Todays Mystery Object?

let's think about it from Einstein’s point of view.

E = mc^2

If light actually had zero mass, that equation would always equal zero!

Wave-particle duality applies?

There is clearly a lot of expertise being brought to bear on this topic, so much so that I am reluctant to dip my toe in the water, but I don't really understand the above post.

I was taught (many years ago ) that this equation meant (in simple terms) that an object of mass "m" contained (or was composed of) energy equal to m multiplied by c squared, where "c" was the speed of light, and that the energy and the mass were interconvertible. In the equation "c squared " is a constant and the mass of the light (if any ) is irrelevant.

I hesitate to say it, but has NDIY got confused and regarded "m" as the mass of the light? Or is it me?

Regards

Rod

My 1st job out of school was as a trainee Instrument Maker with Cambridge Instrument Co. One of my daily tasks was to gently polish the copper wire-clamping bus-bars on the Wheatstone bridge used, among other things, to adjust the resistance of the tiny coils on the indicator arm for these galvanometers. The mature lady who operated the bridge said that her fingers were too delicate for that function and judging by the fineness of the coil wire that she was manipulating, I'll concede that point.
DaveD

To continue with the thread drift. if you allow:

I always wondered what units one has to use for E=mc^2 to get a meaningful result. In SI units it would mean E in Joule, mass in kg, c in metres per second... but is this correct?

Regards, HansR.

Versaboss,

Correct. But smaller units than Joules are often used for atomic and quantum identities.

Rod,

If light has no mass the product of mass and c^2 must be zero - everything multiplied by zero is zero.

We do actually know that light has an energy value and different frequencies (or wavelengths) have different energies. If every photon were assigned as zero mass none of them would have any energy, n’est ce pas?

We know that a huge amount of our energy is transferred from the Sun to the Earth by electromagnetc waves. Mass in the Sun is being converted to electrmagnetic waves? If those waves have energy, they must have mass? Or how else can mc^2 have any value other than zero?

NDIY

I think, with due respect, that you are repeating your original error.

"m" is the mass of the object under consideration, not the mass of the light.

I understand your point about the maths, I think the problem is that you are thinking of "m" as being the mass of the light not the mass of the object, so this is an issue of definition of terms, not maths.

The whole point of the equation is to show the equivalence of the energy "contained" within an object and its mass, and by extension and inference the interchangeability of the two.

For example, in a nuclear reaction, fissile atoms may split releasing energy, and the daughter products will then collectively have slightly less mass than the parents. The "lost" mass has been converted into energy, and this equation enables one to work out how much energy will be released by the "loss" of a given mass of matter. In this example "m" is the lost mass, and E is the energy released. "c squared" remains the constant- that bit we do agree about!

If the object under consideration was "a bit of light" then I would agree with you that if light has no mass then the answer would always be zero, but that was not what Einstein was trying to elucidate when he published his equation.

Or so I have always believed.

Regards

Rod

There lies the theory of duality of light. Does light arrive at the Earth as a pure continuous wave or as photons? Do we, as well as any other macro-sized object, each have a frequency dependent on our mass?

The photo-electric effect appears to rely on ‘packets of energy’ rather than a cumulative supply by a wave of lower frequency. Lots of contradictions, but consider the mass of one photon and try to calculate its mass. Sufficiently low to give a zero result on most pocket calculators!

If the particle theory of light applies, there must be mass included somewhere.

As Dave indicates, there is a mass defect when radio-istopes decay or are atoms are split - it doesn’t just apply to fission in nuclear reactors (but that is clearly more obvious).

We have adopted the speed of light in vacuo as a fixed value. Thought that the energy cannot travel any faster so the frequency is adjusted to comply with the speed limit and Einstein’s theories? Just another take on it? I know it is not cut and dried. Not so very long ago, electrons were the smallest possible particle. Quantum theories have changed all that. We may soon have a few more layers than simple quarks. Where will it stop?

As a matter of interest how do you describe a gamma ray emitted by a single radioactive decay? A continuous wave or a single packet of energy? I don’t claim to know the answers but I do keep an open mind - and occasionally throw a spanner in the works to get some more discussion. Keep it coming!

Kinetic energy is 1/2 mv squared.

If you chose the speed of light as your velocity then v squared becomes a constant because the speed of light never varies.

If the velocity is constant then what you are left with is energy = mass times a constant and you have your conversion.

Simples.

.

The speed of light frequently varies ! [otherwise, how would prisms and lenses work ?]

It's the speed of light in vacuo that [according to dogma] is a constant.

MichaelG.

Edited By Michael Gilligan on 20/07/2018 16:30:33

Not unless you believe in expatriate Russian Oligarch Meerkats who speak English!

Anyway my sorry understanding is that E = ½mv² is only valid at less than 70% light speed. Above that, E=mc² dominates. It's because mass increases with speed.

Nature is fascinating because science doesn't explain what anything is. Taking time as an example, it can be described and measured with enormous accuracy. We understand its effects well. But the what, why and how of time are all unknown.

Natural languages are rotten at expressing facts rigorously. Maths is better but still struggles. Possibly fundamental concepts are beyond human ability to express.

Nothing is ever simples!

Dave

Edit: 0/10 for spelling!

Edited By SillyOldDuffer on 20/07/2018 17:14:30

If I might be a bit nit-picky, the use of a light beam does not increase accuracy (except by being light , ie not heavy, and therefore allowing a less heavy armature assembly) - it actually maximises sensitivity, as the pointer has no inertia, or flywheel effect. So, the response to rapid changes is improved. Later machines on the same lines had a pointer formed of an electron beam for the same reason, and it had to be kept in a glass box (and it was called an oscilloscope).

And how difficult it is to talk about a light pointer made of light without tripping over yourself.

Cheers, Tim

Tim

Interesting, I agree that the use of a beam of light does not of itself lead to greater accuracy, but by the use of mirrors to "fold" the beam, it does allow for a much longer pointer than a conventional (metal?) pointer could conveniently be. The longer pointer allows the measurement to be displayed on a longer, more open scale which can be read to a finer degree than a shorter cramped scale and this can lead to more accurate working?

I also agree with you about the lack of inertia etc but this not always the reason for using a "light pointer.", For example, in the Wheatstone Bridge device mentioned above the unknown resistance is not usually varying rapidly, if at all.

What do you think?

Regards

Rod

I strongly recommend the Royal Institute lectures on particle physics, cosmology, quantum mechanics etc., all interrelated of course. I have been watching them for the last couple of weeks and find them fascinating.