Jump to content
Science Forums

A quantum picture of torsion


Recommended Posts

By the way, what do you think dimensional analysis is, because I've corrected physicists in the past in such things. Telling me I might not be equipped for such things is like telling a horse is not capable of making a long jump. Don't underestimate my capabilities boy, I am just not as wild to think the half baked things you do.

Edited by Dubbelosix
Link to comment
Share on other sites

There! I've even updated the Op for you! I've now given adln exact equality. If Jv is twice the charge, or charge squared, ie. e^2 then we are left with the following exact equality,

B x v ~ (1/2mc² ⋅ ∂U/∂t) 

 

It's approximate because in exact units, Jv is really twice the value of e. An exact equality would be 

 

B x v = (e/2mc² ⋅ ∂U/∂t) 

This is because, Jv is really the same thing as e^2 So we have a remaining charge in the numerator. If you can find three more exact equalities, then I'll concede you know at least something about dimensional analysis!

Edited by Dubbelosix
Link to comment
Share on other sites

2 minutes ago, JeffreysTubes8 said:

Would you please stop turning this into a dick measuring contest. If you can't handle constructive criticism without retaliating you should leave this thread, not me.

No, because you started this whole dick contest when you started talking to me about dimensional analysis when I knew, that you'd no real indepth knowledge of it. So clearly you don't then?

Edited by Dubbelosix
Link to comment
Share on other sites

This tells me, you know nothing, Jon Snow. This isn't about verticle or horizontal movement, a spin and orbit coupling actually describes a curvilinear movement around a fixed axis. Just give it up already, you couldn't do some simple arithmetic which now I will give you an answer. The equalities hold as

Jv = e^2 = Er = Gm^2 = hbar c

(iff) v=c

Edited by Dubbelosix
Link to comment
Share on other sites

Stop telling me what to learn, I know geometry very well. Yes you can combine the two and mechanically, and very roughly explain some movement round a fixed axis, but why be so bothered about such ridiculousness when telleparallel motion is concisely described as it had been in the Op? You blame me of mixing apples amd oranges... Already, your behaviour seems to be fixed round a different common center, one fixated itself on arguments trying to prove in some misplaced way that you are bright enough to path  a cord of significant intelligent discussion in a thread which you are barely capable of understanding. This is my last post to you, I've read enough.

Edited by Dubbelosix
Link to comment
Share on other sites

Anyway, moving on from the weird spell of discussions, the fact that one factor of charge remains in the numerator from a rotation in a magnetic field is very similar to how spinning black holes experience a charge. The motion, induced by the cross product on a magnetic field, does in the picture have encoded in it, not just charge, but the RHS has encoded in it our strict definition of torsion

B x v = (e/2mc² ⋅ ∂U/∂t) 

As exampled in the Op.

Edited by Dubbelosix
Link to comment
Share on other sites

But why? The reason I can only surmise is that relativity has a function in the interpretation, much like how an electrically charged rod with a moving coil only experiences a magnetic field when motion is involved. Similarly, only when the magnetic field interacts with a system involving a motion, with velocity, v,  can it couple to the Gravimagnetic force, and is the primer for the equations of motion involving that giving rise to a Lorentz force e(B x v).

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...