Todays Mystery Object?

[Martin King 2Participant @martinking2]

Hi All,

Just got back from buying a lot of workshop kit and this unusual item was in one of the boxes. I have no idea what it is? Possibly some sort of level???

Seems to have some sort of mirror inside and appears to be very sensitive to movement.

Any thoughts will be most welcome please. Regards, Martin

[Martin King 2Participant @martinking2]

[SillyOldDufferModerator @sillyoldduffer]

I think its a mirror galvanometer for precisely zeroing a Wheatstone Bridge. A beam of light is bounced off the mirror onto a distant screen to greatly amplify the scale of any movement.

The meter is likely to be very sensitive and best practice is to short the terminals out to dampen any movement in storage. It also has a mechanical clamp.

Dave

[SteveWParticipant @stevew54046]

Mirror galvanometer to detect balance point on a Wheatstone bridge? Hint on the picture which shows circuit. Clamp stops the device from swinging free and potentially damaging itself in transport or storage.  SOD beat me to it!

Edited By SteveW on 17/07/2018 18:40:03

[Mike PooleParticipant @mikepoole82104]

It may be the works of a spot reflecting galvanometer the main scale and the rest of the box are missing.

 

Mike

Well while I was looking I was beaten to the answer, but it seems to be three votes for a mirror galvanometer.

Edited By Mike Poole on 17/07/2018 18:47:31

[Neil WyattModerator @neilwyatt]

It's been done.

Clamp/free locks the movement.

By reflecting a narrow beam off the mirror onto a remote scale it provide an incredibly sensitive way of comparing the ratios of two pairs of resistors.

Neil

[Martin King 2Participant @martinking2]

Thanks very much to one and all, what a great place this is!

Cheers, Martin

[paul rushmerParticipant @paulrushmer83015]

If my memory is correct that was made by Cambridge Instruments cam in a bridge !!

Paul

[larneyinParticipant @larneyin]

Used to repair them many years ago

Eyesight was keener and hands steadier then

[Howard LewisParticipant @howardlewis46836]

The wire across the terminals is to short the movement when travelling and minimise pointer movement. Needed, because it is a very sensitive piece of kit. A light beam, reflected off the mirror has no measureable weight or inertia.

A long "pointer" is therefore obtainable for maximum accuracy.

Howard

[Neil WyattModerator @neilwyatt]

Posted by Howard Lewis on 18/07/2018 20:21:18:

A light beam, reflected off the mirror has no measureable weight or inertia.

But it does have momentum

Neil

[peak4Participant @peak4]

Posted by Neil Wyatt on 18/07/2018 20:35:12:

Posted by Howard Lewis on 18/07/2018 20:21:18:

A light beam, reflected off the mirror has no measureable weight or inertia.

But it does have momentum

Neil

As soon as I read this, I thought "Crookes Radiometer" and went to look for a suitable article in order to provide a hyperlink

On reading the above link, I then found that what I'd been taught at school seems to be wrong; I do like learning new things.

Bill

[Ian PParticipant @ianp]

Posted by Neil Wyatt on 18/07/2018 20:35:12:

Posted by Howard Lewis on 18/07/2018 20:21:18:

A light beam, reflected off the mirror has no measureable weight or inertia.

But it does have momentum

Neil

With my pedant hat on… I don't think the light beam has any momentum

Ian P

[Frances IoMParticipant @francesiom58905]

Ian
according to James Clerk Maxwell a light beam does have momentum – eg see https://en.wikipedia.org/wiki/Solar_sail

[not done it yetParticipant @notdoneityet]

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?

[Neil WyattModerator @neilwyatt]

Momentum of a photon = frequency x planck constant

http://www.nasa.gov/mission_pages/tdm/solarsail/index.html

Edited By Neil Wyatt on 19/07/2018 09:45:07

[SillyOldDufferModerator @sillyoldduffer]

Posted by not done it yet on 19/07/2018 06:19:17:

Let’s think about it from Einstein’s point of view.

E = mc^2

…

If only I could!

Can anyone explain how E=mc² is derived?

I understand the root to be evidence from radioactive decay – it was observed that the mass of an unstable element decreases slightly as it emits energy. If Conservation of Mass and Conservation of Energy are both true then the evidence suggests Mass and Energy must be equivalent, perhaps forms of something else. (My brain is starting to overheat!)

I made a promising start on the explanation here. It notes that Energy is measured in Joules, ie kg m² s⁻², which was a step forward for me, but I came unstuck where it says the equation applies only to Invariant Mass, not Relativistic Mass.

Now I'm baffled.

Help!

Dave

[michael pottsParticipant @michaelpotts88182]

The equation E = mc^2 is the first term in an infinite series giving the energy of an object. The second term is

1/2 mv^2 which is the kinetic energy of the object. The other terms are miniscule as all are divided by c^2 and higher powers of c^2. The series derives from the calculation of the kinetic energy of an object. The mass of the object changes as it moves and is calculated by the equation M = M0 / ( 1 – v^2/c^2 )^0.5. M0 is the rest mass of the object, M is the mass of the object moving at a velocity of v, so if the object is moving, its' mass is M0 divided by something that is less than 1 making it more massive, and the moving mass gets larger as the velocity increases. If the velocity of the object reaches the speed of light then the mass is then infinite. In practice if the velocity of the object is less than 10% of the speed of light then normal Newtonian mechanics can be applied with little error.

The mathematics of relativity become very complex very quickly, making an already difficult subject even more impenetrable. Infinite series of terms do not help either.

Radioactive fusion or fission is another issue. Both processes work because the mass of the resultant particle is less than the mass of the starting particle (s). This loss of mass appears as energy, heat,light or kinetic energy of the particles. The amount of energy can be calculated knowing the loss of mass. All the work of measuring the mass of atomic nuclei was carried out after work began to develop atomic weapons during WW 2.

Mike Potts.

[Martin King 2Participant @martinking2]

Blimey, talk about 'thread drift'!

Martin

[SillyOldDufferModerator @sillyoldduffer]

Posted by michael potts on 19/07/2018 10:42:58:

The equation E = mc^2 is the first term in an infinite series giving the energy of an object. The second term is

1/2 mv^2 which is the kinetic energy of the object. The other terms are miniscule as all are divided by c^2 and higher powers of c^2. The series derives from the calculation of the kinetic energy of an object. The mass of the object changes as it moves and is calculated by the equation M = M0 / ( 1 – v^2/c^2 )^0.5. M0 is the rest mass of the object, M is the mass of the object moving at a velocity of v, so if the object is moving, its' mass is M0 divided by something that is less than 1 making it more massive, and the moving mass gets larger as the velocity increases. If the velocity of the object reaches the speed of light then the mass is then infinite. In practice if the velocity of the object is less than 10% of the speed of light then normal Newtonian mechanics can be applied with little error.

The mathematics of relativity become very complex very quickly, making an already difficult subject even more impenetrable. Infinite series of terms do not help either.

Radioactive fusion or fission is another issue. Both processes work because the mass of the resultant particle is less than the mass of the starting particle (s). This loss of mass appears as energy, heat,light or kinetic energy of the particles. The amount of energy can be calculated knowing the loss of mass. All the work of measuring the mass of atomic nuclei was carried out after work began to develop atomic weapons during WW 2.

Mike Potts.

Thanks Mike, interesting to find my belief that E=mc² started from radioactivity is wrong – 'All the work of measuring the mass of atomic nuclei was carried out after work began to develop atomic weapons during WW 2.' I'm never quite sure if what I learned in my youth was over-simplified and explained badly, or if I was too dim to take it in. Sadly for me, the evidence suggests the latter…

Dave

[Cornish JackParticipant @cornishjack]

or, … for Harry Hemsley fans – "What did Horace say, Winnie?"

rgds

Bill

[Rod RenshawParticipant @rodrenshaw28584]

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

[MeunierParticipant @meunier]

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

[VersabossParticipant @versaboss]

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.

[not done it yetParticipant @notdoneityet]

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?

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