# Does a magnetic field have mass?

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11 hours ago, Mordred said:

Good question, it is questions like this that caused modern physics to drop the usage of rest mass. The modern term is the invariant mass.

What’s the difference though?

11 hours ago, Mordred said:

Yes though relativistic mass is replaced by variant mass in modern terminology. Both rest mass and relativistic mass originated in SR treatments where one inertial frame was considered at rest.

Under GR all frames are inertial and no frame is truly at rest. Hence the modern terminology.

To determine the invariant mass of a particle requires extensive experimentation in scattering experiments and particle decays. The CM frame and lab frame is extensively used in these experiments.

Here is an extremely basic oversimplified overview

It would seem then that GR and SR don’t agree on how to quantify mass so how do we actually quantify mass?

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13 minutes ago, MPMin said:

What’s the difference though?

In a practical sense, there is none. If you measure the mass in e.g. a Penning trap, you can determine its speed and subtract the relativistic effects in determining the invariant mass. Effectively you have the rest mass, even though it was never at rest.

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1 hour ago, MPMin said:

It would seem then that GR and SR don’t agree on how to quantify mass so how do we actually quantify mass?

Modern SR and GR now use the same definitions. It was the original definition rest mass that was later changed.

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1 hour ago, MPMin said:

It would seem then that GR and SR don’t agree on how to quantify mass so how do we actually quantify mass?

Not really. SR is just a subset of GR that can be used under certain conditions (when the system can be approximated by an inertial frame of reference). The GR definition if more generic (because it is the general theory of relativity). While the SR definition is a special case of that.

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Keep in mind using rest mass is still acceptable as it's often easier to type than invariant mass so feel free to use the old terminology. Just be aware that it refers to the invariant mass

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Here is a quote from Susskind that may help bring out what Swansont and Mordred have been telling you.

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On 9/17/2019 at 10:03 AM, Mordred said:
On 9/17/2019 at 9:24 AM, MPMin said:

Does relativistic mass contribute to gravitational mass?

Yes though relativistic mass is replaced by variant mass in modern terminology. Both rest mass and relativistic mass originated in SR treatments where one inertial frame was considered at rest.

But if relativistic mass is frame dependent how can it actually contribute gravitational mass? In different frames can an object really have different gravitational mass values?

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All measurements of all types of mass is frame dependent. In order to find a rest mass of a particle for example you have to find a centre of momentum frame where the momentum =0. That requires those scattering experiments I mentioned.

The question you just asked is another one of the those questions that led to the new terminology as the term relativistic mass was often misunderstood by laymen as being the observer dependent mass.

As Swansont mentioned the rest mass term is only valid in its own reference frame other observers in other frames of reference will measure that mass differently as well.

I will give you an example on the LHC  they take protons and accelerate them to near light speeds. From the scattering collisions they produce particles with far higher rest mass than the original two protons. Good example is the Higgs Boson.

The principle of equivalence states one further important detail. Inertia mass=gravitational mass

This means that if you give an object kinetic energy (inertia) it has the same equivalence as giving it gravitational mass

Edited by Mordred

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Ok help me understand this then: if you could trap photons in a perfectly mirrored box, the box would have more mass because of the photons trapped inside, however an emp or magnetic field can pass straight through the box (because we know magnets work through mirrors) does that mean then that an emp doesn’t have relativistic mass?

Further to that, if an incandescent bulb was in the perfectly mirrored box with its own power supply ( assuming the bulb and power supply did not absorb any photons and all the photons produced were perfectly reflected within the box) the magnetic field produced by the filament could pass through box but the photons could not would the box lose mass due to the magnetic field escaping the box?

And if the box could perfectly trap photons, would opening the box after some time release all the trapped photons at once producing an explosion of photons so to speak or would it make no difference and opening the box after some time would look the same as opening the box at any time?

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25 minutes ago, MPMin said:

Ok help me understand this then: if you could trap photons in a perfectly mirrored box, the box would have more mass because of the photons trapped inside, however an emp or magnetic field can pass straight through the box (because we know magnets work through mirrors) does that mean then that an emp doesn’t have relativistic mass?

An EMP has energy. This can be treated as mass (E=mc2) although it is not usually done (just because there is no particular benefit to doing so).

Worth noting that even if you consider the energy of this EM radiation as relativistic mass, calculating the gravitational effect (which you mentioned earlier) is more complicated because gravity depends on more than just mass-energy; you also need to take into account momentum and other things.

Incidentally, if the box is perfectly reflecting (inside) then it is also perfectly opaque from the outside so EM radiation could not pass through it.

30 minutes ago, MPMin said:

Further to that, if an incandescent bulb was in the perfectly mirrored box with its own power supply ( assuming the bulb and power supply did not absorb any photons and all the photons produced were perfectly reflected within the box) the magnetic field produced by the filament could pass through box but the photons could not would the box lose mass due to the magnetic field escaping the box?

If it is a constant current then any magnetic field would be static and not carrying away energy. If it is AC, then any radiation that escapes from the box would reduce the mass of the box.

This seems to be setting up unnecessarily complicated scenarios. If energy escapes then the mass decreases. That is all there is to it.

31 minutes ago, MPMin said:

And if the box could perfectly trap photons, would opening the box after some time release all the trapped photons at once producing an explosion of photons so to speak or would it make no difference and opening the box after some time would look the same as opening the box at any time?

Opening the box would release however much light and heat had accumulated. At some point, the bulb would stop radiating because it would be in equilibrium with the box.

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54 minutes ago, MPMin said:

Ok help me understand this then: if you could trap photons in a perfectly mirrored box, the box would have more mass because of the photons trapped inside, however an emp or magnetic field can pass straight through the box (because we know magnets work through mirrors) does that mean then that an emp doesn’t have relativistic mass?

Why would the EMP pass through it?

54 minutes ago, MPMin said:

Further to that, if an incandescent bulb was in the perfectly mirrored box with its own power supply ( assuming the bulb and power supply did not absorb any photons and all the photons produced were perfectly reflected within the box) the magnetic field produced by the filament could pass through box but the photons could not would the box lose mass due to the magnetic field escaping the box?

As I’ve said before, trying to treat the field separate from the source is problematic. It’s also not a given that the magnetic field would penetrate the box.

54 minutes ago, MPMin said:

And if the box could perfectly trap photons, would opening the box after some time release all the trapped photons at once producing an explosion of photons so to speak or would it make no difference and opening the box after some time would look the same as opening the box at any time?

If you could do this, yes, it would release all the photons at once, but since you can’t do this, it’s a moot point.

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1 hour ago, Strange said:

Incidentally, if the box is perfectly reflecting (inside) then it is also perfectly opaque from the outside so EM radiation could not pass through it.

I tend to disagree with you on this point, a regular mirror reflects around 90% of light but the mirror will not reflect 90% of magnetic field. A perfect reflector of light might not reflect any magnetic radiation at all but instead allow it to pass through.

1 hour ago, swansont said:

Why would the EMP pass through it?

The above is why a magnetic field would pass through it.

1 hour ago, Strange said:

If it is a constant current then any magnetic field would be static and not carrying away energy. If it is AC, then any radiation that escapes from the box would reduce the mass of the box.

Just because a direct current is constant doesn’t mean the magnetic field it produces is static, the magnetic field emanates out from the wire just as light does from a filament wire.

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When was the last time you trapped light without a source? Further more light is part of the EM field.

Edited by Mordred

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36 minutes ago, Mordred said:

When was the last time you trapped light without a source?

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The point is the time scale to trap light with perfect mirrors is extremely brief. You also seem to think light isn't part of the EM spectrum.

Photons are the mediator gauge boson for the EM field this includes mediating the magnetic field though the electric field and magnetic field are one and the same.

So if you have a magnetic field you also have photons though offshell.

Edited by Mordred

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33 minutes ago, Mordred said:

The point is the time scale to trap light with perfect mirrors is extremely brief. You also seem to think light isn't part of the EM spectrum.

I think you might have misunderstood my query. The example I used was actually for collecting photons from a source of light within the box as curiosity while on the subject of photons of . To phrase it differently, can photons be accumulated? Imagine if you could see into the box without losing any photons, would the inside of the box become brighter and brighter as the photons were being emitted inside the box because they couldn’t escape?

54 minutes ago, Mordred said:

Photons are the mediator gauge boson for the EM field this includes mediating the magnetic field though the electric field and magnetic field are one and the same.

So if you have a magnetic field you also have photons though offshell.

Does this also apply to the magnetic fields of magnets?

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No the photons would not accumulate. One thing to realize is that photons are the quantization of the EM field $A_\mu$ the E and B fields are part of the EM field and are in essence different phase polarizations of the same field.

The frequency modes of the EM field  in a given volume gives rise to the photon number density (quanta of each field). These frequency modes also degrade or if you prefer disperse through destructive interference. In QED we can calculate the photon number density via the creation and annihilation operators. Though the formulas take time to understand fully you will immediately see the symalarity between the E and B fields.

$E(r,t)=\frac{i}{2c\sqrt{V}}\sum_k\omega_k[A_k(t)e^{ik\cdot r}-A_k^{*}(t)e^{-ik\cdot r}]$

$B(r,t)=\frac{i}{2c\sqrt{V}}\sum_k k[A_k(t)e^{ik\cdot r}-A_k^{*}(t)e^{-ik\cdot r}]$

now without going into too much detail you should note that only the k for the magnetic field energy and $\Omega_k$ differ in the two above expressions. The latter is the E field energy though in both cases its proper to square those terms through another formula which I won't get into as some care must be taken with the two independent polarizations  E and B of the EM field $A_\mu$

the main point is that in both cases the photon number density of both the E and B fields are both represented in units of quanta, each unit of quanta is a photon so in your box the number density will depend on the frequency modes given off by the light bulb. ( though you can gain a slightly higher number via temperature which is also part of the EM spectrum.)

see here for the polarization waves of the EM field.

Further details see here in terms of EM radiation

Some of the other equations that will relate to the two I provided are contained here under photon polarizations which is linked to the EM radiation link above.

It will help explain the two above equations.

Edited by Mordred

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6 hours ago, MPMin said:

I tend to disagree with you on this point, a regular mirror reflects around 90% of light but the mirror will not reflect 90% of magnetic field. A perfect reflector of light might not reflect any magnetic radiation at all but instead allow it to pass through.

You could have a mirror that only reflected visible light but not other frequencies of radiation (but you didn't say that). In which case, any radiation that escapes will reduce the energy and hence mass of the object.

But you are, again, confusing electromagnetic radiation and a magnetic field. A (static)magnetic field will not carry away energy. A changing magnetic field will generate electromagnetic radiation.

6 hours ago, MPMin said:

Just because a direct current is constant doesn’t mean the magnetic field it produces is static

Yes it does. That is exactly what it means.

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46 minutes ago, Strange said:

But you are, again, confusing electromagnetic radiation and a magnetic field. A (static)magnetic field will not carry away energy. A changing magnetic field will generate electromagnetic radiation.

7 hours ago, MPMin said:

Just because a direct current is constant doesn’t mean the magnetic field it produces is static

Yes it does. That is exactly what it means.

In this context, Is the word static being used to describe something that’s unchanging or not moving at all?

If the static magnetic field produced by a wire carrying a direct current doesn’t carry away any energy is it because the static magnetic field has no energy or the static magnetic field doesn’t move away (emanate) from the wire, neither of these seem to be correct.

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9 minutes ago, MPMin said:

In this context, Is the word static being used to describe something that’s unchanging or not moving at all?

If the static magnetic field produced by a wire carrying a direct current doesn’t carry away any energy is it because the static magnetic field has no energy or the static magnetic field doesn’t move away (emanate) from the wire, neither of these seem to be correct.

A constant current means a current that doesn't change, which results in a magnetic field that doesn't change. The only thing "moving" is the current.

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9 hours ago, MPMin said:

I tend to disagree with you on this point, a regular mirror reflects around 90% of light but the mirror will not reflect 90% of magnetic field.

It wouldn't "reflect" but it depends on the mirror. What properties will make it 100% reflective?

9 hours ago, MPMin said:

A perfect reflector of light might not reflect any magnetic radiation at all but instead allow it to pass through.

9 hours ago, MPMin said:

The above is why a magnetic field would pass through it.

I asked about the EMP. Are you under the impression that these are the same thing?

9 hours ago, MPMin said:

Just because a direct current is constant doesn’t mean the magnetic field it produces is static, the magnetic field emanates out from the wire just as light does from a filament wire.

This claim is based on what, exactly?

6 hours ago, MPMin said:

I think you might have misunderstood my query. The example I used was actually for collecting photons from a source of light within the box as curiosity while on the subject of photons of . To phrase it differently, can photons be accumulated? Imagine if you could see into the box without losing any photons, would the inside of the box become brighter and brighter as the photons were being emitted inside the box because they couldn’t escape?

One of several problems here is that the emitter is inside in this scenario, and will also absorb photons. Another is how you make mirrors that reflect at 100%. You can't actually do it, but if you are going to idealize that you can, the method matters. If it involves the permeability getting very large, then that means any magnetic field is going to be inside the mirror, and not escape it. It's basically how you build magnetic shielding.

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5 hours ago, Strange said:

A constant current means a current that doesn't change, which results in a magnetic field that doesn't change. The only thing "moving" is the current.

I disagree with you on this, the magnetic field emanates out from the wire just as light emanates from a filament wire. Both light and magnetic field travel at the same speed.

3 hours ago, swansont said:

If you take into consideration the context i used; does any mirror reflect a magnetic field or emp the same way it reflects light? Imagine a filament wire carrying a constant direct current, the filament wire will produce light and a magnetic field, imagine the filament wire placed next to a mirror, the mirror will reflect the light but allow the magnetic field to pass through it. This would suggest that the magnetic field is independent of the photons of light.

4 hours ago, swansont said:

I asked about the EMP. Are you under the impression that these are the same thing?

An emp is a segment of a magnetic field.

4 hours ago, swansont said:
Quote

Just because a direct current is constant doesn’t mean the magnetic field it produces is static, the magnetic field emanates out from the wire just as light does from a filament wire.

This claim is based on what, exactly?

It’s based on the fact that the moment the current begins to flow, the magnetic field must begin to emanate out from the wire, if it doesn’t emanate out from the wire then it must instantly appear... which it doesn’t. And if it’s does emanate out from the wire as soon as the current begins to flow then it must continue to emanate while the current is flowing.

4 hours ago, swansont said:

One of several problems here is that the emitter is inside in this scenario, and will also absorb photons. Another is how you make mirrors that reflect at 100%. You can't actually do it, but if you are going to idealize that you can, the method matters. If it involves the permeability getting very large, then that means any magnetic field is going to be inside the mirror, and not escape it. It's basically how you build magnetic shielding.

I did make these propositions hypotheticals only to illustrate a point and I did say the emitter would not absorb any photons.

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39 minutes ago, MPMin said:

I disagree with you on this

Then please provide a reference to support this. (And I should recommend this is moved to Speculations.)

40 minutes ago, MPMin said:

It’s based on the fact that the moment the current begins to flow

That is not a constant current and not a static field.

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1 hour ago, MPMin said:

I disagree with you on this, the magnetic field emanates out from the wire just as light emanates from a filament wire. Both light and magnetic field travel at the same speed.

Based on what?

1 hour ago, MPMin said:

If you take into consideration the context i used; does any mirror reflect a magnetic field or emp the same way it reflects light? Imagine a filament wire carrying a constant direct current, the filament wire will produce light and a magnetic field, imagine the filament wire placed next to a mirror, the mirror will reflect the light but allow the magnetic field to pass through it. This would suggest that the magnetic field is independent of the photons of light.

You don’t reflect a magnetic field. It makes little sense to discuss magnetic field behavior as you do light.

You keep asserting the field would pass through a mirror. But you have provided no justification for this assertion.

1 hour ago, MPMin said:

An emp is a segment of a magnetic field.

No.

EMP stands for electro-magnetic pulse. It’s electromagnetic radiation, aka light, aka photons.

1 hour ago, MPMin said:

It’s based on the fact that the moment the current begins to flow, the magnetic field must begin to emanate out from the wire, if it doesn’t emanate out from the wire then it must instantly appear... which it doesn’t. And if it’s does emanate out from the wire as soon as the current begins to flow then it must continue to emanate while the current is flowing.

Where does the field go as it “emanates” but has reached steady-state? For a wire, the field is circular, centered on the wire? Where is the “sink”?

1 hour ago, MPMin said:

I did make these propositions hypotheticals only to illustrate a point and I did say the emitter would not absorb any photons.

When your scenario violates physical law, it’s impossible to use physical law to derive a valid conclusion.

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