energy between a nucleus and an electron. It only applies to the neutrons and protons.

Yes and i should remind you that e=mc^2. You radiate energy out you lose mass by that amount.

13. Nov 14, 2008

### Phrak

Look Denton. I'm not insulting you. You misunderstand. It's one of these standard terms that that is misnamed and because of it generates confusion.

Wikipedia "Electron binding energy is a measure of the energy required to free electrons from their atomic orbits."

btw, energy is not mass.

14. Nov 14, 2008

### Denton

Ok fair enough, there is a 'binding energy' between the electron and nucleus, however ive only ever heard of the term binding energy used to describe the strong nuclear force in the nucleus.

Just tell me what are you trying to prove? That in a nuclear reaction the products don't liberate mass into energy?