Does an electric field or a magnetic field have mass

passerby

New Member
Messages
12
It has been known for a long time that such fields have mass. I believe it was derived from Maxwell's equations in the late 19th century. A more detailed explanation can be found in the Feynman lectures on physics, volume 2, chapter 28, titled "Electromagnetic mass", and elsewhere online. I'd post a link, but I can't, since I just joined the forum.

The modern view is that of relativity, where E =mc². The older electromagnetic mass derivation was E = (4/3)mc², if I remember correctly.
 

Harte

Senior Member
Messages
4,562
It has been known for a long time that such fields have mass. I believe it was derived from Maxwell's equations in the late 19th century. A more detailed explanation can be found in the Feynman lectures on physics, volume 2, chapter 28, titled "Electromagnetic mass", and elsewhere online. I'd post a link, but I can't, since I just joined the forum.

The modern view is that of relativity, where E =mc². The older electromagnetic mass derivation was E = (4/3)mc², if I remember correctly.
Feynman's lecture doesn't pertain to mass of an EM field, just to the "electromagnetic mass" of a charged particle. A charged particle is not the field it carries. The charge creates the field.
EM fields are governed by photons, which are not charged. However, they do have mass in the form of momentum, which is entirely dependent on wavelength, like I said. An EM field has no mass. If it did, then photons would have a rest mass (invariant mass,) which means photons we observe would have observable mass because of the relativistic mass effect on the invariant mass they carry, along with the momentum they have due to wavelength.

Of course, photons could have such a tiny invariant mass they we can't observe it - yet. However, that is simply speculation. It can't be asserted without evidence.

Harte
 

passerby

New Member
Messages
12
Feynman's lecture doesn't pertain to mass of an EM field, just to the "electromagnetic mass" of a charged particle. A charged particle is not the field it carries. The charge creates the field.
EM fields are governed by photons, which are not charged. However, they do have mass in the form of momentum, which is entirely dependent on wavelength, like I said. An EM field has no mass. If it did, then photons would have a rest mass (invariant mass,) which means photons we observe would have observable mass because of the relativistic mass effect on the invariant mass they carry, along with the momentum they have due to wavelength.

Of course, photons could have such a tiny invariant mass they we can't observe it - yet. However, that is simply speculation. It can't be asserted without evidence.

Harte
I admit I didn't explain it well, as I was in kind of a hurry. If you look at the Wikipedia article on "electric field" & read the section titled "Energy in the electric field", you will find that there is a well-defined energy density in an electromagnetic field equal to:

u = 0.5 * epsilon * E² + 0.5 * B² / mu
Total energy = U = u * V
(V is the volume)

Of course, the E in that equation is not the same as in E = mc². Sometimes we need a bigger alphabet. Anyway, the mass of the EM field is:

m = U / c² = V * ( 0.5 * epsilon * E² + 0.5 * B² / mu) / c²

It's necessarily a small value under ordinary circumstances, due to the c² in the denominator. And I apologize for the ugliness of mathematical equations written in plaintext. I shall attempt to remedy that in the future.
 

Harte

Senior Member
Messages
4,562
Once again, the energy in any EM field is the energy of the photons that govern the field. Photon energy is ENTIRELY based on the wavelength of the photons (light) and has nothing whatsoever to do with the mass of anything that has invariant mass, such as an electron.
An EM field has no mass whatsoever.
You mention that the "U" in your equation is not the same as the "E" in E=mc^2. In fact, it is exactly the same. However, the "m" in Einstein's equation stands for the invariant (rest) mass of an object. Photons are the only objects IN an EM field, and they HAVE no invariant mass (their rest mass is zero.)

What that means is your equation should read:

0 = U / c² = V * ( 0.5 * epsilon * E² + 0.5 * B² / mu) / c²

It should be obvious that the above leads to U = 0, and it does - because the U here is the energy equivalent of the mass of the object.

Harte
 

F Stein

Member
Messages
200
Once again, the energy in any EM field is the energy of the photons that govern the field. Photon energy is ENTIRELY based on the wavelength of the photons (light) and has nothing whatsoever to do with the mass of anything that has invariant mass, such as an electron.
An EM field has no mass whatsoever.
You mention that the "U" in your equation is not the same as the "E" in E=mc^2. In fact, it is exactly the same. However, the "m" in Einstein's equation stands for the invariant (rest) mass of an object. Photons are the only objects IN an EM field, and they HAVE no invariant mass (their rest mass is zero.)

What that means is your equation should read:

0 = U / c² = V * ( 0.5 * epsilon * E² + 0.5 * B² / mu) / c²

It should be obvious that the above leads to U = 0, and it does - because the U here is the energy equivalent of the mass of the object.

Harte
Could a electric magnetic field slow time and if so how many teslas in strength would it have to be?
 

Harte

Senior Member
Messages
4,562
Fields don't exist without charges.
No EM field could dilate time, but if the charge was large enough, the mass of the particles holding it could.
Any particle that can hold a charge has an invariant mass. The charges also have mass, which depends on the wavelength of the particle and in the case of a particle with intrinsic mass, the above equations apply, if there is a relative velocity.

Harte
 

StarLord

Senior Member
Messages
3,187
Yet, STILL, All Things Have "Mass" , "Fuzzy" Equations ..... Ah BE, But Ah Aint, Movin At The Speed Of My TIC TACK.........
 

StarLord

Senior Member
Messages
3,187
Besides, You KNOW They Are Going To Use "Particles" Being SPEWED Out As A "Drive Component" To PUSH Future Space Vehicles......?
 

Top